A 400 mg iron capsule contain 100mg of ferrous fumarate $\left(C_4{H}_2{O}_{4}Fe\right)$.The percentage of Fe is approximately: 

Gram molecular weight if ferrous fumerate $=\left(12\times 4\right)+\left(2\times 1\right)+\left(4\times 16\right)+56=170$ gram per mole 

Therefore 170g of ferrous fumerate contain 56g of Fe

Then 100g of ferrous fumerate contain $\frac{\left(100\times 56\right)}{170}=32.9gofFe$

Then 100mg of ferrous fumerate contain = 32.9 mg of Fe

Therefore, percentage of Fe in the iron capsule = $\left(\frac{32.9}{400}\right)\times 100=8.22$