Determine the empirical formula of an oxide of iron which has $69.9 %$ iron and $30.1 %$ oxygen. $(F e = 55.85 a m u; O = 16.00 a m u)$

F e_{2} O_{3}

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F e_{4} O_{6}

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F e_{8} O_{12}

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`None of the above

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Solution

The correct option is A $F e_{2} O_{3}$
100 g of the compound contains $69.9$ g iron and $30.1$ g oxygen.
The atomic masses of iron and oxygen are $55.85$ amu and $16.00$ amu.

The number of moles of $F e$ present in $100$ g of compound are $\frac{69.9}{55.85} = 1.25$

The number of moles of $O$ present in $100$ g of compound are $\frac{30.1}{16.00} = 1.88$

The mole ratio of $F e$ to $O$ is $\frac{1.25}{1.88} = 0.67 : 1$ or $2 : 3$

Hence, the empirical formula is $F e_{2} O_{3}$
Option A is correct.

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