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

F e_{2} O_{3}

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F e O

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F e_{3} O_{2}

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F e O_{2}

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Solution

The correct option is A $F e_{2} O_{3}$
Given,
$% of iron by mass = 69.9 % % of oxygen by mass = 30.1 %$
$Atomic mass of iron = 55.85 Atomic mass of oxygen = 16.00$
Formula used = Relative moles of atom in compound $= \frac{By mass of atom}{Atomic mass of atom}$
Relative moles of iron in iron oxide $= \frac{by mass of atom}{Atomic mass of atom}$
$= \frac{69.9}{55.85}$
$= 1.25$
Relative moles of oxygen in iron oxide:
$= \frac{by mass of atom}{Atomic mass of atom}$
$= \frac{30.1}{16.00}$
$= 1.88$
Simplest molar ratio of iron to oxygen:
$= 1.25 : 1.88$
$= 1 : 1.5$
$= 2 : 3$
$∴$ The empirical formula of the iron oxide is $F e_{2} O_{3}$

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