Q. 4 g of hydrocarbon on complete combustion gave 12.571 g of $CO_{2}$ and 5.143 g of water. What is the empirical formula of the hydrocarbon?
A) CH
B) $C_{2}H_{3}$
C) $CH_{2}$
D) $CH_{3}$
Explanation
(c)
$\%\ C = \frac{12}{44} \times \frac{12.517}{4.0} \times 100 = 85.7$
$\%\ H = \frac{2}{18} \times \frac{5.143}{4.0} \times 100 = 14.3$
The mole ratio of C to H is $\frac{85.7}{12}:\frac{14.3}{1}$
$= 7.14:14.3 = 1:2 = CH_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.