Question

$\text{0}\text{.246 g}$ of an organic compound on complete combustion gave $\text{0}\text{.198 g}$ of carbon dioxide and $\text{0}\text{.1014 g}$of water, then the percentage composition of carbon and hydrogen in the compound respectively
1) $\text{ }4.58\text{ , 21}\text{.95}$
2) $\text{ 21}.95\text{ , 4}\text{.58}$
3) $\text{ 45}.8\text{ , 2}\text{.195}$
4) $\text{ 2}.195\text{ , 45}\text{.8}$

Answer 1

Hint: The Liebig’s Method is used for the estimation of elements in an organic compound. The organic compound is heated in presence of dry cupric oxide. The percentage of the mass of carbon and hydrogen is determined by taking the ratio of the weight of the $\text{C}{{\text{O}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}\text{O}$ adsorbed to the total weight of the compound taken.

Complete step by step solution:
Carbon and hydrogen in the organic compound are estimated by the Liebig’s Method.
When an organic compound of known mass is strongly heated in presence of dry cupric oxide $\text{(CuO)}$ the carbon $\text{C}$ and hydrogen $\text{H}$in the organic compound quantitatively oxidize to the carbon dioxide $\text{C}{{\text{O}}_{\text{2}}}$and ${{\text{H}}_{\text{2}}}\text{O}$.
Let consider an organic compound having the molecular formula $\text{ }{{\text{C}}_{\text{x}}}{{\text{H}}_{\text{y}}}$ .it undergoes the reaction as follows:
${{\text{C}}_{\text{x}}}{{\text{H}}_{\text{y}}}\text{ + }\left( \text{x+}\dfrac{\text{y}}{\text{4}} \right){{\text{O}}_{\text{2}}}\to \text{ xC}{{\text{O}}_{\text{2}}}\text{ + }\dfrac{\text{y}}{\text{2}}{{\text{H}}_{\text{2}}}\text{O}$
Here we are given with the following data:
$\begin{align}
& \text{weight of unknown compound (}{{\text{C}}_{x}}{{\text{H}}_{y}}\text{) = 0}\text{.246 g} \\
& \text{Weight of C}{{\text{O}}_{\text{2}}}\text{ = 0}\text{.198 g} \\
& \text{weight of }{{\text{H}}_{\text{2}}}\text{O = 0}\text{.1014 g} \\
& \text{To find the }{\scriptstyle{}^{\text{0}}/{}_{\text{0}}}\text{ of carbon and hydrogen} \\
\end{align}$
By knowing the amount of carbon dioxide and water we can determine the weight of the organic compound. The percentage of carbon can be calculated as follows:
$\text{Percentage of carbon=}\dfrac{\text{12}}{\text{44}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{weight of C}{{\text{O}}_{\text{2}}}\text{ }\!\!\times\!\!\text{ 100}}{\text{weight of organic substance}}$
Let’s substitute the values for the weight of compound and the weight of carbon dioxide. We get,
\[\begin{align}
  & \text{ Percentage of carbon=}\dfrac{\text{12}}{\text{44}}\text{ }\!\!\times\!\!\text{ }\dfrac{\left( 0.198\text{ g} \right)\text{ }\!\!\times\!\!\text{ 100}}{\left( 0.246 \right)} \\
 & \Rightarrow \text{Percentage of carbon=}\dfrac{\text{12}}{\text{44}}\text{ }\!\!\times\!\!\text{ }\dfrac{19.8}{\left( 0.246 \right)} \\
 & \Rightarrow \text{Percentage of carbon =}\dfrac{237.6}{10.824} \\
 & \Rightarrow \text{Percentage of carbon = 21}\text{.95 }{\scriptstyle{}^{0}/{}_{0}} \\
\end{align}\]
Let’s calculate the weight percentage for hydrogen. Substitute the values given in the question. We get,
$\begin{align}
  & \text{Percentage of hydrogen=}\dfrac{\text{2}}{\text{18}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{weight of }{{\text{H}}_{\text{2}}}\text{O }\!\!\times\!\!\text{ 100}}{\text{weight of organic substance}} \\
 & \Rightarrow \text{Percentage of hydrogen=}\dfrac{2}{18}\text{ }\!\!\times\!\!\text{ }\dfrac{\left( 0.1014\text{ g} \right)\text{ }\!\!\times\!\!\text{ 100}}{\left( 0.246 \right)} \\
 & \Rightarrow \text{Percentage of carbon=}\dfrac{\text{2}}{18}\text{ }\!\!\times\!\!\text{ }\dfrac{10.14}{\left( 0.246 \right)} \\
 & \Rightarrow \text{Percentage of carbon =}\dfrac{20.28}{4.428} \\
 & \Rightarrow \text{Percentage of carbon = 4}\text{.58 }{\scriptstyle{}^{0}/{}_{0}} \\
\end{align}$
Therefore, the percentage of carbon in the unknown compound is \[\text{21}\text{.95 }{\scriptstyle{}^{0}/{}_{0}}\] and percentage of hydrogen is$\text{4}\text{.58 }{\scriptstyle{}^{0}/{}_{0}}$.

Hence, (B) is the correct option.

Note: To solve such a type of question remember that the amount of carbon dioxide formed is equal to the amount of carbon present in the compound. i.e.
\[\begin{matrix}
\text{Weight of C}{{\text{O}}_{2}} & = & \text{Weight of C} \\
\text{weight of C}{{\text{O}}_{2}}\text{ from the compound} & = & \text{weight of C from the compound (X) } \\
{} & {} & {} \\
\end{matrix}\]