Introduction To Periodic Table
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Types Of Chemical Compounds
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Calculations Involving Moles
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- Moles of Compounds and Elements
- Exercise 1
- Exercise 2
- Calculating Percentage Composition
- Exercise 3
- Exercise 4
- Exercise 5
- Determining Formula from Composition
- Exercise 6
- Exercise 7
- Exercise 8
- Balancing a Chemical Equation for Combustion
- Exercise 9 - Part a
- Exercise 9 - Part b
- Determining Formula From Combustion Analysis
- Exercise 10
- Exercise 11

Inorganic Compounds
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[{"Name":"Introduction To Periodic Table","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introduction To Periodic Table","Duration":"5m 37s","ChapterTopicVideoID":16910,"CourseChapterTopicPlaylistID":154543,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16910.jpeg","UploadDate":"2019-02-20T23:36:40.3130000","DurationForVideoObject":"PT5M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:03.970","Text":"In this video, we will introduce the periodic table of"},{"Start":"00:03.970 ","End":"00:06.869","Text":"the elements and discuss some of its features."},{"Start":"00:06.869 ","End":"00:11.640","Text":"A much more detailed description will be given later in the course."},{"Start":"00:11.640 ","End":"00:16.634","Text":"We\u0027re going to discuss the periodic table of the elements."},{"Start":"00:16.634 ","End":"00:18.870","Text":"Here\u0027s a periodic table of the elements."},{"Start":"00:18.870 ","End":"00:22.905","Text":"It has been color-coded to make it easier to understand."},{"Start":"00:22.905 ","End":"00:25.800","Text":"The first thing we need to note is that"},{"Start":"00:25.800 ","End":"00:28.620","Text":"the elements are arranged according to the value of Z."},{"Start":"00:28.620 ","End":"00:33.105","Text":"Z, you may recall is the atomic number."},{"Start":"00:33.105 ","End":"00:38.344","Text":"It tells us how many protons there are in each element."},{"Start":"00:38.344 ","End":"00:41.585","Text":"We start off with hydrogen,"},{"Start":"00:41.585 ","End":"00:44.620","Text":"which has simply 1 proton,"},{"Start":"00:44.620 ","End":"00:46.675","Text":"so Z is equal to 1."},{"Start":"00:46.675 ","End":"00:50.255","Text":"We jump to the other side with Z equal to 2."},{"Start":"00:50.255 ","End":"00:53.985","Text":"Then lithium, Z equals to 3,"},{"Start":"00:53.985 ","End":"00:58.010","Text":"right up to neon with Z equals 10 and so it goes"},{"Start":"00:58.010 ","End":"01:02.825","Text":"on row after row until we get to the sixth row."},{"Start":"01:02.825 ","End":"01:04.945","Text":"In the sixth row,"},{"Start":"01:04.945 ","End":"01:07.700","Text":"we see this blank space,"},{"Start":"01:07.700 ","End":"01:11.165","Text":"this pink space, which should be 57."},{"Start":"01:11.165 ","End":"01:19.565","Text":"We have in fact 57-71 taken out to make this whole table much less wide."},{"Start":"01:19.565 ","End":"01:23.000","Text":"Because you can understand if we had all of these stuck in here,"},{"Start":"01:23.000 ","End":"01:24.680","Text":"it would be enormously wide."},{"Start":"01:24.680 ","End":"01:30.485","Text":"These are called lanthanides because the very first element is called lanthanum."},{"Start":"01:30.485 ","End":"01:34.190","Text":"Similarly, in the seventh row,"},{"Start":"01:34.190 ","End":"01:36.335","Text":"another group is taken out,"},{"Start":"01:36.335 ","End":"01:38.982","Text":"89-103 is removed,"},{"Start":"01:38.982 ","End":"01:44.545","Text":"and these are called actinides because actinium is the first one."},{"Start":"01:44.545 ","End":"01:48.440","Text":"Let\u0027s look at the structure of the periodic table."},{"Start":"01:48.440 ","End":"01:53.520","Text":"The first thing to notice that there are 18 columns."},{"Start":"01:53.540 ","End":"01:56.790","Text":"We call these columns groups."},{"Start":"01:56.790 ","End":"02:00.080","Text":"We have from group one up to group 18."},{"Start":"02:00.080 ","End":"02:02.900","Text":"We also have rows."},{"Start":"02:02.900 ","End":"02:05.435","Text":"Each row is called a period,"},{"Start":"02:05.435 ","End":"02:08.030","Text":"so we have 7 rows."},{"Start":"02:08.030 ","End":"02:11.540","Text":"Now, we\u0027re going to go over all the properties of"},{"Start":"02:11.540 ","End":"02:17.120","Text":"the various groups to try to understand how they are arranged."},{"Start":"02:17.120 ","End":"02:21.245","Text":"Let\u0027s start off with group 1 in the left-hand side."},{"Start":"02:21.245 ","End":"02:23.570","Text":"Apart from hydrogen, which is quite"},{"Start":"02:23.570 ","End":"02:28.129","Text":"exceptional because it only has one proton and one electron."},{"Start":"02:28.129 ","End":"02:31.880","Text":"The other elements in"},{"Start":"02:31.880 ","End":"02:38.825","Text":"the first group are all metals and they are called alkali metals."},{"Start":"02:38.825 ","End":"02:40.970","Text":"This name is just historical,"},{"Start":"02:40.970 ","End":"02:43.220","Text":"it has no great significance,"},{"Start":"02:43.220 ","End":"02:48.995","Text":"and the properties of these metals change as we go down the group."},{"Start":"02:48.995 ","End":"02:55.195","Text":"The second column, group 2 are also metals."},{"Start":"02:55.195 ","End":"03:02.525","Text":"They are called alkaline earth metals."},{"Start":"03:02.525 ","End":"03:04.570","Text":"Again, a historical name."},{"Start":"03:04.570 ","End":"03:08.715","Text":"Starts with beryllium and goes down to radium."},{"Start":"03:08.715 ","End":"03:16.400","Text":"Now, the central block are called transition metals,"},{"Start":"03:16.400 ","End":"03:19.385","Text":"and you\u0027ll recognize many of them."},{"Start":"03:19.385 ","End":"03:28.015","Text":"Here\u0027s nickel, copper, zinc, silver, gold, platinum."},{"Start":"03:28.015 ","End":"03:31.835","Text":"These are all metals called transition metals."},{"Start":"03:31.835 ","End":"03:34.220","Text":"Now, we got to the final block,"},{"Start":"03:34.220 ","End":"03:36.400","Text":"the block on the right-hand side,"},{"Start":"03:36.400 ","End":"03:40.655","Text":"and here we have metals and non-metals."},{"Start":"03:40.655 ","End":"03:43.580","Text":"Let\u0027s first consider the metals."},{"Start":"03:43.580 ","End":"03:46.325","Text":"The metals are indicated in gray."},{"Start":"03:46.325 ","End":"03:49.985","Text":"Here\u0027s aluminum or as American say aluminum."},{"Start":"03:49.985 ","End":"03:53.760","Text":"These are metals, all the gray ones,"},{"Start":"03:53.760 ","End":"03:59.270","Text":"and in green, we have non-metals like carbon, nitrogen, oxygen."},{"Start":"03:59.270 ","End":"04:01.325","Text":"These are all non-metals."},{"Start":"04:01.325 ","End":"04:03.949","Text":"In-between the metals and non-metals,"},{"Start":"04:03.949 ","End":"04:09.440","Text":"we have the staircase in the darker green."},{"Start":"04:09.440 ","End":"04:13.260","Text":"Now, these are called metalloids."},{"Start":"04:14.080 ","End":"04:18.570","Text":"The intermediate between metals and non-metals."},{"Start":"04:18.570 ","End":"04:21.485","Text":"If we can get them very pure,"},{"Start":"04:21.485 ","End":"04:24.215","Text":"some of them act like semiconductors."},{"Start":"04:24.215 ","End":"04:27.800","Text":"For example, silicon is a well-known semiconductor."},{"Start":"04:27.800 ","End":"04:30.710","Text":"Now, let\u0027s go to group 17,"},{"Start":"04:30.710 ","End":"04:33.005","Text":"here is group 17,"},{"Start":"04:33.005 ","End":"04:36.605","Text":"fluorine, chlorine, bromine, etc."},{"Start":"04:36.605 ","End":"04:39.815","Text":"These are called halogens."},{"Start":"04:39.815 ","End":"04:45.215","Text":"The final group, group 18 are gases."},{"Start":"04:45.215 ","End":"04:50.675","Text":"Gases don\u0027t react with anything else or at least they rarely do."},{"Start":"04:50.675 ","End":"05:00.400","Text":"They are called rare gases or sometimes noble gases."},{"Start":"05:01.480 ","End":"05:05.120","Text":"Here, we\u0027ve learned about the various elements."},{"Start":"05:05.120 ","End":"05:10.220","Text":"We\u0027ve seen that there are many more metals than there are non-metals,"},{"Start":"05:10.220 ","End":"05:14.755","Text":"and the non-metals are all on the right-hand side of the periodic table."},{"Start":"05:14.755 ","End":"05:18.980","Text":"Whereas metals are all over the place going right from the left,"},{"Start":"05:18.980 ","End":"05:22.590","Text":"right up to near the right to the right."},{"Start":"05:23.590 ","End":"05:27.020","Text":"In this video, we\u0027ve introduced the periodic table of"},{"Start":"05:27.020 ","End":"05:30.035","Text":"the elements and discussed some of its features."},{"Start":"05:30.035 ","End":"05:36.810","Text":"In later videos, we\u0027ll discuss the periodic table in much more detail."}],"ID":17657},{"Watched":false,"Name":"Oxidation States","Duration":"8m 38s","ChapterTopicVideoID":16911,"CourseChapterTopicPlaylistID":154543,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16911.jpeg","UploadDate":"2019-02-20T23:38:00.0270000","DurationForVideoObject":"PT8M38S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:05.985","Text":"In this video, we will discuss a useful tool called oxidation states."},{"Start":"00:05.985 ","End":"00:12.465","Text":"These oxidation states are simply useful aids that help us in chemical bookkeeping,"},{"Start":"00:12.465 ","End":"00:15.720","Text":"one shouldn\u0027t try to read too much into them."},{"Start":"00:15.720 ","End":"00:19.785","Text":"In order to make it easy to calculate the oxidation states,"},{"Start":"00:19.785 ","End":"00:21.975","Text":"rules have been devised,"},{"Start":"00:21.975 ","End":"00:25.485","Text":"we\u0027re going to call oxidation states OS,"},{"Start":"00:25.485 ","End":"00:27.165","Text":"just to make it simple."},{"Start":"00:27.165 ","End":"00:33.075","Text":"The first thing we should note is that if there is a conflict between 2 rules,"},{"Start":"00:33.075 ","End":"00:35.595","Text":"the ones that is higher in the list is followed."},{"Start":"00:35.595 ","End":"00:38.400","Text":"For example, we have a conflict retrieved rules 3,"},{"Start":"00:38.400 ","End":"00:42.135","Text":"and 7, we will follow rule 3."},{"Start":"00:42.135 ","End":"00:45.259","Text":"Here is the first rule, if a species,"},{"Start":"00:45.259 ","End":"00:48.290","Text":"a species we mean an isolated atom, a molecule,"},{"Start":"00:48.290 ","End":"00:50.945","Text":"or a formula unit is neutral,"},{"Start":"00:50.945 ","End":"00:52.535","Text":"has no charge,"},{"Start":"00:52.535 ","End":"00:56.615","Text":"the sum of the oxidation states of all the atoms is 0."},{"Start":"00:56.615 ","End":"01:04.250","Text":"In addition, the oxidation states of all the atoms of an element is always 0."},{"Start":"01:04.250 ","End":"01:07.045","Text":"Here are some examples."},{"Start":"01:07.045 ","End":"01:12.565","Text":"In CH_3F, the sum of all the oxidation states of C,"},{"Start":"01:12.565 ","End":"01:14.215","Text":"the 3 hydrogens,"},{"Start":"01:14.215 ","End":"01:17.380","Text":"and the fluorine will be 0."},{"Start":"01:17.380 ","End":"01:21.700","Text":"We\u0027ll use this extensively in what follows."},{"Start":"01:21.700 ","End":"01:25.705","Text":"Now supposing we have free fluorine atoms,"},{"Start":"01:25.705 ","End":"01:29.000","Text":"they will have an oxidation state of 0,"},{"Start":"01:29.000 ","End":"01:32.380","Text":"and if we have fluorine in the gas phase,"},{"Start":"01:32.380 ","End":"01:37.795","Text":"which is F2, each f has an oxidation state of 0."},{"Start":"01:37.795 ","End":"01:42.580","Text":"Here is rule 2, if a species has a charge,"},{"Start":"01:42.580 ","End":"01:44.980","Text":"that\u0027s it\u0027s an ionic species,"},{"Start":"01:44.980 ","End":"01:52.150","Text":"the sum of all the oxidation states of all the atoms is equal to the charge on the ion."},{"Start":"01:52.150 ","End":"01:56.170","Text":"This is best illustrated by some examples."},{"Start":"01:56.170 ","End":"01:59.800","Text":"Here\u0027s a polyatomic ion,"},{"Start":"01:59.800 ","End":"02:04.745","Text":"CO_3^2-, it\u0027s called carbonate,"},{"Start":"02:04.745 ","End":"02:07.590","Text":"and it\u0027s total charge is minus 2."},{"Start":"02:07.590 ","End":"02:10.065","Text":"We write it as 2 minus,"},{"Start":"02:10.065 ","End":"02:12.900","Text":"it just means that the charge is minus 2."},{"Start":"02:12.900 ","End":"02:16.175","Text":"The sum of the oxidation states of the C,"},{"Start":"02:16.175 ","End":"02:19.745","Text":"and the 3 oxygens will be minus 2."},{"Start":"02:19.745 ","End":"02:21.889","Text":"Here is another example."},{"Start":"02:21.889 ","End":"02:23.990","Text":"This time a positive ion,"},{"Start":"02:23.990 ","End":"02:27.515","Text":"in NH_4+, that\u0027s called ammonium,"},{"Start":"02:27.515 ","End":"02:31.700","Text":"is a polyatomic cation because it\u0027s positively charged,"},{"Start":"02:31.700 ","End":"02:33.650","Text":"and the total charge is plus 1,"},{"Start":"02:33.650 ","End":"02:35.390","Text":"that\u0027s a little plus here."},{"Start":"02:35.390 ","End":"02:39.590","Text":"The sum of the oxidation states of the end of nitrogen,"},{"Start":"02:39.590 ","End":"02:43.610","Text":"and the 4 hydrogens will be plus 1 in total."},{"Start":"02:43.610 ","End":"02:45.294","Text":"Here\u0027s the third rule;"},{"Start":"02:45.294 ","End":"02:46.640","Text":"in its compounds,"},{"Start":"02:46.640 ","End":"02:50.210","Text":"the oxidation state of group 1 metals is plus 1,"},{"Start":"02:50.210 ","End":"02:54.155","Text":"the oxidation state of group 2 metals is plus 2."},{"Start":"02:54.155 ","End":"02:56.960","Text":"Here\u0027s an example. Sodium chloride,"},{"Start":"02:56.960 ","End":"02:58.610","Text":"we\u0027ve met this before,"},{"Start":"02:58.610 ","End":"03:05.600","Text":"we remember that it\u0027s an ionic compound composed of Na plus ions,"},{"Start":"03:05.600 ","End":"03:08.360","Text":"and Cl minus ions."},{"Start":"03:08.360 ","End":"03:13.790","Text":"Now, this rule tells us that the oxidation state of Na is plus 1."},{"Start":"03:13.790 ","End":"03:15.980","Text":"We know from previous rules,"},{"Start":"03:15.980 ","End":"03:20.770","Text":"from higher up rules that the total of a neutral species must be 0,"},{"Start":"03:20.770 ","End":"03:23.685","Text":"so that means Cl must be minus 1,"},{"Start":"03:23.685 ","End":"03:28.815","Text":"so that the total of plus 1 and plus-minus 1 will be 0."},{"Start":"03:28.815 ","End":"03:30.930","Text":"In fact, this is a rule,"},{"Start":"03:30.930 ","End":"03:34.160","Text":"the ionic charges are equal to the oxidation states."},{"Start":"03:34.160 ","End":"03:36.140","Text":"Here we see Na has,"},{"Start":"03:36.140 ","End":"03:38.015","Text":"is Na plus 1,"},{"Start":"03:38.015 ","End":"03:41.155","Text":"and the oxidation state of Na is plus 1,"},{"Start":"03:41.155 ","End":"03:44.120","Text":"the ion of Cl is Cl minus,"},{"Start":"03:44.120 ","End":"03:48.320","Text":"and the oxidation state is minus 1, so they\u0027re equal."},{"Start":"03:48.320 ","End":"03:52.250","Text":"The ionic charges are equal to the oxidation states."},{"Start":"03:52.250 ","End":"03:55.580","Text":"Here\u0027s the fourth rule; in its compounds,"},{"Start":"03:55.580 ","End":"03:59.450","Text":"the oxidation state of fluorine is minus 1,"},{"Start":"03:59.450 ","End":"04:01.910","Text":"here\u0027s an example HF."},{"Start":"04:01.910 ","End":"04:03.590","Text":"Now according to the rule,"},{"Start":"04:03.590 ","End":"04:06.815","Text":"the oxidation state of F is minus 1,"},{"Start":"04:06.815 ","End":"04:10.470","Text":"HF, F is minus 1,"},{"Start":"04:10.470 ","End":"04:12.900","Text":"so H must be plus 1."},{"Start":"04:12.900 ","End":"04:14.970","Text":"Because if it is plus 1,"},{"Start":"04:14.970 ","End":"04:16.140","Text":"the sum of plus 1,"},{"Start":"04:16.140 ","End":"04:18.705","Text":"and minus 1 it gives us 0,"},{"Start":"04:18.705 ","End":"04:21.530","Text":"and this is a neutral species."},{"Start":"04:21.530 ","End":"04:23.300","Text":"Here\u0027s the fifth rule."},{"Start":"04:23.300 ","End":"04:27.980","Text":"In its compounds, the oxidation state of hydrogen is plus 1."},{"Start":"04:27.980 ","End":"04:31.085","Text":"Let\u0027s take the example of CH_4,"},{"Start":"04:31.085 ","End":"04:33.785","Text":"CH_4 is methane,"},{"Start":"04:33.785 ","End":"04:35.930","Text":"it\u0027s a gas called methane."},{"Start":"04:35.930 ","End":"04:40.880","Text":"Now the rule tells us that the oxidation state of hydrogen is plus 1,"},{"Start":"04:40.880 ","End":"04:46.590","Text":"so you have CH_4 hydrogen is plus 1 in total,"},{"Start":"04:46.590 ","End":"04:48.480","Text":"that gives us plus 4,"},{"Start":"04:48.480 ","End":"04:51.120","Text":"so carbon must be minus 4."},{"Start":"04:51.120 ","End":"04:53.960","Text":"Now, this row has some exceptions,"},{"Start":"04:53.960 ","End":"04:56.030","Text":"not too many, but some."},{"Start":"04:56.030 ","End":"04:59.705","Text":"For example, lithium hydride."},{"Start":"04:59.705 ","End":"05:01.460","Text":"Now higher up rule,"},{"Start":"05:01.460 ","End":"05:06.860","Text":"rule 3 says that lithium has oxidation state of plus 1,"},{"Start":"05:06.860 ","End":"05:09.830","Text":"so lithium is plus 1,"},{"Start":"05:09.830 ","End":"05:15.860","Text":"so hydrogen must be minus 1 in this case."},{"Start":"05:15.860 ","End":"05:17.660","Text":"Here\u0027s the sixth rule."},{"Start":"05:17.660 ","End":"05:22.460","Text":"In its compounds, the oxidation state of oxygen is minus 2."},{"Start":"05:22.460 ","End":"05:24.800","Text":"Let\u0027s consider CO_2,"},{"Start":"05:24.800 ","End":"05:30.810","Text":"the oxidation state of oxygen is minus 2 according to rule 6."},{"Start":"05:31.400 ","End":"05:36.105","Text":"CO_2, oxygen is minus 2,"},{"Start":"05:36.105 ","End":"05:39.000","Text":"so the total is 2 times minus 2,"},{"Start":"05:39.000 ","End":"05:40.515","Text":"that\u0027s minus 4,"},{"Start":"05:40.515 ","End":"05:44.040","Text":"so carbon must be plus 4,"},{"Start":"05:44.040 ","End":"05:46.835","Text":"so the total is 0."},{"Start":"05:46.835 ","End":"05:53.715","Text":"Now, this rule has an exception, let\u0027s consider H_2O_2."},{"Start":"05:53.715 ","End":"06:00.485","Text":"Rule 5 said that H must have an oxidation state of plus 1."},{"Start":"06:00.485 ","End":"06:06.615","Text":"H_2O_2 hydrogen is plus 1,"},{"Start":"06:06.615 ","End":"06:09.030","Text":"oxygen must be minus 1."},{"Start":"06:09.030 ","End":"06:13.370","Text":"The seventh rule is group 17 atoms of"},{"Start":"06:13.370 ","End":"06:17.314","Text":"oxidation state of minus 1 when combined with the metal;"},{"Start":"06:17.314 ","End":"06:21.140","Text":"group 16 have oxidation state minus 2,"},{"Start":"06:21.140 ","End":"06:25.640","Text":"and Group 15 have oxidation states minus 3."},{"Start":"06:25.640 ","End":"06:27.500","Text":"Here are examples."},{"Start":"06:27.500 ","End":"06:29.345","Text":"Lithium fluoride,"},{"Start":"06:29.345 ","End":"06:31.265","Text":"lithium is a metal."},{"Start":"06:31.265 ","End":"06:38.965","Text":"Fluoride has an oxidation state of minus 1 because it\u0027s in group 17,"},{"Start":"06:38.965 ","End":"06:41.475","Text":"so that fits fine."},{"Start":"06:41.475 ","End":"06:42.945","Text":"Lithium is plus 1,"},{"Start":"06:42.945 ","End":"06:45.210","Text":"fluorine is minus 1,"},{"Start":"06:45.210 ","End":"06:50.745","Text":"oxygen is minus 2 when it\u0027s combined with the metal."},{"Start":"06:50.745 ","End":"06:53.940","Text":"Lithium 2 is plus 2,"},{"Start":"06:53.940 ","End":"06:56.010","Text":"so each lithium is plus 1,"},{"Start":"06:56.010 ","End":"06:58.930","Text":"and has minus 3."},{"Start":"06:58.970 ","End":"07:02.745","Text":"Lithium 3 has to have plus 3,"},{"Start":"07:02.745 ","End":"07:05.760","Text":"that means that each lithium is plus 1,"},{"Start":"07:05.760 ","End":"07:08.825","Text":"so that all fits together nicely."},{"Start":"07:08.825 ","End":"07:10.655","Text":"Now, up to now,"},{"Start":"07:10.655 ","End":"07:14.285","Text":"all the oxidation states have been whole numbers,"},{"Start":"07:14.285 ","End":"07:17.085","Text":"but it\u0027s not always the case,"},{"Start":"07:17.085 ","End":"07:18.855","Text":"let\u0027s take an example."},{"Start":"07:18.855 ","End":"07:22.005","Text":"The example is Fe_3O_4,"},{"Start":"07:22.005 ","End":"07:25.710","Text":"this is the iron ore that is mined."},{"Start":"07:25.710 ","End":"07:30.620","Text":"Now we know that the oxidation state of oxygen is minus 2,"},{"Start":"07:30.620 ","End":"07:35.990","Text":"let\u0027s call the oxidation state of ion of Fe, x."},{"Start":"07:35.990 ","End":"07:43.205","Text":"We can write a simple equation because we know that the total oxidation states must be 0."},{"Start":"07:43.205 ","End":"07:45.830","Text":"We can write 3x"},{"Start":"07:45.830 ","End":"07:54.900","Text":"plus 4 times minus 2 equals 0,"},{"Start":"07:54.920 ","End":"08:03.765","Text":"and from that we can easily deduce that x is equal to 8/3."},{"Start":"08:03.765 ","End":"08:06.705","Text":"Here it\u0027s not a whole number,"},{"Start":"08:06.705 ","End":"08:11.980","Text":"the oxidation state is plus 8 /3,"},{"Start":"08:11.980 ","End":"08:13.655","Text":"not a whole number."},{"Start":"08:13.655 ","End":"08:15.815","Text":"There aren\u0027t too many cases like that,"},{"Start":"08:15.815 ","End":"08:18.125","Text":"but occasionally it occurs."},{"Start":"08:18.125 ","End":"08:23.180","Text":"In this video, we\u0027ve learned how to calculate the oxidation states."},{"Start":"08:23.180 ","End":"08:25.670","Text":"We\u0027re going to use this later on,"},{"Start":"08:25.670 ","End":"08:32.285","Text":"especially when we deal with a reaction called a redox reaction,"},{"Start":"08:32.285 ","End":"08:38.190","Text":"that\u0027s reduction and oxidation, redox reactions."}],"ID":17658},{"Watched":false,"Name":"Exercise 1","Duration":"11m 50s","ChapterTopicVideoID":22851,"CourseChapterTopicPlaylistID":154543,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22851.jpeg","UploadDate":"2020-12-14T07:17:01.1730000","DurationForVideoObject":"PT11M50S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.895","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.895 ","End":"00:07.575","Text":"Indicate the oxidation state of the underlying atom in the following molecules."},{"Start":"00:07.575 ","End":"00:08.910","Text":"Now, before we begin with a,"},{"Start":"00:08.910 ","End":"00:14.910","Text":"we\u0027re going to go over rules for working out the oxidation states of atoms."},{"Start":"00:14.910 ","End":"00:17.865","Text":"Now, if there is a conflict between 2 rules,"},{"Start":"00:17.865 ","End":"00:21.915","Text":"then the one that comes first in the list is the one that is followed."},{"Start":"00:21.915 ","End":"00:24.990","Text":"Now, the first rule states that if we have a neutral species,"},{"Start":"00:24.990 ","End":"00:27.480","Text":"meaning an atom, a molecule,"},{"Start":"00:27.480 ","End":"00:29.590","Text":"or a formula unit."},{"Start":"00:30.140 ","End":"00:36.490","Text":"Now, the oxidation states of all the atoms of an element is always 0."},{"Start":"00:43.430 ","End":"00:50.000","Text":"Now, the second rule states that if the species has a charge species,"},{"Start":"00:50.000 ","End":"00:52.080","Text":"meaning an ion,"},{"Start":"00:58.570 ","End":"01:00.020","Text":"the sum"},{"Start":"01:00.020 ","End":"01:04.145","Text":"of all the oxidation states of all the atoms is equal to the charge of the ion."},{"Start":"01:04.145 ","End":"01:12.480","Text":"The sum of the oxidation states of all the atoms equals the charge of the ion."},{"Start":"01:12.880 ","End":"01:15.560","Text":"Next is Number 3."},{"Start":"01:15.560 ","End":"01:17.960","Text":"Rule Number 3 states that in its compounds,"},{"Start":"01:17.960 ","End":"01:22.950","Text":"the oxidation state of the Group 1 metals is plus 1,"},{"Start":"01:30.200 ","End":"01:39.100","Text":"and the oxidation state of Group 2 metals is plus 2."},{"Start":"01:41.280 ","End":"01:44.665","Text":"Now we\u0027re going to look at the fourth rule."},{"Start":"01:44.665 ","End":"01:47.230","Text":"The fourth rule states that in its compounds,"},{"Start":"01:47.230 ","End":"01:52.790","Text":"the oxidation states of fluorine equals minus 1."},{"Start":"01:52.830 ","End":"01:56.500","Text":"The fifth rule states that in its compounds,"},{"Start":"01:56.500 ","End":"02:01.880","Text":"the oxidation state of hydrogen equals plus 1."},{"Start":"02:02.190 ","End":"02:06.100","Text":"However, we will see that there are exceptions to this rule."},{"Start":"02:06.100 ","End":"02:08.560","Text":"Number 6, in its compounds,"},{"Start":"02:08.560 ","End":"02:15.075","Text":"the oxidation state of oxygen equals minus 2."},{"Start":"02:15.075 ","End":"02:19.070","Text":"Again, we will see that there are exceptions to this rule."},{"Start":"02:19.310 ","End":"02:23.170","Text":"Now, the seventh rule states,"},{"Start":"02:23.630 ","End":"02:28.980","Text":"Group 17 atoms have an oxidation state of minus 1."},{"Start":"02:28.980 ","End":"02:35.790","Text":"Oxidation state of Group 17 atoms"},{"Start":"02:35.790 ","End":"02:41.685","Text":"have an oxidation state of minus 1 when combined with a metal."},{"Start":"02:41.685 ","End":"02:50.500","Text":"Group 16 have an oxidation state of minus 2,"},{"Start":"02:50.500 ","End":"02:54.980","Text":"and Group 15 have an oxidation state of minus 3."},{"Start":"02:56.720 ","End":"02:59.680","Text":"Now, one important thing to note is that"},{"Start":"02:59.680 ","End":"03:02.020","Text":"the oxidation state is not always a whole number."},{"Start":"03:02.020 ","End":"03:04.990","Text":"Now, we\u0027re going to start with a."},{"Start":"03:04.990 ","End":"03:09.200","Text":"In a, we have the dichromate anion."},{"Start":"03:12.780 ","End":"03:18.370","Text":"We want to know the oxidation state of chromium atom."},{"Start":"03:18.370 ","End":"03:21.550","Text":"If we look at our rules, I want to remind you,"},{"Start":"03:21.550 ","End":"03:25.120","Text":"first of all, we see that here we have a charged species."},{"Start":"03:25.120 ","End":"03:28.134","Text":"We know that when we have a charged species,"},{"Start":"03:28.134 ","End":"03:31.450","Text":"just reminding you that rule Number 2 states,"},{"Start":"03:31.450 ","End":"03:33.580","Text":"that the sum of the oxidation states of"},{"Start":"03:33.580 ","End":"03:38.735","Text":"all the atoms in a charged species is equal to the charge of the ion."},{"Start":"03:38.735 ","End":"03:48.830","Text":"The sum of the oxidation states of the atoms equals the charge on the ion."},{"Start":"03:48.830 ","End":"03:52.170","Text":"We\u0027re going to use this in a second."},{"Start":"03:53.530 ","End":"03:56.855","Text":"Now, that\u0027s the first rule we\u0027re going to use."},{"Start":"03:56.855 ","End":"04:01.370","Text":"The second rule we\u0027re going to use is rule Number 6,"},{"Start":"04:01.370 ","End":"04:03.560","Text":"which states that in its compounds,"},{"Start":"04:03.560 ","End":"04:07.970","Text":"the oxidation state of the oxygen equals minus 2."},{"Start":"04:07.970 ","End":"04:09.470","Text":"Now, there are exceptions to this rule."},{"Start":"04:09.470 ","End":"04:13.695","Text":"However, this is not one of the exceptions."},{"Start":"04:13.695 ","End":"04:16.860","Text":"We know that we have 2 chromium atoms,"},{"Start":"04:16.860 ","End":"04:20.665","Text":"so we have 2 times the oxidation state of chromium."},{"Start":"04:20.665 ","End":"04:23.074","Text":"We\u0027re just going to write it like this,"},{"Start":"04:23.074 ","End":"04:26.190","Text":"plus 7 times the oxidation state of the oxygen."},{"Start":"04:26.190 ","End":"04:30.270","Text":"We know that the oxidation state of oxygen is minus 2."},{"Start":"04:30.270 ","End":"04:33.140","Text":"We\u0027re going to write in the beginning oxidation state of oxygen,"},{"Start":"04:33.140 ","End":"04:36.160","Text":"and this equals minus 2,"},{"Start":"04:36.160 ","End":"04:42.340","Text":"since the sum of the oxidation states of all the atoms equals the charge of the ion."},{"Start":"04:42.340 ","End":"04:50.720","Text":"Again, 2 times oxidation state of chromium plus 7 times the oxidation state of oxygen,"},{"Start":"04:50.720 ","End":"04:54.080","Text":"which equals minus 2, equals minus."},{"Start":"04:54.080 ","End":"05:02.405","Text":"Two times the oxidation state of chromium equals 12."},{"Start":"05:02.405 ","End":"05:10.345","Text":"Therefore, the oxidation state of chromium equals 6 or plus 6, is the same."