Question

A $400{\text{ mg}}$ iron capsule contains $100{\text{ mg}}$ of ferrous fumarate ${\left( {CHCOO} \right)_2}Fe$. The percentage of iron in it is approximately ______.
A.$33\% $
B.$25\% $
C.$14\% $
D.$8\% $

Answer 1

Hint:To answer this question, you must find the mass percentage of iron in the given compound. For this, you must know the atomic mass of iron and the molecular mass of ferrous fumarate.

Complete step by step answer:
Ferrous fumarate has the chemical formula ${{\text{C}}_{\text{4}}}{{\text{H}}_{\text{8}}}{\text{Fe}}{{\text{O}}_{\text{4}}}$. Since the atomic masses of carbon, hydrogen, iron and oxygen are known, we can find the molecular mass of ferrous fumarate. It is equal to $169.9{\text{g}}$.
Each mole of ferrous fumarate contains one mole of iron and thus, we can say that,
 $169.9{\text{g}}$ of ferrous fumarate contains $55.9{\text{g}}$ iron.
So $100{\text{g}}$ ferrous fumarate contains $ = 100 \times \dfrac{{55.9}}{{169.9}} = 32.9{\text{ mg}}$ of iron.
Further, each capsule has a weight of $400{\text{ mg}}$ that contains $100{\text{ mg}}$ ferrous fumarate containing $32.9{\text{ mg}}$ iron.
The percentage of iron in the capsule is $ = \dfrac{{32.9}}{{400}} \times 100 = 8.2\% $

Hence, the correct answer is D.

Note:
Stoichiometry of a substance is based upon the very basic laws of chemistry that help us in the better understanding of the concept, namely, the law of conservation of mass, the law of reciprocal proportions, the law of definite proportions (also known as the law of constant composition) and the law of multiple proportions .
Chemicals generally combine in fixed ratios in chemical reactions to form a chemical compound. Since, we know that matter can neither be created nor destroyed, nor can any element transfuse into another during a chemical reaction, hence the amount of each element must be the same before and after the reaction. In other words, the number of atoms of any element in the reactants will always be equal to the number of atoms of that element in the products formed.