A 400 mg iron capsule contain 100mg of ferrous fumarate C4H2O4Fe .The percentage of Fe is approximately:

Gram molecular weight if ferrous fumerate =12×4+2×1+4×16+56=170 gram per mole Therefore 170g of ferrous fumerate contain 56g of FeThen 100g of ferrous fumerate contain 100×56170=32.9 g of FeThen 100mg of ferrous fumerate contain = 32.9 mg of FeTherefore, percentage of Fe in the iron capsule = 32.9400×100=8.22