Question

The ratio of mass percent of Cane H of an organic compound $\left(C_x{H}_y{o}_z\right)$ is 6:1. If one molecule of the above 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 $C_x{H}_y{o}_z$ is:

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

Answer 1

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

When organic compound $-C_x{H}_y{O}_2\to$ molars mass $=12x+y+162$

$\frac{mass\%C}{mass\%H}=\frac{6}{1}$

mass $\%$ of $C=\frac{12x}{12x+yx16z}$ and mass $\%$ of $H=\frac{r}{12x+y+16z}$

$\therefore \frac{12x}{y}=\frac{6}{1}\Rightarrow \frac{x}{y}=\frac{1}{2}----\left(1\right)$

Also, $C_x{H}_y+(x+\frac{y}{4})O_2⟶xC{O}_2+\frac{y}{2}{H}_{2}O$

Noe, $z$ is half of the organ atoms required to better $C_x{H}_y$

$\therefore z_1=\frac{2(x+y/4)}{2}=x\frac{y}{4}$

$\therefore x:y=1:2$ Take $x=p$ and $y=2p$

Then, $z_1=p+\frac{p}{2}=\frac{3p}{2}$

Thus, $x:y:z=p:2p:\frac{3p}{2}=2:4:3$

Required empirical formula $C_2{H}_4{O}_3$

Correct answer $\left(A\right)C_2{H}_4{O}_3$