Question

$10$ mL of gaseous organic compound containing $C,H$ and $O$ only was mixed with $100$ mL of $O_2$ and exploded under identical conditions and then cooled. The volume left after cooling was $90$ mL. On treatment with $KOH$, a contraction of $20$ mL was observed. If vapour density of the compound is $23$, derive molecular formula of the compound.

• A. $C_2{H}_{6}O$
• B. $C_2{H}_{6}OH$
• C. $C_2{H}_{5}O$
• D. $C_2{H}_{5}OH$

Answer 1

Answer: (A), (D)

$C_2{H}_{6}O$
$C_2{H}_{5}OH$

$X_x{H}_y{O}_z+0.5(2x+0.5y-z)O_2\to xC{O}_2+0.5y{H}_{2}O$
The volume of gases is proportional to the number of moles.
$10$ mL of gaseous organic compound containing $C,H$ and $O$ only was mixed with $100$ mL of $O_2$ and exploded under identical conditions and then cooled.
This can be approximated to 10 moles of organic compound and 100 mol of oxygen.
Total number of moles are 110.
On treatment with $KOH$, a contraction of $20$ mL was observed.
This means that 20 moles of $C{O}_2$ were absorbed in KOH.
$x=\frac{20}{10}=2$
The volume left after cooling was $90$ mL.
The volume contraction is $110-90=20$.
This corresponds to water.
Hence, the empirical formula is $C_2{H}_{6}O$ or $C_2{H}_{5}OH$.
Empirical formula weight is 46 g/mol.
Molecular formula weight is $2\times V.D=2\times 23=46$.
Hence, the empirical formula and the molecular formula are same.