How to Determine the Empirical Formula from Combustion Analysis Data?

[shap]

Homework Statement

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?
Answer Options:
C5H6O4
C5H6O5
C5H12O5
C4H6O3
C4H8O3
C3H4O3
C4H4O5
C5H8O4
C4H4O3

Homework Equations

Dimensional Analysis

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
Empirical Formula:CHO<--not an available answer

 

[eli64]

1.1gH20*(1molH2)/18gH2O*(1molH2O/1molH)*(1gH/1molH)=.122gH

how many mol of H in H2O?

 

[shap]

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

 

[eli64]

1.23gC*(1molC/12gC)=.1023molC/.0819mol=1C
.122gH*(1molH/1gH)=.122molH/.0819mol=1H
1.31gO*(1molO/16gO)=.0819molO/.0819mol=1O
Empirical Formula:CHO<--not an available answer

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