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  • Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

1.3    Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

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Given there is an oxide of iron which has 69.9\% iron and 30.1\% dioxygen by mass:

Relative moles of iron in iron oxide:

= \frac{\%\ of\ iron\ by\ mass}{Atomic\ mass\ of\ iron}

= \frac{69.9}{55.85} = 1.25

Relative moles of oxygen in iron oxide:

= \frac{\%\ of\ oxygen\ by\ mass}{Atomic\ mass\ of\ oxygen}

= \frac{30.1}{16.00} = 1.88

The simplest molar ratio of iron to oxygen:

\Rightarrow 1.25:1.88 \Rightarrow 1:1.5 \Rightarrow 2:3

Therefore, the empirical formula of the iron oxide is Fe_{2}O_{3}.

 

Posted by

Divya Prakash Singh

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