What mass (in grams) of iron(iii) oxide contains 58.7 g of iron? iron(iii) oxide is 69.94 % iron by mass?

A substance's percent composition basically reveals to you what number of grams of every constituent component you get per 100 g of said substance. 
For this situation, a percent composition of 
69.94% implies that for each 100 g of iron(III) oxide, 
Fe2O3, you get 69.94 g of iron.

So the solution for this problem would be: 58.7 g Fe = 100 g Fe2O3 / 69.94 g Fe = 83.93 g Fe2O3

The percent composition  (%) of iron (Fe+3) in Fe2O3 equals:
Fe % = Atomic mass of Fe / (Molecular weight of Fe2O3)
∵ Atomic mass of Fe = 55.8 g/mol
and, the atomic mass of Oxygen is 16 g/mol
∴ percent of iron in Fe2O3 = [(2*55.8) / ((3*16) + (2*55.8))] *100 = 69.92 % >>> (1)
And if the mass of the iron is 58.7 g 
∴ mass of Fe2O3 = 58.7 * 100 / 69.92 = 83.95 g >>>> (2)
So, from (2), the mass of iron (III) oxide is 83.95 g
and, from (1), the iron III oxide is 69.92 % iron by mass not 69.94%