Question 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.

From the available data Percentage of iron = 69.9

Percentage of oxygen= 30.1

Total percentage of iron & oxygen= 69.9+30.1= 100%

Step 1 calculation of simplest whole number ratios of the elements

 Element Percentage Atomic mass Atomic ratio Simplest ratio Simplest whole no ratio Fe 69.9 55.84 69.9/55.84=1.25 1.25= 1 2 O 30.1 16 30.1/16 = 1.88 1.88=1.5 3

Step 2  Writing the empirical formula of the compound

The empirical formula of the compound = Fe2 O3

Step 3 determination of molecular formula of the compound

Empirical formula mass = 2 X69.9 + 3 X16=187.8 amu

Molecular mass of oxide= 159.69g/mol(given)

Now we know molecular formula = n x Empirical formula

And n= molecular mass / empirical formula mass= 159.69/187.8 = 0.85 = approx 1

Therefore molecular formula = n x empirical formula

=1 x(Fe2O3) = Fe2O3

The molecular formula of the oxide is Fe2O3