},{"Start":"05:10.345 ","End":"05:17.735","Text":"In a, we found that the oxidation state of chromium in the dichromate anion,"},{"Start":"05:17.735 ","End":"05:21.780","Text":"equals plus 6, so that\u0027s a."},{"Start":"05:21.780 ","End":"05:24.015","Text":"Now, we\u0027re going to go on to b."},{"Start":"05:24.015 ","End":"05:26.490","Text":"If we take a look at b,"},{"Start":"05:26.490 ","End":"05:30.190","Text":"we have the tripotassium phosphate."},{"Start":"05:37.160 ","End":"05:43.170","Text":"We want to find the oxidation state of the phosphorus because that\u0027s the underlined atom."},{"Start":"05:43.360 ","End":"05:46.310","Text":"We\u0027re going to look at a number of rules again."},{"Start":"05:46.310 ","End":"05:49.550","Text":"The first rule reminding you states that,"},{"Start":"05:49.550 ","End":"05:51.425","Text":"if a species is neutral,"},{"Start":"05:51.425 ","End":"05:54.050","Text":"like in this case it\u0027s a neutral species,"},{"Start":"05:54.050 ","End":"06:04.440","Text":"the sum of the oxidation states of all the atoms equals 0."},{"Start":"06:05.050 ","End":"06:09.260","Text":"Another rule we\u0027re going to use here is rule Number 3,"},{"Start":"06:09.260 ","End":"06:11.120","Text":"which states that in its compounds,"},{"Start":"06:11.120 ","End":"06:16.615","Text":"the oxidation state of Group 1 metals is plus 1."},{"Start":"06:16.615 ","End":"06:28.770","Text":"We\u0027re"},{"Start":"06:28.770 ","End":"06:29.100","Text":"talking"},{"Start":"06:29.100 ","End":"06:31.330","Text":"about the potassium here."},{"Start":"06:31.370 ","End":"06:36.995","Text":"Another rule we\u0027re going to use is the same rule we used in a,"},{"Start":"06:36.995 ","End":"06:40.170","Text":"which is rule Number 6,"},{"Start":"06:40.180 ","End":"06:43.130","Text":"which states that in its compounds,"},{"Start":"06:43.130 ","End":"06:49.740","Text":"the oxidation state of oxygen is minus 2."},{"Start":"06:51.970 ","End":"07:02.240","Text":"We have 3 times the oxidation state of potassium plus the oxidation state of phosphorus,"},{"Start":"07:02.240 ","End":"07:04.160","Text":"since we have only 1 plus,"},{"Start":"07:04.160 ","End":"07:12.305","Text":"4 times the oxidation state of oxygen equals 0."},{"Start":"07:12.305 ","End":"07:18.320","Text":"Since the sum of the oxidation states of atoms of a neutral species equals 0."},{"Start":"07:18.320 ","End":"07:20.120","Text":"Now, we\u0027re going to fill these in,"},{"Start":"07:20.120 ","End":"07:22.550","Text":"so 3 times oxidation state of potassium,"},{"Start":"07:22.550 ","End":"07:29.330","Text":"we said that that is plus 1 plus the oxidation state of phosphorus,"},{"Start":"07:29.330 ","End":"07:31.327","Text":"which was what we\u0027re looking for,"},{"Start":"07:31.327 ","End":"07:34.550","Text":"plus 4 times the oxidation state of oxygen,"},{"Start":"07:34.550 ","End":"07:39.480","Text":"which equals minus 2 in this case, equals 0."},{"Start":"07:40.340 ","End":"07:43.940","Text":"The oxidation state of phosphorus, in this case,"},{"Start":"07:43.940 ","End":"07:48.330","Text":"comes to plus 5."},{"Start":"07:48.330 ","End":"07:53.030","Text":"We found that in tripotassium phosphates,"},{"Start":"07:53.030 ","End":"07:57.190","Text":"the oxidation state of the phosphorous equals plus 5,"},{"Start":"07:57.190 ","End":"07:59.925","Text":"so that is our answer for b."},{"Start":"07:59.925 ","End":"08:06.310","Text":"Now, we\u0027re going to go on to c. In c we have the sulfate ion."},{"Start":"08:07.150 ","End":"08:11.020","Text":"Now again, we\u0027re going to use rule Number 2,"},{"Start":"08:11.020 ","End":"08:13.340","Text":"since we have an ion,"},{"Start":"08:13.340 ","End":"08:16.500","Text":"which states that in a charged species,"},{"Start":"08:19.100 ","End":"08:25.800","Text":"the sum of all the oxidation states equals the charge of the ion."},{"Start":"08:28.970 ","End":"08:32.510","Text":"We\u0027re going to use rule Number 6 again,"},{"Start":"08:32.510 ","End":"08:34.700","Text":"which states that in its compounds,"},{"Start":"08:34.700 ","End":"08:40.525","Text":"the oxidation state of the oxygen is usually minus 2,"},{"Start":"08:40.525 ","End":"08:42.500","Text":"except for a number of exceptions."},{"Start":"08:42.500 ","End":"08:45.785","Text":"If we look at the sulfate anion,"},{"Start":"08:45.785 ","End":"08:50.090","Text":"we have the oxidation state of the sulfur plus"},{"Start":"08:50.090 ","End":"08:53.315","Text":"4 times the oxidation state of oxygen"},{"Start":"08:53.315 ","End":"08:58.415","Text":"equals minus 2 because it equals the charge of the ion."},{"Start":"08:58.415 ","End":"09:01.870","Text":"Again, the oxidation state of the sulfur plus 4."},{"Start":"09:01.870 ","End":"09:06.390","Text":"Now, we said that the oxidation state of oxygen equals minus 2,"},{"Start":"09:07.840 ","End":"09:17.615","Text":"so the oxidation state of sulfur comes to plus 6 in our case."},{"Start":"09:17.615 ","End":"09:20.210","Text":"For the sulfate anion,"},{"Start":"09:20.210 ","End":"09:24.400","Text":"the oxidation state of the sulfur equals plus 6,"},{"Start":"09:24.400 ","End":"09:26.780","Text":"and that is our answer for c. Now,"},{"Start":"09:26.780 ","End":"09:29.820","Text":"we\u0027re going to go on to d. In d,"},{"Start":"09:29.820 ","End":"09:32.560","Text":"we have calcium hydride,"},{"Start":"09:34.910 ","End":"09:39.980","Text":"and we want to find the oxidation state of the hydrogen."},{"Start":"09:39.980 ","End":"09:42.290","Text":"Again, we\u0027re going to look at our number of rules."},{"Start":"09:42.290 ","End":"09:44.345","Text":"Again, we have a neutral species."},{"Start":"09:44.345 ","End":"09:51.430","Text":"In a neutral species, the sum of the oxidation states of the atoms equals 0."},{"Start":"09:51.430 ","End":"09:55.725","Text":"Next, we\u0027re going to look at rule Number 3,"},{"Start":"09:55.725 ","End":"09:58.860","Text":"which states that in its compounds,"},{"Start":"09:58.860 ","End":"10:02.760","Text":"the oxidation state of Group 2 metals is plus 2."},{"Start":"10:02.760 ","End":"10:12.710","Text":"We\u0027re"},{"Start":"10:12.710 ","End":"10:13.760","Text":"talking about calcium"},{"Start":"10:13.760 ","End":"10:16.175","Text":"here because that\u0027s a Group 2 metal."},{"Start":"10:16.175 ","End":"10:20.780","Text":"Begin. The oxidation state of calcium plus"},{"Start":"10:20.780 ","End":"10:26.495","Text":"2 times the oxidation state of hydrogen equals 0."},{"Start":"10:26.495 ","End":"10:31.005","Text":"Now again, we know the oxidation state of calcium from rule Number 3, it\u0027s plus 2."},{"Start":"10:31.005 ","End":"10:39.885","Text":"Plus 2 or 2 plus 2 times the oxidation state of hydrogen equals 0."},{"Start":"10:39.885 ","End":"10:44.600","Text":"Two times oxidation state of hydrogen equals minus 2,"},{"Start":"10:44.600 ","End":"10:50.040","Text":"and oxidation state of hydrogen equals minus 1."},{"Start":"10:51.440 ","End":"10:54.015","Text":"Now, just to note,"},{"Start":"10:54.015 ","End":"10:59.090","Text":"rule Number 5 states that in its compounds,"},{"Start":"10:59.090 ","End":"11:01.700","Text":"the oxidation state of hydrogen is plus 1."},{"Start":"11:01.700 ","End":"11:07.050","Text":"However, there are exceptions to this rule. Now remember"},{"Start":"11:19.490 ","End":"11:22.160","Text":"in the beginning, when we went over the rules,"},{"Start":"11:22.160 ","End":"11:25.190","Text":"we said if there is a conflict between 2 rules,"},{"Start":"11:25.190 ","End":"11:28.775","Text":"the first one is the one that is followed."},{"Start":"11:28.775 ","End":"11:33.200","Text":"In our case, rule Number 3 comes before rule Number 5,"},{"Start":"11:33.200 ","End":"11:36.240","Text":"and therefore, we follow rule Number 3."},{"Start":"11:38.350 ","End":"11:46.080","Text":"The oxidation state of hydrogen in calcium hydride equals minus 1."},{"Start":"11:46.080 ","End":"11:49.450","Text":"Now, we\u0027re going to go on d."}],"ID":23671},{"Watched":false,"Name":"Exercise 2","Duration":"5m 52s","ChapterTopicVideoID":22852,"CourseChapterTopicPlaylistID":154543,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22852.jpeg","UploadDate":"2020-12-14T07:17:58.5770000","DurationForVideoObject":"PT5M52S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:02.730","Text":"Now we\u0027re going to go on with"},{"Start":"00:02.730 ","End":"00:10.870","Text":"e. You would have nitrogen dioxide and we want to find the oxidation state of nitrogen."},{"Start":"00:14.270 ","End":"00:16.680","Text":"Again, we\u0027re going to use rule number 1,"},{"Start":"00:16.680 ","End":"00:20.060","Text":"which states that if a species is neutral,"},{"Start":"00:20.060 ","End":"00:23.720","Text":"the sum of the oxidation states of all the atoms is 0,"},{"Start":"00:23.720 ","End":"00:27.845","Text":"so the sum of the oxidation states of all the items equals 0, in this case."},{"Start":"00:27.845 ","End":"00:32.270","Text":"Now, we\u0027re also going to take a look again at rule number 6,"},{"Start":"00:32.270 ","End":"00:38.830","Text":"which states that the oxidation state of oxygen in most of its compounds is minus 2."},{"Start":"00:38.830 ","End":"00:41.660","Text":"Now again, there are exceptions to this rule."},{"Start":"00:41.660 ","End":"00:44.600","Text":"If we take a look at nitrogen dioxide,"},{"Start":"00:44.600 ","End":"00:46.550","Text":"we have the oxidation state of nitrogen,"},{"Start":"00:46.550 ","End":"00:49.755","Text":"which we want to find out plus 2,"},{"Start":"00:49.755 ","End":"00:58.620","Text":"since we have 2 oxygen atoms times the oxidation state of oxygen equals 0,"},{"Start":"00:58.620 ","End":"01:01.035","Text":"since it\u0027s a neutral species."},{"Start":"01:01.035 ","End":"01:04.130","Text":"The oxidation state of nitrogen plus 2 times minus"},{"Start":"01:04.130 ","End":"01:09.870","Text":"2 which is the oxidation state of oxygen in this case, equals 0."},{"Start":"01:09.870 ","End":"01:19.689","Text":"The oxidation state of nitrogen equals plus 4 in nitrogen dioxide."},{"Start":"01:19.689 ","End":"01:25.845","Text":"That is our answer for e. The oxidation state of nitrogen equals plus 4."},{"Start":"01:25.845 ","End":"01:33.430","Text":"Now we\u0027re going to go on to f. In f we have the nitrogen."},{"Start":"01:36.670 ","End":"01:43.070","Text":"Again, we use the first rule which states that the neutral species,"},{"Start":"01:43.070 ","End":"01:51.480","Text":"the sum of the oxidation states equals 0 and that is all we need,"},{"Start":"01:51.480 ","End":"01:56.845","Text":"so 2 times oxidation state of nitrogen equals 0,"},{"Start":"01:56.845 ","End":"02:02.720","Text":"meaning the oxidation state of nitrogen equals 0, in this case."},{"Start":"02:02.720 ","End":"02:08.520","Text":"That is our answer for f. Now we\u0027re going on to g,"},{"Start":"02:08.520 ","End":"02:11.200","Text":"and g we have potassium chloride."},{"Start":"02:12.380 ","End":"02:16.550","Text":"We want to find the oxidation state of the chlorine."},{"Start":"02:16.590 ","End":"02:19.120","Text":"Again, this is a neutral species,"},{"Start":"02:19.120 ","End":"02:22.225","Text":"so we\u0027re going to use rule number 1 in a neutral species."},{"Start":"02:22.225 ","End":"02:30.595","Text":"The sum of the oxidation states of the atoms equals 0."},{"Start":"02:30.595 ","End":"02:36.040","Text":"Another rule we\u0027re going to use that we used before is rule number 3,"},{"Start":"02:36.040 ","End":"02:39.010","Text":"which states that in its compounds,"},{"Start":"02:39.010 ","End":"02:49.305","Text":"the oxidation state of Group 1 metals equals plus 1."},{"Start":"02:49.305 ","End":"02:51.190","Text":"If you use these 2 rules,"},{"Start":"02:51.190 ","End":"02:55.180","Text":"we know that the oxidation state of"},{"Start":"02:55.180 ","End":"03:00.280","Text":"potassium plus the oxidation state of chlorine equals 0,"},{"Start":"03:00.280 ","End":"03:03.844","Text":"and the oxidation state of the potassium equals plus 1,"},{"Start":"03:03.844 ","End":"03:10.614","Text":"so plus 1 plus the oxidation state of chlorine equals 0,"},{"Start":"03:10.614 ","End":"03:14.070","Text":"plus 1 is the same to say 1."},{"Start":"03:14.070 ","End":"03:17.000","Text":"So 1 plus the oxidation state of chlorine equals 0,"},{"Start":"03:17.000 ","End":"03:19.385","Text":"so the oxidation state of chlorine,"},{"Start":"03:19.385 ","End":"03:22.530","Text":"in our case, comes to minus 1."},{"Start":"03:22.670 ","End":"03:28.700","Text":"The oxidation state of chlorine in potassium chloride equals minus 1."},{"Start":"03:28.700 ","End":"03:31.520","Text":"Now if you look at the rules,"},{"Start":"03:31.520 ","End":"03:34.700","Text":"then we stated that in rule number 7,"},{"Start":"03:34.700 ","End":"03:43.190","Text":"we have the Group 17 atoms have"},{"Start":"03:43.190 ","End":"03:52.760","Text":"an oxidation state of minus 1 when they are combined with the metal."},{"Start":"03:52.760 ","End":"03:54.560","Text":"Here, we see that this is true."},{"Start":"03:54.560 ","End":"04:01.710","Text":"However, we didn\u0027t need number 7 for working out the oxidation state of the chlorine."},{"Start":"04:01.750 ","End":"04:06.210","Text":"Now we\u0027re going to go onto h. In h,"},{"Start":"04:06.210 ","End":"04:08.560","Text":"we have nitric acid,"},{"Start":"04:09.020 ","End":"04:14.765","Text":"and again, we\u0027re going to find the oxidation state of the nitrogen."},{"Start":"04:14.765 ","End":"04:23.040","Text":"Again, we\u0027re going to use the number 1 rule since we do have a neutral species."},{"Start":"04:26.360 ","End":"04:31.115","Text":"Again, the sum of the oxidation states of all the atoms equals 0."},{"Start":"04:31.115 ","End":"04:33.890","Text":"The second rule we\u0027re going to use is rule number 5,"},{"Start":"04:33.890 ","End":"04:36.080","Text":"which states that in its compounds,"},{"Start":"04:36.080 ","End":"04:40.295","Text":"the oxidation state of hydrogen equals plus 1."},{"Start":"04:40.295 ","End":"04:45.755","Text":"There are a number of exceptions to this rule that we saw one of the exceptions before."},{"Start":"04:45.755 ","End":"04:49.325","Text":"The next rule we\u0027re going to use is rule number 6,"},{"Start":"04:49.325 ","End":"04:51.350","Text":"which states that in its compounds,"},{"Start":"04:51.350 ","End":"04:54.200","Text":"the oxidation state of oxygen equals minus 2."},{"Start":"04:54.200 ","End":"04:57.330","Text":"Now again, there are exceptions also to this rule."},{"Start":"04:57.820 ","End":"05:05.903","Text":"We have the oxidation state of hydrogen plus the oxidation state of nitrogen plus"},{"Start":"05:05.903 ","End":"05:13.290","Text":"3 times the oxidation state of oxygen equals 0 since we have a neutral species."},{"Start":"05:13.290 ","End":"05:14.870","Text":"Now the oxidation state of hydrogen,"},{"Start":"05:14.870 ","End":"05:16.055","Text":"we said is plus 1,"},{"Start":"05:16.055 ","End":"05:20.330","Text":"so we have 1 plus oxidation state of nitrogen,"},{"Start":"05:20.330 ","End":"05:22.490","Text":"which is what we\u0027re working out."},{"Start":"05:22.490 ","End":"05:25.250","Text":"Plus 3 times the oxidation state of oxygen,"},{"Start":"05:25.250 ","End":"05:30.515","Text":"which equals minus 2 in our case, equals 0."},{"Start":"05:30.515 ","End":"05:38.105","Text":"The oxidation state of nitrogen equals plus 5 in this case."},{"Start":"05:38.105 ","End":"05:39.800","Text":"Again, in nitric acid,"},{"Start":"05:39.800 ","End":"05:44.240","Text":"the oxidation state of nitrogen equals plus 5,"},{"Start":"05:44.240 ","End":"05:50.120","Text":"and that is our answer for h. These are our final answers."},{"Start":"05:50.120 ","End":"05:52.770","Text":"Thank you very much for watching."}],"ID":23672}],"Thumbnail":null,"ID":154543},{"Name":"Types Of Chemical Compounds","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Molecular Compounds","Duration":"9m 3s","ChapterTopicVideoID":16912,"CourseChapterTopicPlaylistID":154544,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16912.jpeg","UploadDate":"2019-02-20T23:38:21.5570000","DurationForVideoObject":"PT9M3S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.650","Text":"In previous videos, we said that compounds are composed of molecules,"},{"Start":"00:04.650 ","End":"00:06.899","Text":"and elements are composed of atoms."},{"Start":"00:06.899 ","End":"00:10.320","Text":"In this video, we will discuss the chemical formulas used to"},{"Start":"00:10.320 ","End":"00:14.010","Text":"represent compounds and some elements."},{"Start":"00:14.010 ","End":"00:17.340","Text":"The simplest formulas tell us which atoms are"},{"Start":"00:17.340 ","End":"00:21.030","Text":"present and the relative numbers of each element."},{"Start":"00:21.030 ","End":"00:22.889","Text":"Let\u0027s take an example."},{"Start":"00:22.889 ","End":"00:26.040","Text":"The example we\u0027re going to discuss is butane."},{"Start":"00:26.040 ","End":"00:27.975","Text":"In each butane molecule,"},{"Start":"00:27.975 ","End":"00:31.995","Text":"there are 4 carbon atoms and 10 hydrogen atoms."},{"Start":"00:31.995 ","End":"00:38.505","Text":"The simplest way of writing this C_4H_10."},{"Start":"00:38.505 ","End":"00:41.955","Text":"We call this a molecular formula."},{"Start":"00:41.955 ","End":"00:45.845","Text":"Because it tells us that in each molecule of butane,"},{"Start":"00:45.845 ","End":"00:51.920","Text":"there are 4 carbons 4C, and 10 hydrogens."},{"Start":"00:51.920 ","End":"00:55.025","Text":"Now often when we perform an experiment,"},{"Start":"00:55.025 ","End":"00:56.675","Text":"all we can find out,"},{"Start":"00:56.675 ","End":"00:59.465","Text":"is the ratio of carbon to hydrogen."},{"Start":"00:59.465 ","End":"01:07.475","Text":"The ratio of carbon to hydrogen is 2 carbons to every 5 hydrogens."},{"Start":"01:07.475 ","End":"01:10.460","Text":"So we write C_2H_5,"},{"Start":"01:10.460 ","End":"01:13.790","Text":"so the ratio is 2 to 5."},{"Start":"01:13.790 ","End":"01:17.720","Text":"We wouldn\u0027t be able to find out from a simple experiment where there was"},{"Start":"01:17.720 ","End":"01:24.460","Text":"C_2H_5 or twice as much C_4H_10 which really is."},{"Start":"01:24.460 ","End":"01:27.890","Text":"This is the best we can do, the simple ratio."},{"Start":"01:27.890 ","End":"01:30.469","Text":"It\u0027s called the empirical formula."},{"Start":"01:30.469 ","End":"01:38.225","Text":"Now sometimes we want more information than just the simple molecular formula."},{"Start":"01:38.225 ","End":"01:41.735","Text":"This is called the condensed structural formula."},{"Start":"01:41.735 ","End":"01:45.320","Text":"Condensed because it\u0027s a shortened version of"},{"Start":"01:45.320 ","End":"01:47.774","Text":"the full structural formula"},{"Start":"01:47.774 ","End":"01:53.080","Text":"and structural because it gives us information about the structure."},{"Start":"01:53.080 ","End":"01:55.610","Text":"What information do we get from this?"},{"Start":"01:55.610 ","End":"02:01.595","Text":"We see the first carbon is associated with 3 hydrogens as the last one."},{"Start":"02:01.595 ","End":"02:04.475","Text":"The intermediate ones, the middle ones,"},{"Start":"02:04.475 ","End":"02:07.160","Text":"are associated with only 2 hydrogens."},{"Start":"02:07.160 ","End":"02:10.640","Text":"Each carbon is associated with only 2 hydrogens."},{"Start":"02:10.640 ","End":"02:14.375","Text":"But it doesn\u0027t give us any more information than that."},{"Start":"02:14.375 ","End":"02:17.705","Text":"Now we can draw the full structural formula,"},{"Start":"02:17.705 ","End":"02:20.360","Text":"which gives us much more information."},{"Start":"02:20.360 ","End":"02:24.245","Text":"In the plane of the page, we have H,"},{"Start":"02:24.245 ","End":"02:30.440","Text":"C, C, C, C, H, like zigzag."},{"Start":"02:30.440 ","End":"02:33.140","Text":"That\u0027s the plane of the molecule."},{"Start":"02:33.140 ","End":"02:37.955","Text":"If we look at the first carbon in addition to the hydrogen that\u0027s in the plane,"},{"Start":"02:37.955 ","End":"02:40.760","Text":"there\u0027s hydrogen that comes out of the plane."},{"Start":"02:40.760 ","End":"02:44.570","Text":"That\u0027s this symbol tells us that comes out of the plane,"},{"Start":"02:44.570 ","End":"02:47.120","Text":"and hydrogen going into the plane that\u0027s"},{"Start":"02:47.120 ","End":"02:50.765","Text":"the symbol that tells us hydrogen going into the plane."},{"Start":"02:50.765 ","End":"02:56.000","Text":"We have, this is out of the plane and this is into the plane."},{"Start":"02:56.000 ","End":"02:59.104","Text":"Again the same, the next carbon."},{"Start":"02:59.104 ","End":"03:02.255","Text":"Then we have the C in the last carbon,"},{"Start":"03:02.255 ","End":"03:05.210","Text":"1 in the plane, 1 going out to the plane,"},{"Start":"03:05.210 ","End":"03:06.740","Text":"1 going into the plane."},{"Start":"03:06.740 ","End":"03:11.570","Text":"In the 2 intermediate carbons or 2 middle carbons are only 2 hydrogens;"},{"Start":"03:11.570 ","End":"03:12.980","Text":"1 coming out of the plane,"},{"Start":"03:12.980 ","End":"03:14.585","Text":"1 going into the plane."},{"Start":"03:14.585 ","End":"03:16.835","Text":"Often when we study chemistry,"},{"Start":"03:16.835 ","End":"03:22.925","Text":"we can use a set of models called ball-and-stick models."},{"Start":"03:22.925 ","End":"03:25.865","Text":"Here\u0027s butane in a ball-and-stick model."},{"Start":"03:25.865 ","End":"03:31.685","Text":"The carbons are gray and the hydrogens are white."},{"Start":"03:31.685 ","End":"03:35.870","Text":"This gives us perhaps more indication of how the molecule actually looks."},{"Start":"03:35.870 ","End":"03:39.140","Text":"The advantage of such a model is that we can turn"},{"Start":"03:39.140 ","End":"03:42.770","Text":"it around and look at it from all different directions."},{"Start":"03:42.770 ","End":"03:46.055","Text":"Another possibility is to use a space-filling model,"},{"Start":"03:46.055 ","End":"03:49.670","Text":"which are calculated using computer simulations."},{"Start":"03:49.670 ","End":"03:53.855","Text":"We met such a model where we discussed the structure of water."},{"Start":"03:53.855 ","End":"03:57.560","Text":"What we had there was a large bowl, red,"},{"Start":"03:57.560 ","End":"04:01.415","Text":"large red ball into the cutting oxygen,"},{"Start":"04:01.415 ","End":"04:08.600","Text":"and 2 smaller hydrogens touching the large red ball."},{"Start":"04:08.600 ","End":"04:13.510","Text":"These are hydrogens and this is oxygen."},{"Start":"04:13.540 ","End":"04:16.835","Text":"This is called a space-filling model."},{"Start":"04:16.835 ","End":"04:18.875","Text":"Let\u0027s discuss another example."},{"Start":"04:18.875 ","End":"04:21.995","Text":"The second example is acetic acid,"},{"Start":"04:21.995 ","End":"04:24.250","Text":"which we know is vinegar."},{"Start":"04:24.250 ","End":"04:28.900","Text":"The molecular formula is C_2H_4O_2."},{"Start":"04:28.900 ","End":"04:30.620","Text":"That means there are 2 carbons,"},{"Start":"04:30.620 ","End":"04:34.520","Text":"4 hydrogens, and 2 oxygens in the molecule."},{"Start":"04:34.520 ","End":"04:41.285","Text":"The empirical formula, That\u0027s the formula we obtain experimentally, is CH_2O."},{"Start":"04:41.285 ","End":"04:44.470","Text":"It\u0027s like the molecular formula divided by 2."},{"Start":"04:44.470 ","End":"04:46.380","Text":"C_2 becomes C,"},{"Start":"04:46.380 ","End":"04:50.425","Text":"H_4 becomes H_2, O_2 becomes O."},{"Start":"04:50.425 ","End":"04:55.550","Text":"This is the best we can achieve from many, many experiments."},{"Start":"04:55.550 ","End":"04:59.870","Text":"Now the condensed structural formula of acetic acid is"},{"Start":"04:59.870 ","End":"05:07.420","Text":"CH_3COOH or as it\u0027s often written, CH_3CO_2H."},{"Start":"05:08.500 ","End":"05:13.925","Text":"That tells us that there are 3 hydrogens attached to 1 carbon."},{"Start":"05:13.925 ","End":"05:17.645","Text":"2 oxygens are attached to 1 carbon,"},{"Start":"05:17.645 ","End":"05:22.235","Text":"the other carbon, and an H is attached to 1 of the oxygens."},{"Start":"05:22.235 ","End":"05:26.645","Text":"If we want more information or more precise information."},{"Start":"05:26.645 ","End":"05:28.775","Text":"If you use a structural formula,"},{"Start":"05:28.775 ","End":"05:32.555","Text":"here\u0027s a structural formula of acetic acid."},{"Start":"05:32.555 ","End":"05:39.086","Text":"On the plane of the molecule is CH,"},{"Start":"05:39.086 ","End":"05:43.010","Text":"CC, and CO with another OH,"},{"Start":"05:43.010 ","End":"05:45.230","Text":"that\u0027s all in the plane of the molecule."},{"Start":"05:45.230 ","End":"05:46.806","Text":"In addition on the plane,"},{"Start":"05:46.806 ","End":"05:53.330","Text":"this is what we\u0027ll later learn as a double bond C attached to O by a double bond."},{"Start":"05:53.330 ","End":"05:55.379","Text":"You see these double lines,"},{"Start":"05:55.379 ","End":"05:57.430","Text":"has a double bond."},{"Start":"06:00.440 ","End":"06:06.540","Text":"We have an H going into the plane of"},{"Start":"06:06.540 ","End":"06:13.645","Text":"the paper and a CH going out of the plane of the paper."},{"Start":"06:13.645 ","End":"06:17.065","Text":"Here we have this CH_3,"},{"Start":"06:17.065 ","End":"06:19.120","Text":"just like we had a butane,"},{"Start":"06:19.120 ","End":"06:24.355","Text":"CO double bond, and COH."},{"Start":"06:24.355 ","End":"06:30.160","Text":"This gives us much more information of the structure of the molecule."},{"Start":"06:30.160 ","End":"06:35.230","Text":"Here we have a ball-and-stick model of citric acid."},{"Start":"06:35.230 ","End":"06:39.490","Text":"The oxygens are red, is 2 oxygens."},{"Start":"06:39.490 ","End":"06:41.450","Text":"The carbons are gray,"},{"Start":"06:41.450 ","End":"06:46.530","Text":"so carbons, and the hydrogens are white."},{"Start":"06:46.900 ","End":"06:53.435","Text":"We have the first carbon associated with 3 hydrogens as 1, 2, 3."},{"Start":"06:53.435 ","End":"07:02.554","Text":"The last oxygen associates with 1 hydrogen and the carbon is attached to 1 carbon,"},{"Start":"07:02.554 ","End":"07:07.020","Text":"1 oxygen, and 1 OH."},{"Start":"07:08.510 ","End":"07:11.795","Text":"This gives us a lot of information."},{"Start":"07:11.795 ","End":"07:14.360","Text":"Now, in previous videos,"},{"Start":"07:14.360 ","End":"07:17.825","Text":"we said that elements consist of atoms."},{"Start":"07:17.825 ","End":"07:20.075","Text":"This is generally true,"},{"Start":"07:20.075 ","End":"07:27.290","Text":"but sometimes elements have more than 1 atom in each molecule."},{"Start":"07:27.290 ","End":"07:31.565","Text":"Some elements consists of molecules of identical atoms."},{"Start":"07:31.565 ","End":"07:34.160","Text":"Let\u0027s consider some examples."},{"Start":"07:34.160 ","End":"07:36.785","Text":"For example, O_2,"},{"Start":"07:36.785 ","End":"07:39.185","Text":"which is a gas, oxygen gas,"},{"Start":"07:39.185 ","End":"07:43.805","Text":"nitrogen gas into chlorine gas, Cl_2,"},{"Start":"07:43.805 ","End":"07:45.920","Text":"fluorine gas F_2,"},{"Start":"07:45.920 ","End":"07:50.505","Text":"all consists of 2 atoms in each molecule."},{"Start":"07:50.505 ","End":"07:53.620","Text":"Oxygen will be O."},{"Start":"07:53.800 ","End":"07:58.910","Text":"Some solids such as sulfur or phosphorus,"},{"Start":"07:58.910 ","End":"08:02.179","Text":"also consists of several atoms."},{"Start":"08:02.179 ","End":"08:09.350","Text":"For example, S_8 has 8 sulfur atoms in molecule of sulfur,"},{"Start":"08:09.350 ","End":"08:15.040","Text":"and 4 phosphorus atoms in a molecule of phosphorus."},{"Start":"08:15.040 ","End":"08:20.225","Text":"Another concept we need to know is that of allotropes."},{"Start":"08:20.225 ","End":"08:23.810","Text":"Some elements have more than 1 possible structure."},{"Start":"08:23.810 ","End":"08:26.600","Text":"For example, oxygen can be O_2,"},{"Start":"08:26.600 ","End":"08:29.390","Text":"which is ordinary oxygen that we breathe,"},{"Start":"08:29.390 ","End":"08:33.300","Text":"and O_3, which is ozone."},{"Start":"08:34.010 ","End":"08:40.745","Text":"Now we know the ozone protects us from ultraviolet light coming from the sun."},{"Start":"08:40.745 ","End":"08:44.059","Text":"It\u0027s quite different from the artery oxygen,"},{"Start":"08:44.059 ","End":"08:48.305","Text":"artery oxygen is O_2 and ozone is O_3."},{"Start":"08:48.305 ","End":"08:50.660","Text":"These are called allotropes,"},{"Start":"08:50.660 ","End":"08:55.520","Text":"I will come across quite a few examples."},{"Start":"08:55.520 ","End":"09:02.760","Text":"In this video, we discussed the various types of formulas for molecular compounds."