How do you solve for volume to volume conversions and empirical formula?

Question

What is the empirical formula of a compound containing 60.0% sulfur and 40.0% oxygen by mass?

Answer 1

Assume 100.0 g of the sample. This gives

mass of S in compound = 60% x 100.0 g = 60.0 g

mass of O in compound = 40% x 100.0 g = 40.0 g

Convert g to mol using their MW.

mol of S = 60.0 g x (1 mol / 32.06 g) = 1.87 mol S

mol of O = 40.0 g x (1 mol / 16.00 g) = 2.50 mol O

Divide moles of each element by the lowest # of moles (S = 1.87 mol) to give the mole ratio of each element present in the compound.

S = 1.87 / /1.87 = 1.00 mol S

O = 2.50 / 1.87 = 1.33 mol O

Mole ratio must be a whole #!

Since # of moles of O = 1.33 (or 1 1/3), need to multiply it by 3 to give a whole #.

O = 1.33 mol x 3 = 3.99 mol ~ 4 mol O

Since O was multiplied by 3, need to multiply moles of S by 3 as well to maintain the 60%:40% ratio.

S = 1.00 mol x 3 = 3.00 mol ~ 3 mol S

This gives the empirical formula as S₃O₄.