Question

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 1

Mass percent of iron (Fe) = \(69.9\)% (Given)
Mass percent of oxygen (O) = \(30.1\)% (Given)
Number of moles of iron present in the oxide = \(\frac {69.90}{55.85}\) = \(1.25\)
Number of moles of oxygen present in the oxide = \(\frac {30.1}{16.0}\) = \(1.88\)
Ratio of iron to oxygen in the oxide,
= \(1.25 : 1.88\)
= \(\frac {1.25}{1.25} : \frac {1.88}{1.25}\)
= \(1 : 1.5\)
= \(2 : 3\)
The empirical formula of the oxide is \(Fe_2O_3\).
Empirical formula mass of \(Fe_2O_3 = [2(55.85) + 3(16.00)]\ g\)
Molar mass of \(Fe_2O_3 = 159.69\ g\)
∴ \(n = \frac {\text {Molar\ mass}}{\text {Emperical\ formula\ mass }}\)

\(n = \frac {159.69\ g}{159.7\ g}\)
\(n = 0.999 \)
\(n = 1\) (approx)
Molecular formula of a compound is obtained by multiplying the empirical formula with \(n\).
Thus, the empirical formula of the given oxide is \(Fe_2O_3\) and \(n\) is \(1\).

Hence, the molecular formula of the oxide is \(Fe_2O_3\).