00:01
We are asked to perform a combustion analysis in order to determine the empirical formula, just empirical, for a compound that contains only carbon, hydrogen, and oxygen.
00:16
We are told that we produce 2 .086 grams of co2, and we produce 1 .134 grams of water.
00:30
Okay, let's begin.
00:31
My first order of business is to find grams of carbon, and i will do that by a percent composition analysis, in which i use the molar mass of carbon dioxide and the atomic mass of carbon.
00:47
And this will get me 2 .086 times 12 .01 divided by 44 .01.
00:55
0 .56925 moles of carbon.
01:03
Next, i'll use 2 .02 grams of hydrogen and 18 .02 grams of water.
01:15
So, 1 .134 times 2 .02 divided by 18 .02 is 0 .12712 grams of hydrogen.
01:34
My mass of oxygen will be equal to 1 .200 grams minus 0 .56925 grams minus 0 .12712 grams.
01:55
1 .2 minus 0 .56925 minus 0 .12712.
02:07
And this will equal 0 .50363 grams of o.
02:13
So, i've got 0 .560360 grams of o.
02:22
So, now i've got these three values here, here, and here.
02:28
There's a little poem we use that goes, percent of mass, mass to mole, divide by small, multiply till whole.
02:36
I'm already at, i've done my percent to mass, this should have been mass up here.
02:43
I don't have to do my mass, i'm already at mass.
02:47
Percent to mass, now i need to do mass to mole.
02:50
To do the mass to mole, i'm going to divide each of these by the molar mass to only one sig fig, or one decimal place for each of these.
03:00
So, this will be 0 .12712 moles of h...