Question

$12 ml$ of a gaseous hydrocarbon was exploded with $50ml$ of Oxygen. The volume measured after the explosion was $32 ml$. After treatment with KOH the volume diminished to $8ml$. Determine the formula of the hydrocarbon.

Answer 1

At first we will write what is given in the question then we will write the equation of explosion. It must be balanced. Then we will find the volume of Oxygen required to react with hydrocarbon. Then we will find out what happens when the KOH is used. In the end we will find the formula of hydrocarbon.

Complete solution:
Step1: The hydrocarbon volume is $12 ml$.
Volume of Oxygen is $50 ml$.
The volume after the explosion is $32 ml$.
The volume left after KOH is $8 ml$.
Step2: When the hydrocarbon reacts with the oxygen it creates heat and the carbon di oxygen water. It is a combustion reaction. The reaction is given below. Let us assume the chemical formula of hydrocarbon is
${C_x}{H_y} + (x + \dfrac{y}{4}){O_2} \to xC{O_2} + \dfrac{y}{2}{H_2}O$
The volume of oxygen required to react with the hydrocarbon is
$(x + \dfrac{y}{4}) volume of {C_x}{H_y}$
=$\left( {x + \dfrac{y}{4}} \right) \times 12 - - - - - - 1$
Step3. When the KOH is passed through the solution the carbon dioxide get absorbed. The volume left is only of Oxygen which is $8ml$.
$\\{ 50 - 12(x + \dfrac{y}{4})\\} + 12x = 32$
On solving we get that the
$y = \dfrac{{50 - 32}}{3} = 6$
Now we will write equation with only oxygen left
$50 - 12(x + \dfrac{y}{4}) = 8$
On solving we get
$2x = 32 - 8 = 2$
Step4. So the value of x will be 2 and the value of y is 6 so the formula of hydrocarbon would be ${C_2}{H_6}$.

Note: The volume of ${H_2}O$ is not considered because the volume is often measured because it is measured below ${100^ \circ }C$. At that temperature the water is liquid so volume of it is not counted.