Question

A hydrocarbon contains 10.5 g of carbon per gram of hydrogen; 1 litre of the vapour of the hydrocarbon at 127°C and 1 atmosphere pressure weighs 2.8g. Find the molecular formula.

Answer 1

Calculation of empirical formula.

Let wt. of hydrogen = 1 g 

Wt. of carbon = 10.5 g 

Moles of C : Moles of H = \(\frac{10.5}{12} : \frac 11\)

= 0.875 : 1

= 0.875 x 8:1 x 8 

= 7 moles: 8 moles 

= 7 atoms: 8 atoms 

Empirical formula of the hydrocarbon = C₇H₈

Calculation of mol. wt. and mol. formula

We know that, 

\(PV = \frac{m}{M} \times RT\)

Here, P = 1 atmosphere, V = 1 litre, T = 127 + 273 = 400 A, R= 0.082 litre atm./degree/mole, m = 2.8g, M = ?

\(\therefore M = \frac {m \times R \times T}{P\times V}\)

\(= \frac{2.8 \times 0.082 \times 400}{1 \times 1}\)

\(= 91.8\)

\(n = \frac{\text{Mol. wt.}}{\text{Emp. formula wt.}}\)

\(= \frac{91.8}{84 + 8}\)

\(= 1\) (app.)

Hence mol. formula (C₇H₈)₁ = C₇H₈