Question

$4 g$ of hydrocarbon $\left( C _{ x } H _{ y }\right)$ on complete combustion gave $12 g$ of $CO _{2} .$ What is the empirical formula of the hydrocarbon?

• A. $CH _{3}$
• B. $C _{4} H _{9}$
• C. $CH$
• D. $C _{3} H _{8}$

Answer 1

Answer: (D)

Per cent of carbon in compound

$\frac{12}{44} \times \frac{\text { weight of } CO _{2}}{\text { weight of compound }} \times 100$

$\frac{12}{44} \times \frac{12}{4} \times 100=81.81 \%$

$\%$ of $H =100-81.81=18.19 \%$



Hence; empirical formula of the compound $= C_{3}H_{8} $