00:01
Alright, so our goal is to determine an empirical formula.
00:03
We know that we have carbon, hydrogen, oxygen, and nitrogen.
00:07
So let's start with the combustion.
00:13
So we have 3 .39 grams of co2, and we know that in carbon dioxide there's 12 .01 grams of carbon for every 44 .01 grams, which is the molar mass of co2.
00:31
So this will tell us the amount of carbon.
00:35
And then similarly, we have 1 .74 grams of h2o.
00:40
From that we can get the amount of hydrogen.
00:42
So there's 2 .02 grams of hydrogen in every 18 .02 grams of water, just using a ratio of the masses.
01:00
So let's do 3 .39 times 12 .01 divided by 44 .01.
01:05
For the first one is 0 .925 grams of carbon.
01:12
Then we have 1 .74 times 2 .02 divided by 18 .02 comes out to 0 .195 grams of hydrogen.
01:23
Then for the nitrogen, we can consider the ammonia experiment.
01:27
So we have 7 .94 grams of the compound produced 2 .28 grams of ammonia.
01:46
So 7 .94 grams of our compound, which i'll just call x, produced 2 .28 grams of nh3.
01:56
So if we had 2 .28 grams of our sample of x, how much nh3 would that be? because we want to keep it the same, because this is for a 2 .28 gram sample.
02:11
So we want to find the composition.
02:16
Okay, so 7 .94 times a question mark is equal to 2 .28 squared.
02:23
If we multiply, then divide by 7 .94.
02:27
So we get 0 .655 grams of nh3 in 2 .28 grams of x.
02:41
And then if we do, let's see, there's 14 .01 grams of nitrogen in every 17 .04 grams of ammonia.
02:54
So then we'll get the amount of nitrogen is in 2 .28 grams of our sample.
03:04
So this comes out to 0 .538 grams of nitrogen.
03:13
Then lastly, for the amount of oxygen, if we take the overall mass and we subtract out the mass of carbon, which is 0 .925, subtract out the mass of hydrogen, and subtract out the mass of nitrogen, the only thing left is oxygen...