}],"ID":17663},{"Watched":false,"Name":"Ionic Compounds","Duration":"4m 29s","ChapterTopicVideoID":16913,"CourseChapterTopicPlaylistID":154544,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16913.jpeg","UploadDate":"2019-02-20T23:39:06.0170000","DurationForVideoObject":"PT4M29S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:02.265","Text":"In the previous video,"},{"Start":"00:02.265 ","End":"00:04.965","Text":"we discussed molecular compounds."},{"Start":"00:04.965 ","End":"00:08.070","Text":"In molecular compounds, the basic unit is the molecule."},{"Start":"00:08.070 ","End":"00:11.565","Text":"A few points about these molecular compounds."},{"Start":"00:11.565 ","End":"00:14.550","Text":"First of all we need to note that the atoms that form"},{"Start":"00:14.550 ","End":"00:18.045","Text":"these molecules are usually non-metals."},{"Start":"00:18.045 ","End":"00:20.880","Text":"The second thing we need to note is that"},{"Start":"00:20.880 ","End":"00:23.835","Text":"the bonds between the atoms are called covalent bonds."},{"Start":"00:23.835 ","End":"00:30.630","Text":"Single bonds and double bonds we saw are called covalent bonds."},{"Start":"00:30.630 ","End":"00:34.095","Text":"We\u0027ll meet them a great deal later in the course."},{"Start":"00:34.095 ","End":"00:39.710","Text":"In this video, we\u0027re going to talk about compounds of a different sort."},{"Start":"00:39.710 ","End":"00:42.980","Text":"These are called ionic compounds."},{"Start":"00:42.980 ","End":"00:47.495","Text":"Instead of being composed of non-metals,"},{"Start":"00:47.495 ","End":"00:51.200","Text":"they are composed of metals and non-metals."},{"Start":"00:51.200 ","End":"00:56.525","Text":"The simplest examples of these compounds are composed of a metal and a non-metal."},{"Start":"00:56.525 ","End":"01:00.650","Text":"The simplest example, the one that we\u0027re very familiar with,"},{"Start":"01:00.650 ","End":"01:03.560","Text":"sodium chloride, that\u0027s common salt,"},{"Start":"01:03.560 ","End":"01:05.885","Text":"which we write as NaCl."},{"Start":"01:05.885 ","End":"01:09.890","Text":"In a few moments, we\u0027ll see why we write it NaCl."},{"Start":"01:09.890 ","End":"01:15.260","Text":"Now what do we know about the constituents of sodium chloride? First sodium."},{"Start":"01:15.260 ","End":"01:17.045","Text":"Sodium, that\u0027s Na,"},{"Start":"01:17.045 ","End":"01:19.510","Text":"is an alkali metal,"},{"Start":"01:19.510 ","End":"01:22.635","Text":"and that\u0027s in group 1 of the periodic table."},{"Start":"01:22.635 ","End":"01:28.344","Text":"The first column in the periodic table has sodium in it."},{"Start":"01:28.344 ","End":"01:30.080","Text":"Sodium is a metal."},{"Start":"01:30.080 ","End":"01:31.744","Text":"What about chlorine?"},{"Start":"01:31.744 ","End":"01:38.600","Text":"Chlorine, which we write as Cl is a non-metal and it\u0027s in group 17 of the periodic table."},{"Start":"01:38.600 ","End":"01:43.850","Text":"Remember the periodic table has 18 columns and this is one before the end,"},{"Start":"01:43.850 ","End":"01:46.655","Text":"group 17 of the periodic table."},{"Start":"01:46.655 ","End":"01:51.095","Text":"Chlorine is an example of what we call halogens."},{"Start":"01:51.095 ","End":"01:53.870","Text":"Sodium chloride, when the two are together,"},{"Start":"01:53.870 ","End":"01:59.645","Text":"is a cubic crystal of Na plus and Cl minus ions."},{"Start":"01:59.645 ","End":"02:06.365","Text":"They\u0027re attracted due to electric forces between a positive charge and a negative charge."},{"Start":"02:06.365 ","End":"02:09.920","Text":"These attract very strongly positive and negative."},{"Start":"02:09.920 ","End":"02:13.190","Text":"Let\u0027s look at the crystal, here\u0027s a crystal."},{"Start":"02:13.190 ","End":"02:15.770","Text":"You see here is Na plus,"},{"Start":"02:15.770 ","End":"02:20.404","Text":"which is small, and Cl minus which is larger."},{"Start":"02:20.404 ","End":"02:22.325","Text":"It\u0027s like a cube,"},{"Start":"02:22.325 ","End":"02:25.140","Text":"we can draw it like a cube."},{"Start":"02:25.510 ","End":"02:29.520","Text":"This is top square of the cube."},{"Start":"02:33.680 ","End":"02:35.935","Text":"Here we have a cube."},{"Start":"02:35.935 ","End":"02:37.810","Text":"Now for every Na plus,"},{"Start":"02:37.810 ","End":"02:39.895","Text":"there\u0027s 1 Cl minus."},{"Start":"02:39.895 ","End":"02:43.210","Text":"We can count in any cube how many there are."},{"Start":"02:43.210 ","End":"02:49.999","Text":"It\u0027s the same Na plus ions as there are Cl minus ions."},{"Start":"02:51.510 ","End":"02:56.350","Text":"The ratio of Na plus the Cl minus is 1:1."},{"Start":"02:56.350 ","End":"02:59.465","Text":"Now how are we going to write this?"},{"Start":"02:59.465 ","End":"03:02.845","Text":"We use what\u0027s called a formula unit."},{"Start":"03:02.845 ","End":"03:07.030","Text":"That\u0027s the smallest possible electrically neutral units in the crystal."},{"Start":"03:07.030 ","End":"03:12.510","Text":"The smallest possible unit for NaCl is 1Na plus,"},{"Start":"03:12.510 ","End":"03:14.855","Text":"and 1 Cl minus."},{"Start":"03:14.855 ","End":"03:18.510","Text":"Rewrite the formula unit NaCl."},{"Start":"03:18.510 ","End":"03:23.685","Text":"They don\u0027t bother to write Na plus Cl minus just NaCl."},{"Start":"03:23.685 ","End":"03:28.125","Text":"It\u0027s completely neutral because there\u0027s 1 positive charge,"},{"Start":"03:28.125 ","End":"03:30.580","Text":"and 1 negative charge."},{"Start":"03:31.040 ","End":"03:36.125","Text":"There are other ionic crystals that are not 1:1,"},{"Start":"03:36.125 ","End":"03:39.685","Text":"and here\u0027s an example, calcium chloride."},{"Start":"03:39.685 ","End":"03:45.635","Text":"The calcium is in the second column of the periodic table."},{"Start":"03:45.635 ","End":"03:49.670","Text":"That\u0027s in the second column."},{"Start":"03:49.670 ","End":"03:55.230","Text":"It\u0027s iron is Ca_2 plus."},{"Start":"03:55.230 ","End":"03:56.780","Text":"In order to balance it,"},{"Start":"03:56.780 ","End":"04:01.045","Text":"we need not just 1 Cl minus, but another one."},{"Start":"04:01.045 ","End":"04:02.610","Text":"We need 2 of them."},{"Start":"04:02.610 ","End":"04:04.140","Text":"Then we have 2 minuses,"},{"Start":"04:04.140 ","End":"04:07.875","Text":"and here we have 2 plus and that together that\u0027s 0."},{"Start":"04:07.875 ","End":"04:10.400","Text":"This is electrically neutral."},{"Start":"04:10.400 ","End":"04:13.320","Text":"We write it Ca,"},{"Start":"04:13.320 ","End":"04:18.065","Text":"Cl_2, that\u0027s completely electrically neutral."},{"Start":"04:18.065 ","End":"04:24.605","Text":"In this video, we\u0027ve discussed compounds consist of ions rather than molecules."},{"Start":"04:24.605 ","End":"04:28.290","Text":"We call them ionic compounds."}],"ID":17664}],"Thumbnail":null,"ID":154544},{"Name":"Calculations Involving Moles","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Moles of Compounds and Elements","Duration":"5m 18s","ChapterTopicVideoID":16914,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16914.jpeg","UploadDate":"2019-02-20T23:38:41.0930000","DurationForVideoObject":"PT5M18S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.795","Text":"In previous videos we talked about moles of atoms."},{"Start":"00:03.795 ","End":"00:08.399","Text":"Here we\u0027re going to talk about moles of compounds and elements."},{"Start":"00:08.399 ","End":"00:14.505","Text":"We learned that the molar mass of C^12 is exactly 12 grams."},{"Start":"00:14.505 ","End":"00:17.040","Text":"We learned that the molar mass of oxygen,"},{"Start":"00:17.040 ","End":"00:19.515","Text":"taking into account the various isotopes,"},{"Start":"00:19.515 ","End":"00:24.945","Text":"is 15.999 grams, which is almost 16 grams."},{"Start":"00:24.945 ","End":"00:28.785","Text":"Now we\u0027re going to learn how to calculate the molar mass of compounds."},{"Start":"00:28.785 ","End":"00:32.550","Text":"If we have an empirical formula, or molecular formula,"},{"Start":"00:32.550 ","End":"00:34.395","Text":"or a formula unit,"},{"Start":"00:34.395 ","End":"00:37.920","Text":"all of which we\u0027ve defined in previous videos."},{"Start":"00:37.920 ","End":"00:40.610","Text":"We can find it\u0027s molar mass by adding"},{"Start":"00:40.610 ","End":"00:45.175","Text":"the molar masses of all the atoms that constitute the formula."},{"Start":"00:45.175 ","End":"00:47.520","Text":"Let\u0027s take an example."},{"Start":"00:47.520 ","End":"00:52.075","Text":"The first example we consider is NaCl,"},{"Start":"00:52.075 ","End":"00:53.974","Text":"which is sodium chloride,"},{"Start":"00:53.974 ","End":"00:55.925","Text":"which is ordinary salt."},{"Start":"00:55.925 ","End":"00:58.400","Text":"You may remember in the previous video,"},{"Start":"00:58.400 ","End":"01:02.340","Text":"we said that this was an ionic solid,"},{"Start":"01:04.580 ","End":"01:12.420","Text":"consisting of Na plus cations and Cl minus anions."},{"Start":"01:12.420 ","End":"01:14.805","Text":"Now the molar mass of NaCl,"},{"Start":"01:14.805 ","End":"01:17.700","Text":"consists of the molar mass of sodium,"},{"Start":"01:17.700 ","End":"01:20.070","Text":"plus the molar mass of chlorine."},{"Start":"01:20.070 ","End":"01:24.455","Text":"If we add these 2 we should get the molar mass of sodium chloride."},{"Start":"01:24.455 ","End":"01:27.675","Text":"This is its formula unit NaCl."},{"Start":"01:27.675 ","End":"01:35.900","Text":"If we add 22.99 for sodium and 35.45 grams for chlorine,"},{"Start":"01:35.900 ","End":"01:38.480","Text":"we get 58.44 grams."},{"Start":"01:38.480 ","End":"01:42.890","Text":"That\u0027s the molar mass of the formula unit, NaCl."},{"Start":"01:42.890 ","End":"01:45.110","Text":"Now let\u0027s take another example."},{"Start":"01:45.110 ","End":"01:50.150","Text":"Second example we\u0027re going to talk about is acetic acid,"},{"Start":"01:50.150 ","End":"01:55.010","Text":"CH_3CO_2H, which if we add up all the carbons,"},{"Start":"01:55.010 ","End":"01:56.330","Text":"and all the hydrogens,"},{"Start":"01:56.330 ","End":"01:58.470","Text":"we get C_2O_2 and H_4."},{"Start":"02:00.350 ","End":"02:05.525","Text":"That\u0027s the molecular formula of acetic acid."},{"Start":"02:05.525 ","End":"02:09.325","Text":"What the molar mass, you have to take twice the mass of carbon,"},{"Start":"02:09.325 ","End":"02:11.330","Text":"and twice the mass of oxygen,"},{"Start":"02:11.330 ","End":"02:13.475","Text":"because there are 2 carbons and 2 oxygens,"},{"Start":"02:13.475 ","End":"02:15.800","Text":"and 4 times the mass of hydrogen,"},{"Start":"02:15.800 ","End":"02:17.515","Text":"those are 4 hydrogens."},{"Start":"02:17.515 ","End":"02:26.580","Text":"It\u0027s 2 times 12.01 plus 2 times 16.00 plus 4 times 1.008,"},{"Start":"02:26.580 ","End":"02:36.315","Text":"and altogether, that makes 60.05 grams So the molar mass of acetic acid is 60.05 grams,"},{"Start":"02:36.315 ","End":"02:40.835","Text":"and it\u0027s very important always to say molar mass of what?"},{"Start":"02:40.835 ","End":"02:43.385","Text":"I want to explain why it\u0027s important."},{"Start":"02:43.385 ","End":"02:46.718","Text":"Let\u0027s consider the molar mass of oxygen."},{"Start":"02:46.718 ","End":"02:48.735","Text":"Now the molar mass of oxygen,"},{"Start":"02:48.735 ","End":"02:52.195","Text":"the element, is 16.00 grams."},{"Start":"02:52.195 ","End":"02:57.110","Text":"But the formula of oxygen gas isn\u0027t just O, it\u0027s O_2."},{"Start":"02:57.110 ","End":"03:04.020","Text":"So that, the molar mass of O_2 is twice as much as 16.00 grams,"},{"Start":"03:04.020 ","End":"03:07.260","Text":"and that\u0027s 32.00 grams."},{"Start":"03:07.260 ","End":"03:09.240","Text":"You can see how important is,"},{"Start":"03:09.240 ","End":"03:10.815","Text":"to say whether we\u0027re talking about O,"},{"Start":"03:10.815 ","End":"03:16.425","Text":"16.00 grams, or whether we\u0027re talking about O_2,"},{"Start":"03:16.425 ","End":"03:19.225","Text":"where it\u0027s 32.00 grams."},{"Start":"03:19.225 ","End":"03:22.310","Text":"Now sometimes we\u0027re asked the number of molecules in a mole."},{"Start":"03:22.310 ","End":"03:24.965","Text":"You\u0027ll recall, that in every mole,"},{"Start":"03:24.965 ","End":"03:27.380","Text":"there are Avogadro\u0027s number of units."},{"Start":"03:27.380 ","End":"03:29.330","Text":"Let\u0027s take an example."},{"Start":"03:29.330 ","End":"03:31.760","Text":"Once again, consider O_2."},{"Start":"03:31.760 ","End":"03:37.125","Text":"Now Avogadro\u0027s number, 6.022 times 10^23."},{"Start":"03:37.125 ","End":"03:44.780","Text":"There are 6.022 times 10^23 molecules in 1 mole of oxygen gas."},{"Start":"03:44.780 ","End":"03:47.585","Text":"We can use the conversion factor,"},{"Start":"03:47.585 ","End":"03:54.620","Text":"1 mole of O_2 divided by 6.022 times 10^23 molecules of O_2,"},{"Start":"03:54.620 ","End":"03:57.205","Text":"and these are totally equivalent,"},{"Start":"03:57.205 ","End":"04:00.935","Text":"the ratio is 1, this conversion factor."},{"Start":"04:00.935 ","End":"04:03.275","Text":"Let\u0027s take an example."},{"Start":"04:03.275 ","End":"04:07.175","Text":"How many molecules are in 12 grams of O_2?"},{"Start":"04:07.175 ","End":"04:10.210","Text":"What we want to do is to go from grams,"},{"Start":"04:10.210 ","End":"04:16.060","Text":"to moles, and from moles to molecules."},{"Start":"04:16.060 ","End":"04:21.000","Text":"The number of molecules is 12.00 grams of"},{"Start":"04:21.000 ","End":"04:29.280","Text":"oxygen times 1 mole of oxygen is gas is equivalent to 32.00 grams of oxygen,"},{"Start":"04:29.280 ","End":"04:32.170","Text":"so this is a conversion factor."},{"Start":"04:32.170 ","End":"04:36.830","Text":"Then we want to convert moles to molecules."},{"Start":"04:36.830 ","End":"04:41.840","Text":"So 6.022 times 10^23 molecules of O_2,"},{"Start":"04:41.840 ","End":"04:45.805","Text":"and 1 mole of oxygen is now a conversion factor."},{"Start":"04:45.805 ","End":"04:47.870","Text":"If we multiply this out,"},{"Start":"04:47.870 ","End":"04:50.870","Text":"grams goes with grams of O_2,"},{"Start":"04:50.870 ","End":"04:54.095","Text":"moles of O_2, goes with moles of O_2,"},{"Start":"04:54.095 ","End":"04:56.600","Text":"we\u0027re left with molecules of O_2,"},{"Start":"04:56.600 ","End":"05:03.355","Text":"and that\u0027s 2.258 times 10^23 molecules of O_2."},{"Start":"05:03.355 ","End":"05:12.365","Text":"Now we know precisely how many molecules of O_2 there are in 12 grams of oxygen."},{"Start":"05:12.365 ","End":"05:14.600","Text":"In this video, we learned how to calculate"},{"Start":"05:14.600 ","End":"05:18.060","Text":"the molar mass and the number of molecules in a mole."}],"ID":17659},{"Watched":false,"Name":"Exercise 1","Duration":"6m 31s","ChapterTopicVideoID":18721,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/18721.jpeg","UploadDate":"2019-04-28T06:42:08.9430000","DurationForVideoObject":"PT6M31S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:03.419","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.419 ","End":"00:06.210","Text":"Calculate the total number of a,"},{"Start":"00:06.210 ","End":"00:08.850","Text":"atoms in 1 molecule of methionine,"},{"Start":"00:08.850 ","End":"00:12.435","Text":"C_5H_11(NO_2)S, b,"},{"Start":"00:12.435 ","End":"00:20.790","Text":"atoms in 0.00304 mole hexanol CH_3(CH_2)_4CH_2OH and c,"},{"Start":"00:20.790 ","End":"00:28.065","Text":"fluorine atoms in 11.45 mole halothane, C_2HBrClF_3."},{"Start":"00:28.065 ","End":"00:32.580","Text":"We\u0027re going to start in a. You have to calculate the atoms in 1 molecule of methionine."},{"Start":"00:32.580 ","End":"00:37.530","Text":"If we look at methionine we see first of all that we have 5 carbon atoms."},{"Start":"00:37.530 ","End":"00:39.965","Text":"In a, we have 5 atoms,"},{"Start":"00:39.965 ","End":"00:44.820","Text":"which are for the carbon plus 11 hydrogen atoms,"},{"Start":"00:44.820 ","End":"00:46.570","Text":"so 11 atoms,"},{"Start":"00:46.570 ","End":"00:52.275","Text":"which originate from the hydrogen plus 1 nitrogen atom,"},{"Start":"00:52.275 ","End":"00:57.120","Text":"so that\u0027s 1 atom plus 2 oxygen atoms,"},{"Start":"00:57.120 ","End":"01:02.370","Text":"so that\u0027s another 2 atoms plus 1 sulfur atom,"},{"Start":"01:02.370 ","End":"01:06.160","Text":"so plus another atom."},{"Start":"01:06.160 ","End":"01:09.690","Text":"Altogether these equal 20 atoms."},{"Start":"01:11.240 ","End":"01:13.925","Text":"In a we have 20 atoms."},{"Start":"01:13.925 ","End":"01:15.350","Text":"Now we\u0027re going to go on to b."},{"Start":"01:15.350 ","End":"01:19.355","Text":"In b we\u0027re asked to calculate the number of atoms in"},{"Start":"01:19.355 ","End":"01:24.950","Text":"0.00304 mole of hexanol, which we have here."},{"Start":"01:24.950 ","End":"01:27.650","Text":"First of all, we\u0027re going to calculate the number of molecules of"},{"Start":"01:27.650 ","End":"01:31.345","Text":"hexanol and then we\u0027ll go on to calculate the number of atoms."},{"Start":"01:31.345 ","End":"01:35.510","Text":"To calculate the number of moles we\u0027re going to take the following equation,"},{"Start":"01:35.510 ","End":"01:40.590","Text":"n, which is the number of moles equals N divided by N_A."},{"Start":"01:40.590 ","End":"01:42.360","Text":"N_A is Avogadro\u0027s number,"},{"Start":"01:42.360 ","End":"01:43.490","Text":"and N could be atoms,"},{"Start":"01:43.490 ","End":"01:45.630","Text":"or molecules, or ions and so on."},{"Start":"01:45.630 ","End":"01:49.750","Text":"In this question it\u0027s going to be molecules because we have molecules of hexanol."},{"Start":"01:49.750 ","End":"01:57.990","Text":"Therefore the Avogadro number is going to be molecules per mole in our case."},{"Start":"01:57.990 ","End":"02:00.680","Text":"We want to calculate the number of molecules therefore,"},{"Start":"02:00.680 ","End":"02:04.025","Text":"we\u0027re going to multiply both sides by Avogadro\u0027s number."},{"Start":"02:04.025 ","End":"02:06.815","Text":"N, number of molecules of hexanol,"},{"Start":"02:06.815 ","End":"02:12.020","Text":"equals number of moles of hexanol times Avogadro\u0027s number."},{"Start":"02:12.020 ","End":"02:15.950","Text":"The number of moles of hexanol are given,"},{"Start":"02:15.950 ","End":"02:20.590","Text":"0.00304 moles times Avogadro\u0027s number."},{"Start":"02:20.590 ","End":"02:21.810","Text":"We\u0027re going to write that down here,"},{"Start":"02:21.810 ","End":"02:30.880","Text":"times 6.022 times 10^23 molecules per mole."},{"Start":"02:30.880 ","End":"02:35.155","Text":"Because again, the mole is of the molecule hexanol not"},{"Start":"02:35.155 ","End":"02:39.680","Text":"of a certain atom so we\u0027re calculating the molecules."},{"Start":"02:39.680 ","End":"02:50.005","Text":"The moles cancel out and we get 1.83 times 10^21 molecules of hexanol."},{"Start":"02:50.005 ","End":"02:52.400","Text":"Now the next step is to take these molecules and"},{"Start":"02:52.400 ","End":"02:56.075","Text":"multiply by the number of atoms in 1 molecule of hexanol."},{"Start":"02:56.075 ","End":"02:59.375","Text":"We\u0027re going to calculate the number of atoms in hexanol."},{"Start":"02:59.375 ","End":"03:03.935","Text":"The number of atoms in hexanol equals,"},{"Start":"03:03.935 ","End":"03:05.360","Text":"we can see if we count the carbons,"},{"Start":"03:05.360 ","End":"03:07.150","Text":"we have 4 carbons here,"},{"Start":"03:07.150 ","End":"03:11.910","Text":"plus 1, plus 1 so that comes to 6 carbon atoms."},{"Start":"03:11.910 ","End":"03:18.230","Text":"We have 6 atoms from our carbon plus, we look at our hydrogens,"},{"Start":"03:18.230 ","End":"03:21.780","Text":"we have 8 hydrogens here plus 3,"},{"Start":"03:21.780 ","End":"03:24.090","Text":"which is 11, plus 2,"},{"Start":"03:24.090 ","End":"03:27.045","Text":"which is 13 plus 1, which comes to 14."},{"Start":"03:27.045 ","End":"03:30.450","Text":"That\u0027s another 14 atoms plus,"},{"Start":"03:30.450 ","End":"03:32.640","Text":"we have 1 oxygen atom,"},{"Start":"03:32.640 ","End":"03:37.035","Text":"so another 1 atom and this comes to 21 atoms."},{"Start":"03:37.035 ","End":"03:39.425","Text":"These are the number of atoms we have in our molecule."},{"Start":"03:39.425 ","End":"03:41.600","Text":"Now we\u0027re going to take the number of molecules times"},{"Start":"03:41.600 ","End":"03:44.266","Text":"the number of atoms we have in every molecule,"},{"Start":"03:44.266 ","End":"03:46.112","Text":"so that\u0027s atoms per molecule,"},{"Start":"03:46.112 ","End":"03:48.560","Text":"and we\u0027re going to multiply one by the other."},{"Start":"03:48.560 ","End":"03:52.970","Text":"The number of molecules is 1.83 times"},{"Start":"03:52.970 ","End":"04:00.890","Text":"10^21 molecules times 21 atoms per molecule."},{"Start":"04:00.890 ","End":"04:06.570","Text":"Is 21 atoms per molecule in every molecule."},{"Start":"04:07.030 ","End":"04:16.625","Text":"The molecules cancel out and we get 3.84 times 10^22 atoms of hexanol."},{"Start":"04:16.625 ","End":"04:18.980","Text":"That is our answer for b."},{"Start":"04:18.980 ","End":"04:24.620","Text":"In c, we have to calculate fluorine in atoms in 11.45 mole of halothane."},{"Start":"04:24.620 ","End":"04:27.200","Text":"First of all, we\u0027re going to calculate the moles of"},{"Start":"04:27.200 ","End":"04:29.885","Text":"the fluorine and then go on to the atoms."},{"Start":"04:29.885 ","End":"04:38.370","Text":"The moles of fluorine equal the moles of halothane times,"},{"Start":"04:38.370 ","End":"04:43.130","Text":"if we look at our molecule we can see that for every 1 molecule of halothane we have"},{"Start":"04:43.130 ","End":"04:48.671","Text":"3 fluorine atoms meaning that for 1 mole of halothane we have 3 moles of fluorine,"},{"Start":"04:48.671 ","End":"04:56.800","Text":"so this is times 3 mole of fluorine for every 1 mole of halothane."},{"Start":"04:58.810 ","End":"05:04.085","Text":"This equals, the moles of halothane are 11.45 times"},{"Start":"05:04.085 ","End":"05:10.120","Text":"3 moles of fluorine for every 1 mole of halothane."},{"Start":"05:13.030 ","End":"05:20.765","Text":"The moles of halothane cancel out and we get 34.35 moles of fluorine."},{"Start":"05:20.765 ","End":"05:23.615","Text":"You want to calculate atoms of fluorine."},{"Start":"05:23.615 ","End":"05:25.480","Text":"We\u0027re going to use the following equation,"},{"Start":"05:25.480 ","End":"05:29.090","Text":"n number of moles equals N divided by N_A."},{"Start":"05:29.090 ","End":"05:32.600","Text":"N_A is Avogadro\u0027s number and N could be the number of atoms,"},{"Start":"05:32.600 ","End":"05:33.920","Text":"molecules, ions, so on,"},{"Start":"05:33.920 ","End":"05:35.555","Text":"depending on the question."},{"Start":"05:35.555 ","End":"05:38.680","Text":"Here we\u0027re going to calculate the number of atoms."},{"Start":"05:38.680 ","End":"05:42.390","Text":"Avogadro\u0027s number is going to be the atoms per mole."},{"Start":"05:42.390 ","End":"05:44.510","Text":"When we calculate the number of atoms we\u0027re going to"},{"Start":"05:44.510 ","End":"05:46.370","Text":"multiply both sides by Avogadro\u0027s number."},{"Start":"05:46.370 ","End":"05:50.240","Text":"N, the number of atoms of the fluorine equal"},{"Start":"05:50.240 ","End":"05:55.410","Text":"the number of moles of fluorine times Avogadro\u0027s number."},{"Start":"05:55.460 ","End":"06:00.990","Text":"This equals 34.35 moles of"},{"Start":"06:00.990 ","End":"06:08.890","Text":"fluorine times 6.022 times 10^23 atoms per mole."},{"Start":"06:09.500 ","End":"06:14.700","Text":"The moles cancel out and we get"},{"Start":"06:14.700 ","End":"06:23.445","Text":"2.07 times 10^25 atoms of fluorine."},{"Start":"06:23.445 ","End":"06:27.320","Text":"Our answer is 2.07 times 10^25 atoms of fluorine."},{"Start":"06:27.320 ","End":"06:28.955","Text":"That is our final answer."},{"Start":"06:28.955 ","End":"06:31.770","Text":"Thank you very much for watching."}],"ID":23663},{"Watched":false,"Name":"Exercise 2","Duration":"7m 37s","ChapterTopicVideoID":18720,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/18720.jpeg","UploadDate":"2019-04-28T06:39:16.4170000","DurationForVideoObject":"PT7M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.180","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.180 ","End":"00:11.610","Text":"Trinitrotoluene, TNT has a molecular formula, CH_3C_6H_2NO_2 3 times."},{"Start":"00:11.610 ","End":"00:15.285","Text":"Determine a, its molecular mass, b,"},{"Start":"00:15.285 ","End":"00:19.920","Text":"the number of moles of hydrogen atoms per mole of TNT,"},{"Start":"00:19.920 ","End":"00:24.300","Text":"c, the number of grams of carbon per mole of TNT,"},{"Start":"00:24.300 ","End":"00:29.985","Text":"and d, the number of carbon atoms in 8.57 mole TNT."},{"Start":"00:29.985 ","End":"00:33.855","Text":"We\u0027re going to start with a and find the molecular mass of TNT."},{"Start":"00:33.855 ","End":"00:42.025","Text":"The molecular mass of TNT equals,"},{"Start":"00:42.025 ","End":"00:45.015","Text":"we can see that in TNT we have 7 carbons,"},{"Start":"00:45.015 ","End":"00:48.890","Text":"so it\u0027s 7 times the atomic mass of carbon."},{"Start":"00:51.140 ","End":"00:54.720","Text":"Plus we have 5 hydrogens,"},{"Start":"00:54.720 ","End":"01:02.805","Text":"so plus 5 times the atomic mass of hydrogen plus,"},{"Start":"01:02.805 ","End":"01:05.385","Text":"you can see that we have 3 nitrogens,"},{"Start":"01:05.385 ","End":"01:15.060","Text":"so 3 times the atomic mass of nitrogen plus,"},{"Start":"01:15.060 ","End":"01:18.225","Text":"we can see that we have 6 oxygens,"},{"Start":"01:18.225 ","End":"01:25.850","Text":"so 6 times the atomic mass of oxygen."},{"Start":"01:25.850 ","End":"01:29.030","Text":"Because we can see it\u0027s 2 times 3 equals 6."},{"Start":"01:29.030 ","End":"01:35.360","Text":"This equals 7 times 12.01u,"},{"Start":"01:35.360 ","End":"01:37.700","Text":"which is the atomic mass of carbon,"},{"Start":"01:37.700 ","End":"01:41.435","Text":"plus 5 times 1.01u,"},{"Start":"01:41.435 ","End":"01:45.005","Text":"so it\u0027s 5 hydrogens times the atomic mass of hydrogen,"},{"Start":"01:45.005 ","End":"01:51.590","Text":"plus 3 times the atomic mass of nitrogen,"},{"Start":"01:51.590 ","End":"01:55.230","Text":"which equals 14.01u,"},{"Start":"01:56.120 ","End":"02:04.145","Text":"plus 6 times the atomic mass of oxygen, which equals 16u."},{"Start":"02:04.145 ","End":"02:07.890","Text":"This comes to 227.15u."},{"Start":"02:10.280 ","End":"02:13.963","Text":"That\u0027s the molecular mass of the TNT,"},{"Start":"02:13.963 ","End":"02:15.905","Text":"and that is our final answer for a."},{"Start":"02:15.905 ","End":"02:17.900","Text":"Now we\u0027re going to go on to b."},{"Start":"02:17.900 ","End":"02:23.735","Text":"In b, we have to find the number of moles of hydrogen atoms per mole of TNT."},{"Start":"02:23.735 ","End":"02:28.760","Text":"Now in b we have to find the number of moles of hydrogen atoms per mole of TNT."},{"Start":"02:28.760 ","End":"02:31.415","Text":"If we look at the TNT,"},{"Start":"02:31.415 ","End":"02:35.035","Text":"we can see again that it\u0027s CH_3C_6H_2,"},{"Start":"02:35.035 ","End":"02:36.720","Text":"and NO_2 3 times."},{"Start":"02:36.720 ","End":"02:39.020","Text":"For every 1 mole of TNT,"},{"Start":"02:39.020 ","End":"02:41.870","Text":"we can see that we have 5 moles of hydrogen since we have"},{"Start":"02:41.870 ","End":"02:45.940","Text":"5 hydrogen oxygen atoms for every 1 molecule of TNT."},{"Start":"02:45.940 ","End":"02:52.560","Text":"The moles of the hydrogen equal the moles of TNT times 5 moles"},{"Start":"02:52.560 ","End":"03:00.040","Text":"of hydrogen for every 1 mole of TNT."},{"Start":"03:00.040 ","End":"03:05.780","Text":"Since we\u0027re interested in the number of moles of hydrogen atoms per 1 mole of TNT,"},{"Start":"03:05.780 ","End":"03:10.080","Text":"this is 1 mole of"},{"Start":"03:10.080 ","End":"03:19.030","Text":"TNT times 5 moles of hydrogen per 1 mole of TNT."},{"Start":"03:19.030 ","End":"03:21.380","Text":"The moles of TNT cancel out,"},{"Start":"03:21.380 ","End":"03:25.130","Text":"and our answer is 5 moles of hydrogen."},{"Start":"03:25.130 ","End":"03:29.035","Text":"The answer to b is 5 moles of hydrogen."},{"Start":"03:29.035 ","End":"03:33.425","Text":"In c we have to calculate the number of grams of carbon per mole of TNT."},{"Start":"03:33.425 ","End":"03:37.295","Text":"First we\u0027re going to find the moles of carbon and go on to the grams."