The molecular formula of the oxide of iron in which mass percentage of iron and oxygen are 69.0 and 30.1 respectively ?
The molecular formula of the oxide of iron in which mass percentage of iron and oxygen are 69.0 and 30.1 respectively ?
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(55.85) + 3(16.00)] =159.7
Molecular mass of oxide= 159.69g/mol(given)
Now we know molecular formula = n x Empirical formula
And n= molecular mass / empirical formula mass= 159.69/159.7 = 0.999 = approx 1
Therefore molecular formula = n x empirical formula
=1 x(Fe2O3) = Fe2O3
The molecular formula of the oxide is Fe2O3
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(55.85) + 3(16.00)] =159.7
Molecular mass of oxide= 159.69g/mol(given)
Now we know molecular formula = n x Empirical formula
And n= molecular mass / empirical formula mass= 159.69/159.7 = 0.999 = approx 1
Therefore molecular formula = n x empirical formula
=1 x(Fe2O3) = Fe2O3
The molecular formula of the oxide is Fe2O3
