Question

A 2.20 g-sample of a compound containing carbon, hydrogen, and oxygen is burned and it produces 4.61 g CO2 and 0.94 g H2O. What is the empirical formula of this compound? (i.e., C2H4O)

I found the mass of Carbon
(4.61/44g)*(1 mol C/1 mol CO2)=.1047

I found the mass of Hydrogen
(.94/18g)*(2 mol H/1 mol H20)=.1044

2.20-.1047-.1044=1.99

Then I divided all the numbers by the lowest which is .1044. I keep getting CH019?
What am I doing everything wrong?

Answer 1

<b>The problem is that you didn't quite follow my instructions.</b>

I found the mass of Carbon
(4.61/44g)*(1 mol C/1 mol CO2)=.1047
<b>What you have done here is to calculate MOLES C, not grams C. Multiply by 12 to get grams C. </b>

I found the mass of Hydrogen
(.94/18g)*(2 mol H/1 mol H20)=.1044
<b>Same thing here. You have calculated MOLES H. Convert this number to grams H atoms</b>

2.20-.1047-.1044=1.99
<b>Now you are subtracting 2.20 GRAMS - mols C - mols H and you can't do that. You should go back and calculate grams C and grams H so when you subtract from 2.20 g sample, you will get grams O instead of some other fictitious number. </b>
Then I divided all the numbers by the lowest which is .1044. I keep getting CH019?
What am I doing everything wrong?