Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69•9 and 30•1 respectively

Since 69.9% and 30.1% of iron and oxygen are present in 100 grams of iron oxide.

Step I - Calculation of number of moles

Number of moles of iron present in 100 gram of iron oxide = $\frac{weightingrams\left(iron\right)}{molecularweightofiron}$ =$\frac{69.9}{55.8}=1.25mole$

Number of moles of oxygen present in 100 gram of iron oxide = $\frac{weightingrams\left(oxygen\right)}{molecularweightofoxygen}=\frac{30.1}{16}=1.88moles$

Step II - Calculation of ratio of number of moles of iron to number of moles of oxygen

Ratio of iron to oxygen = $\frac{1.25}{1.88}=\frac{1}{1.5}$

On multiplying the given ration with 2 we get $\frac{2}{3}$

Hence the empirical formula of iron oxide is ${Fe}_2{O}_3$

Step III - Calculation of n

Empirical mass of ${Fe}_2{O}_3=2\times 55.85+3\times 16=159.7$

Molar mass of ${Fe}_2{O}_3=159.69$

$n=\frac{molarmass}{Empiricalformulamass}=\frac{159.69}{159.7}=1$