A 7.417 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 10.52 grams of CO2 and 6.460 grams of H2O are produced. In a separate experiment, the molar mass is found to be 62.07 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Empirical formula: CH₃O

Molecular formula: C₂H₆O₂

Explanation:

All carbon is converted in CO2 and all hydrogen in H2O. Thus, we can find mass of each element and, by subtracting, the mass of Oxygen:

Mass C:

10.52g CO2 * (1mol / 44g) = 0.239 moles CO2 = Moles C * (12.01g / mol) =

2.87g C

Mass H:

6.460g H2O * (1mol / 18g) = 0.359 moles H2O * 2 = 0.718 moles H * (1g / mol) =  

0.718g H

Mass O:

7.417g Sample - 2.87g C - 0.718g H =

3.829g O

Now, we can determine empirical formula defined as the simples whole number ratio of atoms present in a compound with moles of each atom:

Moles C: 0.239 moles C

Moles H: 0.718 moles H

Moles O: 3.829g * (1mol / 16g) = 0.239 moles O

The ratio of moles of each atom divided in moles of oxygen:

C: 0.239 mol / 0.239 mol = 1

H: 0.718 mol / 0.239 mol = 3

C: 0.239 mol / 0.239 mol = 1

The molar mass of CH3O is:

12.01g/mol + 1g/mol*3 + 16g/mol =

31.01g/mol

That means molecular formula is twice empirical formula because its molecular mass is 31.01g/mol * 2 = 62.02g/mol ≈ 62.07g/mol