Complete combustion of 0.86g of compound ‘X’ gives 2.64g of CO₂ and 1.26g of H₂O. The minimum molecular mass of ‘X’ can have

$\mathrm{C}={\mathrm{n}}_{{\mathrm{CO}}_2}=\frac{2.64}{44}=0.06$
$\Rightarrow \mathrm{mass} \mathrm{of} \mathrm{C}=0.72$
Moles of H = 2 x moles of H2O
$=2\times \frac{1.26}{18}=0.14$
mass of H = 0.14
Compound does not contains oxygen,
So,  $\mathrm{EF}\to {\mathrm{C}}_{0.06}{\mathrm{H}}_{0.14}\Rightarrow {\mathrm{C}}_3{\mathrm{H}}_7$
Lowest molecular mass of ‘X = 43