question_answer 16)

Determine 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.

Answer:

  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 = \[F{{e}_{2}}{{O}_{3}}\]