## 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.Given that the molar mass of the oxide is 159.69 g mol-1

### Q.8:- 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.Given that the molar mass of the oxide is 159.69 g mol-1

% of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]

Atomic mass of iron = 55.85 amu
Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88
Simplest molar ratio = 1.25/1.25 : 1.88/1.25
⇒  1 : 1.5 = 2 : 3

∴ The empirical formula of the iron oxide is Fe2O3.
Mass of Fe2O3 = (2×55.85) + (3×16.00) = 159.7 g mol-1
n = Molar mass/Empirical formula mass = 159.7/159.6 = 1(approx)

Thus, Molecular formula is same as Empirical Formula i.e.  Fe2O3.