Question

Complete the following calculations.

If the empirical formal of a compound is $CH$ and it has a vapour density of $13$, find the molecular formula of the compound?

Answer 1

Given that the empirical formula is $CH$.

V.D. = 13

Now, Mol. Mass = 2 $\times$ V.D. = 2$\times$13 = 26

Empirical formula weight of $CH$ = 12+1 = 13.

Now,

$n=\frac{Mol.mass}{Empiricalformulamass}$

$=\frac{26}{13}=2$

Molecular formula = (Empirical formula)${}_n$

=$(CH{)}_n$ = $C_2{H}_2$.