},{"Start":"03:37.295 ","End":"03:39.400","Text":"If we look at TNT,"},{"Start":"03:39.400 ","End":"03:42.690","Text":"we can say that for every 1 molecule of TNT,"},{"Start":"03:42.690 ","End":"03:44.430","Text":"we have 7 carbon atoms,"},{"Start":"03:44.430 ","End":"03:48.140","Text":"meaning that from 1 mole of TNT we\u0027re going to have 7 moles of carbon."},{"Start":"03:48.140 ","End":"03:54.570","Text":"The moles of carbon equal the moles of TNT times 7 moles of"},{"Start":"03:54.570 ","End":"04:01.850","Text":"carbon for every 1 mole of TNT."},{"Start":"04:01.850 ","End":"04:06.230","Text":"In our case, we\u0027re talking about 1 mole of TNT because it\u0027s for a mole of TNT,"},{"Start":"04:06.230 ","End":"04:09.200","Text":"so the moles of TNT equal 1 mole of"},{"Start":"04:09.200 ","End":"04:14.180","Text":"TNT times 7 moles"},{"Start":"04:14.180 ","End":"04:19.700","Text":"of carbon divided by 1 mole of TNT."},{"Start":"04:19.700 ","End":"04:22.580","Text":"The moles of TNT cancel out,"},{"Start":"04:22.580 ","End":"04:26.715","Text":"and we get 7 moles of carbon."},{"Start":"04:26.715 ","End":"04:29.045","Text":"Now we\u0027re going to calculate the mass of carbon."},{"Start":"04:29.045 ","End":"04:31.280","Text":"It\u0027s just a reminder, n,"},{"Start":"04:31.280 ","End":"04:32.870","Text":"number of moles equals m,"},{"Start":"04:32.870 ","End":"04:35.375","Text":"which is the mass divided by the molar mass."},{"Start":"04:35.375 ","End":"04:39.050","Text":"We\u0027re looking for the mass so we\u0027re going to multiply both sides by the molar mass."},{"Start":"04:39.050 ","End":"04:46.590","Text":"The mass of carbon equals number of moles of carbon times the molar mass of carbon."},{"Start":"04:47.080 ","End":"04:50.210","Text":"This equals the number of moles is 7,"},{"Start":"04:50.210 ","End":"04:54.230","Text":"so 7 moles of carbon times the molar mass of carbon,"},{"Start":"04:54.230 ","End":"05:00.455","Text":"which is taken from the periodic table of elements and equals 12.01 grams per mole."},{"Start":"05:00.455 ","End":"05:03.380","Text":"The moles cancel out, and we\u0027re left with grams,"},{"Start":"05:03.380 ","End":"05:08.480","Text":"so this equals 84.07 grams."},{"Start":"05:08.480 ","End":"05:14.045","Text":"In c, our final answer is 84.07 grams of carbon."},{"Start":"05:14.045 ","End":"05:16.670","Text":"Now we\u0027re going to go on to d. In d,"},{"Start":"05:16.670 ","End":"05:21.065","Text":"we have to find the number of carbon atoms in 8.57 moles of TNT."},{"Start":"05:21.065 ","End":"05:23.450","Text":"Again, we will first calculate the number of moles of"},{"Start":"05:23.450 ","End":"05:26.690","Text":"carbon and then go on to the number of atoms of carbon."},{"Start":"05:26.690 ","End":"05:32.435","Text":"The number of moles of carbon equals the number of moles of TNT times, again,"},{"Start":"05:32.435 ","End":"05:34.160","Text":"for every 1 mole of TNT,"},{"Start":"05:34.160 ","End":"05:35.570","Text":"we have 7 moles of carbon,"},{"Start":"05:35.570 ","End":"05:41.260","Text":"so is time 7 moles of carbon divided by 1 mole of TNT."},{"Start":"05:42.040 ","End":"05:46.040","Text":"This time we\u0027re talking about 8.57 moles of TNT."},{"Start":"05:46.040 ","End":"05:51.206","Text":"The moles of TNT is 8.57 moles of"},{"Start":"05:51.206 ","End":"05:59.230","Text":"TNT times 7 moles of carbon for every 1 mole of TNT."},{"Start":"05:59.360 ","End":"06:09.750","Text":"The moles of TNT cancel out and this equals 59.99 moles of carbon."},{"Start":"06:09.750 ","End":"06:11.640","Text":"Now we have the moles of carbon,"},{"Start":"06:11.640 ","End":"06:14.075","Text":"we\u0027re going to calculate the number of atoms."},{"Start":"06:14.075 ","End":"06:16.880","Text":"The number of moles,"},{"Start":"06:16.880 ","End":"06:23.950","Text":"which is n, equal N divided by N_a."},{"Start":"06:23.950 ","End":"06:28.280","Text":"N_a is the Avogadro\u0027s number and N in our case, is going to be atoms."},{"Start":"06:28.280 ","End":"06:31.280","Text":"However, remember that it can generally be molecules,"},{"Start":"06:31.280 ","End":"06:33.215","Text":"ions, and so on."},{"Start":"06:33.215 ","End":"06:35.990","Text":"Avogadro number, since we\u0027re talking about atoms,"},{"Start":"06:35.990 ","End":"06:39.120","Text":"is going to be in atoms per mole."},{"Start":"06:39.160 ","End":"06:41.640","Text":"We\u0027re looking for the number of atoms,"},{"Start":"06:41.640 ","End":"06:44.675","Text":"so we\u0027re going to multiply both sides by the Avogadro\u0027s number."},{"Start":"06:44.675 ","End":"06:48.890","Text":"N, which is the number atoms of carbon equals number"},{"Start":"06:48.890 ","End":"06:53.139","Text":"of moles of carbon times the Avogadro number."},{"Start":"06:53.139 ","End":"06:54.860","Text":"This equals the moles of carbon,"},{"Start":"06:54.860 ","End":"07:01.160","Text":"which is 59.99 moles times the Avogadro number,"},{"Start":"07:01.160 ","End":"07:05.585","Text":"which is 6.022 times 10^23."},{"Start":"07:05.585 ","End":"07:08.910","Text":"Here we\u0027re talking about atoms per mole."},{"Start":"07:10.750 ","End":"07:13.860","Text":"The moles cancel out,"},{"Start":"07:14.350 ","End":"07:24.210","Text":"and we get 3.61 times 10^25 atoms."},{"Start":"07:24.430 ","End":"07:32.450","Text":"Our final answer for d is 3.61 times 10^25 carbon atoms."},{"Start":"07:32.450 ","End":"07:34.505","Text":"That is our final answer."},{"Start":"07:34.505 ","End":"07:37.140","Text":"Thank you very much for watching."}],"ID":23660},{"Watched":false,"Name":"Calculating Percentage Composition","Duration":"4m 19s","ChapterTopicVideoID":16915,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16915.jpeg","UploadDate":"2019-02-20T23:39:13.4730000","DurationForVideoObject":"PT4M19S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.385","Text":"In a previous video,"},{"Start":"00:02.385 ","End":"00:05.850","Text":"we defined the composition in the following way."},{"Start":"00:05.850 ","End":"00:10.620","Text":"We said that the composition of a sample of matter is a quantity of"},{"Start":"00:10.620 ","End":"00:15.465","Text":"each component in the sample. Let\u0027s take water."},{"Start":"00:15.465 ","End":"00:17.280","Text":"One of the earlier videos,"},{"Start":"00:17.280 ","End":"00:19.905","Text":"we said that the percentage composition of water,"},{"Start":"00:19.905 ","End":"00:25.820","Text":"H_2_O, is 89% oxygen, 11% hydrogen."},{"Start":"00:25.820 ","End":"00:30.770","Text":"In this video, we\u0027ll see how we can calculate such quantities."},{"Start":"00:30.770 ","End":"00:33.725","Text":"How can we calculate the percentage composition?"},{"Start":"00:33.725 ","End":"00:36.890","Text":"The first thing we have to do is to learn how to"},{"Start":"00:36.890 ","End":"00:40.870","Text":"convert moles of molecules to moles of atoms."},{"Start":"00:40.870 ","End":"00:46.670","Text":"We know from the molecular formula of water, H_2_O,"},{"Start":"00:46.670 ","End":"00:52.880","Text":"that there are 2 hydrogen atoms and 1 oxygen atom in every molecule of H_2_O."},{"Start":"00:52.880 ","End":"00:56.840","Text":"If we now multiply this by Avogadro\u0027s number,"},{"Start":"00:56.840 ","End":"01:00.125","Text":"we can see that in N_A molecules in H_2_O,"},{"Start":"01:00.125 ","End":"01:05.590","Text":"there are 2N_A_H atoms and N_A_O atoms."},{"Start":"01:05.590 ","End":"01:08.055","Text":"We know that N_A of anything,"},{"Start":"01:08.055 ","End":"01:10.935","Text":"Avogadro\u0027s number of anything is a mole,"},{"Start":"01:10.935 ","End":"01:20.615","Text":"so we can say that 1 mole of water contains 2 moles of hydrogen and 1 mole of oxygen."},{"Start":"01:20.615 ","End":"01:23.134","Text":"You probably think this is rather strange."},{"Start":"01:23.134 ","End":"01:28.939","Text":"How can we take 1 mole and get from it 2 plus 1 moles?"},{"Start":"01:28.939 ","End":"01:31.490","Text":"It seems illogical."},{"Start":"01:31.490 ","End":"01:34.010","Text":"But if we consider this previous slide,"},{"Start":"01:34.010 ","End":"01:36.635","Text":"we\u0027ll see that it\u0027s really very reasonable."},{"Start":"01:36.635 ","End":"01:40.550","Text":"This is something we\u0027re going to use over and over and over again."},{"Start":"01:40.550 ","End":"01:47.250","Text":"Now how can we calculate the percentage composition of water again?"},{"Start":"01:47.250 ","End":"01:51.650","Text":"Now we know that the molar mass of water is"},{"Start":"01:51.650 ","End":"01:56.320","Text":"twice the molar mass of hydrogen plus the molar mass of oxygen."},{"Start":"01:56.320 ","End":"02:00.770","Text":"That\u0027s 2 times 1 plus for the hydrogen plus"},{"Start":"02:00.770 ","End":"02:05.480","Text":"16 for the oxygen gives me a total of 18 grams."},{"Start":"02:05.480 ","End":"02:12.730","Text":"1 mole of H_2_O weighs or has a mass of 18 grams."},{"Start":"02:12.730 ","End":"02:15.755","Text":"Now what\u0027s the percentage composition?"},{"Start":"02:15.755 ","End":"02:18.320","Text":"We have 2 hydrogens,"},{"Start":"02:18.320 ","End":"02:20.380","Text":"that\u0027s 2 times 1,"},{"Start":"02:20.380 ","End":"02:25.408","Text":"1 is the mass of a single hydrogen atom,"},{"Start":"02:25.408 ","End":"02:27.440","Text":"divided by 18,"},{"Start":"02:27.440 ","End":"02:32.600","Text":"that\u0027s the mass of a water molecule, molar mass."},{"Start":"02:32.600 ","End":"02:35.765","Text":"How can we calculate the percentage composition?"},{"Start":"02:35.765 ","End":"02:41.280","Text":"The percentage of hydrogen is the molar mass of hydrogen,"},{"Start":"02:41.280 ","End":"02:44.250","Text":"which is 1, multiplied by 2,"},{"Start":"02:44.250 ","End":"02:49.565","Text":"because they have 2 hydrogens in every molecule,"},{"Start":"02:49.565 ","End":"02:53.345","Text":"divided by 18, which is the molar mass of H_2_O,"},{"Start":"02:53.345 ","End":"02:54.935","Text":"so that\u0027s a ratio."},{"Start":"02:54.935 ","End":"02:59.930","Text":"It\u0027s a fraction and in order to get a fraction into a percentage,"},{"Start":"02:59.930 ","End":"03:01.940","Text":"we multiply by 100%."},{"Start":"03:01.940 ","End":"03:06.530","Text":"When we do this multiplication, we get 11%."},{"Start":"03:06.530 ","End":"03:09.380","Text":"Now let\u0027s look at the percentage of oxygen."},{"Start":"03:09.380 ","End":"03:15.320","Text":"We know the mass of 1 oxygen atom is 16 and the mass of H_2_O was 18."},{"Start":"03:15.320 ","End":"03:17.749","Text":"This gives us a ratio,"},{"Start":"03:17.749 ","End":"03:19.385","Text":"gives us a fraction."},{"Start":"03:19.385 ","End":"03:23.720","Text":"We multiply by 100% to get a percentage,"},{"Start":"03:23.720 ","End":"03:25.220","Text":"and we do the calculation,"},{"Start":"03:25.220 ","End":"03:29.190","Text":"we\u0027ll get that that\u0027s 89%."},{"Start":"03:29.260 ","End":"03:37.205","Text":"We see that the percentage of hydrogen is 11% and the percentage of oxygen is 89%,"},{"Start":"03:37.205 ","End":"03:39.930","Text":"just as we claimed before."},{"Start":"03:40.570 ","End":"03:44.060","Text":"We could have done this a slightly different way."},{"Start":"03:44.060 ","End":"03:49.918","Text":"If we know the percent of hydrogen is 11%,"},{"Start":"03:49.918 ","End":"03:54.380","Text":"and there\u0027s only hydrogen and oxygen in this particular molecule,"},{"Start":"03:54.380 ","End":"03:57.670","Text":"we can say that the percentage of oxygen is 100%,"},{"Start":"03:57.670 ","End":"03:59.210","Text":"that\u0027s the total,"},{"Start":"03:59.210 ","End":"04:01.714","Text":"minus percentage of hydrogen."},{"Start":"04:01.714 ","End":"04:06.035","Text":"That\u0027s 100 minus 11%, which is 89%."},{"Start":"04:06.035 ","End":"04:08.600","Text":"We get precisely the same answer,"},{"Start":"04:08.600 ","End":"04:10.975","Text":"either doing it this way or this way."},{"Start":"04:10.975 ","End":"04:13.745","Text":"It doesn\u0027t really matter which way you choose."},{"Start":"04:13.745 ","End":"04:16.280","Text":"In this video, we\u0027ve seen how we can calculate"},{"Start":"04:16.280 ","End":"04:20.010","Text":"the percentage composition from a chemical formula."}],"ID":17660},{"Watched":false,"Name":"Exercise 3","Duration":"2m 33s","ChapterTopicVideoID":18724,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/18724.jpeg","UploadDate":"2019-04-28T06:42:52.0170000","DurationForVideoObject":"PT2M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.504","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.504 ","End":"00:07.215","Text":"Determine the mass percent of hydrogen in the hydrocarbon hexane C_6H_14."},{"Start":"00:07.215 ","End":"00:10.810","Text":"The mass percent of an element"},{"Start":"00:12.470 ","End":"00:18.000","Text":"equals the number of atoms of the element for formula unit."},{"Start":"00:18.000 ","End":"00:21.345","Text":"That\u0027s the number of"},{"Start":"00:21.345 ","End":"00:29.160","Text":"atoms of the element for formula unit,"},{"Start":"00:29.160 ","End":"00:37.690","Text":"times the molar mass of the element,"},{"Start":"00:38.710 ","End":"00:43.530","Text":"divided by the molar mass"},{"Start":"00:45.260 ","End":"00:52.595","Text":"of the compound, times 100 percent."},{"Start":"00:52.595 ","End":"00:57.020","Text":"We want to calculate the mass percent of hydrogen in hexane."},{"Start":"00:57.020 ","End":"01:05.075","Text":"The mass percent of hydrogen equals the number of atoms of hydrogen in hexane."},{"Start":"01:05.075 ","End":"01:07.625","Text":"We can see is 14 hydrogen atoms."},{"Start":"01:07.625 ","End":"01:11.990","Text":"That\u0027s 14 times the molar mass of hydrogen,"},{"Start":"01:11.990 ","End":"01:15.750","Text":"which is 1.01 grams per mole,"},{"Start":"01:16.300 ","End":"01:20.870","Text":"divided by the molar mass of hexane."},{"Start":"01:20.870 ","End":"01:25.110","Text":"We\u0027re going to give them a calculate the molar mass of hexane, which is C_6H_14."},{"Start":"01:26.690 ","End":"01:30.680","Text":"This equals 6 because we have 6 carbon atoms,"},{"Start":"01:30.680 ","End":"01:34.370","Text":"times the molar mass of carbon plus 14."},{"Start":"01:34.370 ","End":"01:36.395","Text":"Since we have 14 hydrogen atoms,"},{"Start":"01:36.395 ","End":"01:38.675","Text":"times the molar mass of hydrogen."},{"Start":"01:38.675 ","End":"01:42.530","Text":"This equals 6 times 12.01 grams per"},{"Start":"01:42.530 ","End":"01:51.555","Text":"mole plus 14 times 1.01 grams per mole."},{"Start":"01:51.555 ","End":"01:55.640","Text":"The molar masses were taken from the periodic table of elements."},{"Start":"01:55.640 ","End":"02:00.360","Text":"Now this equals 86.2 grams per mole."},{"Start":"02:00.430 ","End":"02:04.115","Text":"That\u0027s the molar mass of hexane."},{"Start":"02:04.115 ","End":"02:06.410","Text":"The molar mass of hexane."},{"Start":"02:06.410 ","End":"02:09.120","Text":"I\u0027m going to put in the denominator."},{"Start":"02:10.270 ","End":"02:14.240","Text":"Now the units cancel out grams per mole divided by grams per mole."},{"Start":"02:14.240 ","End":"02:17.435","Text":"All this is times 100 percent."},{"Start":"02:17.435 ","End":"02:22.800","Text":"This equals 16.40 percent."},{"Start":"02:22.800 ","End":"02:29.945","Text":"The mass percent of hydrogen in hexane equals 16.40 percent."},{"Start":"02:29.945 ","End":"02:31.490","Text":"That is our final answer."},{"Start":"02:31.490 ","End":"02:34.170","Text":"Thank you very much for watching."}],"ID":23664},{"Watched":false,"Name":"Exercise 4","Duration":"6m 37s","ChapterTopicVideoID":18722,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/18722.jpeg","UploadDate":"2019-04-28T06:40:33.7630000","DurationForVideoObject":"PT6M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.495","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.495 ","End":"00:09.120","Text":"Determine the mass percent of each of the elements in methionine, C_5 H_11NO_2S."},{"Start":"00:09.120 ","End":"00:11.880","Text":"The mass percent of"},{"Start":"00:11.880 ","End":"00:18.440","Text":"an element equals the number of atoms of the element or a formula unit."},{"Start":"00:18.440 ","End":"00:29.625","Text":"It\u0027s the number of atoms of the element times the molar mass of the element"},{"Start":"00:29.625 ","End":"00:33.930","Text":"divided by the molar mass of"},{"Start":"00:33.930 ","End":"00:42.925","Text":"the compound and all of this is multiplied by 100 percent."},{"Start":"00:42.925 ","End":"00:45.035","Text":"Let\u0027s start with the carbon."},{"Start":"00:45.035 ","End":"00:48.065","Text":"See the percent of carbon equals,"},{"Start":"00:48.065 ","End":"00:53.205","Text":"first of all, we can see that there are 5 carbon atoms in methionine."},{"Start":"00:53.205 ","End":"00:56.240","Text":"It equals 5,"},{"Start":"00:56.240 ","End":"00:57.665","Text":"that\u0027s the number of atoms,"},{"Start":"00:57.665 ","End":"00:59.750","Text":"times the molar mass of carbon,"},{"Start":"00:59.750 ","End":"01:03.755","Text":"which is 12.01 grams per mole."},{"Start":"01:03.755 ","End":"01:10.805","Text":"This we need to divide by the molar mass of methionine."},{"Start":"01:10.805 ","End":"01:14.125","Text":"Now we\u0027re going to calculate the molar mass of methionine."},{"Start":"01:14.125 ","End":"01:21.975","Text":"The molar mass of methionine equals,"},{"Start":"01:21.975 ","End":"01:24.060","Text":"you can see that we have 5 carbons,"},{"Start":"01:24.060 ","End":"01:27.493","Text":"so it\u0027s 5 times the molar mass of carbon."},{"Start":"01:27.493 ","End":"01:30.335","Text":"We can see that we have 11 hydrogens,"},{"Start":"01:30.335 ","End":"01:37.130","Text":"plus 11 times the molar mass of hydrogen plus 1 nitrogen,"},{"Start":"01:37.130 ","End":"01:39.640","Text":"molar mass of nitrogen,"},{"Start":"01:39.640 ","End":"01:41.820","Text":"plus we have 2 oxygens,"},{"Start":"01:41.820 ","End":"01:49.050","Text":"so 2 times the molar mass of oxygen plus the molar mass of sulfur."},{"Start":"01:50.060 ","End":"02:00.510","Text":"This equals 5 times the molar mass of carbon is 12.01 grams per mole,"},{"Start":"02:00.510 ","End":"02:05.660","Text":"plus 11 times the molar mass of hydrogen,"},{"Start":"02:05.660 ","End":"02:09.409","Text":"1.01 grams per mole,"},{"Start":"02:09.409 ","End":"02:12.380","Text":"plus the molar mass of nitrogen,"},{"Start":"02:12.380 ","End":"02:17.930","Text":"which is 14.01 grams per mole,"},{"Start":"02:17.930 ","End":"02:23.480","Text":"plus 2 times the molar mass of oxygen,"},{"Start":"02:23.480 ","End":"02:26.880","Text":"which is 16 grams per mole,"},{"Start":"02:27.180 ","End":"02:32.290","Text":"plus the molar mass of sulfur,"},{"Start":"02:32.290 ","End":"02:39.220","Text":"which equals 32.07 grams per mole."},{"Start":"02:39.220 ","End":"02:47.080","Text":"This equals 149.24 grams per mole."},{"Start":"02:47.080 ","End":"02:53.405","Text":"The molar mass of methionine equals 149.24 grams per mole."},{"Start":"02:53.405 ","End":"02:56.380","Text":"Again, we\u0027re going back to the percent of carbon."},{"Start":"02:56.380 ","End":"03:00.425","Text":"We said that it\u0027s the number of atoms in methionine, which is 5,"},{"Start":"03:00.425 ","End":"03:03.440","Text":"times the molar mass of carbon,"},{"Start":"03:03.440 ","End":"03:05.390","Text":"which is 12.01 grams per mole,"},{"Start":"03:05.390 ","End":"03:08.975","Text":"divided by the molar mass of methionine,"},{"Start":"03:08.975 ","End":"03:14.510","Text":"which is 149.24 grams per"},{"Start":"03:14.510 ","End":"03:20.655","Text":"mole and all of this is times 100 percent."},{"Start":"03:20.655 ","End":"03:23.600","Text":"These grams per mole cancel out,"},{"Start":"03:23.600 ","End":"03:28.615","Text":"and this equals 40.24 percent."},{"Start":"03:28.615 ","End":"03:34.185","Text":"That\u0027s the mass percent of carbon, 40.24 percent."},{"Start":"03:34.185 ","End":"03:37.880","Text":"Now we\u0027re going to calculate the mass percent of hydrogen."},{"Start":"03:37.880 ","End":"03:41.585","Text":"The mass percent of hydrogen equals"},{"Start":"03:41.585 ","End":"03:46.640","Text":"again the number of atoms of hydrogen in methionine, which is 11,"},{"Start":"03:46.640 ","End":"03:49.700","Text":"times the molar mass of hydrogen,"},{"Start":"03:49.700 ","End":"03:54.020","Text":"which is 1.01 grams per mole,"},{"Start":"03:54.020 ","End":"03:57.770","Text":"divided by the molar mass of methionine,"},{"Start":"03:57.770 ","End":"04:01.715","Text":"which we calculated and equals 149.24"},{"Start":"04:01.715 ","End":"04:09.210","Text":"grams per mole times 100 percent."},{"Start":"04:09.580 ","End":"04:16.560","Text":"The grams per mole cancels out and we get 7.44 percent."},{"Start":"04:18.520 ","End":"04:22.429","Text":"That\u0027s the mass percent of hydrogen."},{"Start":"04:22.429 ","End":"04:25.060","Text":"Now we\u0027ll calculate the mass percent of nitrogen."},{"Start":"04:25.060 ","End":"04:28.490","Text":"The mass percent of nitrogen equals 1,"},{"Start":"04:28.490 ","End":"04:34.640","Text":"since we have 1 nitrogen atom in methionine times 14.01 grams per mole,"},{"Start":"04:34.640 ","End":"04:38.030","Text":"which is the molar mass of nitrogen,"},{"Start":"04:38.030 ","End":"04:41.945","Text":"divided by the molar mass of methionine,"},{"Start":"04:41.945 ","End":"04:52.410","Text":"which equals 149.24 grams per mole, times 100 percent."},{"Start":"04:52.660 ","End":"04:55.355","Text":"The grams per mole cancels out,"},{"Start":"04:55.355 ","End":"04:59.975","Text":"and this equals 9.39 percent."},{"Start":"04:59.975 ","End":"05:03.530","Text":"That\u0027s the mass percent of nitrogen."},{"Start":"05:03.530 ","End":"05:07.340","Text":"Now we will go on to calculate the mass percent of oxygen."},{"Start":"05:07.340 ","End":"05:12.380","Text":"The mass percent of oxygen equals the number of atoms of oxygen and methionine,"},{"Start":"05:12.380 ","End":"05:17.420","Text":"which are 2, that\u0027s 2 atoms times the molar mass of oxygen,"},{"Start":"05:17.420 ","End":"05:20.280","Text":"which is 16 grams per mole,"},{"Start":"05:20.320 ","End":"05:24.260","Text":"divided by the molar mass of methionine,"},{"Start":"05:24.260 ","End":"05:26.930","Text":"which equals 149.24 grams per"},{"Start":"05:26.930 ","End":"05:36.915","Text":"mole and all of this is times 100 percent."},{"Start":"05:36.915 ","End":"05:41.560","Text":"This comes to 21.44 percent."},{"Start":"05:41.560 ","End":"05:47.680","Text":"The mass percent of oxygen in methionine is 21.44 percent."},{"Start":"05:47.680 ","End":"05:49.575","Text":"Now, we\u0027ll go on to the sulfur."},{"Start":"05:49.575 ","End":"05:53.400","Text":"The mass percent of sulfur equals 1,"},{"Start":"05:53.400 ","End":"05:56.420","Text":"since we have 1 sulfur atom in methionine,"},{"Start":"05:56.420 ","End":"05:58.640","Text":"times the molar mass of sulfur,"},{"Start":"05:58.640 ","End":"06:03.085","Text":"which is 32.07 grams per mole,"},{"Start":"06:03.085 ","End":"06:06.920","Text":"divided by the molar mass of methionine,"},{"Start":"06:06.920 ","End":"06:09.680","Text":"which is 149.24 grams per"},{"Start":"06:09.680 ","End":"06:18.040","Text":"mole times 100 percent."},{"Start":"06:18.710 ","End":"06:28.045","Text":"The grams per mole cancel out and we get 21.49 percent."},{"Start":"06:28.045 ","End":"06:31.130","Text":"I just want to remind you that the molar masses of the elements"},{"Start":"06:31.130 ","End":"06:34.145","Text":"that we used were all taken from the periodic table of elements."},{"Start":"06:34.145 ","End":"06:35.825","Text":"That is our final answer."},{"Start":"06:35.825 ","End":"06:38.520","Text":"Thank you very much for watching."}],"ID":23665},{"Watched":false,"Name":"Exercise 5","Duration":"5m 16s","ChapterTopicVideoID":18723,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/18723.jpeg","UploadDate":"2019-04-28T06:42:18.1470000","DurationForVideoObject":"PT5M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.392","Text":"Hi, we\u0027re going to calculate the percent composition of hexanol,"},{"Start":"00:04.392 ","End":"00:07.275","Text":"CH_3, (CH_2)4 CH_2OH."},{"Start":"00:07.275 ","End":"00:12.513","Text":"First, we need to calculate the mass percent of the elements in hexanol."},{"Start":"00:12.513 ","End":"00:19.930","Text":"Again, CH_3, (CH_2)4, CH_2OH."},{"Start":"00:21.110 ","End":"00:23.310","Text":"If we look at the carbons,"},{"Start":"00:23.310 ","End":"00:24.390","Text":"we see that we have 4,"},{"Start":"00:24.390 ","End":"00:27.780","Text":"5, 6 carbons, we have C_6."},{"Start":"00:27.780 ","End":"00:30.690","Text":"If we look at the hydrogens, we have 8 from here,"},{"Start":"00:30.690 ","End":"00:33.768","Text":"so it\u0027s 8 plus 3, which is 11."},{"Start":"00:33.768 ","End":"00:36.345","Text":"Another 2 and another 1."},{"Start":"00:36.345 ","End":"00:42.270","Text":"In all, we have 14 hydrogens and we have 1 oxygen."},{"Start":"00:42.270 ","End":"00:44.920","Text":"Hexanol is C_6H_14O."},{"Start":"00:44.920 ","End":"00:47.135","Text":"That\u0027s just to simplify it for us."},{"Start":"00:47.135 ","End":"00:50.270","Text":"Now we\u0027re going to calculate the mass percent of the carbon, hydrogen,"},{"Start":"00:50.270 ","End":"00:54.305","Text":"and oxygen and then we\u0027ll know the percent composition of hexanol."},{"Start":"00:54.305 ","End":"01:01.985","Text":"The mass percent equals the number of atoms of the element in a formula unit,"},{"Start":"01:01.985 ","End":"01:06.174","Text":"times the molar mass of the element,"},{"Start":"01:06.174 ","End":"01:12.172","Text":"divided by the molar mass of the whole compound"},{"Start":"01:12.172 ","End":"01:18.760","Text":", times 100 percent."},{"Start":"01:18.760 ","End":"01:22.700","Text":"Let\u0027s first calculate the molar mass of hexanol."},{"Start":"01:22.700 ","End":"01:24.530","Text":"The molar mass of hexanol,"},{"Start":"01:24.530 ","End":"01:28.880","Text":"which is C_6H_14O equals,"},{"Start":"01:28.880 ","End":"01:30.230","Text":"we can see that we have 6 carbons,"},{"Start":"01:30.230 ","End":"01:33.504","Text":"so it\u0027s 6 times the molar mass of carbon,"},{"Start":"01:33.504 ","End":"01:35.640","Text":"plus 14 hydrogens,"},{"Start":"01:35.640 ","End":"01:38.570","Text":"14 times the molar mass of hydrogen,"},{"Start":"01:38.570 ","End":"01:40.553","Text":"plus the molar mass of oxygen,"},{"Start":"01:40.553 ","End":"01:42.280","Text":"since we only have 1."},{"Start":"01:42.280 ","End":"01:45.560","Text":"Now this equals 6 times the molar mass of carbon,"},{"Start":"01:45.560 ","End":"01:49.180","Text":"which is 12.01 grams per mole,"},{"Start":"01:49.180 ","End":"01:53.070","Text":"plus 14 times the molar mass of hydrogen,"},{"Start":"01:53.070 ","End":"01:56.620","Text":"which is 1.01 grams per mole,"},{"Start":"01:57.460 ","End":"02:00.260","Text":"plus the molar mass of oxygen,"},{"Start":"02:00.260 ","End":"02:02.910","Text":"which is 16 grams per mole."},{"Start":"02:04.730 ","End":"02:10.270","Text":"This equals 102.2 grams per mole."},{"Start":"02:10.570 ","End":"02:14.120","Text":"That\u0027s the molar mass of the hexanol."},{"Start":"02:14.120 ","End":"02:19.055","Text":"Now we\u0027re going to calculate the mass percent of the carbon."},{"Start":"02:19.055 ","End":"02:21.665","Text":"The mass percent of the carbon equals,"},{"Start":"02:21.665 ","End":"02:23.970","Text":"again, we said the number of atoms."},{"Start":"02:23.970 ","End":"02:27.130","Text":"We can see that we have 6 carbon atoms in hexanol,"},{"Start":"02:27.130 ","End":"02:30.440","Text":"it\u0027s going to be 6, times the molar mass of the carbon,"},{"Start":"02:30.440 ","End":"02:33.390","Text":"which is 12.01 grams per mole,"},{"Start":"02:34.820 ","End":"02:38.825","Text":"divided by the molar mass of hexanol,"},{"Start":"02:38.825 ","End":"02:47.920","Text":"which equals 102.2 grams per mole, times 100 percent."},{"Start":"02:47.920 ","End":"02:55.405","Text":"The grams per mole cancel out and this equals 70.51 percent."},{"Start":"02:55.405 ","End":"03:00.620","Text":"The mass percent of the carbon equals 70.51 percent."},{"Start":"03:00.620 ","End":"03:04.730","Text":"Now we\u0027re going to go on and calculate the mass percent of hydrogen."},{"Start":"03:04.730 ","End":"03:08.570","Text":"The mass percent of hydrogen equals the number of hydrogen atoms in hexanol,"},{"Start":"03:08.570 ","End":"03:12.950","Text":"which is 14, times the molar mass of hydrogen,"},{"Start":"03:12.950 ","End":"03:16.710","Text":"which is 1.01 grams per mole,"},{"Start":"03:16.710 ","End":"03:19.640","Text":"divided by the molar mass of hexanol,"},{"Start":"03:19.640 ","End":"03:22.740","Text":"which is 102.2 grams per mole,"},{"Start":"03:24.650 ","End":"03:28.390","Text":"and all of this is times 100 percent."},{"Start":"03:28.390 ","End":"03:30.560","Text":"Grams per mole cancel out,"},{"Start":"03:30.560 ","End":"03:36.540","Text":"and the mass percent of hydrogen equals 13.84 percent."