Question

The ratio of mass percent of $C$ and $H$ of an organic compound $\left(C_x{H}_Y{O}_Z\right)$ is $6:1$. If one molecule of the given compound $\left(C_X{H}_Y{O}_Z\right)$ contains half as much oxygen as required to burn one molecule of compound $C_X{H}_Y$ completely to $C{O}_2$ and $H_{2}O$. The empirical formula of compound $C_X{H}_Y{O}_Z$ is:

• A. $C_3{H}_6{O}_3$
• B. $C_2{H}_{4}O$
• C. $C_3{H}_4{O}_2$
• D. $C_2{H}_4{O}_3$

Answer 1

The correct option is D $C_2{H}_4{O}_3$

Given: Ratio of masses of % of $C$ & $H$ in $C_x{H}_y{O}_2$ is $6:1$

$\therefore$ Ratio of mole % of $C$ & $H$ will be $1:2$

$\therefore \frac{12x}{y}=\frac{6}{1}$

${C_x{H}_y}_{\left(g\right)}+(x+y/4){O_2}_{\left(g\right)}\to {xC{O}_2}_{\left(g\right)}+\frac{y}{2}{H}_2{O}_{\left(l\right)}$

Number of oxygen atoms in $C_x{H}_y{O}_2=2$

Number of oxygen atoms required for combustion of $C_x{H}_y$ is $(x+\frac{y}{4})\times 2=2x+\frac{y}{2}$

So, $Z=\frac{1}{2}(2x+\frac{y}{2})$

$\Rightarrow Z=x+y/4$

$=x+\frac{2x}{4}=\frac{3x}{2}$

$x:2x:\frac{3x}{2}\Rightarrow 2x:4x:3x\Rightarrow 2:4:3$

Hence, Empirical formula= $C_2{H}_4{O}_3$