00:01
Let's go through how to find the empirical formula in a combustion reaction.
00:06
So we're given 0 .1 grams of some molecule, and we're told it makes this much carbon dioxide and this much water vapor.
00:16
So this carbon here, it's going to be converted into carbon dioxide.
00:24
And it's the carbon and the carbon dioxide.
00:27
And this hydrogen here is going to be converted to this hydrogen.
00:31
In water vapor.
00:33
So we need to find the mass of carbon and the mass of hydrogen.
00:38
So we'll go here and we'll find out how much carbon we have.
00:43
So we have 0 .1783 grams of carbon dioxide.
00:49
We'll divide that by the atomic mass of the entire carbon dioxide molecule.
00:55
That's going to be 44.
00:57
And then we multiply that number by the mass of just the carbon because that's all we care about is the carbon we'll do that and when we do that we see that point 1783 divided by 44 times 12 is 0 .0486 so we have 0 .04 86 grams of carbon i've already got carbon over here so i don't want to run it to us alright, so for this hydrogen, we're going to take the .0734, and we divide it by the mass of water.
01:55
And then we need to multiply it by how many parts we have of hydrogen.
02:00
So hydrogen is one atomic mass, but we have two molecules.
02:07
So we're going to add a 2 there.
02:10
So we do 0 .0734 divided by 18 times 2.
02:17
And it gives you 0 .00815.
02:26
0 .00815 grams.
02:28
All right.
02:29
So this is how much carbon we have.
02:31
This is how much hydrogen have.
02:32
So we can find out how much oxygen we have since we're given the mass of this sample.
02:38
So the oxygen is just going to be present.
02:42
Let's line it up here.
02:43
It's just going to be present in the remaining amount over here...