How Do You Determine the Empirical Formula from Combustion Data?

[rudders93]

Homework Statement

A 2.203g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32g of water and the carbon was oxidised to 3.23g of CO₂. Find the empirical formula of the compound

2. The attempt at a solution
So I found the number of moles of H₂O that were reacted using the n = m/Mr formula (so m(H₂O) = 1.32g, Mr(H₂O) = 18g/mol therefore n(H₂O) = 0.0733mol).

Also, I found then number of CO₂ the same way, and came out nicely that n(CO₂) = 0.0733mol

Here was my first problem. The ratio of n(H₂O) : n(CO₂) = 0.0733 : 0.0733, but since there are two H atoms in H₂O does that mean there is the equivalent of 0.1466mol of H₂O (ie: 2 * n(H₂O)) or does it mean there are 0.03665mol of H₂O (through: 0.5 * n(H₂O)). Which one is it and why?

The second problem I had is this. The answer says that the empirical formula of the compound is: CH₂O. What I don't see is how they found the number of O. Also, I'm confused as to how they knew there was O in the compound, does Organic mean it has O in it? Because I always thought that organic meant it had carbon, and that was the only prerequisite?

Thanks for your help!

 

[chemisttree]

That analysis only calculates C and H. Anything left over must be due to O.

1. Find the # grams of CO2 due only to carbon itself. (moles CO2/FW CO2)*(12grams C/ 1 mole C)

2. Repeat #1 using H2O.

3. Sum result of 1 & 2. Is it equal to 2.203 g? If not, what is missing?

4. Calculate moles of O.

5. Calculate empirical formula given results in 1, 2 & 4.