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 160.9?

• A. $FeO$
• B. $Fe_{3}O_{4}$
• C. $Fe_{2}O_{3}$
• D. $FeO_{2}$

Answer 1

Answer: (C)

$Fe_{2}O_{3}$

Answer 2

(3) : For element Fe, mole of atoms = $\frac{69.9}{56}$=$1.25$

For element O, mole of atoms = $\frac{30.1}{16} = 1.88$

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 = ${Fe}_{2}O_{3}$

Molecular mass of $Fe_{2}O_{3}$=160

n=$\frac{Molecularmass}{Empiricalformulamass} = \frac{160}{160} = 1$

Molecular formula $= Fe_{2}O_{3}$