Question

4 g of hydrocarbon $({{C}_{x}}{{H}_{y}})$ on complete combustion gave 12 g of $\text{C}{{\text{O}}_{\text{2}}}\text{.}$ What is the empirical formula of the hydrocarbon? (C=12,H=1)

• A. $C{{H}_{3}}$
• B. ${{C}_{4}}{{H}_{9}}$
• C. $CH$
• D. ${{C}_{3}}{{H}_{g}}$

Answer 1

Answer: (D)

${{C}_{3}}{{H}_{g}}$

Answer 2

% of carbon in compound $=\frac{12}{44}\times \frac{\text{weight of C}{{\text{O}}_{\text{2}}}}{\text{weight of comp}\text{.}}\times 100$ $=\frac{12}{44}\times \frac{12}{4}\times 100$ $=81.81$ % of $H=100-81.81=18.19$
Element
$\frac{\text{percentage}}{\text{at}\text{. wt}\text{.}}$
Simple ratio
C
$\frac{81.81}{12}=6.81$
$\frac{6.81}{6.81}=1.0\times 3=3$
H
$\frac{18.19}{1}=18.19$
$\frac{18.19}{6.81}=2.67\times 3=8$
Hence, empirical formula of the compound $={{C}_{3}}{{H}_{8}}$