A compound is found to contain 50.05 percent sulfur and 49.95 percent oxygen by mass. What is the empirical formula for this compound?

Assuming 100 g of compound
50.05 g S x 1 mol S/32 g S = 1.56 moles S present
49.95 g O x 1 mol O/16 g O = 3.12 moles O present
mole ratio of O to S is 3.12/1/56 = 2 to 1
Empirical formula = SO2