• # question_answer 11) Determine the molecular formula of an oxide of iron in which the mass percentage of iron and oxygen are 69.9 and 30.1 respectively.

Element % Atomic mass Relative no. of atoms Simplest ratio Whole no. ratio Iron 69.9 56 $\frac{69.9}{56}=1.248$ $\frac{1.248}{1.248}=1$ $1\times 2=2$ Oxygen 30.1 16 $\frac{30.1}{16}=1.88$ $\frac{1.88}{1.248}=1.5$ $1.5\times 2=3$   Formula of compound = $F{{e}_{2}}{{O}_{3}}$