Question

In a compound C, H and N are present in the ratio 9:1:3.5 by weight. Molecular mass of the compound is 108. What is the molecular formula of the compound?

Answer 1

The ratios of the weights of the elements present in the compound are given.

Let us assume that the weight of:

C = 9x
H = 1x
N = 3.5x

therefore,
9x+1x+3.5x = 108
x = 8

Mass of C in the compound= 9x = 9*8 = 72 g
Mass of H in the compound= 1x = 8 g
Mass of N in the compound= 3.5x = 3.5*8 = 28 g

We can calculate number of moles from the mass of the element in the compound,

Number of moles of C in the compound = 72/12 = 6
no. of moles of H in the compound = 8/1= 8
no. of moles of N in the compound = 28/14 = 2

Ratios between the number of moles of each element,

Divide the number of moles by the smallest value:

Ratio of C:N = 3

Ratio of H:N = 4

Ration of N:N = 1

Therefore, Empirical formula is C₃H₄N₁

Empirical formula mass = (12*3) + (1*4) + (14) = 54

Number of empirical formula units = 108/54 = 2

Therefore, the molecular formula can be derived by multiplying the empirical formula by 2

= C₆H₂N₂