Ii) An organic liquid having carbon, hydrogen, nitrogen and oxygen was found to

contain C = 41.37%; H = 5.75% ; N= 16.09 % and the rest oxygen. Calculate

the Empirical formula

(6 marks)

C3H5NO2

Explanation:

C = 41.37%;

H = 5.75% ;

N= 16.09 %;

O= (100 - 41.37 - 5.75 - 16.09)% = 36.79 %.

In 100 g of substance we have

C = 41.37 g;

H = 5.75 g ;

N= 16.09 g;

O = 36.79 g.

Molar mass (C) = 12 g/mol;

Molar mass(H) =1 g/mol;

Molar mass(N)= 14 g/mol;

Molar mass(O) = 16 g/mol.

C = 41.37 g* 1 mol/12g = 3.4475 mol;

H = 5.75 g *1 mol/1g = 5.75 mol;

N= 16.09 g*1mol/14g = 1.1493 mol;

O = 36.79 g* 1mol/16g = 2.2994 mol.

The Empirical formula shows ratio of moles of elements in the substance , so

C : H : N : O = 3.4475 mol : 5.75 mol : 1.1493 mol : 2.2994 mol =

= (3.4475 mol /1.1493 mol) : (5.75 mol/1.1493 mol) : (1.1493 mol /1.1493 mol) :   :( 2.2994 mol/1.1493 mol) = 3 : 5 : 1 : 2

C : H : N : O = 3 : 5 : 1 : 2

C3H5NO2