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
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Respected teacher, You have taken the percentage of iron instead of the molar mass of iron while calculating the Empirical Formula mass.Thank you for making available such detailed answers to us.
In NCERT the molecular mass of oxide is not given....that's why I was in trouble to find the answer...thank's for this information..