0:02 hi everyone in this last video I just
0:05 wanted to do one more quick example of
0:07 combustion analysis in this example we
0:10 have a compound that contains carbon
0:13 hydrogen and also oxygen so when 4.18
0:16 grams of this compound is burned we're
0:19 able to collect 2.5 grams of water and
0:22 6.13 grams of carbon dioxide we're told
0:25 the molar mass of the substance we want
0:26 to figure out what the molecular formula
0:29 is of the compound so just as we did in
0:32 the previous video we want to use the
0:34 mass of the water and carbon dioxide
0:36 collected to figure out how many moles
0:39 of hydrogen and carbon were in our
0:41 original compound
0:43 so that's where we're going to start to
0:45 find the moles of carbon
0:48 we use the mass of carbon dioxide that
0:50 was collected in the combustion analysis
0:57 as we did previously we'll want to use
0:59 the molar mass of carbon dioxide in
1:02 order to convert this to moles the molar
1:05 mass is 44.01 grams for every one mole
1:08 of carbon dioxide
1:10 so grams of carbon dioxide cancel and we
1:12 have moles and then we write the
1:14 stoichiometric factor based upon the
1:16 molecular formula of CO2 to figure out
1:19 how many moles of carbon this is we know
1:22 that for every one mole of CO2 we have
1:25 only one mole of carbon
1:28 so this will give us the moles of carbon
1:29 that were collected in the carbon
1:34 dioxide this is 0.139 moles
1:37 we can do the same thing with the water
1:40 to figure out moles of hydrogen
1:42 so starting with the mass of water which
1:46 is 2.50 grams we're able to figure out
1:49 how many moles of Water by dividing this
1:52 by the molar mass of water which is
1:56 18.02 grams for every one mole
1:58 of H2O
2:01 grams of water cancel and then we can
2:03 use the stoichiometric factor based on
2:05 the molecular formula of water to know
2:08 that this for every one mole of water we
2:12 have two moles of hydrogen
2:14 doing this calculation we find that we have
2:16 have
2:20 0.277 moles of hydrogen
2:22 now when we got to this point in the
2:24 last problem we were able to take the
2:26 molar ratio to determine the empirical
2:29 formula right the numbers of carbon and
2:30 hydrogen that should be in our empirical
2:33 formula however if we know that we have
2:40 that is made up of carbon hydrogen and
2:44 oxygen we can subtract the mass of the
2:47 carbon and the mass of the hydrogen and
2:49 what should be left would be the mass of
2:52 the oxygen
2:55 so let's use these numbers of moles to
2:58 figure out the mass of hydrogen and the
3:01 mass of carbon the mass of carbon we can
3:10 and using the molar mass of carbon to
3:12 figure out the mass so here we would say
3:15 for every one mole of carbon right the
3:16 mass would be
3:20 12.011 grams
3:24 so this gives me a mass of 1.67 grams of
3:27 carbon we can do the same thing for the
3:29 moles of hydrogen from
3:35 0.277 moles of hydrogen I can use the
3:37 molar mass of hydrogen where every one
3:41 mole of hydrogen is
3:44 1.0079 grams
3:46 so this tells me that the mass of
3:48 hydrogen is
3:54 all right so now if we want to figure
3:57 out the mass of oxygen we take our
4:05 we subtract the mass of carbon which we
4:09 found is 1.67 grams and we subtract the
4:11 mass of hydrogen which is
4:14 0.279 grams
4:18 so the mass of oxygen should be
4:22 2.231 grams
4:24 and just as we did before for carbon and
4:26 hydrogen we'll need to know how many
4:30 moles this is so the moles of oxygen
4:32 would be
4:39 and we use the molar mass of oxygen
4:43 where one mole of oxygen is
4:47 16.00 grams
4:50 grams cancel and this tells me that the
4:59 and as we learned previously if we want
5:00 to figure out what our empirical formula
5:03 is we can write our empirical formula
5:05 with the number of moles that we've calculated
5:14 and then we can divide all of these
5:16 numbers by the smallest one so the
5:27 and doing these calculations we find
5:31 that the empirical formula should be c h
5:35 two o now our problem told us that the
5:38 molar mass of the substance is
5:41 180.156 and they want to know the
5:43 molecular formula we just found the
5:45 empirical formula in order to determine
5:48 the molecular formula we would want to
5:58 and divide it by the mass of our
6:03 empirical formula so this is 30. 03
6:05 grams per mole
6:07 so this is the mass of our empirical
6:11 formula ch2o when I do this calculation
6:13 it's going to tell me the number by
6:15 which I need to multiply all of my
6:18 subscripts and so this calculation comes
6:21 out to 5 which means that to arrive at
6:24 my molecular formula for the compound I
6:26 would need to multiply all my subscripts
6:29 by 5. so this becomes C5 h10
6:31 h10 o5
