# Combustion Train Analysis

1. Oct 28, 2007

### shap

1. The problem statement, all variables and given/known data
When 2.66 grams of a compound containing only carbon, hydrogen, and oxygen is burned completely, 4.50 grams of CO2 and 1.10 grams of H2O are produced. What is the empirical formula of the compound?
C5H6O4
C5H6O5
C5H12O5
C4H6O3
C4H8O3
C3H4O3
C4H4O5
C5H8O4
C4H4O3

2. Relevant equations
Dimensional Analysis

3. The attempt at a solution
4.5gCO2*(1molCO2/44gCO2)*(1molCO2/1molC)*(12gC/1molC)=1.23gC
1.1gH20*(1molH2)/18gH2O*(1molH2O/1molH)*(1gH/1molH)=.122gH
2.66g-1.23g-.122g=1.308gO

1.23gC*(1molC/12gC)=.1023molC/.0819mol=1C
.122gH*(1molH/1gH)=.122molH/.0819mol=1H
1.31gO*(1molO/16gO)=.0819molO/.0819mol=1O

2. Oct 28, 2007

### eli64

how many mol of H in H2O?

3. Oct 28, 2007

### shap

Sorry that was a typo, in my calculation I used 2 mol of H in H2O.

4. Oct 28, 2007

### eli64

you can't round down to 1. Keep the decimals and multiply all mole numbers by the same factor until you get whole numbers