I can't remember combustion analysis

[1MileCrash]

63.8 mg of a C - H - O compound produces 145.0 mg of CO2 and 59.38 mg of H2O. Find empirical formula.

I THOUGHT that I simply had to find the moles of each compound as a ratio, but in doing that I get an answer that isn't an option.

I calculate moles C as (12/44)(.145) = .0395 grams carbon = .00329 moles carbon
Moles O as .10546 grams + .052 grams = .0097 moles oxygen
Moles H as .00329 moles H

Or CO3H...

What am I doing wrong here? I can't remember this crap.

 

[Borek]

Combustion products contain oxygen both from the compound and from the air.

Check hydrogen moles, you look for H, not for H₂.

Check what is the mass of 0.00329 moles C, corrected moles of H and 0.0097 moles of O. Compare with the sample mass. Can you think about a way of finding amount of oxygen in the sample?

 

[1MileCrash]

Hows this look?

.145 g CO2 = .003295 Moles C = .0395 g
.05938 g H2O = .00659 Moles H = .0065 g

Grams oxygen = .0638 - .0395 - .0065 = .0178 g = .00106 moles oxygen

= C3H6O

 

[Borek]

OK now.