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  4. 400 mg of a capsule contains 100 mg of ferrous fumarate. The percentage of iron present in the capsule is approximately.

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

JamiaJamia 2015

Solution:

Molecular formula of ferrous fumarate is $ H{{e}^{+}}=\frac{1}{\lambda }={{R}_{H}}.{{Z}^{2}}\left[ \frac{1}{n_{2}^{2}}-\frac{1}{n_{2}^{2}} \right] $ , molar mass is 170 $ ={{R}_{H}}{{(2)}^{2}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right] $ 100 mg of $ =4{{R}_{H}}\times \frac{3}{16}=\frac{3}{4}{{R}_{H}} $ contains $ \text{s }>\text{ p}>\text{ d}>\text{ f} $ $ CsCl $ Total Fe in 400 mg of capsule $ C{{l}^{-}} $ $ \text{C}{{\text{s}}^{+}} $ percentage of Fe in capsule $ \frac{1}{8}\times 8=1 $