How to Determine the Empirical Formula from Combustion Analysis Data?

AI Thread Summary
To determine the empirical formula from combustion analysis data, the mass of carbon, hydrogen, and oxygen must be calculated from the products of combustion. In this case, burning 2.66 grams of the compound produced 4.50 grams of CO2 and 1.10 grams of H2O, leading to 1.23 grams of carbon, 0.122 grams of hydrogen, and 1.308 grams of oxygen. The mole ratios were calculated, resulting in a formula of CHO, which is not among the provided answer options. The discussion emphasizes the importance of not rounding down mole values and suggests multiplying by a common factor to achieve whole numbers. Ultimately, the correct empirical formula must be derived from accurate mole calculations.
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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
 
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shap said:
1.1gH20*(1molH2)/18gH2O*(1molH2O/1molH)*(1gH/1molH)=.122gH

how many mol of H in H2O?
 
Sorry that was a typo, in my calculation I used 2 mol of H in H2O.
 
shap said:
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
 
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