Question

A carbon compound containing carbon and oxygen has molar mass equal to 288. On analysis it is found to contain 50% by mass of each element. Therefore, molecular formula of the compound is:

Answer 1

Element	Atomic Mass	Relative no. of atoms	Simplest Ratio	Whole no. Ratio
$C$ (50%)	$12$	$4.166$	$1.33$	$1.33\times 3=4$
$O$ (50%)	$16$	$3.125$	$1$	$1\times 3=3$

$\therefore$ Empirical formula= $C_4{O}_3$

Mass= $(4\times 12)+(3\times 16)=96$

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}=\frac{288}{96}=3$

$\therefore$ Molecular formula= $n\times$ Empirical formula

$=3\left(C_4{O}_3\right)$

$=C_{12}{O}_9$ .