Question

The ratio of mass percent of C and H in 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 it is required to burn one molecule of a compound $C_x{H}_y$ completely to $C{O}_2$ and $H_{2}O$, then the empirical formula of compound $C_x{H}_y{O}_z$ is:

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

Answer 1

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

Given,
the ratio of mass percent of C and H in $\left(C_x{H}_y{O}_z\right)$ is 6:1
So, the ratio of mole percent of C and H in $\left(C_x{H}_y{O}_z\right)$ is 1:2.From this deduction, option (d) can be eliminated.
Now the amount of oxygen required to burn $C_x{H}_y$,
$C_x{H}_y+(x+\frac{y}{4})O_2\to xC{O}_2+\frac{y}{2}{H}_{2}O$
If z is the half of the oxygen atoms required to completely burn $C_x{H}_y$, then
$z=\frac{2\times (x+\frac{y}{4})}{2}=x+\frac{y}{4}$
In option (a), x=2, y=4, So z=3
In option (b), x=3, y=6, So z=4.5
So, the empirical formula of the compound $C_x{H}_y{O}_z$ is $C_2{H}_4{O}_3$