},{"Start":"03:38.860 ","End":"03:42.680","Text":"Now we will calculate the mass percent of oxygen."},{"Start":"03:42.680 ","End":"03:46.305","Text":"The mass percent of oxygen equals,"},{"Start":"03:46.305 ","End":"03:48.480","Text":"we can see that we have 1 oxygen atom,"},{"Start":"03:48.480 ","End":"03:49.925","Text":"so it\u0027s 1,"},{"Start":"03:49.925 ","End":"03:51.920","Text":"times the molar mass of oxygen,"},{"Start":"03:51.920 ","End":"03:57.289","Text":"which is 16 grams per mole and divided by the molar mass of hexanol,"},{"Start":"03:57.289 ","End":"04:05.495","Text":"which we calculated and it\u0027s 102.2 grams per mole, times 100 percent."},{"Start":"04:05.495 ","End":"04:11.935","Text":"The grams per mole cancel out and this equals 15.66 percent."},{"Start":"04:11.935 ","End":"04:15.790","Text":"We were asked to find the percent composition of hexanol."},{"Start":"04:15.790 ","End":"04:19.400","Text":"The percent composition of hexanol"},{"Start":"04:28.550 ","End":"04:34.810","Text":"is 70.51 percent carbon,"},{"Start":"04:36.770 ","End":"04:40.620","Text":"13.84 percent hydrogen,"},{"Start":"04:40.620 ","End":"04:45.940","Text":"and 15.66 percent oxygen."},{"Start":"04:48.830 ","End":"04:51.845","Text":"Now one more thing I want to say,"},{"Start":"04:51.845 ","End":"04:56.780","Text":"just if we take all the mass percent of the carbon, of the hydrogen,"},{"Start":"04:56.780 ","End":"05:02.370","Text":"and the oxygen in a compound and we sum them up,"},{"Start":"05:02.370 ","End":"05:05.654","Text":"they equal 100 percent."},{"Start":"05:05.654 ","End":"05:12.495","Text":"70.51 plus 13.84 plus 15.66 percent comes to 100 percent."},{"Start":"05:12.495 ","End":"05:14.253","Text":"That is our final answer."},{"Start":"05:14.253 ","End":"05:16.750","Text":"Thank you very much for watching."}],"ID":23662},{"Watched":false,"Name":"Determining Formula from Composition","Duration":"3m 33s","ChapterTopicVideoID":16916,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16916.jpeg","UploadDate":"2019-02-20T23:39:38.4700000","DurationForVideoObject":"PT3M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:02.415","Text":"In the previous video,"},{"Start":"00:02.415 ","End":"00:06.615","Text":"we learned how to calculate the percentage composition from a chemical formula."},{"Start":"00:06.615 ","End":"00:09.269","Text":"We now learned how to do the reverse process,"},{"Start":"00:09.269 ","End":"00:13.005","Text":"how to calculate the empirical formula from the composition."},{"Start":"00:13.005 ","End":"00:17.505","Text":"We\u0027re going to calculate the empirical formula from percentage composition."},{"Start":"00:17.505 ","End":"00:21.195","Text":"Once again, we\u0027re going to take the example of water."},{"Start":"00:21.195 ","End":"00:27.765","Text":"We know from the previous video that water is 89% oxygen and 11% hydrogen."},{"Start":"00:27.765 ","End":"00:31.905","Text":"Let\u0027s take a sample of water with mass 100 grams."},{"Start":"00:31.905 ","End":"00:36.120","Text":"You can take all sorts of masses but this is the simplest thing to do."},{"Start":"00:36.120 ","End":"00:38.280","Text":"Here we have a little table,"},{"Start":"00:38.280 ","End":"00:40.635","Text":"oxygen with 89 grams,"},{"Start":"00:40.635 ","End":"00:42.795","Text":"hydrogen is 11 grams."},{"Start":"00:42.795 ","End":"00:46.280","Text":"The second step is to calculate the number of moles of O and"},{"Start":"00:46.280 ","End":"00:49.865","Text":"H by dividing the mass by molar atomic mass,"},{"Start":"00:49.865 ","End":"00:51.470","Text":"16 grams for oxygen,"},{"Start":"00:51.470 ","End":"00:53.395","Text":"1 gram for hydrogen."},{"Start":"00:53.395 ","End":"01:00.335","Text":"When we do that, we get 89 grams of oxygen divided by 16 grams per mole."},{"Start":"01:00.335 ","End":"01:03.950","Text":"That\u0027s the molar mass of oxygen."},{"Start":"01:03.950 ","End":"01:07.640","Text":"That gives us 5.56 moles."},{"Start":"01:07.640 ","End":"01:13.190","Text":"For hydrogen, we have 11 grams divided by the molar mass of hydrogen,"},{"Start":"01:13.190 ","End":"01:17.500","Text":"1 gram per mole and that gives us 11 moles."},{"Start":"01:17.500 ","End":"01:19.760","Text":"What we found so far,"},{"Start":"01:19.760 ","End":"01:29.365","Text":"we found that the ratio of oxygen to hydrogen is 5.56-11."},{"Start":"01:29.365 ","End":"01:34.415","Text":"We don\u0027t usually write formulas like that, with such numbers."},{"Start":"01:34.415 ","End":"01:37.129","Text":"We need small whole numbers."},{"Start":"01:37.129 ","End":"01:41.300","Text":"The empirical formula has the smallest whole numbers possible."},{"Start":"01:41.300 ","End":"01:46.925","Text":"One easy way of getting to that is to divide by the smallest number."},{"Start":"01:46.925 ","End":"01:49.055","Text":"That\u0027s step 3."},{"Start":"01:49.055 ","End":"01:51.560","Text":"Divide by the smallest number of moles,"},{"Start":"01:51.560 ","End":"01:56.440","Text":"and the result is dimensionless because we\u0027re dividing moles by moles."},{"Start":"01:56.440 ","End":"02:01.080","Text":"We have 5.56 divide by 5.56 gives us 1."},{"Start":"02:01.080 ","End":"02:05.415","Text":"11 divide by 5.56 is approximately 2."},{"Start":"02:05.415 ","End":"02:09.465","Text":"This tells us the ratio of O to H is 1-2."},{"Start":"02:09.465 ","End":"02:12.310","Text":"The empirical formula is H_2."},{"Start":"02:12.310 ","End":"02:16.160","Text":"Now such a calculation just gives us the empirical formula."},{"Start":"02:16.160 ","End":"02:18.290","Text":"If we want to know the molecular formula,"},{"Start":"02:18.290 ","End":"02:20.255","Text":"we need more information."},{"Start":"02:20.255 ","End":"02:24.545","Text":"How do we calculate the molecular formula from the empirical formula?"},{"Start":"02:24.545 ","End":"02:28.835","Text":"The extra information that\u0027s required is the molar mass of water."},{"Start":"02:28.835 ","End":"02:31.990","Text":"Supposing it was 18 grams per mole,"},{"Start":"02:31.990 ","End":"02:35.115","Text":"then the molecular formula would be H_2O."},{"Start":"02:35.115 ","End":"02:36.465","Text":"Why H_2O?"},{"Start":"02:36.465 ","End":"02:43.590","Text":"Because 2 hydrogen weighs 2 times 1 and 1 oxygen weighs 1 times 16."},{"Start":"02:43.590 ","End":"02:46.220","Text":"The total is 18 grams."},{"Start":"02:46.220 ","End":"02:52.040","Text":"However, if by some chance it was 36 grams per mole,"},{"Start":"02:52.040 ","End":"02:55.790","Text":"then that is twice as much as 18 grams per mole."},{"Start":"02:55.790 ","End":"03:00.950","Text":"The molecular formula would be doubled, would be H_4O_2."},{"Start":"03:00.950 ","End":"03:04.565","Text":"Notice that we always keep the same ratio."},{"Start":"03:04.565 ","End":"03:06.545","Text":"The ratio here is 2-1."},{"Start":"03:06.545 ","End":"03:09.570","Text":"The ratio here is also 2-1."},{"Start":"03:12.260 ","End":"03:15.890","Text":"Now we have the molecular formula."},{"Start":"03:15.890 ","End":"03:23.120","Text":"Now we know that in fact the molar mass of water is 18 grams per mole."},{"Start":"03:23.120 ","End":"03:27.040","Text":"The molecular form is indeed H_2O."},{"Start":"03:27.040 ","End":"03:29.270","Text":"In this video, we learned how to calculate"},{"Start":"03:29.270 ","End":"03:33.240","Text":"the empirical formula from the percentage composition."}],"ID":17661},{"Watched":false,"Name":"Exercise 6","Duration":"4m 41s","ChapterTopicVideoID":18725,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/18725.jpeg","UploadDate":"2019-04-28T06:40:35.1300000","DurationForVideoObject":"PT4M41S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.535","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.535 ","End":"00:06.690","Text":"Determine the empirical formula of a carbon-hydrogen compound which consists of"},{"Start":"00:06.690 ","End":"00:11.730","Text":"82.63 percent carbon and 17.37 percent hydrogen by mass."},{"Start":"00:11.730 ","End":"00:16.215","Text":"Now, we\u0027re asked to find the empirical formula of a carbon-hydrogen compounds."},{"Start":"00:16.215 ","End":"00:19.785","Text":"The empirical formula is the simplest formula for a compound."},{"Start":"00:19.785 ","End":"00:23.730","Text":"It shows the types of atoms present and their relative numbers."},{"Start":"00:23.730 ","End":"00:29.010","Text":"The subscripts in empirical formula are reduced to the simplest whole number ratio."},{"Start":"00:29.010 ","End":"00:31.035","Text":"Let\u0027s take an example."},{"Start":"00:31.035 ","End":"00:33.670","Text":"If we take the compound P_4H_10,"},{"Start":"00:34.370 ","End":"00:40.780","Text":"we can see that the subscripts are 4 and 10 and their ratio is 4:10."},{"Start":"00:40.780 ","End":"00:46.505","Text":"This also equals, if we divide the 4 and 10 by 2 equals 2:5."},{"Start":"00:46.505 ","End":"00:50.033","Text":"That\u0027s the smallest whole number ratio that is possible."},{"Start":"00:50.033 ","End":"00:53.570","Text":"We can\u0027t divide by another number and get small whole numbers."},{"Start":"00:53.570 ","End":"01:00.670","Text":"P_4H_10, its empirical formula equals P_2H_5."},{"Start":"01:00.670 ","End":"01:02.915","Text":"Now we\u0027re going to continue to the question."},{"Start":"01:02.915 ","End":"01:08.327","Text":"We have to determine the empirical formula of a carbon-hydrogen compound."},{"Start":"01:08.327 ","End":"01:10.730","Text":"We know the mass percent of carbon,"},{"Start":"01:10.730 ","End":"01:13.025","Text":"and the mass percent of hydrogen."},{"Start":"01:13.025 ","End":"01:16.190","Text":"There are a couple of steps. We\u0027re going to start with the first step."},{"Start":"01:16.190 ","End":"01:20.360","Text":"The first step is to assume that we have 100 grams in our sample."},{"Start":"01:20.360 ","End":"01:22.580","Text":"If we had 100 grams and our sample,"},{"Start":"01:22.580 ","End":"01:32.510","Text":"we\u0027re going to have 82.63 grams of carbon and 17.37 grams of hydrogen."},{"Start":"01:32.510 ","End":"01:36.245","Text":"That\u0027s our first step. The second step"},{"Start":"01:36.245 ","End":"01:39.875","Text":"is to convert the masses we have in the first step into moles."},{"Start":"01:39.875 ","End":"01:43.280","Text":"Remember that n, number of moles, equals m,"},{"Start":"01:43.280 ","End":"01:46.250","Text":"which is the mass, divided by the molar mass."},{"Start":"01:46.250 ","End":"01:48.980","Text":"The number of moles of carbon equal the mass of"},{"Start":"01:48.980 ","End":"01:52.520","Text":"carbon divided by the molar mass of carbon."},{"Start":"01:52.520 ","End":"01:57.500","Text":"The mass of carbon equals 82.63 grams,"},{"Start":"01:57.500 ","End":"01:59.780","Text":"divided by the molar mass of carbon,"},{"Start":"01:59.780 ","End":"02:04.105","Text":"which is 12.01 grams per mole."},{"Start":"02:04.105 ","End":"02:08.645","Text":"The molar mass of carbon is taken from the periodic table of elements."},{"Start":"02:08.645 ","End":"02:15.000","Text":"This equals 6.88 moles of carbon."},{"Start":"02:15.000 ","End":"02:17.360","Text":"Now here, if we look at our units,"},{"Start":"02:17.360 ","End":"02:19.370","Text":"we have grams divided by grams per mole."},{"Start":"02:19.370 ","End":"02:23.090","Text":"I just want to remind you that when you\u0027re dividing by a fraction,"},{"Start":"02:23.090 ","End":"02:25.070","Text":"we have grams divided by grams per mole,"},{"Start":"02:25.070 ","End":"02:30.170","Text":"which is a fraction, it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"02:30.170 ","End":"02:33.575","Text":"It\u0027s the same as grams times mole per grams."},{"Start":"02:33.575 ","End":"02:36.630","Text":"Grams cancel out and we\u0027re left with moles."},{"Start":"02:36.630 ","End":"02:38.615","Text":"Our units are mole."},{"Start":"02:38.615 ","End":"02:40.865","Text":"Now we\u0027re going to calculate the moles of hydrogen."},{"Start":"02:40.865 ","End":"02:46.820","Text":"The moles of hydrogen equals the mass of hydrogen divided by the molar mass of hydrogen."},{"Start":"02:46.820 ","End":"02:49.280","Text":"In this equals the mass of hydrogen,"},{"Start":"02:49.280 ","End":"02:52.380","Text":"which is 17.37 grams,"},{"Start":"02:52.990 ","End":"02:55.880","Text":"divided by the molar mass of hydrogen,"},{"Start":"02:55.880 ","End":"02:58.950","Text":"which is 1.01 grams per mole."},{"Start":"02:59.200 ","End":"03:03.870","Text":"This comes to 17.20 mole."},{"Start":"03:03.910 ","End":"03:08.480","Text":"Now we know the moles of carbon and the moles of hydrogen."},{"Start":"03:08.480 ","End":"03:11.195","Text":"The next step, step number 3,"},{"Start":"03:11.195 ","End":"03:14.780","Text":"is to write a formula based on the moles we just found."},{"Start":"03:14.780 ","End":"03:17.840","Text":"Carbon is going to be 6.88,"},{"Start":"03:17.840 ","End":"03:21.445","Text":"and hydrogen is 17.20."},{"Start":"03:21.445 ","End":"03:24.195","Text":"The next step, step number 4,"},{"Start":"03:24.195 ","End":"03:28.775","Text":"is to take our subscripts and divide them by the smallest subscript."},{"Start":"03:28.775 ","End":"03:31.175","Text":"The smallest subscript is 6.88."},{"Start":"03:31.175 ","End":"03:35.325","Text":"We\u0027re going to divide both these subscripts by 6.88."},{"Start":"03:35.325 ","End":"03:39.180","Text":"In carbon, 6.88 divided by 6.88 equals 1,"},{"Start":"03:39.180 ","End":"03:48.390","Text":"and then hydrogen 17.20 divided by 6.88 equals 2.5."},{"Start":"03:48.390 ","End":"03:50.490","Text":"I\u0027m just going to put that there."},{"Start":"03:50.490 ","End":"03:52.200","Text":"It\u0027s hydrogen 2.5."},{"Start":"03:52.200 ","End":"03:53.460","Text":"Now our last step,"},{"Start":"03:53.460 ","End":"03:55.090","Text":"step number 5,"},{"Start":"03:55.090 ","End":"04:01.325","Text":"is to multiply all our subscripts by a small whole number to make them integral."},{"Start":"04:01.325 ","End":"04:03.545","Text":"Now we\u0027re going to multiply these by 2,"},{"Start":"04:03.545 ","End":"04:06.130","Text":"because 1 times 2 is going to give us 2,"},{"Start":"04:06.130 ","End":"04:09.070","Text":"and 2.5 times 2 is going to give us 5."},{"Start":"04:09.070 ","End":"04:13.460","Text":"Now note that I took the smallest whole number because I could\u0027ve taken,"},{"Start":"04:13.460 ","End":"04:16.850","Text":"for example, for just going to write that here."},{"Start":"04:16.850 ","End":"04:20.370","Text":"If I multiply 1 by 4 it\u0027s going to give me C_4 and"},{"Start":"04:20.370 ","End":"04:24.295","Text":"the hydrogen 2.5 times 4 is going to give me 10."},{"Start":"04:24.295 ","End":"04:26.765","Text":"But this will not be the empirical formula"},{"Start":"04:26.765 ","End":"04:30.440","Text":"because it is not the smallest whole number ratio."},{"Start":"04:30.440 ","End":"04:33.080","Text":"C_2H_5 is a smallest whole number ratio,"},{"Start":"04:33.080 ","End":"04:35.329","Text":"and therefore that\u0027s our empirical formula."},{"Start":"04:35.329 ","End":"04:38.220","Text":"Empirical formula is C_2H_5,"},{"Start":"04:38.220 ","End":"04:39.635","Text":"that\u0027s our final answer."},{"Start":"04:39.635 ","End":"04:42.270","Text":"Thank you very much for watching."}],"ID":23658},{"Watched":false,"Name":"Exercise 7","Duration":"5m 14s","ChapterTopicVideoID":18726,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/18726.jpeg","UploadDate":"2019-04-28T06:42:43.7330000","DurationForVideoObject":"PT5M14S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.045","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.045 ","End":"00:07.320","Text":"Determine the empirical formula of a carbon-hydrogen oxygen compound with"},{"Start":"00:07.320 ","End":"00:11.970","Text":"39.99 percent carbon and 6.73 percent hydrogen by mass."},{"Start":"00:11.970 ","End":"00:16.020","Text":"The empirical formula of a compound is the simplest formula for a compound."},{"Start":"00:16.020 ","End":"00:19.635","Text":"It shows the types of atoms present and the relative numbers."},{"Start":"00:19.635 ","End":"00:21.900","Text":"For example, if we take P_4H_10,"},{"Start":"00:24.500 ","End":"00:33.030","Text":"the empirical formula equals P_2H_5 because this is the simplest formula of the compound."},{"Start":"00:33.030 ","End":"00:34.440","Text":"In an empirical formula,"},{"Start":"00:34.440 ","End":"00:40.430","Text":"the subscripts in the formula are reduced to their simplest whole number ratio,"},{"Start":"00:40.430 ","End":"00:42.220","Text":"meaning if we take our subscripts and P_4H_10,"},{"Start":"00:42.220 ","End":"00:45.465","Text":"the ratio is 4-10."},{"Start":"00:45.465 ","End":"00:49.870","Text":"We can divide the 4 and the 10 by 2, we get 2-5."},{"Start":"00:49.870 ","End":"00:51.800","Text":"That\u0027s our simplest whole number ratio."},{"Start":"00:51.800 ","End":"00:57.365","Text":"Therefore, P_4H_10 will be P_2H_5, the empirical formula."},{"Start":"00:57.365 ","End":"01:02.120","Text":"Now we know that our compound is a carbon-hydrogen oxygen compound,"},{"Start":"01:02.120 ","End":"01:05.570","Text":"and we know that our mass percent of carbon is 39.99 percent,"},{"Start":"01:05.570 ","End":"01:08.615","Text":"and our mass percent of hydrogen is 6.73 percent."},{"Start":"01:08.615 ","End":"01:11.690","Text":"However, we were not given the mass percent of oxygen."},{"Start":"01:11.690 ","End":"01:14.540","Text":"The sum of all of the mass percent has to give us"},{"Start":"01:14.540 ","End":"01:17.905","Text":"100 percent because it\u0027s a carbon-hydrogen oxygen compound."},{"Start":"01:17.905 ","End":"01:22.720","Text":"The mass percent of oxygen equals 100 percent,"},{"Start":"01:22.720 ","End":"01:26.355","Text":"minus the mass percent of carbon,"},{"Start":"01:26.355 ","End":"01:29.285","Text":"minus the mass percent of hydrogen."},{"Start":"01:29.285 ","End":"01:36.120","Text":"This equals 100 percent minus 39.99 percent,"},{"Start":"01:36.120 ","End":"01:40.065","Text":"which is the carbon, minus 6.73 percent,"},{"Start":"01:40.065 ","End":"01:46.395","Text":"which is the hydrogen, and this comes to 53.28 percent."},{"Start":"01:46.395 ","End":"01:49.290","Text":"That\u0027s the mass percent of oxygen."},{"Start":"01:49.290 ","End":"01:53.585","Text":"Now we have the mass percent of carbon, hydrogen, and oxygen."},{"Start":"01:53.585 ","End":"01:57.240","Text":"The first step is to assume we have 100 gram sample."},{"Start":"01:57.240 ","End":"01:58.310","Text":"In our 100 gram sample,"},{"Start":"01:58.310 ","End":"02:02.730","Text":"we have 39.99 grams of carbon,"},{"Start":"02:02.730 ","End":"02:06.430","Text":"we have 6.73 grams of hydrogen,"},{"Start":"02:07.370 ","End":"02:13.015","Text":"and 53.28 grams of oxygen."},{"Start":"02:13.015 ","End":"02:14.915","Text":"That\u0027s our first step."},{"Start":"02:14.915 ","End":"02:18.350","Text":"In our second step, we\u0027re going to convert these masses into moles."},{"Start":"02:18.350 ","End":"02:20.390","Text":"We\u0027re going to use the equation n,"},{"Start":"02:20.390 ","End":"02:21.950","Text":"which is the number of moles equals m,"},{"Start":"02:21.950 ","End":"02:25.030","Text":"which is the mass divided by the molar mass."},{"Start":"02:25.030 ","End":"02:31.190","Text":"Number of moles of carbon equals the mass of carbon divided by the molar mass of carbon."},{"Start":"02:31.190 ","End":"02:38.780","Text":"This equals 39.99 grams divided by the molar mass of carbon,"},{"Start":"02:38.780 ","End":"02:41.740","Text":"which is 12.01 grams per mole."},{"Start":"02:41.740 ","End":"02:45.530","Text":"The molar mass is taken from the periodic table of elements."},{"Start":"02:45.530 ","End":"02:50.720","Text":"This equals 3.33 moles of carbon."},{"Start":"02:50.720 ","End":"02:53.120","Text":"Now we\u0027re going to calculate the moles of hydrogen."},{"Start":"02:53.120 ","End":"02:58.220","Text":"That\u0027s equals the mass of hydrogen divided by the molar mass of hydrogen."},{"Start":"02:58.220 ","End":"03:05.660","Text":"The mass of hydrogen equals 6.73 grams divided by the molar mass of hydrogen,"},{"Start":"03:05.660 ","End":"03:09.080","Text":"which is 1.01 grams per mole,"},{"Start":"03:09.080 ","End":"03:12.600","Text":"and this equals 6.66 mole."},{"Start":"03:12.600 ","End":"03:14.690","Text":"Now I just want to remind you we had this"},{"Start":"03:14.690 ","End":"03:18.590","Text":"also in calculating the moles of carbon and the moles of hydrogen."},{"Start":"03:18.590 ","End":"03:22.055","Text":"If we look at our units, we have grams divide by grams per mole."},{"Start":"03:22.055 ","End":"03:25.835","Text":"We\u0027re dividing by a fraction dividing by grams per mole."},{"Start":"03:25.835 ","End":"03:28.025","Text":"When you divide by a fraction,"},{"Start":"03:28.025 ","End":"03:31.760","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction,"},{"Start":"03:31.760 ","End":"03:36.755","Text":"so grams divided by grams per mole equals grams times mole per grams."},{"Start":"03:36.755 ","End":"03:39.380","Text":"The grams cancel out and we\u0027re left with mole."},{"Start":"03:39.380 ","End":"03:42.050","Text":"Therefore our units are mole."},{"Start":"03:42.050 ","End":"03:45.110","Text":"We\u0027re going on to calculate the moles of oxygen,"},{"Start":"03:45.110 ","End":"03:50.240","Text":"and this equals the mass of oxygen divided by the molar mass of oxygen."},{"Start":"03:50.240 ","End":"03:54.815","Text":"This equals the mass of oxygen is 53.28 grams,"},{"Start":"03:54.815 ","End":"03:57.200","Text":"divided by the molar mass of oxygen,"},{"Start":"03:57.200 ","End":"04:04.560","Text":"which is 16 grams per mole and this equals 3.33 mole."},{"Start":"04:04.940 ","End":"04:10.100","Text":"This third step is to write a formula based on the moles we have calculated."},{"Start":"04:10.100 ","End":"04:13.650","Text":"That\u0027s C_3.33, H_6.66, and O_3.33."},{"Start":"04:18.880 ","End":"04:25.140","Text":"The fourth step is to divide these subscripts by the smallest subscript."},{"Start":"04:25.140 ","End":"04:27.041","Text":"The smallest subscript here is 3.33."},{"Start":"04:27.041 ","End":"04:30.270","Text":"We\u0027re going to divide all of these subscripts by 3.33,"},{"Start":"04:30.270 ","End":"04:34.900","Text":"so 3.33 divided by 3.33 equals 1."},{"Start":"04:34.900 ","End":"04:36.590","Text":"If we go on to the hydrogen,"},{"Start":"04:36.590 ","End":"04:39.965","Text":"6.66 divided by 3.33 equals 2,"},{"Start":"04:39.965 ","End":"04:44.120","Text":"and oxygen 3.33 divide by 333 is going to give us 1."},{"Start":"04:44.120 ","End":"04:46.130","Text":"Now when we have a chemical formula,"},{"Start":"04:46.130 ","End":"04:48.250","Text":"we don\u0027t have to actually write the 1."},{"Start":"04:48.250 ","End":"04:51.960","Text":"This equals CH_2O."},{"Start":"04:52.030 ","End":"04:55.670","Text":"Since we already have whole numbers here,"},{"Start":"04:55.670 ","End":"04:58.370","Text":"1, 2, and 1, we don\u0027t have to do step 5."},{"Start":"04:58.370 ","End":"05:00.740","Text":"Step 5 is when we don\u0027t have whole numbers and then we have to"},{"Start":"05:00.740 ","End":"05:03.440","Text":"multiply by a small whole numbers to give us integers."},{"Start":"05:03.440 ","End":"05:05.320","Text":"But in this case, we have whole numbers,"},{"Start":"05:05.320 ","End":"05:07.250","Text":"so we don\u0027t have to multiply by anything."},{"Start":"05:07.250 ","End":"05:11.315","Text":"In this case, CH_2O is our empirical formula."},{"Start":"05:11.315 ","End":"05:12.710","Text":"That is our final answer."},{"Start":"05:12.710 ","End":"05:15.120","Text":"Thank you very much for watching."}],"ID":23661},{"Watched":false,"Name":"Exercise 8","Duration":"7m 51s","ChapterTopicVideoID":18727,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/18727.jpeg","UploadDate":"2019-04-28T06:44:05.3070000","DurationForVideoObject":"PT7M51S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.519","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.519 ","End":"00:07.110","Text":"A compound has a percent composition by mass of 74.03 percent carbon,"},{"Start":"00:07.110 ","End":"00:09.360","Text":"7.47 percent hydrogen,"},{"Start":"00:09.360 ","End":"00:11.100","Text":"8.64 percent nitrogen,"},{"Start":"00:11.100 ","End":"00:13.590","Text":"and 9.86 percent oxygen."},{"Start":"00:13.590 ","End":"00:16.140","Text":"Determine the molecular formula of the compound."},{"Start":"00:16.140 ","End":"00:20.190","Text":"The molecular mass of the compound is 324.5 u."},{"Start":"00:20.190 ","End":"00:23.460","Text":"In order to determine the molecular formula of the compound,"},{"Start":"00:23.460 ","End":"00:26.175","Text":"we must first determine its empirical formula."},{"Start":"00:26.175 ","End":"00:29.010","Text":"The empirical formula is the simplest formula for a compound"},{"Start":"00:29.010 ","End":"00:32.430","Text":"and it shows the types of atom present and their relative numbers."},{"Start":"00:32.430 ","End":"00:36.040","Text":"For example, if we take P_4H_10,"},{"Start":"00:36.040 ","End":"00:38.855","Text":"if we look at the ratio between the subscripts,"},{"Start":"00:38.855 ","End":"00:41.450","Text":"we have 4 and 10, so it\u0027s 4-10."},{"Start":"00:41.450 ","End":"00:45.440","Text":"If we divide both the 4 and 10 by 2, we get 2.5."},{"Start":"00:45.440 ","End":"00:48.230","Text":"This is our simplest whole-number ratio."},{"Start":"00:48.230 ","End":"00:51.680","Text":"The empirical formula is the simplest formula for a compound,"},{"Start":"00:51.680 ","End":"00:53.360","Text":"meaning the subscripts in"},{"Start":"00:53.360 ","End":"00:56.795","Text":"an empirical formula are reduced to their simplest whole-number ratio."},{"Start":"00:56.795 ","End":"00:58.100","Text":"Instead of P_4H_10,"},{"Start":"00:58.100 ","End":"01:02.075","Text":"we\u0027re going to have P_2H_5 in our empirical formula."},{"Start":"01:02.075 ","End":"01:05.150","Text":"In order to find the empirical formula, there are a number of steps."},{"Start":"01:05.150 ","End":"01:10.085","Text":"The first step is to assume we have a sample of 100 grams in our compounds."},{"Start":"01:10.085 ","End":"01:13.830","Text":"We have 74.03 grams of carbon,"},{"Start":"01:14.290 ","End":"01:18.510","Text":"7.47 grams of hydrogen,"},{"Start":"01:19.810 ","End":"01:24.090","Text":"8.64 grams of nitrogen,"},{"Start":"01:25.990 ","End":"01:30.899","Text":"and 9.86 grams of oxygen."},{"Start":"01:31.970 ","End":"01:37.130","Text":"Our second step is to take these masses and convert them into moles."},{"Start":"01:37.130 ","End":"01:39.530","Text":"For this purpose, we\u0027re going to use the equation n,"},{"Start":"01:39.530 ","End":"01:41.060","Text":"the number of moles equals m,"},{"Start":"01:41.060 ","End":"01:43.340","Text":"which is mass divided by the molar mass."},{"Start":"01:43.340 ","End":"01:44.360","Text":"The number of moles,"},{"Start":"01:44.360 ","End":"01:47.030","Text":"we\u0027re going to start with the carbon of carbon equals the mass of"},{"Start":"01:47.030 ","End":"01:50.000","Text":"carbon divided by the molar mass of carbon."},{"Start":"01:50.000 ","End":"01:56.720","Text":"This equals 74.03 grams divided by the molar mass of carbon,"},{"Start":"01:56.720 ","End":"02:00.450","Text":"which is 12.01 grams per mole,"},{"Start":"02:00.450 ","End":"02:04.735","Text":"and this equals 6.16 moles of carbon."},{"Start":"02:04.735 ","End":"02:08.870","Text":"Just a reminder, when you\u0027re dividing by a fraction if we look at our units here,"},{"Start":"02:08.870 ","End":"02:11.140","Text":"we have grams divided by grams per mole."},{"Start":"02:11.140 ","End":"02:16.010","Text":"Dividing by a fraction equals the same as multiplying by the reciprocal of the fraction."},{"Start":"02:16.010 ","End":"02:18.230","Text":"Grams divided by grams per mole equals grams,"},{"Start":"02:18.230 ","End":"02:20.735","Text":"times mole per grams."},{"Start":"02:20.735 ","End":"02:23.855","Text":"The grams cancel out and we\u0027re left with mole."},{"Start":"02:23.855 ","End":"02:30.474","Text":"The moles of hydrogen equal the mass of hydrogen divided by the molar mass of hydrogen."},{"Start":"02:30.474 ","End":"02:34.100","Text":"This equals 7.47 grams,"},{"Start":"02:34.540 ","End":"02:39.145","Text":"divided by 1.01 grams per mole,"},{"Start":"02:39.145 ","End":"02:43.835","Text":"and this equals 7.40 moles of hydrogen."},{"Start":"02:43.835 ","End":"02:46.295","Text":"Now we\u0027re going to calculate the moles of nitrogen."},{"Start":"02:46.295 ","End":"02:54.200","Text":"This equals 8.64 grams divided by the molar mass of nitrogen,"},{"Start":"02:54.200 ","End":"02:58.820","Text":"which equals 14.01 grams per mole,"},{"Start":"02:58.820 ","End":"03:03.620","Text":"and this equals 0.62 moles of nitrogen."