Q. An organic compound has an empirical formula (${CH}_{2}O$) its vapour density is 45. The molecular formula of the compound is
A) ${CH}_{2}O$
B) ${\ C}_{2}H_{5}O$
C) $C_{2}H_{2}O$
D) $C_{3}H_{6}O_{3}$
Explanation
(d)
Mol. wt.=$2 \times$ vapour density
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2 \times 45 = 90$
Empirical formula weight=12+2+16=30
$\therefore n = \frac{mol.wt}{empirical\ formula\ wt.}$
$\ \ \ \ \ \ \ = \frac{90}{30} = 3$
$\therefore$ Molecular formula of the compound
$= \left( CH_{2}O \right)_{3}$
$= C_{3}H_{6}O_{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.