Question

$500$ml of a gaseous hydrocarbon burnt in excess of O gave $2.5$ lt. of ${\text{C}}{{\text{O}}_{\text{2}}}$and $3.0$lt. of water vapour under same conditions. Molecular formula of the hydrocarbon is—
(1) ${{\text{C}}_{\text{4}}}{{\text{H}}_{\text{8}}}$
(2) ${{\text{C}}_{\text{4}}}{{\text{H}}_{10}}$
(3) ${{\text{C}}_5}{{\text{H}}_{10}}$
(4) ${{\text{C}}_5}{{\text{H}}_{12}}$

Answer 1

Hint: We will write the general reaction for the combustion. We will determine the relation between volume and number of moles so, we can replace the volume with number of moles. The number of moles of carbon dioxide will give the number of carbon atoms in hydrocarbon.

Complete Step by step answer: Combustion reactions are defined as the reaction in which the chemical substance reacts with oxygen and forms carbon dioxide and water and release the energy.
The equation for the combustion of hydrocarbon is represented as follows:
${{\text{C}}_n}{{\text{H}}_{2n}}\,{\text{ + }}\,\,{{\text{O}}_{\text{2}}}\,\mathop \to \limits^\Delta \,{\text{C}}{{\text{O}}_2}\, + \,{{\text{H}}_{\text{2}}}{\text{O}}$
$500$ml of a gaseous hydrocarbon burnt in excess of O gave $2.5$ lt. of ${\text{C}}{{\text{O}}_{\text{2}}}$and $3.0$lt. of water vapour under same conditions. So, first we will convert the volume of hydrocarbon from ml to lt as follows:
${\text{1000}}\,{\text{ml}}\,{\text{ = }}\,{\text{1}}\,{\text{lt}}$
${\text{500}}\,{\text{ml}}\,{\text{ = }}\,0.2\,{\text{lt}}$
So,
${{\text{C}}_n}{{\text{H}}_{2n}}\,{\text{ + }}\,\,{{\text{O}}_{\text{2}}}\,\mathop \to \limits^\Delta \,{\text{C}}{{\text{O}}_2}\, + \,{{\text{H}}_{\text{2}}}{\text{O}}$
${\text{0}}{\text{.5}}\,{\text{L}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{2}}{\text{.5L}}\,\,\,\,\,{\text{3}}{\text{.0L}}$
The ideal has equation is as follows:
${\text{pV}}\,{\text{ = nRT}}$
All the species are reacting at same condition so, we can say temperature and pressure is constant. R is gas constant.
So, according to the ideal gas equation, volume is directly proportional to the number of moles. So, we replace the volume with number of moles.
${{\text{C}}_n}{{\text{H}}_{2n}}\,{\text{ + }}\,\,{{\text{O}}_{\text{2}}}\,\mathop \to \limits^\Delta \,{\text{C}}{{\text{O}}_2}\, + \,{{\text{H}}_{\text{2}}}{\text{O}}$
${\text{0}}{\text{.5}}\,{\text{mole}}\,\,\,\,\,\,\,\,\,{\text{2}}{\text{.5mole}}\,\,\,\,\,{\text{3}}{\text{.0}}\,{\text{mole}}$
Multiply the integer with two to convert into whole numbers.
${\text{1}}\,{\text{mole}}\,\,\,\,\,\,\,{\text{5mole}}\,\,\,\,\,6\,{\text{mole}}$
So, we can write the combustion reaction as,
${\text{1}}{{\text{C}}_n}{{\text{H}}_{2n}}\,{\text{ + }}\,\,{{\text{O}}_{\text{2}}}\,\mathop \to \limits^\Delta \,5\,{\text{C}}{{\text{O}}_2}\, + \,6\,{{\text{H}}_{\text{2}}}{\text{O}}$
So, the value of n in hydrocarbon is $5$because we are obtaining five mole carbon dioxide.
On placing $5$for n is hydrocarbon formula,
${{\text{C}}_5}{{\text{H}}_{2 \times 5}}$
${{\text{C}}_5}{{\text{H}}_{10}}$
So, the molecular formula of the hydrocarbon is${{\text{C}}_5}{{\text{H}}_{10}}$.

Therefore, option (C) ${{\text{C}}_5}{{\text{H}}_{10}}$is correct.

Note: The general formula for hydrocarbon is${{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n}}}}$. Where n is the number of carbons atoms. The number of carbon atoms represents the mole of products produced by one mole of hydrocarbon. The combustion reaction for ${{\text{C}}_5}{{\text{H}}_{10}}$can be written as follows:
${{\text{C}}_5}{{\text{H}}_{10}}\,{\text{ + }}\,\,\frac{{15}}{2}{{\text{O}}_{\text{2}}}\,\mathop \to \limits^\Delta \,5\,{\text{C}}{{\text{O}}_2}\, + \,5\,{{\text{H}}_2}{\text{O}}$