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