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 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 $CO _{2}$ and $H _{2} O$. The empirical formula of compound $C _{ X } H _{ Y } O _{ Z }$ is

• A. $C_3H_6O_3$
• B. $C_2H_4O$
• C. $C_3H_4O_2$
• D. $C_2H_4O_3$

Answer 1

Answer: (D)

$C_2H_4O_3$

Answer 2

So, $X=1, Y=2$ Equation for combustion of $C _{ x } H _{ Y }$ $C _{ X } H _{ Y }+\left( X +\frac{ Y }{4}\right) O _{2} \longrightarrow XCO _{2}+\frac{ Y }{2} H _{2} O$ Oxygen atoms required $=2\left( X +\frac{ Y }{4}\right)$ As per information, $2\left(X+\frac{Y}{4}\right)=2 Z$ $\Rightarrow\left(1+\frac{2}{4}\right)=Z$ $\Rightarrow Z =1.5$ Molecule can be written $C _{ x } H _{ Y } O _{ z }$ $C _{1} H _{2} O _{3 / 2}$ $\Rightarrow C _{2} H _{4} O _{3}$