Complete the following Empirical Formula & Molecular Formula problem:

1.) A compound containing 63.15% C, 5.30% H, and 31.55% O. (Assume 100 g sample)

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 isC₂H₂O