An unknown compound contains only C , H , and O . Combustion of 3.10 g of this compound produced 6.19 g CO2 and 2.54 g H2O . What is the empirical formula of the unknown compound?

Answer 1

We assume that upon combustion of the unknown that all of the carbon (C) ends up in CO2, and all of the hydrogen (H) ends up in H2O. The remaining mass would then be oxygen (O).

mols C = 6.19 g CO2 x 1 mol CO2 / 44 g = 0.141 mols C x 12 g C/mol = 1.69 g C

mols H = 2.54 g H2O x 1 mol H2O / 18 g x 2 mol H / mol H2O = 0.282 mols H x 1 g H/mol = 0.282 g H

mols O = 3.10 g - 1.69 g - 0.282 g = 1.14 g O x 1 mol O/16 g = 0.0711 mols O

Summary of moles:

mols C = 0.141

mols H = 0.282

mols O = 0.0711

Dividing all by 0.0711 in an attempt to get all whole numbers, we have...

1.98 mols C ~ 2 mols C

3.97 mols H ~ 4 mols H

1 mol O

Empirical formula = C₂H₄O