Question

$400mg$ capsule contains $100mg$ of ferrous fumarate. The percentage of iron present in the capsule approximately is: (Mol mass of $Fe\left(C_4{H}_2{O}_4\right)=170$)

• A. $8.2$%
• B. $25$%
• C. $16$%
• D. $50$%

Answer 1

Answer: (A)

$8.2$%

Molar mass of ferrous fumarate $\left(C_4{H}_{2}Fe{O}_4\right)=169.9$ $g$

Atomic mass of Iron $\left(Fe\right)=55.8$

$\therefore 169.9$ $g$ of $C_4{H}_{2}Fe{O}_4\to 55.8$ $g$ of $Fe$

$\therefore 100$ $mg$ of $C_4{H}_{2}Fe{O}_4$ contain:

$=\frac{55.8g}{169.9g}\times 100$ $mg$

$=32.8$ $mg$

Amount of $C_4{H}_{2}Fe{O}_4$ in $400$ $mg$ capsule $=100$ $mg$

$\therefore$ Amount of iron in $400$ $mg$ capsule $=32.8$ $mg$

Percentage of iron in $400$ $mg$ capusle $=100\times \frac{AmountofIronincapsule}{Totalmassofcapsule}$

$=100\times \frac{32.8mg}{400mg}$

$=8.2\%$