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.
We have given,
Mass percent of iron (Fe) =69.9%
Mass percent of oxygen (O)=30.1%
Number of moles of iron present in the oxide =69.90/55.85 =1.25
Number of moles of oxygen present in the oxide=30.1/16.0
=1.88
Ratio of iron to oxygen in the oxide,
1.25:1.88
=1:1.5
=2:3
Therefore the empirical formula of the oxide is Fe2O3.
Empirical formula mass of Fe2O3= [2(55.85) +3(16)]g
Molar mass of Fe2O3=159.69g
Thus, n= Molar mass/Empirical Formula mass =159.69/159.7g
=0.999 = 1(approx.)
Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus, the empirical formula of the given oxide is Fe2O3 and n is 1.
Hence, the molecular formula of the oxide is Fe2O3.
