Q. 5.6 g of an organic compound on burning with excess of oxygen gave 17.6g of $CO_{2}$ and 7.2 g of $H_{2}O$. The organic compound is
A) $C_{6}H_{6}$
B) $C_{4}H_{8}$
C) $C_{3}H_{8}$
D) $CH_{3}COOH$
Explanation
(b)
Organic compound $\overset{\ \ \ \ \ \lbrack O\rbrack\ \ \ }{\rightarrow}CO_{2} + H_{2}O$
17.6 g 7.2g
$\% of\ C = \frac{12}{44} \times \frac{17.6}{5.6} \times 100 = 85.7\%$
$\%\ of\ H = \frac{2}{18} \times \frac{7.2}{5.6} \times 100 = 14.28\%$
[table] Element | Percentage | Relative no. of atoms | Simplest ratio ; C | 85.7 | 85.7/12=7.14 | 7.14/7.14=1 ; H | 14.28 | 14.28/1=14.28 | 14.28/7.14=2 [/table]
Hence, empirical formula of compound of $= CH_{2}$
∴ Molecular formula of compound =$C_{4}H_{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.