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 is C₂H₂O

It’s b. mass i searched it up