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Question

An oxide of iron contains $69.9\%$ iron, its empirical formula, is :

(Given : Molar mass of Fe and O are 56 and $16\text{ }g{\text{ }mol}^{- 1}$ respectively.)

Ans. (2)

Sol. [Key Concept: Convert the given mass percentages of Fe and O to moles, take the simplest whole-number ratio ($2:3$), giving ${Fe}_{2}O_{3}$.]

Fe Oxygen
Moles : $$\frac{69.9}{56}$$ $$\frac{30.1}{16}$$
Molar Ratio : 1 $$:$$ 1.5
Molar Ratio : 2 $$:$$ 3

Options

A.FeO
B.${Fe}_{2}O_{3}$
C.${Fe}_{3}O_{4}$
D.${FeO}_{3}$

Solution

Sol. [Key Concept: Convert the given mass percentages of Fe and O to moles, take the simplest whole-number ratio ($2:3$), giving ${Fe}_{2}O_{3}$.]

Fe Oxygen
Moles : $$\frac{69.9}{56}$$ $$\frac{30.1}{16}$$
Molar Ratio : 1 $$:$$ 1.5
Molar Ratio : 2 $$:$$ 3

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