Question

A gaseous hydrocarbon on combustion gives 0.72g of water and 3.08 g of $C{O_2}$. The empirical formula of the hydrocarbon is:

{a.{\text{ }}-{C_3}{H_4}} \\\ {b.- {\text{ }}{C_6}{H_5}} \\\ {c.{\text{ }}-{C_7}{H_8}} \\\ {d.- {\text{ }}{C_2}{H_4}} \end{array}$$

Answer 1

In combustion, hydrocarbon reacts with oxygen to give energy. Other products of this reaction are carbon dioxide and water. General representation for such reaction is:
${C_x}{H_Y}{\text{ }} + {\text{ }}(x + \dfrac{y}{4}){O_2}\; \to {\text{ }}\;xC{O_2}\; + {\text{ }}2y{H_2}O$

Complete answer:
Step-1:
First of all find the given identities and what to find out.
Given: Amount of water obtained: 0.72 g
Amount of carbon dioxide obtained: 3.08 g
Empirical formula of hydrocarbon: ?
Step-2
Find the number of moles of water and carbon dioxide obtained:
Moles of ${H_2}O$ obtained=$\dfrac{W}{{MW}}$
= $\dfrac{{0.72}}{{18}}$
= 0.04 mole
Moles of$\;C{O_2}$obtained​=$\dfrac{W}{{MW}}$
= $\dfrac{{3.08}}{{44}}$
= 0.07 mole
Step-3
Now apply general representation equation and find x, y
${C_x}{H_Y}{\text{ }} + {\text{ }}(x + \dfrac{y}{4}){O_2}\; \to {\text{ }}\;xC{O_2}\; + {\text{ }}2y{H_2}O$
2y= moles of water obtained
2y=0.04
$\dfrac{y}{2}$ =0.04
y= 0.08 …..(i)
x= moles of carbon dioxide obtained
x=0.07 …. (ii)
Use the value of x and y from equation (i) and (ii) and try to make formula of hydrocarbon
Here y= 0.08 and x=0.07
${C_x}{H_Y} = {C_{0.07}}{H_{0.08}}$
Or we can write ${C_7}{H_8}$.

Hence, the empirical formula of the hydrocarbon is ${C_7}{H_8}$i.e. option (C).

Note:
Direct prediction of molecular formula is not possible. We have to convert the grams of product into mole first then frame the required hydrocarbon.