If an iron oxide has $69.9$ mass $\%$ of Fe (molar mass $=56.0 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and $30.1$ mass $\%$ of oxygen, its molecular formula will be

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Aicharya · Accepted Answer

The iron oxide has $69.9 \%$ iron and $30.1 \%$ dioxygen by mass.
Thus, $100 \mathrm{~g}$ of iron oxide contains $69.9\mathrm{~g}$ iron and $30.1 \mathrm{~g}$ dioxygen.
The number of moles of iron present in $100 \mathrm{~g}$ of iron oxide are $\frac{69.9}{55.8}=1.25$.
The number of moles of dioxygen present in $100 \mathrm{~g}$ of iron oxide are $\frac{30.1}{32}=$ $0.94 .$
The ratio of the number of oxygen atoms to the number of iron atoms present in one formula unit of iron oxide is $\frac{2 \times 0.94}{1.25}=1.5: 1=3: 2$.
Hence, the formula of the iron oxide is $\mathrm{Fe}_{2} \mathrm{O}_{3}$.

If an iron oxide has $69.9$ mass $\%$ of Fe (molar mass $=56.0 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and $30.1$…

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