Question

An organic compound has an empirical formula $ C{{H}_{2}}O, $ its vapour density is $45$. The molecular formula of the compound is:

• A. $ C{{H}_{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 $ vap. 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} $
$= C _{3} H _{6} O _{3}$