chemistry question, please help?

Question

what are the empirical and molecular formulas for a compound that is 58.84% barium 13.74% sulfur and the rest oxygen, and a molar mass of 233.39 g/mol?

Answer 1

First you have to divide each %age by the relative atomic mass of each element:
Ba = 58.84/137
= 0.429
S = 13.74/32
= 0.429
O = 27.42/16
= 1.714

Now you divide the answers you got by the smallest one
Ba = 0.429/0.429
= 1
S = 0.429/0.429
= 1
O = 1.714/0.429
= 4
So the epirical formula = BaSO4

For the molecular formula first fined the RMM of the empirical formula
Ba + S + 4O = 137 + 32 + (16 x4)
= 233
since this is the same as the molar mass the molecular formula will be the same as the empirical formula
BaSO4

Answer 2

100 - 58.84 - 13.74 = 27.42
Molecular mass 233.39
Atomic masses
Ba 137.3
S 32.01
O 16
0.5884*233.39 = 137.326676 ≈ 1*137.3
0.1374*233.39 = 32.067786 ≈ 1*32.01
0.2742*233.39 = 63.995538 ≈ 4*16
The empirical and molecular formulas are the same.