},{"Start":"03:03.620 ","End":"03:09.380","Text":"The moles of oxygen equal the mass of oxygen divided by the molar mass of oxygen,"},{"Start":"03:09.380 ","End":"03:17.150","Text":"and this equals 9.86 grams divided by the molar mass of oxygen,"},{"Start":"03:17.150 ","End":"03:19.820","Text":"which is 16 grams per mole."},{"Start":"03:19.820 ","End":"03:24.545","Text":"All of the molar masses of the elements were taken from the periodic table of elements."},{"Start":"03:24.545 ","End":"03:27.650","Text":"This equals 0.62 mole."},{"Start":"03:27.650 ","End":"03:31.107","Text":"Now we calculated the moles of the elements."},{"Start":"03:31.107 ","End":"03:33.920","Text":"The next step is to write a formula based on these moles."},{"Start":"03:33.920 ","End":"03:35.795","Text":"We have C,"},{"Start":"03:35.795 ","End":"03:39.920","Text":"6.16, hydrogen,"},{"Start":"03:39.920 ","End":"03:44.840","Text":"7.40, nitrogen,"},{"Start":"03:44.840 ","End":"03:53.000","Text":"0.62, and oxygen, 0.62."},{"Start":"03:53.000 ","End":"03:59.705","Text":"Now step number 4 is to take our subscripts and divide them by the smallest subscript."},{"Start":"03:59.705 ","End":"04:01.417","Text":"Our smallest subscript is 0.62,"},{"Start":"04:01.417 ","End":"04:05.300","Text":"we\u0027re going to divide all of our subscripts by 0.62."},{"Start":"04:05.300 ","End":"04:11.958","Text":"Carbon 6.16 divided by 0.62 gives us 9.94."},{"Start":"04:11.958 ","End":"04:19.260","Text":"Hydrogen, 7.40 divided by 0.62 gives us 11.94."},{"Start":"04:19.260 ","End":"04:22.330","Text":"Nitrogen, we have 1,"},{"Start":"04:22.330 ","End":"04:24.750","Text":"since 0.62 divided by 0.62 is 1,"},{"Start":"04:24.750 ","End":"04:26.750","Text":"and oxygen we also have 1,"},{"Start":"04:26.750 ","End":"04:28.730","Text":"since it\u0027s also 0.62."},{"Start":"04:28.730 ","End":"04:30.920","Text":"Now we can see that the subscripts of"},{"Start":"04:30.920 ","End":"04:33.290","Text":"the carbon and hydrogen are very close to whole numbers,"},{"Start":"04:33.290 ","End":"04:35.550","Text":"so we\u0027re going to just round them off."},{"Start":"04:35.550 ","End":"04:37.050","Text":"Instead of C_9.94,"},{"Start":"04:37.050 ","End":"04:42.185","Text":"we\u0027re going to write C_10H_12 instead of 11.94."},{"Start":"04:42.185 ","End":"04:44.055","Text":"Then we\u0027re going to write N and O,"},{"Start":"04:44.055 ","End":"04:47.635","Text":"since we don\u0027t write the number 1 in the formula."},{"Start":"04:47.635 ","End":"04:51.560","Text":"So C_10H_12NO is our empirical formula."},{"Start":"04:51.560 ","End":"04:54.515","Text":"Just to note and a reminder at this stage,"},{"Start":"04:54.515 ","End":"04:57.080","Text":"if our subscripts are not all whole numbers,"},{"Start":"04:57.080 ","End":"05:00.320","Text":"we have to multiply them by a small whole number in"},{"Start":"05:00.320 ","End":"05:04.175","Text":"order to get integers for our empirical formula."},{"Start":"05:04.175 ","End":"05:06.470","Text":"But since here we have all whole numbers,"},{"Start":"05:06.470 ","End":"05:08.008","Text":"we don\u0027t have to do anything,"},{"Start":"05:08.008 ","End":"05:12.425","Text":"and this is actual our empirical formula C_10H_12NO."},{"Start":"05:12.425 ","End":"05:14.975","Text":"Now we have to find the molecular formula."},{"Start":"05:14.975 ","End":"05:16.970","Text":"In order to find the molecular formula,"},{"Start":"05:16.970 ","End":"05:20.569","Text":"we\u0027re going to use the molecular mass of the compound,"},{"Start":"05:20.569 ","End":"05:24.170","Text":"which is 324.5 u."},{"Start":"05:24.170 ","End":"05:30.320","Text":"The molecular mass of"},{"Start":"05:30.320 ","End":"05:36.395","Text":"our compound equals 324.5 u."},{"Start":"05:36.395 ","End":"05:39.530","Text":"Now in order to find our molecular formula,"},{"Start":"05:39.530 ","End":"05:43.445","Text":"we\u0027re going to first find the empirical formula mass."},{"Start":"05:43.445 ","End":"05:54.075","Text":"Our empirical formula mass equals,"},{"Start":"05:54.075 ","End":"05:55.770","Text":"if we look, we have 10 carbons,"},{"Start":"05:55.770 ","End":"05:58.910","Text":"so it\u0027s 10 times the atomic mass of carbon,"},{"Start":"05:58.910 ","End":"06:03.560","Text":"which is 12.01 u plus 12 hydrogen,"},{"Start":"06:03.560 ","End":"06:06.080","Text":"so it\u0027s 12 times the atomic mass of hydrogen as well,"},{"Start":"06:06.080 ","End":"06:10.065","Text":"1.01 u plus nitrogen,"},{"Start":"06:10.065 ","End":"06:12.675","Text":"that\u0027s 14.01 u,"},{"Start":"06:12.675 ","End":"06:15.240","Text":"plus oxygen, which is 16 u."},{"Start":"06:15.240 ","End":"06:20.785","Text":"This comes to 162.23 u."},{"Start":"06:20.785 ","End":"06:22.490","Text":"The empirical formula mass,"},{"Start":"06:22.490 ","End":"06:27.430","Text":"which is the mass of our empirical formula, equals 162.23 u."},{"Start":"06:27.430 ","End":"06:30.200","Text":"Now we\u0027re going to take the molecular mass we were given,"},{"Start":"06:30.200 ","End":"06:36.605","Text":"which is 324.5 u and divide it by our empirical formula mass."},{"Start":"06:36.605 ","End":"06:44.110","Text":"The molecular mass we were given divided by"},{"Start":"06:44.110 ","End":"06:47.550","Text":"our empirical formula mass"},{"Start":"06:53.510 ","End":"06:58.060","Text":"equals 324.5 u,"},{"Start":"06:59.720 ","End":"07:05.500","Text":"divided by 162.23 u,"},{"Start":"07:07.880 ","End":"07:10.240","Text":"and this equals 2."},{"Start":"07:10.240 ","End":"07:14.645","Text":"Therefore, the molecular formula is twice our empirical formula."},{"Start":"07:14.645 ","End":"07:21.690","Text":"Again, our empirical formula is C_10H_12NO."},{"Start":"07:21.690 ","End":"07:25.205","Text":"Now we know that our molecular formula is 2 times our empirical formula,"},{"Start":"07:25.205 ","End":"07:34.249","Text":"meaning that our molecular formula equals,"},{"Start":"07:34.249 ","End":"07:36.265","Text":"instead of C_10, we have C_20,"},{"Start":"07:36.265 ","End":"07:37.980","Text":"instead of H_12,"},{"Start":"07:37.980 ","End":"07:40.095","Text":"we have H_24,"},{"Start":"07:40.095 ","End":"07:44.020","Text":"instead of N, we have N_2 and O_2."},{"Start":"07:44.020 ","End":"07:47.915","Text":"That\u0027s our molecular formula of our compound."},{"Start":"07:47.915 ","End":"07:52.300","Text":"That is our final answer and thank you very much for watching."}],"ID":23659},{"Watched":false,"Name":"Balancing a Chemical Equation for Combustion","Duration":"6m 25s","ChapterTopicVideoID":22844,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22844.jpeg","UploadDate":"2020-12-10T05:40:35.3070000","DurationForVideoObject":"PT6M25S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.860 ","End":"00:03.870","Text":"In this video, we\u0027re going to write"},{"Start":"00:03.870 ","End":"00:08.145","Text":"the chemical equation for combustion of an organic molecule."},{"Start":"00:08.145 ","End":"00:14.160","Text":"An organic molecule is a molecule that containing at least the carbon,"},{"Start":"00:14.160 ","End":"00:16.110","Text":"hydrogen, and oxygen."},{"Start":"00:16.110 ","End":"00:21.510","Text":"Combustion means that we\u0027re going to burn it in excess of oxygen."},{"Start":"00:21.510 ","End":"00:26.115","Text":"That means enough oxygen for the substance to completely burn."},{"Start":"00:26.115 ","End":"00:31.860","Text":"The first thing to note is that when we burn an organic compound containing C, H,"},{"Start":"00:31.860 ","End":"00:36.270","Text":"and O, the products are only water,"},{"Start":"00:36.270 ","End":"00:40.380","Text":"that\u0027s H_2O, and carbon dioxide CO_2."},{"Start":"00:40.380 ","End":"00:42.910","Text":"Nothing else is obtained."},{"Start":"00:42.910 ","End":"00:45.649","Text":"Before we write the chemical equation,"},{"Start":"00:45.649 ","End":"00:51.445","Text":"we\u0027re going to remember something we learned before in Dalton\u0027s laws."},{"Start":"00:51.445 ","End":"00:56.165","Text":"One of Dalton\u0027s law stated that each chemical element is composed of"},{"Start":"00:56.165 ","End":"01:01.595","Text":"atoms that cannot be created or destroyed in a chemical reaction."},{"Start":"01:01.595 ","End":"01:05.120","Text":"In other words, we have to have the same number of"},{"Start":"01:05.120 ","End":"01:09.350","Text":"atoms of the same type before and after the reaction."},{"Start":"01:09.350 ","End":"01:14.435","Text":"For example, if we had 6 carbons before the reaction,"},{"Start":"01:14.435 ","End":"01:18.415","Text":"after the reaction, there would still be 6 carbons."},{"Start":"01:18.415 ","End":"01:21.250","Text":"Not more, and not less."},{"Start":"01:21.250 ","End":"01:24.290","Text":"Let\u0027s consider a chemical equation."},{"Start":"01:24.290 ","End":"01:26.690","Text":"How do we write a chemical equation?"},{"Start":"01:26.690 ","End":"01:28.685","Text":"Let us take an example."},{"Start":"01:28.685 ","End":"01:32.510","Text":"We\u0027re going to consider the combustion of glucose."},{"Start":"01:32.510 ","End":"01:35.365","Text":"Glucose is C_6H_12O_6,"},{"Start":"01:35.365 ","End":"01:38.900","Text":"that\u0027s its molecular formula."},{"Start":"01:38.900 ","End":"01:42.080","Text":"Glucose, of course, is a form of sugar."},{"Start":"01:42.080 ","End":"01:44.120","Text":"When we write a chemical reaction,"},{"Start":"01:44.120 ","End":"01:45.582","Text":"we write the reactants,"},{"Start":"01:45.582 ","End":"01:50.850","Text":"that\u0027s all the substances that react on the left-hand side and the products,"},{"Start":"01:50.850 ","End":"01:53.480","Text":"that\u0027s all the substances that are created,"},{"Start":"01:53.480 ","End":"01:57.140","Text":"that are produced in the chemical reaction on the right-hand side."},{"Start":"01:57.140 ","End":"01:59.195","Text":"For this particular reaction,"},{"Start":"01:59.195 ","End":"02:04.144","Text":"the reactants are glucose and oxygen gas,"},{"Start":"02:04.144 ","End":"02:06.665","Text":"which remember is O_2,"},{"Start":"02:06.665 ","End":"02:08.702","Text":"and the products are carbon dioxide,"},{"Start":"02:08.702 ","End":"02:12.740","Text":"that\u0027s CO_2 and water, which is H_2O."},{"Start":"02:12.740 ","End":"02:15.785","Text":"Now this chemical reaction is incomplete,"},{"Start":"02:15.785 ","End":"02:20.910","Text":"because it is not balanced, it\u0027s unbalanced."},{"Start":"02:22.390 ","End":"02:28.270","Text":"That means that we don\u0027t have the same number of carbons on each side,"},{"Start":"02:28.270 ","End":"02:30.330","Text":"or the same number of hydrogens,"},{"Start":"02:30.330 ","End":"02:32.580","Text":"or indeed the same number of oxygens."},{"Start":"02:32.580 ","End":"02:35.345","Text":"We have to balance this equation."},{"Start":"02:35.345 ","End":"02:38.060","Text":"Let\u0027s first consider the carbon atoms."},{"Start":"02:38.060 ","End":"02:39.320","Text":"On the left-hand side,"},{"Start":"02:39.320 ","End":"02:41.735","Text":"we see that there are 6 carbon atoms."},{"Start":"02:41.735 ","End":"02:44.375","Text":"On the right-hand side only 1."},{"Start":"02:44.375 ","End":"02:49.805","Text":"We need to multiply the carbon on the right-hand side by 6."},{"Start":"02:49.805 ","End":"02:53.915","Text":"We do that by multiplying the whole of CO_2 by 6."},{"Start":"02:53.915 ","End":"02:56.116","Text":"We can\u0027t just multiply part of it,"},{"Start":"02:56.116 ","End":"03:00.440","Text":"we have to multiply the whole of CO_2 by 6."},{"Start":"03:00.440 ","End":"03:04.129","Text":"That gives us C_6H_12O_6,"},{"Start":"03:04.129 ","End":"03:05.820","Text":"glucose plus O_2,"},{"Start":"03:05.820 ","End":"03:09.299","Text":"giving 6 molecules,"},{"Start":"03:09.299 ","End":"03:13.215","Text":"or moles of CO_2 plus H_2O."},{"Start":"03:13.215 ","End":"03:17.075","Text":"Now we have to balance hydrogen atoms."},{"Start":"03:17.075 ","End":"03:19.340","Text":"If we look at the equation,"},{"Start":"03:19.340 ","End":"03:24.735","Text":"we see that there are 12 hydrogens here in glucose,"},{"Start":"03:24.735 ","End":"03:26.985","Text":"and only 2 in water."},{"Start":"03:26.985 ","End":"03:29.935","Text":"If we multiply water by 6,"},{"Start":"03:29.935 ","End":"03:37.655","Text":"will have a total of 12 hydrogen atoms on the right and 12 hydrogen atoms on the left."},{"Start":"03:37.655 ","End":"03:39.790","Text":"Here\u0027s 6."},{"Start":"03:39.790 ","End":"03:44.360","Text":"Now our equation is C_6H_12O_6 plus O_2,"},{"Start":"03:44.360 ","End":"03:48.285","Text":"giving 6CO_2 plus 6H_2O."},{"Start":"03:48.285 ","End":"03:50.805","Text":"We can read this equation in 2 ways."},{"Start":"03:50.805 ","End":"03:52.950","Text":"Always we can read it in 2 ways,"},{"Start":"03:52.950 ","End":"03:58.515","Text":"1 molecule of glucose plus 1 molecule of O_2,"},{"Start":"03:58.515 ","End":"04:03.065","Text":"giving 6 molecules of CO_2 and 6 molecules of water."},{"Start":"04:03.065 ","End":"04:09.480","Text":"Or we can write this as 1 mole of glucose and 1 mole of oxygen,"},{"Start":"04:09.480 ","End":"04:13.595","Text":"giving 6 moles of carbon dioxide and 6 moles of water."},{"Start":"04:13.595 ","End":"04:16.445","Text":"However, the equation is still unbalanced,"},{"Start":"04:16.445 ","End":"04:20.605","Text":"because we have different numbers of oxygen on each side."},{"Start":"04:20.605 ","End":"04:22.845","Text":"Let\u0027s look at the oxygen."},{"Start":"04:22.845 ","End":"04:27.755","Text":"In glucose there are 6 oxygen atoms."},{"Start":"04:27.755 ","End":"04:30.395","Text":"In carbon dioxide, there are 2,"},{"Start":"04:30.395 ","End":"04:32.179","Text":"but we have 6 molecules,"},{"Start":"04:32.179 ","End":"04:34.895","Text":"so 6 times 2,"},{"Start":"04:34.895 ","End":"04:36.365","Text":"and we have water,"},{"Start":"04:36.365 ","End":"04:39.035","Text":"which is 6 times 1,"},{"Start":"04:39.035 ","End":"04:43.400","Text":"because in each H_2O molecule there\u0027s only 1 oxygen atom."},{"Start":"04:43.400 ","End":"04:48.815","Text":"In total we have here 12 plus 6 is equal to 18."},{"Start":"04:48.815 ","End":"04:52.235","Text":"The left-hand side, we only have 6."},{"Start":"04:52.235 ","End":"04:54.980","Text":"In order to balance the equation,"},{"Start":"04:54.980 ","End":"04:57.980","Text":"we need another 12 oxygen atoms."},{"Start":"04:57.980 ","End":"05:02.130","Text":"Now, each oxygen molecule has 2 oxygens,"},{"Start":"05:02.130 ","End":"05:11.210","Text":"so we need 6 times 2 in order to get to a total of 18 on the left and 18 on the right,"},{"Start":"05:11.210 ","End":"05:14.315","Text":"so we need to write 6 here."},{"Start":"05:14.315 ","End":"05:23.675","Text":"Here\u0027s our equation balanced C_6H_12O_6 plus 6O_2,"},{"Start":"05:23.675 ","End":"05:28.690","Text":"giving us 6CO_2 plus 6H_2O."},{"Start":"05:28.690 ","End":"05:30.735","Text":"As we learned before,"},{"Start":"05:30.735 ","End":"05:32.629","Text":"we can read this in 2 ways,"},{"Start":"05:32.629 ","End":"05:35.750","Text":"one relating to the number of molecules,"},{"Start":"05:35.750 ","End":"05:38.515","Text":"and one relating to the number of moles."},{"Start":"05:38.515 ","End":"05:45.515","Text":"1 mole of this plus 6 moles of oxygen gives me 6 moles of carbon dioxide,"},{"Start":"05:45.515 ","End":"05:48.310","Text":"and 6 moles of water."},{"Start":"05:48.310 ","End":"05:53.650","Text":"This calculation in terms of moles is called stoichiometry."},{"Start":"06:01.580 ","End":"06:07.230","Text":"We\u0027ll use it very widely in the later videos."},{"Start":"06:08.980 ","End":"06:15.515","Text":"This video we saw how to balance a chemical equation for combustion."},{"Start":"06:15.515 ","End":"06:19.730","Text":"In the next video, we\u0027ll show that we can determine the empirical formula"},{"Start":"06:19.730 ","End":"06:24.840","Text":"of an unknown organic compound from combustion analysis."}],"ID":23655},{"Watched":false,"Name":"Exercise 9 - Part a","Duration":"3m 32s","ChapterTopicVideoID":22848,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22848.jpeg","UploadDate":"2020-12-13T04:26:12.1430000","DurationForVideoObject":"PT3M32S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.910","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.910 ","End":"00:07.050","Text":"Write balanced equations to represent the complete combustion of: 1,"},{"Start":"00:07.050 ","End":"00:11.265","Text":"butane, C_4H_10; 2, sucrose C_12H_22O_11."},{"Start":"00:11.265 ","End":"00:13.960","Text":"Butane, a hydrocarbon,"},{"Start":"00:15.470 ","End":"00:19.750","Text":"when it is burnt in excess oxygen,"},{"Start":"00:21.260 ","End":"00:25.990","Text":"we get carbon dioxide and water."},{"Start":"00:26.240 ","End":"00:29.550","Text":"We have to balance this equation."},{"Start":"00:29.550 ","End":"00:34.695","Text":"We\u0027re going to divide this into our reactant side and our product side."},{"Start":"00:34.695 ","End":"00:39.765","Text":"In our reactant side, we have 4 carbons as you can see,"},{"Start":"00:39.765 ","End":"00:42.760","Text":"we have 10 hydrogens,"},{"Start":"00:43.490 ","End":"00:46.930","Text":"and we have 2 oxygens."},{"Start":"00:47.330 ","End":"00:51.430","Text":"In our product side, we have 1 carbon,"},{"Start":"00:51.980 ","End":"00:57.060","Text":"2 hydrogens, and we have 3 oxygens,"},{"Start":"00:57.060 ","End":"01:02.480","Text":"2 from the carbon dioxide and 1 from water, that\u0027s 3 oxygens."},{"Start":"01:02.480 ","End":"01:05.560","Text":"In our first step, we\u0027ll look at our carbons and we\u0027ll balance them."},{"Start":"01:05.560 ","End":"01:09.190","Text":"We see that we have 4 on our reactant side and 1 on our product side."},{"Start":"01:09.190 ","End":"01:13.140","Text":"Therefore, we\u0027re going to multiply the carbon dioxide by 4,"},{"Start":"01:13.140 ","End":"01:15.280","Text":"and then we\u0027re going to have 4 carbons."},{"Start":"01:15.280 ","End":"01:18.350","Text":"You\u0027re going to change this 1 to 4."},{"Start":"01:18.490 ","End":"01:24.470","Text":"We also see that now we have 8 oxygens from the carbon dioxide and 1 from the water,"},{"Start":"01:24.470 ","End":"01:26.810","Text":"all in all we have 9 oxygens."},{"Start":"01:26.810 ","End":"01:29.330","Text":"We\u0027re going to change the oxygen to 9."},{"Start":"01:29.330 ","End":"01:33.470","Text":"Now our carbons are balanced and we\u0027ll go on to balance our hydrogens."},{"Start":"01:33.470 ","End":"01:35.555","Text":"We have 10 on our reactant side,"},{"Start":"01:35.555 ","End":"01:37.365","Text":"and we have 2 on our product side."},{"Start":"01:37.365 ","End":"01:40.910","Text":"You have to multiply the water by 5,"},{"Start":"01:40.910 ","End":"01:43.865","Text":"because we have 2 hydrogens here in the water,"},{"Start":"01:43.865 ","End":"01:46.220","Text":"times 5 will give us 10 hydrogens."},{"Start":"01:46.220 ","End":"01:49.470","Text":"Now we\u0027re going to write 10 hydrogens, because again,"},{"Start":"01:49.470 ","End":"01:50.895","Text":"2 times 5 is 10,"},{"Start":"01:50.895 ","End":"01:54.845","Text":"and we can see that we have 10 hydrogens on our reactant side and on our product side."},{"Start":"01:54.845 ","End":"01:58.385","Text":"But also, as we can see now we have 5 oxygens from the water,"},{"Start":"01:58.385 ","End":"02:01.870","Text":"plus 8 oxygens from our carbon dioxide."},{"Start":"02:01.870 ","End":"02:08.105","Text":"We have now 13 oxygens because it\u0027s 5 from the water plus 8 from the carbon dioxide."},{"Start":"02:08.105 ","End":"02:10.775","Text":"We\u0027re going to change our oxygen to 13."},{"Start":"02:10.775 ","End":"02:14.090","Text":"Now the only difference is in"},{"Start":"02:14.090 ","End":"02:18.830","Text":"our oxygens because our carbons are balanced and our hydrogens are balanced,"},{"Start":"02:18.830 ","End":"02:20.795","Text":"and now we have to balance the oxygens."},{"Start":"02:20.795 ","End":"02:24.170","Text":"We can see that we have 13 in O on our product side,"},{"Start":"02:24.170 ","End":"02:27.365","Text":"and on our reactant side, we only have 2."},{"Start":"02:27.365 ","End":"02:30.005","Text":"If we take our oxygen, our O_2,"},{"Start":"02:30.005 ","End":"02:34.125","Text":"and multiply it by 13 divide by 2,"},{"Start":"02:34.125 ","End":"02:36.570","Text":"it\u0027s actually 2 times 13 divided by 2."},{"Start":"02:36.570 ","End":"02:38.405","Text":"We\u0027re going to have 13 oxygens."},{"Start":"02:38.405 ","End":"02:41.075","Text":"Now we have 13 oxygens."},{"Start":"02:41.075 ","End":"02:43.460","Text":"You can see that our carbons are balanced,"},{"Start":"02:43.460 ","End":"02:44.570","Text":"our hydrogens are balanced,"},{"Start":"02:44.570 ","End":"02:46.105","Text":"and our oxygens are balanced too."},{"Start":"02:46.105 ","End":"02:48.455","Text":"At this point, we have a balanced equation."},{"Start":"02:48.455 ","End":"02:53.000","Text":"Now you can leave it this way or if you don\u0027t want a fraction in your balanced equation,"},{"Start":"02:53.000 ","End":"02:55.520","Text":"you can multiply everything by 2."},{"Start":"02:55.520 ","End":"02:57.710","Text":"We\u0027re just going to do that right here."},{"Start":"02:57.710 ","End":"02:59.585","Text":"If we multiply everything by 2,"},{"Start":"02:59.585 ","End":"03:08.065","Text":"we have 2 butane plus 13 oxygen,"},{"Start":"03:08.065 ","End":"03:18.940","Text":"because it\u0027s times 2, gives us 8 carbon dioxide plus 10 water."},{"Start":"03:24.520 ","End":"03:28.040","Text":"That\u0027s our balanced equation for 1, for butane."},{"Start":"03:28.040 ","End":"03:30.005","Text":"We\u0027ll do number 2 in the next video."},{"Start":"03:30.005 ","End":"03:32.430","Text":"Thank you very much for watching."}],"ID":23667},{"Watched":false,"Name":"Exercise 9 - Part b","Duration":"3m 7s","ChapterTopicVideoID":22849,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22849.jpeg","UploadDate":"2020-12-13T04:26:40.5800000","DurationForVideoObject":"PT3M7S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:02.925","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.925 ","End":"00:06.600","Text":"Write balanced equations to represent a complete combustion of: 1,"},{"Start":"00:06.600 ","End":"00:10.860","Text":"butane, C_4H_10; 2, sucrose C_12H_22O_11."},{"Start":"00:10.860 ","End":"00:12.750","Text":"We balanced butane in previous video."},{"Start":"00:12.750 ","End":"00:16.120","Text":"Now we\u0027re going on to sucrose."},{"Start":"00:16.340 ","End":"00:22.360","Text":"When sucrose is burned in excess oxygen,"},{"Start":"00:22.850 ","End":"00:27.090","Text":"it yields carbon dioxide and water."},{"Start":"00:27.090 ","End":"00:29.460","Text":"Now we\u0027re going to balance this equation."},{"Start":"00:29.460 ","End":"00:33.525","Text":"We\u0027re going to divide this into our reactant side and our product side."},{"Start":"00:33.525 ","End":"00:34.980","Text":"We can see that on our reactant side,"},{"Start":"00:34.980 ","End":"00:37.380","Text":"we have 12 carbons,"},{"Start":"00:37.380 ","End":"00:46.450","Text":"22 hydrogens, and we have 13 oxygens."},{"Start":"00:47.150 ","End":"00:51.575","Text":"11 from the sucrose and 2 from the oxygen."},{"Start":"00:51.575 ","End":"00:54.260","Text":"We\u0027ll look at our product side, we have 1 carbon,"},{"Start":"00:54.260 ","End":"00:59.399","Text":"2 hydrogens, and 3 oxygens."},{"Start":"00:59.560 ","End":"01:02.480","Text":"The first step, we\u0027re going to balance our carbons."},{"Start":"01:02.480 ","End":"01:06.395","Text":"We have 12 carbons on the reactant side and we have 1 on our product side."},{"Start":"01:06.395 ","End":"01:09.455","Text":"We\u0027re going to add 12 to the carbon dioxide."},{"Start":"01:09.455 ","End":"01:13.079","Text":"That way, we\u0027ll have 12 carbons."},{"Start":"01:13.990 ","End":"01:16.505","Text":"Now our carbons are balanced,"},{"Start":"01:16.505 ","End":"01:18.065","Text":"now we have to look at our oxygen,"},{"Start":"01:18.065 ","End":"01:21.890","Text":"because now we have 24 from the carbon dioxide because it\u0027s 2 times 12,"},{"Start":"01:21.890 ","End":"01:25.280","Text":"so that\u0027s 24 oxygens plus 1 in our water."},{"Start":"01:25.280 ","End":"01:27.695","Text":"We have 25 oxygens now."},{"Start":"01:27.695 ","End":"01:29.930","Text":"That\u0027s 25. Next,"},{"Start":"01:29.930 ","End":"01:31.280","Text":"we\u0027re going to balance our hydrogens."},{"Start":"01:31.280 ","End":"01:33.050","Text":"We see that in our reactant side,"},{"Start":"01:33.050 ","End":"01:36.320","Text":"we have 22 hydrogens and on our product side, we only have 2."},{"Start":"01:36.320 ","End":"01:37.610","Text":"In order to fix this,"},{"Start":"01:37.610 ","End":"01:40.505","Text":"we\u0027re going to add 11 to the water."},{"Start":"01:40.505 ","End":"01:42.290","Text":"This way, we can see we have hydrogen,"},{"Start":"01:42.290 ","End":"01:43.610","Text":"we have 2 times 11."},{"Start":"01:43.610 ","End":"01:46.745","Text":"That\u0027s 22 just like in our reactant side."},{"Start":"01:46.745 ","End":"01:48.380","Text":"You\u0027re just going to add it to here."},{"Start":"01:48.380 ","End":"01:50.179","Text":"Now we have 22 hydrogens,"},{"Start":"01:50.179 ","End":"01:52.024","Text":"and let\u0027s look at our oxygens."},{"Start":"01:52.024 ","End":"02:00.015","Text":"We have 24 oxygens from our carbon dioxide plus 11 oxygens from our water."},{"Start":"02:00.015 ","End":"02:02.510","Text":"This gives us 35 oxygens."},{"Start":"02:02.510 ","End":"02:04.445","Text":"I\u0027m going to change that to 35."},{"Start":"02:04.445 ","End":"02:08.179","Text":"Now we have to look at our reactant side in order to balance our oxygens."},{"Start":"02:08.179 ","End":"02:10.309","Text":"Because our carbons are already balanced,"},{"Start":"02:10.309 ","End":"02:13.625","Text":"our hydrogens are balanced and now we\u0027re going on to the oxygen."},{"Start":"02:13.625 ","End":"02:15.440","Text":"We can see that in our product side,"},{"Start":"02:15.440 ","End":"02:17.390","Text":"we have 35 oxygen in all,"},{"Start":"02:17.390 ","End":"02:19.505","Text":"in our reactant side, we have 13."},{"Start":"02:19.505 ","End":"02:21.320","Text":"Now, we can see that on the reactant side,"},{"Start":"02:21.320 ","End":"02:26.580","Text":"we have 11 oxygens from our sucrose and then we have oxygen."},{"Start":"02:26.580 ","End":"02:29.660","Text":"In order to make the oxygens 35,"},{"Start":"02:29.660 ","End":"02:34.520","Text":"what we\u0027re going to do is we\u0027re just going to put a 12 here in front of the oxygen."},{"Start":"02:34.520 ","End":"02:36.410","Text":"That way, from our oxygen,"},{"Start":"02:36.410 ","End":"02:39.990","Text":"we have 24 because it\u0027s 12 times 2,"},{"Start":"02:39.990 ","End":"02:42.480","Text":"plus 11 oxygens from our sucrose."},{"Start":"02:42.480 ","End":"02:49.290","Text":"24 plus 11 is going to give us our 35. That\u0027s 35."},{"Start":"02:49.290 ","End":"02:51.300","Text":"Now we can see that our carbons are balanced,"},{"Start":"02:51.300 ","End":"02:52.700","Text":"our hydrogens are balanced,"},{"Start":"02:52.700 ","End":"02:54.185","Text":"and our oxygen is balanced."},{"Start":"02:54.185 ","End":"02:56.460","Text":"Everything is balanced."},{"Start":"03:00.970 ","End":"03:03.590","Text":"That\u0027s our final answer for number 2,"},{"Start":"03:03.590 ","End":"03:05.555","Text":"the complete combustion of sucrose."},{"Start":"03:05.555 ","End":"03:08.460","Text":"Thank you very much for watching."}],"ID":23668},{"Watched":false,"Name":"Determining Formula From Combustion Analysis","Duration":"11m 27s","ChapterTopicVideoID":16917,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16917.jpeg","UploadDate":"2019-02-20T23:40:45.0770000","DurationForVideoObject":"PT11M27S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.780","Text":"In this video, we will show that we can determine"},{"Start":"00:03.780 ","End":"00:10.110","Text":"the empirical formula of an unknown organic compound from combustion analysis."},{"Start":"00:10.110 ","End":"00:13.665","Text":"So we\u0027re going to calculate the formula from combustion."},{"Start":"00:13.665 ","End":"00:15.854","Text":"Let\u0027s consider an example."},{"Start":"00:15.854 ","End":"00:22.260","Text":"Supposing we have 1.8 grams of an unknown organic compound containing C, H,"},{"Start":"00:22.260 ","End":"00:29.760","Text":"and O, and it\u0027s completely burnt in a furnace in the presence of oxygen."},{"Start":"00:29.760 ","End":"00:38.110","Text":"It produces 2.64 grams of CO_2 and 1.08 grams of water, H_2O."},{"Start":"00:38.110 ","End":"00:42.604","Text":"We\u0027re going to use this information in order"},{"Start":"00:42.604 ","End":"00:48.035","Text":"to find the empirical formula of the unknown organic compound."