What is the empirical formula of a compound composed of 72.36 % Fe and 27.64 % O by mass?

Answer 1

Assume 100 g of the compound

72.36 g of Fe and 27.64 g O

Change gram to moles

72.36 g Fe x 1 mol Fe/55.85 g Fe = 1.296 mol

27.64 g O x 1 mol O/16.00 g = 1.727 mol

calculate mole ratio of O to Fe (divide by smaller number) 1.727 mol/1.269 mol = 1.333mol O for 1 mol Fe

Since we want integer numbers we need to multiply each number by 3

1.333 x 3 = 4 mol O

1 x 3 = 3 mol Fe

The formula is Fe3O4