00:01
So for this problem, we are given some percent composition of a certain compound.
00:05
We know that the molar mass of the final compound, and we want to find molecular formula.
00:12
So in general, our approach is going to be find the empirical formula for the compound, and then relate that to the molecular formula of the final compound via the molar mass.
00:22
So first, we want to find the empirical formula.
00:26
So we know that the compound is 31.
00:32
0 .51 % carbon, 4 .77 % hydrogen, or 37 ,0 .85 % oxygen, 8 .29 % nitrogen, and 13 .60 % sodium.
01:00
So, hypothetically, if the compound was 100 grams, then this would correspond.
01:06
To 35 .51 grams of carbon, 4 .77 grams of hydrogen, 37 .85 grams of oxygen, 8 .29 grams of nitrogen, and 13 .60 grams of sodium.
01:26
Now, once again, the compound obviously isn't going to necessarily weigh 100 grams, but we hypothetically assume it is because that helps us build proportion that we can, then use to find the empirical formula.
01:39
So if we have 35 .51 grams of carbon, we want to convert that into moles of carbon, and we can do that by dividing by the molar mass.
01:50
We know that the molar mass of one mole of carbon is 12 .01 grams, and when we multiply that out, we get 2 .96 moles of carbon.
02:01
Same thing for hydrogen.
02:02
You know that it's a molar mass is 1 .008.
02:07
We multiply that out.
02:09
We get 4 .73 moles of hydrogen.
02:17
Oxygen.
02:18
We know that in every mole of oxygen.
02:22
16 .00 grams.
02:24
We multiply that out.
02:26
We get 2 .37 moles of oxygen.
02:33
And then we do nitrogen.
02:35
We know that in every mole of nitrogen, we know that in every mole of nitrogen, that there's 14 .01 grams.
02:42
That's a molar mass of one mole of nitrogen.
02:44
And we get 0 .59 moles in nitrogen.
02:51
We need the same thing to sodium.
02:53
We know that the molar mass of sodium is 22 and 9 grams.
02:59
We multiply that out, we get 0 .59 moles of sodium.
03:08
So now that we have the relative proportions of the moles of each, we can then derive the empirical formula.
03:16
So we're just going to use sodium as a base because that is the lowest.
03:23
It's the same as nitrogen...