},{"Start":"00:48.035 ","End":"00:51.935","Text":"First, let\u0027s write the equation for combustion."},{"Start":"00:51.935 ","End":"00:54.920","Text":"We studied how to find the equation for combustion,"},{"Start":"00:54.920 ","End":"01:01.130","Text":"how to balance the equation for combustion of a known compound in a previous video."},{"Start":"01:01.130 ","End":"01:04.625","Text":"Here, our compound is unknown, so we know,"},{"Start":"01:04.625 ","End":"01:07.200","Text":"we can write C_x,"},{"Start":"01:07.210 ","End":"01:10.445","Text":"H_y, O_z, where x, y,"},{"Start":"01:10.445 ","End":"01:13.025","Text":"and z are unknowns,"},{"Start":"01:13.025 ","End":"01:18.215","Text":"which we would determine by calculation plus"},{"Start":"01:18.215 ","End":"01:27.085","Text":"oxygen giving x molecules of CO_2 and y over 2 molecules of H_2O."},{"Start":"01:27.085 ","End":"01:29.465","Text":"This equation as written,"},{"Start":"01:29.465 ","End":"01:35.510","Text":"is balanced with respect to carbon and balanced with respect to hydrogen,"},{"Start":"01:35.510 ","End":"01:38.480","Text":"but not balanced with respect to oxygen."},{"Start":"01:38.480 ","End":"01:45.680","Text":"Let\u0027s check that it\u0027s really balanced with respect to carbon and hydrogen. First carbon."},{"Start":"01:45.680 ","End":"01:52.980","Text":"There are x atoms of carbon here and x atoms of carbon here on the right-hand side,"},{"Start":"01:52.980 ","End":"01:55.560","Text":"so carbon is balanced."},{"Start":"01:55.560 ","End":"01:58.620","Text":"Next, let\u0027s look at hydrogen."},{"Start":"01:58.620 ","End":"02:03.410","Text":"We have y atoms of hydrogen on the left-hand side,"},{"Start":"02:03.410 ","End":"02:06.770","Text":"that\u0027s y, and on the right-hand side,"},{"Start":"02:06.770 ","End":"02:10.355","Text":"we have y over 2 molecules of water."},{"Start":"02:10.355 ","End":"02:21.410","Text":"But each water molecule has 2 hydrogens so y over 2, multiplied by 2=y."},{"Start":"02:21.410 ","End":"02:25.070","Text":"We have y on the left and y on the right."},{"Start":"02:25.070 ","End":"02:27.230","Text":"Before we can proceed,"},{"Start":"02:27.230 ","End":"02:32.620","Text":"we need to calculate the molar mass of carbon dioxide and of water."},{"Start":"02:32.620 ","End":"02:35.570","Text":"Now we\u0027ve learned how to do this in previous videos,"},{"Start":"02:35.570 ","End":"02:37.259","Text":"but just as to remind you,"},{"Start":"02:37.259 ","End":"02:42.740","Text":"the molar mass of carbon dioxide is 12 grams for carbon,"},{"Start":"02:42.740 ","End":"02:51.305","Text":"and 2 times 16 grams for the 2 oxygens giving a total of 44.0 grams."},{"Start":"02:51.305 ","End":"02:55.780","Text":"Now, the molar mass of water is 2 times 1.01,"},{"Start":"02:55.780 ","End":"02:58.800","Text":"1.01 is the molar mass of hydrogen,"},{"Start":"02:58.800 ","End":"03:04.250","Text":"we have 2 hydrogens plus 16.0 for the oxygen."},{"Start":"03:04.250 ","End":"03:09.770","Text":"That gives us a total of 18.0 grams for water."},{"Start":"03:09.770 ","End":"03:13.580","Text":"The first thing we\u0027re going to do is to determine the number of moles of"},{"Start":"03:13.580 ","End":"03:19.225","Text":"carbon dioxide and carbon that are produced."},{"Start":"03:19.225 ","End":"03:22.400","Text":"Now we can turn the mass of carbon dioxide,"},{"Start":"03:22.400 ","End":"03:26.510","Text":"that\u0027s 2.64 grams of carbon dioxide into"},{"Start":"03:26.510 ","End":"03:31.840","Text":"moles of carbon dioxide by using the conversion factor,"},{"Start":"03:31.840 ","End":"03:38.540","Text":"1 mole of carbon dioxide is equivalent to 44.0 grams of carbon dioxide."},{"Start":"03:38.540 ","End":"03:40.727","Text":"That\u0027s what we calculated above,"},{"Start":"03:40.727 ","End":"03:43.565","Text":"so this is the conversion factor."},{"Start":"03:43.565 ","End":"03:50.535","Text":"When we multiply, we can cancel the grams of carbon dioxide at the top and bottom,"},{"Start":"03:50.535 ","End":"03:53.120","Text":"and we\u0027re left with moles of carbon dioxide."},{"Start":"03:53.120 ","End":"03:55.355","Text":"When we work this out,"},{"Start":"03:55.355 ","End":"03:56.630","Text":"we get that we"},{"Start":"03:56.630 ","End":"04:03.440","Text":"have 0.06"},{"Start":"04:03.440 ","End":"04:07.545","Text":"moles of carbon dioxide."},{"Start":"04:07.545 ","End":"04:10.595","Text":"Now, in every mole of carbon dioxide,"},{"Start":"04:10.595 ","End":"04:13.055","Text":"there\u0027s 1 mole of carbon."},{"Start":"04:13.055 ","End":"04:18.949","Text":"The moles of carbon are precisely equal to the moles of carbon dioxide,"},{"Start":"04:18.949 ","End":"04:23.075","Text":"and that\u0027s equal to 0.06 moles."},{"Start":"04:23.075 ","End":"04:27.560","Text":"This gives us the value of x."},{"Start":"04:28.010 ","End":"04:33.725","Text":"The number of moles of carbon has to be the same on the right and the left."},{"Start":"04:33.725 ","End":"04:36.800","Text":"If it\u0027s 0.06 on the right,"},{"Start":"04:36.800 ","End":"04:39.680","Text":"it will be 0.06 on the left,"},{"Start":"04:39.680 ","End":"04:41.675","Text":"and that\u0027s the value of x,"},{"Start":"04:41.675 ","End":"04:44.265","Text":"so now we know what x is."},{"Start":"04:44.265 ","End":"04:49.430","Text":"Now, in addition to finding out the number of moles of carbon,"},{"Start":"04:49.430 ","End":"04:50.560","Text":"we need its mass,"},{"Start":"04:50.560 ","End":"04:52.295","Text":"we\u0027ll need this later."},{"Start":"04:52.295 ","End":"04:55.640","Text":"The mass of carbon is the number of moles of carbon,"},{"Start":"04:55.640 ","End":"04:58.340","Text":"0.06 moles of carbon."},{"Start":"04:58.340 ","End":"05:02.060","Text":"Again, we have a conversion factor which converts"},{"Start":"05:02.060 ","End":"05:07.269","Text":"1 mole of carbon to 12.0 grams of carbon."},{"Start":"05:07.269 ","End":"05:11.210","Text":"We have this, we multiply this out,"},{"Start":"05:11.210 ","End":"05:14.630","Text":"and we get 0.72 grams."},{"Start":"05:14.630 ","End":"05:18.545","Text":"This is the mass of carbon."},{"Start":"05:18.545 ","End":"05:25.685","Text":"Now we\u0027re going to determine the number of moles of H_2O and H produced."},{"Start":"05:25.685 ","End":"05:32.230","Text":"Now, we can find the number of moles of water produced because we know"},{"Start":"05:32.230 ","End":"05:39.761","Text":"the mass of water produced so it was 1.08 grams of water,"},{"Start":"05:39.761 ","End":"05:42.514","Text":"that\u0027s what they told us in the question."},{"Start":"05:42.514 ","End":"05:49.340","Text":"We can convert that to moles of water by"},{"Start":"05:49.340 ","End":"05:55.700","Text":"using the conversion factor that 1 mole is equivalent to 18.0 grams water."},{"Start":"05:55.700 ","End":"05:58.130","Text":"That\u0027s what we calculated above,"},{"Start":"05:58.130 ","End":"05:59.480","Text":"the grams water,"},{"Start":"05:59.480 ","End":"06:06.735","Text":"grams water cancel and we\u0027re left with 0.06 moles."},{"Start":"06:06.735 ","End":"06:12.664","Text":"We have 0.06 moles of water produced."},{"Start":"06:12.664 ","End":"06:17.075","Text":"Now we have to convert that to the moles of hydrogen."},{"Start":"06:17.075 ","End":"06:21.380","Text":"The first thing we note is that in every water molecule,"},{"Start":"06:21.380 ","End":"06:24.140","Text":"there are 2 hydrogens."},{"Start":"06:24.140 ","End":"06:26.345","Text":"That means for every mole of water,"},{"Start":"06:26.345 ","End":"06:28.400","Text":"there are 2 moles of hydrogen,"},{"Start":"06:28.400 ","End":"06:35.503","Text":"so the moles of hydrogen can be calculated by using this conversion factor,"},{"Start":"06:35.503 ","End":"06:40.910","Text":"2 moles of hydrogen is equivalent to 1 mole of water."},{"Start":"06:40.910 ","End":"06:45.500","Text":"When we multiply this by the moles of water,"},{"Start":"06:45.500 ","End":"06:48.035","Text":"we will get the moles of hydrogen."},{"Start":"06:48.035 ","End":"06:52.440","Text":"We have 2, the moles of water we\u0027ve already calculated,"},{"Start":"06:52.440 ","End":"06:58.820","Text":"0.06 multiply 2 times 0.06,"},{"Start":"06:58.820 ","End":"07:02.045","Text":"we get 0.12 moles."},{"Start":"07:02.045 ","End":"07:06.050","Text":"That\u0027s the number of moles of hydrogen."},{"Start":"07:06.050 ","End":"07:10.160","Text":"Since we have the same number of moles on the right and the left,"},{"Start":"07:10.160 ","End":"07:14.705","Text":"that must be equivalent to the value of y."},{"Start":"07:14.705 ","End":"07:17.285","Text":"Now we found x and y,"},{"Start":"07:17.285 ","End":"07:19.195","Text":"we need to find z."},{"Start":"07:19.195 ","End":"07:21.030","Text":"Before we can do that,"},{"Start":"07:21.030 ","End":"07:23.720","Text":"we have to find the mass of hydrogen."},{"Start":"07:23.720 ","End":"07:28.130","Text":"The mass of hydrogen is 0.12 moles of hydrogen."},{"Start":"07:28.130 ","End":"07:32.250","Text":"We know the conversion factor that"},{"Start":"07:32.250 ","End":"07:36.880","Text":"1 mole of hydrogen is equivalent to 1.01 grams of hydrogen."},{"Start":"07:36.880 ","End":"07:38.770","Text":"The moles cancel."},{"Start":"07:38.770 ","End":"07:43.225","Text":"We\u0027re left with grams of hydrogen is 0.12 grams."},{"Start":"07:43.225 ","End":"07:45.880","Text":"Now we can determine the value of z."},{"Start":"07:45.880 ","End":"07:51.295","Text":"Now we know that the mass of the sample is 1.8 grams that was given to us."},{"Start":"07:51.295 ","End":"07:55.750","Text":"We\u0027ve calculated the mass of carbon is 0.72 grams,"},{"Start":"07:55.750 ","End":"07:58.720","Text":"and the mass of hydrogen is 0.12 grams,"},{"Start":"07:58.720 ","End":"08:03.040","Text":"so we can find out the mass of oxygen,"},{"Start":"08:03.040 ","End":"08:07.480","Text":"because the substance contains only C, H, and O."},{"Start":"08:07.480 ","End":"08:11.569","Text":"The mass of oxygen is 1.8 grams,"},{"Start":"08:11.569 ","End":"08:13.335","Text":"that\u0027s the total mass,"},{"Start":"08:13.335 ","End":"08:20.705","Text":"minus 0.72 grams of carbon minus 0.12 grams of hydrogen."},{"Start":"08:20.705 ","End":"08:25.744","Text":"That gives us 0.96 grams of oxygen."},{"Start":"08:25.744 ","End":"08:31.100","Text":"We can convert that to the moles of oxygen because we"},{"Start":"08:31.100 ","End":"08:37.040","Text":"know that 1 mole of oxygen is equivalent to 16.0 grams of oxygen."},{"Start":"08:37.040 ","End":"08:42.305","Text":"Multiply this by the mass of oxygen,"},{"Start":"08:42.305 ","End":"08:46.700","Text":"0.96 grams of oxygen times the conversion factor."},{"Start":"08:46.700 ","End":"08:49.280","Text":"The grams of oxygen cancel."},{"Start":"08:49.280 ","End":"08:53.300","Text":"We\u0027re left with 0.06 moles of oxygen."},{"Start":"08:53.300 ","End":"08:57.250","Text":"Now that gives us the number of moles of oxygen."},{"Start":"08:57.370 ","End":"09:00.565","Text":"Now we can write the empirical formula."},{"Start":"09:00.565 ","End":"09:03.520","Text":"We know that x is 0.06,"},{"Start":"09:03.520 ","End":"09:09.280","Text":"y is 0.12, and z is 0.06."},{"Start":"09:09.280 ","End":"09:12.115","Text":"This is the empirical formula,"},{"Start":"09:12.115 ","End":"09:14.380","Text":"but it\u0027s not in a very convenient form."},{"Start":"09:14.380 ","End":"09:17.875","Text":"We know that we prefer whole numbers."},{"Start":"09:17.875 ","End":"09:20.770","Text":"If we divide by the smallest number,"},{"Start":"09:20.770 ","End":"09:26.800","Text":"divide by 0.06, divide by 0.06,"},{"Start":"09:26.800 ","End":"09:30.535","Text":"and again divide by 0.06,"},{"Start":"09:30.535 ","End":"09:32.040","Text":"we get C_1,"},{"Start":"09:32.040 ","End":"09:33.300","Text":"we don\u0027t write the 1,"},{"Start":"09:33.300 ","End":"09:35.355","Text":"H_2, O_1,"},{"Start":"09:35.355 ","End":"09:37.110","Text":"but again, we don\u0027t write the 1,"},{"Start":"09:37.110 ","End":"09:43.300","Text":"so the empirical formula is CH_2O."},{"Start":"09:45.070 ","End":"09:48.809","Text":"That\u0027s the empirical formula."},{"Start":"09:51.700 ","End":"09:55.580","Text":"Now, is this the molecular formula?"},{"Start":"09:55.580 ","End":"09:59.224","Text":"As before, as we saw in previous videos,"},{"Start":"09:59.224 ","End":"10:02.690","Text":"we need to know the molecular formula,"},{"Start":"10:02.690 ","End":"10:09.740","Text":"the molecular mass, in order to find out whether this is indeed the molecular formula."},{"Start":"10:09.740 ","End":"10:14.809","Text":"First, let\u0027s calculate the molar mass of the empirical formula."},{"Start":"10:14.809 ","End":"10:20.580","Text":"The molar mass of the empirical formula is 12 for the carbon,"},{"Start":"10:20.580 ","End":"10:23.139","Text":"1 for each of the hydrogen,"},{"Start":"10:23.139 ","End":"10:24.660","Text":"it\u0027s 2 times 1,"},{"Start":"10:24.660 ","End":"10:27.365","Text":"and 16 for the oxygen."},{"Start":"10:27.365 ","End":"10:29.255","Text":"This is just an approximation."},{"Start":"10:29.255 ","End":"10:31.849","Text":"That gives us 30 grams."},{"Start":"10:31.849 ","End":"10:39.155","Text":"Now, if we were told and this is the case that the compound we\u0027re considering,"},{"Start":"10:39.155 ","End":"10:45.260","Text":"that the molar mass is 180 grams per mole,"},{"Start":"10:45.800 ","End":"10:53.705","Text":"we would immediately say that this is 6 times the molar mass of the empirical formula."},{"Start":"10:53.705 ","End":"10:56.075","Text":"So in order to get the molecular formula,"},{"Start":"10:56.075 ","End":"10:58.385","Text":"we have to multiply by 6,"},{"Start":"10:58.385 ","End":"11:01.380","Text":"so we get C_6H_12O_6."},{"Start":"11:05.470 ","End":"11:08.750","Text":"If you remember from previous video,"},{"Start":"11:08.750 ","End":"11:14.310","Text":"that is indeed the molecular formula of glucose."},{"Start":"11:14.650 ","End":"11:19.190","Text":"In this video, we have shown how to balance equations and"},{"Start":"11:19.190 ","End":"11:24.240","Text":"how to obtain the empirical formula from combustion data."}],"ID":17662},{"Watched":false,"Name":"Exercise 10","Duration":"10m 19s","ChapterTopicVideoID":22847,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22847.jpeg","UploadDate":"2020-12-13T04:25:39.9400000","DurationForVideoObject":"PT10M19S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.165","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.165 ","End":"00:05.820","Text":"A 3.244 gram sample of phenol,"},{"Start":"00:05.820 ","End":"00:09.570","Text":"C_6H_5OH is burned in an excess of oxygen."},{"Start":"00:09.570 ","End":"00:12.885","Text":"What masses of carbon dioxide and water should be obtained?"},{"Start":"00:12.885 ","End":"00:14.730","Text":"In the first step, we\u0027re going to take our phenol."},{"Start":"00:14.730 ","End":"00:16.620","Text":"We\u0027re going to write and balance an equation."},{"Start":"00:16.620 ","End":"00:18.660","Text":"We have phenol which is C_6."},{"Start":"00:18.660 ","End":"00:20.790","Text":"We see that we have 6 hydrogens,"},{"Start":"00:20.790 ","End":"00:22.110","Text":"so we\u0027re just going to write H_6,"},{"Start":"00:22.110 ","End":"00:24.780","Text":"just going to be more comfortable, O."},{"Start":"00:24.780 ","End":"00:27.720","Text":"We know that it is burned in an excess of oxygen,"},{"Start":"00:27.720 ","End":"00:29.950","Text":"so we\u0027re going to add oxygen."},{"Start":"00:30.590 ","End":"00:34.830","Text":"We have an organic compound burned in oxygen."},{"Start":"00:34.830 ","End":"00:39.280","Text":"We\u0027re going to get carbon dioxide and water as our products."},{"Start":"00:40.760 ","End":"00:43.445","Text":"Now we\u0027re going to balance this equation."},{"Start":"00:43.445 ","End":"00:46.610","Text":"We have our reactant side and our product side."},{"Start":"00:46.610 ","End":"00:49.220","Text":"On our reactant side, we can see that we have 6 carbons,"},{"Start":"00:49.220 ","End":"00:55.500","Text":"6 hydrogens, and we have 3 oxygens,"},{"Start":"00:55.500 ","End":"00:58.785","Text":"1 is from the phenol and 2 are from the oxygen."},{"Start":"00:58.785 ","End":"01:00.365","Text":"We have 3 oxygens."},{"Start":"01:00.365 ","End":"01:03.990","Text":"On our product side, we can see we have 1 carbon,"},{"Start":"01:04.010 ","End":"01:09.000","Text":"2 hydrogens, and 3 oxygens,"},{"Start":"01:09.000 ","End":"01:11.490","Text":"2 from the carbon dioxide and 1 from the water."},{"Start":"01:11.490 ","End":"01:13.250","Text":"That\u0027s 3 oxygens in all."},{"Start":"01:13.250 ","End":"01:15.980","Text":"We\u0027re going to start by balancing our carbons."},{"Start":"01:15.980 ","End":"01:20.435","Text":"We see that we have 6 on our reactant side and 1 on our product side."},{"Start":"01:20.435 ","End":"01:22.030","Text":"We\u0027re going to balance that."},{"Start":"01:22.030 ","End":"01:25.385","Text":"We\u0027re going to multiply the carbon dioxide by 6."},{"Start":"01:25.385 ","End":"01:29.190","Text":"That way, our number of carbons changes to 6."},{"Start":"01:29.190 ","End":"01:33.230","Text":"Now we have 6 carbons and we can also see that we change the number of oxygens."},{"Start":"01:33.230 ","End":"01:37.100","Text":"Now we have 12 oxygens from the carbon dioxide because it\u0027s 2 times 6,"},{"Start":"01:37.100 ","End":"01:39.620","Text":"so that\u0027s 12 oxygens, plus 1."},{"Start":"01:39.620 ","End":"01:41.635","Text":"In all we have 13 oxygens."},{"Start":"01:41.635 ","End":"01:43.790","Text":"We\u0027re just going to write 13 oxygen."},{"Start":"01:43.790 ","End":"01:45.500","Text":"Then we\u0027re going to balance our hydrogens."},{"Start":"01:45.500 ","End":"01:47.660","Text":"We can see that on our reactant side we have"},{"Start":"01:47.660 ","End":"01:50.120","Text":"6 hydrogens and on our product side we have 2."},{"Start":"01:50.120 ","End":"01:53.060","Text":"We have to multiply our water by 3."},{"Start":"01:53.060 ","End":"01:58.660","Text":"That way we are going to get 6 hydrogens because we have 2 times 3, so that\u0027s 6."},{"Start":"01:58.660 ","End":"02:01.520","Text":"We\u0027re going to change our hydrogens to 6."},{"Start":"02:01.520 ","End":"02:04.625","Text":"We see that we also changed our oxygens."},{"Start":"02:04.625 ","End":"02:09.650","Text":"We have 12 oxygens from the carbon dioxide and we have another 3 from the water."},{"Start":"02:09.650 ","End":"02:11.780","Text":"In all we have 15 oxygens."},{"Start":"02:11.780 ","End":"02:13.970","Text":"That\u0027s 15 oxygens."},{"Start":"02:13.970 ","End":"02:16.880","Text":"Now we have to balance our oxygen because in our product side we have"},{"Start":"02:16.880 ","End":"02:21.515","Text":"15 and in our reactant side we only have 3 at this point."},{"Start":"02:21.515 ","End":"02:24.505","Text":"We said that we have 1 oxygen from our phenol."},{"Start":"02:24.505 ","End":"02:28.400","Text":"What is left to do is to multiply our oxygen by 7."},{"Start":"02:28.400 ","End":"02:30.440","Text":"That way we\u0027re going to get 2 times 7,"},{"Start":"02:30.440 ","End":"02:32.495","Text":"14 oxygens plus 1."},{"Start":"02:32.495 ","End":"02:34.630","Text":"It\u0027s going to give us 15 oxygens."},{"Start":"02:34.630 ","End":"02:37.320","Text":"Now we can see that both sides are balanced."},{"Start":"02:37.320 ","End":"02:40.340","Text":"Now that we have a balanced equation,"},{"Start":"02:40.340 ","End":"02:45.460","Text":"we have to calculate the masses of the carbon dioxide and the water that will be obtained"},{"Start":"02:45.460 ","End":"02:50.990","Text":"if 3.244 grams of phenol is burnt in an excess of oxygen."},{"Start":"02:50.990 ","End":"02:54.170","Text":"First of all, we\u0027re going to calculate the moles of the phenol,"},{"Start":"02:54.170 ","End":"02:57.080","Text":"then go on to calculate the moles of the carbon dioxide and"},{"Start":"02:57.080 ","End":"03:00.485","Text":"the water and convert these moles into masses."},{"Start":"03:00.485 ","End":"03:03.110","Text":"In order to calculate the number of moles of the phenol,"},{"Start":"03:03.110 ","End":"03:04.325","Text":"we\u0027re going to use the equation,"},{"Start":"03:04.325 ","End":"03:06.080","Text":"n, number of moles equals m,"},{"Start":"03:06.080 ","End":"03:08.780","Text":"which is the mass divided by the molar mass."},{"Start":"03:08.780 ","End":"03:12.560","Text":"Now we know the mass of the phenol is 3.244 grams."},{"Start":"03:12.560 ","End":"03:14.660","Text":"We have to calculate the molar mass."},{"Start":"03:14.660 ","End":"03:16.490","Text":"The molar mass of phenol,"},{"Start":"03:16.490 ","End":"03:20.975","Text":"which is C_6H_6O equals"},{"Start":"03:20.975 ","End":"03:25.440","Text":"6 times the molar mass of carbon since we have 6 carbons in phenol,"},{"Start":"03:25.440 ","End":"03:28.475","Text":"plus 6 times the molar mass of hydrogen,"},{"Start":"03:28.475 ","End":"03:30.155","Text":"since we have 6 hydrogens,"},{"Start":"03:30.155 ","End":"03:31.790","Text":"plus the molar mass of oxygen,"},{"Start":"03:31.790 ","End":"03:33.365","Text":"since we only have 1 oxygen."},{"Start":"03:33.365 ","End":"03:39.515","Text":"This equals 6 times 12.01 grams per mole,"},{"Start":"03:39.515 ","End":"03:41.450","Text":"which is the molar mass of carbon,"},{"Start":"03:41.450 ","End":"03:45.425","Text":"plus 6 times 1.01 grams per mole,"},{"Start":"03:45.425 ","End":"03:47.510","Text":"which is the molar mass of hydrogen,"},{"Start":"03:47.510 ","End":"03:49.280","Text":"plus 16 grams per mole,"},{"Start":"03:49.280 ","End":"03:51.635","Text":"which is the molar mass of oxygen."},{"Start":"03:51.635 ","End":"03:56.850","Text":"This equals 94.12 grams per mole."},{"Start":"03:58.270 ","End":"04:01.870","Text":"The molar masses are taken from a periodic table of elements."},{"Start":"04:01.870 ","End":"04:03.320","Text":"Now that we have, our molar mass,"},{"Start":"04:03.320 ","End":"04:05.420","Text":"we can calculate the moles of the phenol."},{"Start":"04:05.420 ","End":"04:07.175","Text":"The number of moles of phenol,"},{"Start":"04:07.175 ","End":"04:09.755","Text":"which is C_6H_6O,"},{"Start":"04:09.755 ","End":"04:14.315","Text":"equals the mass of phenol divided by the molar mass of phenol."},{"Start":"04:14.315 ","End":"04:23.540","Text":"This equals 3.244 grams divided by the molar mass which we calculated now,"},{"Start":"04:23.540 ","End":"04:27.600","Text":"94.12 grams per mole."},{"Start":"04:28.150 ","End":"04:33.500","Text":"This comes to 0.035 mole."},{"Start":"04:33.500 ","End":"04:35.945","Text":"Now just a quick reminder,"},{"Start":"04:35.945 ","End":"04:37.490","Text":"when we\u0027re dividing by a fraction,"},{"Start":"04:37.490 ","End":"04:38.960","Text":"if we look at our units,"},{"Start":"04:38.960 ","End":"04:41.690","Text":"we have grams divided by grams per mole,"},{"Start":"04:41.690 ","End":"04:43.740","Text":"which is a fraction."},{"Start":"04:44.870 ","End":"04:50.930","Text":"It\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"04:50.930 ","End":"04:54.930","Text":"It\u0027s the same as grams times mole per grams."},{"Start":"04:55.060 ","End":"04:58.425","Text":"Grams cancel out and we\u0027re left with mole."},{"Start":"04:58.425 ","End":"05:00.400","Text":"Therefore our units are moles."},{"Start":"05:00.400 ","End":"05:02.485","Text":"Now we have the number of moles of phenol."},{"Start":"05:02.485 ","End":"05:04.730","Text":"From here we\u0027re going to calculate the number of moles of"},{"Start":"05:04.730 ","End":"05:08.390","Text":"the carbon dioxide and convert it to the mass of carbon dioxide."},{"Start":"05:08.390 ","End":"05:13.955","Text":"Then we\u0027ll calculate the number of moles of water and convert it to the mass of water."},{"Start":"05:13.955 ","End":"05:17.630","Text":"If we look at our equation, we can see that for every 1 mole of phenol that reacts,"},{"Start":"05:17.630 ","End":"05:19.985","Text":"we get 6 moles of carbon dioxide."},{"Start":"05:19.985 ","End":"05:24.200","Text":"Number of moles of carbon dioxide equal the number of moles of phenol,"},{"Start":"05:24.200 ","End":"05:26.330","Text":"which is C_6H_6O,"},{"Start":"05:26.330 ","End":"05:32.220","Text":"times 6 moles of carbon dioxide for"},{"Start":"05:32.220 ","End":"05:38.535","Text":"every 1 mole of phenol, C_6H_6O."},{"Start":"05:38.535 ","End":"05:42.450","Text":"Now the number of moles of the phenol are 0.035."},{"Start":"05:42.450 ","End":"05:45.600","Text":"It\u0027s 0.035 moles of"},{"Start":"05:45.600 ","End":"05:56.190","Text":"phenol times 6 moles carbon dioxide for every 1 mole of phenol."},{"Start":"05:58.360 ","End":"06:08.620","Text":"The moles of phenol cancel out and we get 0.21 moles carbon dioxide."},{"Start":"06:08.750 ","End":"06:12.365","Text":"The next step is to convert these moles into mass."},{"Start":"06:12.365 ","End":"06:14.840","Text":"For this purpose, we\u0027re going to use the equation n,"},{"Start":"06:14.840 ","End":"06:17.075","Text":"the number of moles equals m,"},{"Start":"06:17.075 ","End":"06:20.475","Text":"mass, divided by the molar mass."},{"Start":"06:20.475 ","End":"06:22.575","Text":"We want to calculate the mass."},{"Start":"06:22.575 ","End":"06:27.425","Text":"The mass equals the number of moles times the molar mass."},{"Start":"06:27.425 ","End":"06:33.635","Text":"The number of moles, we calculated it\u0027s 0.21 moles carbon dioxide,"},{"Start":"06:33.635 ","End":"06:36.140","Text":"and now we have to calculate the molar mass of carbon dioxide."},{"Start":"06:36.140 ","End":"06:39.785","Text":"The molar mass of carbon dioxide equals,"},{"Start":"06:39.785 ","End":"06:42.800","Text":"we have 1 carbon so we have the molar mass of carbon"},{"Start":"06:42.800 ","End":"06:45.830","Text":"plus 2 times the molar mass of oxygen,"},{"Start":"06:45.830 ","End":"06:47.300","Text":"since we have 2 oxygens."},{"Start":"06:47.300 ","End":"06:50.000","Text":"This equals 12.01 grams per mole,"},{"Start":"06:50.000 ","End":"06:52.085","Text":"which is the molar mass of carbon,"},{"Start":"06:52.085 ","End":"06:55.870","Text":"plus 2 times 16 grams per mole,"},{"Start":"06:55.870 ","End":"06:58.460","Text":"which is the molar mass of oxygen."},{"Start":"06:58.460 ","End":"07:03.020","Text":"This equals 44.01 grams per mole."},{"Start":"07:03.020 ","End":"07:05.690","Text":"That\u0027s the molar mass of carbon dioxide."},{"Start":"07:05.690 ","End":"07:09.349","Text":"Again, we\u0027re going to calculate the mass of carbon dioxide,"},{"Start":"07:09.349 ","End":"07:11.420","Text":"which equals the number of moles of"},{"Start":"07:11.420 ","End":"07:14.135","Text":"carbon dioxide times the molar mass of carbon dioxide."},{"Start":"07:14.135 ","End":"07:22.430","Text":"The number of moles is 0.21 mole times the molar mass of carbon dioxide,"},{"Start":"07:22.430 ","End":"07:28.620","Text":"which equals 44.01 grams per mole."},{"Start":"07:28.620 ","End":"07:36.895","Text":"Moles cancel out and we get 9.24 grams of carbon dioxide."},{"Start":"07:36.895 ","End":"07:40.490","Text":"That\u0027s the mass of carbon dioxide that we found."},{"Start":"07:40.490 ","End":"07:43.550","Text":"Now we will go on to find the mass of the water."},{"Start":"07:43.550 ","End":"07:45.050","Text":"Now going back to our equation,"},{"Start":"07:45.050 ","End":"07:47.630","Text":"we can see that for every 1 mole of phenol which reacts,"},{"Start":"07:47.630 ","End":"07:49.865","Text":"we get 3 moles of water."},{"Start":"07:49.865 ","End":"07:55.100","Text":"The number of moles of water equal the number of moles of"},{"Start":"07:55.100 ","End":"08:01.170","Text":"phenol C_6H_6O times 3 moles of water,"},{"Start":"08:01.370 ","End":"08:07.560","Text":"for every 1 mole of phenol C_6H_6O."},{"Start":"08:07.560 ","End":"08:11.320","Text":"The moles of the phenol are 0.035 mole,"},{"Start":"08:15.110 ","End":"08:19.035","Text":"times 3 moles of water,"},{"Start":"08:19.035 ","End":"08:22.510","Text":"for every 1 mole of phenol."},{"Start":"08:24.090 ","End":"08:34.490","Text":"The moles of phenol cancel out and we get 0.105 mole of water."},{"Start":"08:34.710 ","End":"08:37.720","Text":"Now we\u0027re going to calculate the mass of the water."},{"Start":"08:37.720 ","End":"08:39.970","Text":"For this purpose, we use the equation n,"},{"Start":"08:39.970 ","End":"08:43.795","Text":"the number of moles equals the mass divided by the molar mass."},{"Start":"08:43.795 ","End":"08:46.570","Text":"Now the mass of water,"},{"Start":"08:46.570 ","End":"08:48.970","Text":"just going to multiply both sides by the molar mass,"},{"Start":"08:48.970 ","End":"08:54.620","Text":"equals the number of moles of water times the molar mass of water."},{"Start":"08:55.700 ","End":"08:59.735","Text":"We know the number of moles of water because we calculated them."