Home NCERT Solutions NCERT Exemplar CBSE Sample Papers NCERT Books Class 12 Class 10
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.

Answer

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

Popular Questions of Class 11th chemistry

Recently Viewed Questions of Class 11th chemistry

10 Comment(s) on this Question

Write a Comment: