Question

an organic compound on analysis gave the following % composition C= 57.8% H= 3.6 the result is oxygen. the vapour density of the compound was found to be 83. find the molecular formula of the compound

Answer 1

Given,

C = 57.8%, H = 3.6% and O = 100 – ( 57.8 + 3.6)%.= 38.6%

Therefore, in 100 g of compound,

C = 57.8 g

H = 3.6 g

O = 38.6 g

Moles of C = 57.8/12 = 4.8

Moles of H = 3.6/1 = 3.6

Moles of O = 38.6/16 = 2.4

Ratio of moles of C, H, and O = 4.8 : 3.6 : 2.4

Dividing by 2.4

C : H : O = 2: 1.5 :1

Multiplying by 2 to get whole number ratio

C : H : O = 4 : 3: 2

Therefore, empirical formula of the compound = C₄H₃O₂

Empirical formula mass = 12 x 4 + 1 x 3 + 2 x 16 = 83 u

Vapour density = 83

Now,

Molecular mass = 2 x Vapour density

= 2 x 83 u

And

n = Molecular mass / empirical formula mass

= 2 x 83 / 83

=2

Hence, the molecular formula = (C₄H₃O₂)₂

= C₈H₆O₄