},{"Start":"08:59.735 ","End":"09:02.570","Text":"Now the molar mass of water we\u0027re just going to calculate here."},{"Start":"09:02.570 ","End":"09:05.150","Text":"The molar mass of water equals,"},{"Start":"09:05.150 ","End":"09:07.235","Text":"we can see that we have 2 hydrogens and 1 oxygen."},{"Start":"09:07.235 ","End":"09:10.085","Text":"It\u0027s 2 times the molar mass of hydrogen,"},{"Start":"09:10.085 ","End":"09:14.895","Text":"plus the molar mass of oxygen."},{"Start":"09:14.895 ","End":"09:21.330","Text":"This equals 2 times 1.01 grams per mole,"},{"Start":"09:21.330 ","End":"09:24.100","Text":"plus 16 grams per mole."},{"Start":"09:25.550 ","End":"09:30.640","Text":"This equals 18.02 grams per mole."},{"Start":"09:31.060 ","End":"09:34.350","Text":"Again, the mass of water"},{"Start":"09:34.850 ","End":"09:45.265","Text":"equals number of moles of water times the molar mass of water."},{"Start":"09:45.265 ","End":"09:47.140","Text":"This equals the number of moles,"},{"Start":"09:47.140 ","End":"09:50.000","Text":"which is 0.105 mole,"},{"Start":"09:51.090 ","End":"09:53.530","Text":"times the molar mass of water,"},{"Start":"09:53.530 ","End":"09:57.860","Text":"which is 18.02 grams per mole."},{"Start":"10:00.030 ","End":"10:03.050","Text":"Moles cancel out."},{"Start":"10:03.090 ","End":"10:09.860","Text":"This equals 1.89 grams of water."},{"Start":"10:10.680 ","End":"10:15.655","Text":"The mass of water equals 1.89 grams."},{"Start":"10:15.655 ","End":"10:17.635","Text":"That is our final answer."},{"Start":"10:17.635 ","End":"10:20.329","Text":"Thank you very much for watching."}],"ID":23666},{"Watched":false,"Name":"Exercise 11","Duration":"5m 56s","ChapterTopicVideoID":22853,"CourseChapterTopicPlaylistID":154545,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22853.jpeg","UploadDate":"2020-12-14T07:39:14.5500000","DurationForVideoObject":"PT5M56S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:03.345","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.345 ","End":"00:08.040","Text":"Anhydrous copper sulfate can be used to dry liquids in which it is insoluble."},{"Start":"00:08.040 ","End":"00:11.760","Text":"The copper sulfate is converted to copper sulfate pentahydrate,"},{"Start":"00:11.760 ","End":"00:14.115","Text":"which can be filtered off from the liquid."},{"Start":"00:14.115 ","End":"00:18.675","Text":"What is the minimum mass of anhydrous copper sulfate"},{"Start":"00:18.675 ","End":"00:24.075","Text":"needed to remove 14.2 grams of water from a tank full of gasoline?"},{"Start":"00:24.075 ","End":"00:28.710","Text":"As you can see, we\u0027re given the mass of the water."},{"Start":"00:28.710 ","End":"00:33.555","Text":"From the mass of the water, we\u0027re going to calculate the number of moles of the water."},{"Start":"00:33.555 ","End":"00:35.220","Text":"Then we\u0027ll calculate the number of moles of"},{"Start":"00:35.220 ","End":"00:39.270","Text":"the copper sulfate and convert these moles into the mass."},{"Start":"00:39.270 ","End":"00:43.280","Text":"In order to convert the mass of the water into moles,"},{"Start":"00:43.280 ","End":"00:44.540","Text":"we\u0027re going to use the equation,"},{"Start":"00:44.540 ","End":"00:46.460","Text":"n number of moles equal m,"},{"Start":"00:46.460 ","End":"00:49.890","Text":"which is the mass divided by the molar mass."},{"Start":"00:50.720 ","End":"00:53.055","Text":"We know the mass of the water,"},{"Start":"00:53.055 ","End":"00:57.240","Text":"14.2 grams and now we have to calculate the molar mass of the water."},{"Start":"00:57.590 ","End":"01:02.644","Text":"The molar mass of water equals 2 times the molar mass of hydrogen,"},{"Start":"01:02.644 ","End":"01:04.280","Text":"since we have 2 hydrogens and water,"},{"Start":"01:04.280 ","End":"01:06.710","Text":"plus the molar mass of oxygen,"},{"Start":"01:06.710 ","End":"01:08.750","Text":"since we have 1 oxygen in water."},{"Start":"01:08.750 ","End":"01:13.250","Text":"This equals 2 times 1.01 grams per mole,"},{"Start":"01:13.250 ","End":"01:15.905","Text":"which is the molar mass of hydrogen,"},{"Start":"01:15.905 ","End":"01:18.380","Text":"plus the molar mass of oxygen,"},{"Start":"01:18.380 ","End":"01:20.910","Text":"which is 16 grams per mole."},{"Start":"01:21.520 ","End":"01:26.700","Text":"This equals 18.02 grams per mole."},{"Start":"01:27.550 ","End":"01:31.730","Text":"The molar masses were taken from the periodic table of elements."},{"Start":"01:31.730 ","End":"01:33.635","Text":"Now we have the molar mass of water,"},{"Start":"01:33.635 ","End":"01:35.120","Text":"and we have the mass of water."},{"Start":"01:35.120 ","End":"01:37.610","Text":"We can calculate the moles of the water."},{"Start":"01:37.610 ","End":"01:40.235","Text":"The number of moles of water,"},{"Start":"01:40.235 ","End":"01:42.260","Text":"equal the mass of water,"},{"Start":"01:42.260 ","End":"01:44.360","Text":"divided by the molar mass of water,"},{"Start":"01:44.360 ","End":"01:48.420","Text":"which equals 14.2 grams,"},{"Start":"01:49.070 ","End":"01:53.770","Text":"divided by 18.02 grams per mole."},{"Start":"01:59.390 ","End":"02:06.490","Text":"This equals 0.788 moles of water."},{"Start":"02:07.330 ","End":"02:10.265","Text":"Now note the number of moles of water."},{"Start":"02:10.265 ","End":"02:14.510","Text":"We\u0027re going to calculate the number of moles of the copper sulfate."},{"Start":"02:14.510 ","End":"02:16.475","Text":"If we look at our hydrate,"},{"Start":"02:16.475 ","End":"02:18.980","Text":"we can see that for every 1 mole of copper sulfate,"},{"Start":"02:18.980 ","End":"02:21.395","Text":"we have 5 moles of water."},{"Start":"02:21.395 ","End":"02:30.290","Text":"Therefore, the number of moles of copper sulfate equal the number of moles of water,"},{"Start":"02:30.290 ","End":"02:40.180","Text":"times 1 mole of copper sulfate for every 5 moles of water."},{"Start":"02:42.060 ","End":"02:45.670","Text":"Now we have the moles of the water in the denominator."},{"Start":"02:45.670 ","End":"02:48.460","Text":"This way, the moles of the water will cancel out."},{"Start":"02:48.460 ","End":"02:53.480","Text":"This equals the moles of water equal 0.7088 mole."},{"Start":"02:58.400 ","End":"03:11.190","Text":"Again, times 1 mole of copper sulfate for every 5 moles of water."},{"Start":"03:11.190 ","End":"03:20.360","Text":"The moles of water cancel out and this equals 0.158 moles of copper sulfate."},{"Start":"03:21.360 ","End":"03:25.494","Text":"Now we know the moles of copper sulfate,"},{"Start":"03:25.494 ","End":"03:27.925","Text":"which is 0.158 moles."},{"Start":"03:27.925 ","End":"03:30.880","Text":"All we have to do is convert these moles into the mass."},{"Start":"03:30.880 ","End":"03:32.770","Text":"Again, we\u0027re going to use the equation,"},{"Start":"03:32.770 ","End":"03:34.840","Text":"n number of moles equal,"},{"Start":"03:34.840 ","End":"03:37.495","Text":"which is the mass divided by the molar mass."},{"Start":"03:37.495 ","End":"03:40.405","Text":"Now we want to calculate the mass of a copper sulfate."},{"Start":"03:40.405 ","End":"03:44.330","Text":"The mass of copper sulfate equals,"},{"Start":"03:44.330 ","End":"03:47.680","Text":"we can see that if we multiply both sides by the molar mass."},{"Start":"03:47.680 ","End":"03:50.530","Text":"The mass of copper sulfate equal the moles of"},{"Start":"03:50.530 ","End":"03:56.750","Text":"copper sulfate times the molar mass of copper sulfate."},{"Start":"03:59.390 ","End":"04:02.450","Text":"Now we calculated the moles of copper sulfate."},{"Start":"04:02.450 ","End":"04:06.140","Text":"Now we need to calculate the molar mass of copper sulfate."},{"Start":"04:06.930 ","End":"04:13.765","Text":"The molar mass of copper sulfate equals the molar mass of copper,"},{"Start":"04:13.765 ","End":"04:17.649","Text":"plus the molar mass of sulfur,"},{"Start":"04:17.649 ","End":"04:21.250","Text":"plus 4, times the molar mass of oxygen."},{"Start":"04:21.250 ","End":"04:23.230","Text":"Because we see that we have 1 copper,"},{"Start":"04:23.230 ","End":"04:26.290","Text":"1 sulfur, and 4 oxygens and copper sulfate."},{"Start":"04:26.290 ","End":"04:31.180","Text":"This equals 63.55 grams per"},{"Start":"04:31.180 ","End":"04:39.200","Text":"mole plus 32.07 grams per mole,"},{"Start":"04:40.020 ","End":"04:44.480","Text":"plus 4 times 16 grams per mole."},{"Start":"04:48.360 ","End":"04:55.620","Text":"This equals 159.62 grams per mole."},{"Start":"04:55.620 ","End":"04:59.670","Text":"That\u0027s the molar mass of the copper sulfate."},{"Start":"05:00.310 ","End":"05:04.835","Text":"Now we\u0027re going to go back to this equation to that mass of copper sulfate."},{"Start":"05:04.835 ","End":"05:07.320","Text":"The mass of copper sulfate"},{"Start":"05:10.150 ","End":"05:18.090","Text":"equals the number of moles of copper sulfate times the molar mass of copper sulfate."},{"Start":"05:18.430 ","End":"05:28.400","Text":"Number of moles of copper sulfate is 0.158 mole times the molar mass of copper sulfate,"},{"Start":"05:28.400 ","End":"05:33.900","Text":"which is 159.62 grams per mole."},{"Start":"05:34.280 ","End":"05:45.240","Text":"The moles cancel out and this equals 25.22 grams of copper sulfate."},{"Start":"05:46.030 ","End":"05:51.665","Text":"The mass of copper sulfate equals 25.22 grams."},{"Start":"05:51.665 ","End":"05:53.885","Text":"That is our final answer."},{"Start":"05:53.885 ","End":"05:56.400","Text":"Thank you very much for watching."}],"ID":23673}],"Thumbnail":null,"ID":154545},{"Name":"Inorganic Compounds","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Naming inorganic compounds 1","Duration":"7m 2s","ChapterTopicVideoID":22846,"CourseChapterTopicPlaylistID":154546,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22846.jpeg","UploadDate":"2020-12-10T05:44:09.1670000","DurationForVideoObject":"PT7M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.700","Text":"In this video and in the next one,"},{"Start":"00:02.700 ","End":"00:07.500","Text":"we\u0027ll learn about the names and formulas of inorganic compounds."},{"Start":"00:07.500 ","End":"00:12.870","Text":"Let\u0027s begin with a binary compounds of metals and nonmetals."},{"Start":"00:12.870 ","End":"00:17.775","Text":"A binary compound involves 2 elements, that\u0027s the binary."},{"Start":"00:17.775 ","End":"00:22.350","Text":"Binary compounds of a metal and non-metal are ionic compounds,"},{"Start":"00:22.350 ","End":"00:27.700","Text":"and they consist of a metal cation and a non-metal anion."},{"Start":"00:27.770 ","End":"00:31.940","Text":"We write the name of the metal first and then the name of"},{"Start":"00:31.940 ","End":"00:35.370","Text":"the non-metal with the suffix ide."},{"Start":"00:35.370 ","End":"00:38.520","Text":"Instead of chlorine, we\u0027ll have chloride."},{"Start":"00:38.520 ","End":"00:45.380","Text":"We should note throughout and always check that the compound is electrically neutral."},{"Start":"00:45.380 ","End":"00:49.010","Text":"Many metal ions have only one possibility,"},{"Start":"00:49.010 ","End":"00:52.285","Text":"that means there\u0027s only one oxidation state."},{"Start":"00:52.285 ","End":"00:56.580","Text":"These include lithium plus, sodium plus,"},{"Start":"00:56.580 ","End":"00:58.620","Text":"potassium plus, rubidium plus,"},{"Start":"00:58.620 ","End":"01:01.200","Text":"cesium plus, magnesium 2 plus,"},{"Start":"01:01.200 ","End":"01:02.460","Text":"calcium 2 plus,"},{"Start":"01:02.460 ","End":"01:03.810","Text":"strontium 2 plus,"},{"Start":"01:03.810 ","End":"01:05.235","Text":"barium 2 plus,"},{"Start":"01:05.235 ","End":"01:07.140","Text":"aluminum 3 plus,"},{"Start":"01:07.140 ","End":"01:08.760","Text":"zinc 2 plus,"},{"Start":"01:08.760 ","End":"01:12.520","Text":"silver plus, and gold plus."},{"Start":"01:12.520 ","End":"01:17.105","Text":"However, there are other metals that have more than one possibility."},{"Start":"01:17.105 ","End":"01:20.359","Text":"That means there\u0027s more than one oxidation state."},{"Start":"01:20.359 ","End":"01:23.955","Text":"For example, chromium can be 2 plus or 3 plus,"},{"Start":"01:23.955 ","End":"01:26.190","Text":"iron can be 2 plus or 3 plus,"},{"Start":"01:26.190 ","End":"01:28.635","Text":"cobalt can be 2 plus or 3 plus,"},{"Start":"01:28.635 ","End":"01:31.830","Text":"copper can be plus or 2 plus,"},{"Start":"01:31.830 ","End":"01:36.299","Text":"mercury can have 2 mercury atoms together,"},{"Start":"01:36.299 ","End":"01:43.605","Text":"Hg_2 and that\u0027s 2 plus or just one mercury atom Hg 2 plus."},{"Start":"01:43.605 ","End":"01:47.625","Text":"Tin can be 2 plus or 4 plus,"},{"Start":"01:47.625 ","End":"01:51.290","Text":"and lead can be 2 plus or 4 plus."},{"Start":"01:51.290 ","End":"01:54.650","Text":"Now there are far fewer nonmetal ions."},{"Start":"01:54.650 ","End":"01:55.910","Text":"We have H minus,"},{"Start":"01:55.910 ","End":"01:58.460","Text":"that\u0027s called hydride, F minus,"},{"Start":"01:58.460 ","End":"02:00.845","Text":"fluoride, Cl minus,"},{"Start":"02:00.845 ","End":"02:03.635","Text":"chloride, Br minus, bromide,"},{"Start":"02:03.635 ","End":"02:06.100","Text":"I minus, iodide,"},{"Start":"02:06.100 ","End":"02:07.451","Text":"O 2 minus,"},{"Start":"02:07.451 ","End":"02:10.325","Text":"that\u0027s oxide, S^2 minus,"},{"Start":"02:10.325 ","End":"02:14.450","Text":"sulfide, N^3 minus nitrite."},{"Start":"02:14.450 ","End":"02:16.490","Text":"Let\u0027s take some examples."},{"Start":"02:16.490 ","End":"02:18.920","Text":"Here\u0027s one that you\u0027re very familiar with."},{"Start":"02:18.920 ","End":"02:23.930","Text":"NaCl is sodium chloride, MgO,"},{"Start":"02:23.930 ","End":"02:29.785","Text":"magnesium oxide, Al_2O_3 aluminum oxide."},{"Start":"02:29.785 ","End":"02:33.270","Text":"We can check that this is the correct formula."},{"Start":"02:33.270 ","End":"02:35.055","Text":"Al is 3 plus,"},{"Start":"02:35.055 ","End":"02:39.330","Text":"that\u0027s 2 times 3 plus and oxygen is 2 minus,"},{"Start":"02:39.330 ","End":"02:42.290","Text":"that\u0027s 3 times 2 minus,"},{"Start":"02:42.290 ","End":"02:46.430","Text":"so we have plus 6 plus minus 6,"},{"Start":"02:46.430 ","End":"02:47.735","Text":"which gives us 0."},{"Start":"02:47.735 ","End":"02:51.625","Text":"So 6 plus minus 6 giving us 0."},{"Start":"02:51.625 ","End":"02:56.545","Text":"Now, when we have the possibility of more than one oxidation state,"},{"Start":"02:56.545 ","End":"03:01.900","Text":"then we have to distinguish between the different oxidation states in the name."},{"Start":"03:01.900 ","End":"03:05.815","Text":"For example, FeO, we write iron II."},{"Start":"03:05.815 ","End":"03:08.860","Text":"We use the II oxide."},{"Start":"03:08.860 ","End":"03:13.615","Text":"That means iron has a plus 2 oxidation state."},{"Start":"03:13.615 ","End":"03:16.850","Text":"If it\u0027s Fe_2O_3,"},{"Start":"03:16.850 ","End":"03:18.750","Text":"that\u0027s iron III oxide,"},{"Start":"03:18.750 ","End":"03:22.790","Text":"that means iron has a plus 3 oxidation state."},{"Start":"03:22.790 ","End":"03:29.320","Text":"Remember for ions, the oxidation state and the ionic charge are the same,"},{"Start":"03:29.320 ","End":"03:35.205","Text":"so it\u0027s Fe_2 plus an oxidation state plus 2."},{"Start":"03:35.205 ","End":"03:37.695","Text":"Where we have Cu_2O,"},{"Start":"03:37.695 ","End":"03:39.530","Text":"that\u0027s copper I oxide,"},{"Start":"03:39.530 ","End":"03:43.390","Text":"that means copper has the oxidation state of plus 1,"},{"Start":"03:43.390 ","End":"03:48.320","Text":"and CuO, that\u0027s copper II oxide,"},{"Start":"03:48.320 ","End":"03:52.280","Text":"that means copper has the oxidation state of plus 2."},{"Start":"03:52.280 ","End":"03:56.690","Text":"Now we go on to the binary compounds of 2 nonmetals."},{"Start":"03:56.690 ","End":"04:04.660","Text":"These are molecular compounds consisting of 2 nonmetals before we add ionic compounds."},{"Start":"04:04.660 ","End":"04:08.750","Text":"We write the name of the element with the positive oxidation state"},{"Start":"04:08.750 ","End":"04:12.665","Text":"first and then the name of the negative oxidation state,"},{"Start":"04:12.665 ","End":"04:16.685","Text":"and we again have the suffix ide."},{"Start":"04:16.685 ","End":"04:21.530","Text":"Now, very often an element has several oxidation states,"},{"Start":"04:21.530 ","End":"04:24.715","Text":"and then we use prefixes."},{"Start":"04:24.715 ","End":"04:27.000","Text":"Here\u0027s a list of them."},{"Start":"04:27.000 ","End":"04:28.890","Text":"Greek if it come from Greek."},{"Start":"04:28.890 ","End":"04:30.090","Text":"Mono, di,"},{"Start":"04:30.090 ","End":"04:31.650","Text":"tri, tetra, penta,"},{"Start":"04:31.650 ","End":"04:35.050","Text":"hexa, hepta, octa, nona, and deca."},{"Start":"04:35.050 ","End":"04:36.800","Text":"I will have some examples,"},{"Start":"04:36.800 ","End":"04:39.245","Text":"for example, carbon monoxide."},{"Start":"04:39.245 ","End":"04:42.860","Text":"Here the carbon has the oxidation state of plus 2."},{"Start":"04:42.860 ","End":"04:44.755","Text":"We don\u0027t write usually,"},{"Start":"04:44.755 ","End":"04:48.965","Text":"mono carbon monoxide, we just leave it as carbon."},{"Start":"04:48.965 ","End":"04:51.050","Text":"Here is carbon dioxide,"},{"Start":"04:51.050 ","End":"04:52.415","Text":"the 2 oxygens,"},{"Start":"04:52.415 ","End":"04:55.820","Text":"and the oxidation state of carbon is plus 4."},{"Start":"04:55.820 ","End":"04:57.590","Text":"If we\u0027re given a formula,"},{"Start":"04:57.590 ","End":"05:00.515","Text":"we can always check what the oxidation state is,"},{"Start":"05:00.515 ","End":"05:06.020","Text":"because oxygen is minus 2 and carbon therefore must be plus 4."},{"Start":"05:06.020 ","End":"05:11.135","Text":"Now nitrogen has a lot of possible oxidation states and here are a few examples."},{"Start":"05:11.135 ","End":"05:15.245","Text":"Dinitrogen oxide, that\u0027s N_2O."},{"Start":"05:15.245 ","End":"05:19.985","Text":"The oxidation state of N must be plus 1."},{"Start":"05:19.985 ","End":"05:23.240","Text":"NO_2 nitrogen dioxide."},{"Start":"05:23.240 ","End":"05:27.370","Text":"The oxidation state of N must be plus 4."},{"Start":"05:27.370 ","End":"05:31.755","Text":"N_2O_5 dinitrogen pentoxide."},{"Start":"05:31.755 ","End":"05:35.690","Text":"The oxidation state here of N is plus 5."},{"Start":"05:35.690 ","End":"05:37.970","Text":"We can check if that\u0027s correct."},{"Start":"05:37.970 ","End":"05:39.350","Text":"There are 2 nitrogens,"},{"Start":"05:39.350 ","End":"05:43.380","Text":"so it\u0027s 2 times 5 and 5 oxygens,"},{"Start":"05:43.380 ","End":"05:46.380","Text":"so that\u0027s 5 times minus 2."},{"Start":"05:46.380 ","End":"05:48.180","Text":"That gives us 10 plus,"},{"Start":"05:48.180 ","End":"05:50.895","Text":"minus 10 total of 0."},{"Start":"05:50.895 ","End":"05:52.580","Text":"Now we get to phosphorus."},{"Start":"05:52.580 ","End":"05:56.570","Text":"We can have phosphorus trichloride, PCl_3,"},{"Start":"05:56.570 ","End":"06:01.850","Text":"where the oxidation state of the phosphorus is plus 3 and PCl_5,"},{"Start":"06:01.850 ","End":"06:08.510","Text":"that\u0027s phosphorous pentachloride, where the oxidation states of phosphorus is plus 5."},{"Start":"06:08.510 ","End":"06:11.585","Text":"Now we go into binary acids."},{"Start":"06:11.585 ","End":"06:19.645","Text":"Binary acid is a compound consisting of hydrogen and a non-metal in an aqueous solution."},{"Start":"06:19.645 ","End":"06:22.220","Text":"We write the prefix hydro,"},{"Start":"06:22.220 ","End":"06:27.020","Text":"followed by the name of the non-metal with the suffix ic."},{"Start":"06:27.020 ","End":"06:29.560","Text":"Here\u0027s some examples."},{"Start":"06:29.560 ","End":"06:33.030","Text":"HF is hydrofluoric acid,"},{"Start":"06:33.030 ","End":"06:36.965","Text":"HBr is hydrobromic acid, HCl,"},{"Start":"06:36.965 ","End":"06:38.460","Text":"with which you\u0027re all familiar,"},{"Start":"06:38.460 ","End":"06:40.245","Text":"is hydrochloric acid,"},{"Start":"06:40.245 ","End":"06:43.070","Text":"HI is hydroiodic acid,"},{"Start":"06:43.070 ","End":"06:46.235","Text":"H_2S is hydrosulfuric acid."},{"Start":"06:46.235 ","End":"06:50.720","Text":"You have to remember all of these are in aqueous media,"},{"Start":"06:50.720 ","End":"06:52.595","Text":"they\u0027re all in water."},{"Start":"06:52.595 ","End":"06:57.875","Text":"In this video, we talked about the names of inorganic compounds,"},{"Start":"06:57.875 ","End":"07:01.800","Text":"and we\u0027ll continue this in the next video."}],"ID":23656},{"Watched":false,"Name":"Naming inorganic compounds 2","Duration":"6m 17s","ChapterTopicVideoID":23177,"CourseChapterTopicPlaylistID":154546,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/23177.jpeg","UploadDate":"2021-01-01T03:14:48.5130000","DurationForVideoObject":"PT6M17S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"In the previous video,"},{"Start":"00:01.740 ","End":"00:04.725","Text":"we talked about naming inorganic compounds."},{"Start":"00:04.725 ","End":"00:09.705","Text":"We talked about binary compounds that consists of only 2 elements."},{"Start":"00:09.705 ","End":"00:15.975","Text":"In this video, we will talk about inorganic compounds consist of more than 2 elements."},{"Start":"00:15.975 ","End":"00:20.355","Text":"I\u0027m going to begin by naming polyatomic ions."},{"Start":"00:20.355 ","End":"00:22.695","Text":"These are ions, usually anions,"},{"Start":"00:22.695 ","End":"00:25.185","Text":"consisting of more than 1 element,"},{"Start":"00:25.185 ","End":"00:28.440","Text":"and they\u0027ll often contain oxygen."},{"Start":"00:28.440 ","End":"00:31.140","Text":"There\u0027s only one very common cation,"},{"Start":"00:31.140 ","End":"00:35.150","Text":"and that\u0027s NH_ 4 plus that\u0027s called ammonium."},{"Start":"00:35.150 ","End":"00:40.010","Text":"We should recall that NH_4 is 1 group,"},{"Start":"00:40.010 ","End":"00:44.065","Text":"and the plus relates to the whole collection of atoms."},{"Start":"00:44.065 ","End":"00:46.895","Text":"Now, what about the anions?"},{"Start":"00:46.895 ","End":"00:48.980","Text":"The whole collection of these, and again,"},{"Start":"00:48.980 ","End":"00:52.459","Text":"the names are unfortunately not very trivial."},{"Start":"00:52.459 ","End":"00:59.140","Text":"CH_3CO_2 minus acetate, CO_3^2 minus carbonate,"},{"Start":"00:59.140 ","End":"01:02.580","Text":"HCO_3 minus hydrogen carbonate,"},{"Start":"01:02.580 ","End":"01:05.375","Text":"or sometimes called bicarbonate,"},{"Start":"01:05.375 ","End":"01:11.540","Text":"CRO_4^2 minus chromate, CR_2O_7^2 minus dichromate,"},{"Start":"01:11.540 ","End":"01:16.225","Text":"CN minus cyanide, OH minus hydroxide,"},{"Start":"01:16.225 ","End":"01:21.425","Text":"NO_2 minus nitrite, NO_3 minus nitrate."},{"Start":"01:21.425 ","End":"01:26.420","Text":"When there are 2 possible oxidation states as here,"},{"Start":"01:26.420 ","End":"01:31.580","Text":"then one of the lower oxidation state here is plus 3, is called ite,"},{"Start":"01:31.580 ","End":"01:39.155","Text":"and the one with a higher oxidation state here it\u0027s plus 5, is called ate."},{"Start":"01:39.155 ","End":"01:44.595","Text":"MnO_4 minus permanganate, PO_4^3 minus phosphate,"},{"Start":"01:44.595 ","End":"01:47.715","Text":"HPO_4^2 minus hydrogen phosphate,"},{"Start":"01:47.715 ","End":"01:52.155","Text":"H_2PO_4 minus dihydrogen phosphate,"},{"Start":"01:52.155 ","End":"01:56.990","Text":"SO_3 minus sulfite, HSO_3 minus"},{"Start":"01:56.990 ","End":"02:02.795","Text":"hydrogen sulfite or bisulfite, SO_4^2 minus sulfate."},{"Start":"02:02.795 ","End":"02:09.810","Text":"Again, ite for the lower oxidation state and ate for the higher one."},{"Start":"02:09.810 ","End":"02:18.875","Text":"HSO_4 minus hydrogen sulfate or bisulfate, S_2O_3^2 minus thiosulfate."},{"Start":"02:18.875 ","End":"02:24.935","Text":"Unfortunately, names are not very obvious and you just have to learn them."},{"Start":"02:24.935 ","End":"02:26.450","Text":"Here\u0027s some more."},{"Start":"02:26.450 ","End":"02:32.600","Text":"This is a series where we have 4 different oxidation states for chlorine."},{"Start":"02:32.600 ","End":"02:38.345","Text":"ClO minus, that\u0027s plus 1, ClO_2 minus,"},{"Start":"02:38.345 ","End":"02:41.640","Text":"plus 3, ClO_3 minus,"},{"Start":"02:41.640 ","End":"02:43.890","Text":"plus 5, ClO_4,"},{"Start":"02:43.890 ","End":"02:46.090","Text":"minus, plus 7."},{"Start":"02:46.090 ","End":"02:48.995","Text":"The names are hypochlorite,"},{"Start":"02:48.995 ","End":"02:51.905","Text":"hypo and ite at the end,"},{"Start":"02:51.905 ","End":"02:57.420","Text":"chlorite, chlorate, and perchlorate."},{"Start":"02:57.420 ","End":"02:59.335","Text":"Per and ate."},{"Start":"02:59.335 ","End":"03:01.180","Text":"Here\u0027s some examples."},{"Start":"03:01.180 ","End":"03:06.030","Text":"Sodium acetate, NaCH_3CO_2,"},{"Start":"03:06.030 ","End":"03:15.790","Text":"consists of Na plus, and CH_3CO_2 minus."},{"Start":"03:16.280 ","End":"03:22.770","Text":"Na_3PO_4, sodium phosphate consists of Na plus"},{"Start":"03:22.770 ","End":"03:30.680","Text":"3 ions like that, and PO_4^3 minus."},{"Start":"03:30.680 ","End":"03:32.815","Text":"Some more examples."},{"Start":"03:32.815 ","End":"03:35.160","Text":"Potassium permanganate,"},{"Start":"03:35.160 ","End":"03:44.835","Text":"KMnNO_4 consists of K plus MnO_4 minus."},{"Start":"03:44.835 ","End":"03:48.825","Text":"Na_2CO_3, that\u0027s sodium carbonate,"},{"Start":"03:48.825 ","End":"03:56.070","Text":"that\u0027s Na plus 2 of those, and CO_3^2 minus."},{"Start":"03:56.070 ","End":"04:01.040","Text":"Now, there are some acids which contain"},{"Start":"04:01.040 ","End":"04:06.935","Text":"oxygen and they\u0027re often related to anions that we\u0027ve talked about already."},{"Start":"04:06.935 ","End":"04:10.000","Text":"Oxo acids usually contain H,"},{"Start":"04:10.000 ","End":"04:13.325","Text":"O, and one other element."},{"Start":"04:13.325 ","End":"04:18.275","Text":"Here\u0027s a list of acids and their associated anions,"},{"Start":"04:18.275 ","End":"04:21.120","Text":"which we\u0027ve already met above."},{"Start":"04:21.390 ","End":"04:27.630","Text":"HNO_2 is called nitrous acid and HNO_3 nitric acid."},{"Start":"04:27.630 ","End":"04:33.890","Text":"Us the lower oxidation state and ate the higher oxidation state."},{"Start":"04:33.890 ","End":"04:37.330","Text":"The associated anions or nitrite,"},{"Start":"04:37.330 ","End":"04:39.210","Text":"us goes with ite,"},{"Start":"04:39.210 ","End":"04:42.760","Text":"nitric acid goes with nitrates."},{"Start":"04:42.760 ","End":"04:44.585","Text":"Ic goes with ate,"},{"Start":"04:44.585 ","End":"04:46.415","Text":"us goes with ite."},{"Start":"04:46.415 ","End":"04:50.730","Text":"Sulfuric acid, ic going with ate,"},{"Start":"04:50.730 ","End":"04:54.635","Text":"SO_4^2 minus is the associated ion,"},{"Start":"04:54.635 ","End":"05:00.480","Text":"phosphoric acid goes with phosphate."},{"Start":"05:00.480 ","End":"05:03.060","Text":"Again, ic and ate."},{"Start":"05:03.060 ","End":"05:12.265","Text":"Again we have the series of acids containing chlorine with different oxidation states."},{"Start":"05:12.265 ","End":"05:14.820","Text":"We have HClO;"},{"Start":"05:14.820 ","End":"05:19.775","Text":"that\u0027s hypo at the beginning and us at the end."},{"Start":"05:19.775 ","End":"05:22.100","Text":"That goes with ClO minus,"},{"Start":"05:22.100 ","End":"05:25.315","Text":"which is hypo at the beginning and ite the end."},{"Start":"05:25.315 ","End":"05:27.320","Text":"Us goes with ite."},{"Start":"05:27.320 ","End":"05:30.260","Text":"HClO_2 is chlorous acid,"},{"Start":"05:30.260 ","End":"05:35.510","Text":"the associated ion is ClO_2 minus with chlorite."},{"Start":"05:35.510 ","End":"05:38.525","Text":"Again, us goes with ite."},{"Start":"05:38.525 ","End":"05:43.050","Text":"HClO_3 is chloric acid,"},{"Start":"05:43.050 ","End":"05:46.370","Text":"and the associated anion is ClO minus."},{"Start":"05:46.370 ","End":"05:48.665","Text":"So ic goes with ate."},{"Start":"05:48.665 ","End":"05:53.960","Text":"Then HClO_4 is perchloric acid,"},{"Start":"05:53.960 ","End":"05:59.485","Text":"and that goes with ClO_4 minus perchlorate."},{"Start":"05:59.485 ","End":"06:01.545","Text":"Ic again with ate,"},{"Start":"06:01.545 ","End":"06:04.855","Text":"and we have per at the beginning."},{"Start":"06:04.855 ","End":"06:09.725","Text":"These are some examples of the names of"},{"Start":"06:09.725 ","End":"06:15.900","Text":"inorganic compounds that consist of more than 2 elements."}],"ID":23998}],"Thumbnail":null,"ID":154546}]

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