Question

$400mg$ of a capsule contains $100mg$ of ferrous fumarate. The percentage of iron present in the capsule is approximately:

• A. $8.2\%$
• B. $25\%$
• C. $16\%$
• D. unpredicatable

Answer 1

Answer: (A)

$8.2\%$

The molecular formula of ferrous fumarate is $C_4{H}_{2}Fe{O}_4$. Its molecular weight is $4\left(12\right)+2\left(1\right)+55.9+4\left(16\right)=169.9g/mol$. The atomic mass of iron is 55.9 g/mol.

One molecule of ferrous fumarate $C_4{H}_{2}Fe{O}_4$ contains one iron atom.

100 mg of ferrous fumarate $=100mg\times \frac{55.9g/mol}{169.9g/mol}=32.9mg$

Thus, 100 mg of ferrous sulphate corresponds to 32.9 mg of iron.

The percentage of iron present in 400 mg of capsule is $\frac{32.9mg\times 100}{400mg}=8.2\%$.