Question

Choose the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are $69.9$ and $30.1$ respectively and its molecular mass is $160$

• A. $FeO$
• B. $F{e}_3{O}_4$
• C. $F{e}_2{O}_3$
• D. $Fe{O}_2$

Answer 1

Answer: (C)

For element $Fe$, mole of atoms $=\frac{69.9}{56}=1.25$
For element $O$, mole of atoms $=30.116=1.88$
Mole ratio of $Fe=\frac{1.25}{1.25}=1$,
Mole ratio of $O=\frac{1.88}{1.25}=1.5$
Simplest whole number ratio of $Fe$ and $O=2,3$
Empirical formula of compound $=F{e}_2{O}_3$
Molecular mass of $F{e}_2{O}_3=160$
$n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}=\frac{160}{160}=1$
Molecular formula $=F{e}_2{O}_3$