An 85.00 g sample of an unknown compound contains 24.26 g molybdenum, 24.33 g sulfur, and the rest is oxygen. How many moles of molybdenum are present in the compound

A) 0.253 mol B) 0.7603 mol C) 2.276 mol D) MoS₃O₉ or Mo(SO₃)₃ E) molybdenum(VI) sulfite

Explanation:

For all the Mole problems you want to find divide grams of the element in compound by the molar mass of the element.

24.26 g Mo / 95.95 g = 0.2528

24.33 g S / 32.06 g = 0.7589 mol S

36.41 g O / 16 g = 2.276 mol O

S /  Mo =  0.7587 mol  / 0.2529 mol  =  3/ 1

O / Mo =  2.276 mol  / 0.2529 mol  =  9/ 1

The empirical formula of the compound is MoS₃O₉ or Mo(SO₃)₃.