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. $C_2{H}_4{O}_2$
• D. $C_3{H}_6{O}_3$

Answer 1

Answer: (D)

$C_3{H}_6{O}_3$

The empirical formula is $C{H}_{2}O$. Therefore, the empirical weight $=$ $30$ g.

Given, the molecular weight = $2\times V.D$ = $2\times 45=90$ g.

$x=\frac{\text{Molecular weight}}{\text{Empirical weight}}=\frac{90}{30}=3$

$\text{Molecular Formula}=(\text{Empirical Formula}{)}_x$

$\therefore \text{Molecular Formula}=(C{H}_{2}O{)}_3=C_3{H}_6{O}_3$