63.15% C ; 5.30% H; 31.55% O
1) Assume 100 g sample
63.15% C * 100 g = 63.15g
5.30% H * 100 g = 5.30g
31.55% O * 100g = 31.55g
2) Convert mass to moles using their atomic weights
63.15 g * 1 mol C / 12.0107 g C = 5.2870 mol C
5.30 g * 1mol H / 1.0079 g H = 5.2585 mol H
31.55 g * 1mol O / 15.9994 O = 1.9719 mol O
3) Divide each quantity by the smallest number of moles
5.2870 mol C / 1.9719 mol = 2.6812 C = 2 C
5.2584 mol H / 1.9719 mol = 2.6667 H = 2 H
1.9719 mol O / 1.9719 mol = 1.000 O = 1 O
Empirical Formula is C₂H₂O
If the problem is silent, the molecular weight is equivalent to the empirical weight.
To get the molecular formula, divide the molecular weight by the empirical weight to get the multiple.
Molecular Weight is not mentioned thus it is equivalent to Empirical Weight which is:
Atom Number in Molecule Atomic Weight Total Mass
C 2 12.0107 24.0214
H 2 1.0079 2.0158
O 1 15.9994 15.9994
Total weight is 42.0366
Molecular weight / Empirical Weight = Multiple to be multiplied to the Empirical Formula
42.0366 / 42.0366 = 1
Molecular Formula is C₂H₂O