Q. 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
Solution:
Element
Relative mass
Relative mole
Simplest whole number ratio
C
6
$\frac{6}{12}=0.5$
1
H
1
$\frac{1}{1}=1$
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
$C _{ x } H _{ Y } O _{ z }$
$C _{1} H _{2} O _{3 / 2}$
$\Rightarrow C _{2} H _{4} O _{3}$
| Element | Relative mass | Relative mole | Simplest whole number ratio |
|---|---|---|---|
| C | 6 | $\frac{6}{12}=0.5$ | 1 |
| H | 1 | $\frac{1}{1}=1$ | 2 |