Question

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. $C_{3}H_{6}O_{3}$

Answer 1

Answer: (D)

Mol. wt. $=2 \times$ vapour density
$=2 \times 45=90$
Empirical formula weight
$=12+2+16=30$
$\therefore n =\frac{\text { mol. wt. }}{\text { empirical formula wt. }}$
$=\frac{90}{30}=3$
$\therefore $ Molecular formula of the compound
$=\left( CH _{2} O \right)_{3}$
$= C _{3} H _{6} O _{3}$