A 15.267 g sample of a sulfide of gold was found to contain 12.272 g of gold, and the rest, sulfur. Determine the empirical formula of this sulfide of gold

Explanation:

gold sulfide = 15.267 g

gold = 12.272 g

sulphur = 15.267 - 12.272 = 2.995 g

moles of gold = 12.272 / 196.96 = .0623 moles

moles of sulphur = 2.995 / 32 = .0936 moles

ratio of moles of gold and sulphur in the compound = .0623 : .0936

= 2 : 3

empirical formula  Au₂S₃

the answer is b & c i believe : )