M FeO = A Fe + A O = 55.84 + 16 = 71.84 g/mol
71.84 g/mol FeO ......55.84 g Fe
100 g FeO ......................a g Fe
a= 100 x 55.84 / 71.84 = 77.72% Fe
M Fe2O3 = 2 x A Fe + 3 x A O = 2 x 55.84 + 3 x 16 = 159.68 g/mol
159.68 g/mol Fe2O3 ... 111.68 g Fe
100 g Fe2O3 ................b g Fe
b = 100 x 111.68 / 159.68 = 69.93 % Fe
M Fe3O4 = 3 x A Fe + 4 x A O = 3 x 55.84 + 4 x 16 = 231.52 g/mol
231.52 g/mol Fe3O4 .......... 167.52 g Fe
100 g Fe3O4 ................. .....c g Fe
c= 100 x 167.52 / 231.52 = 72.35 % Fe
