Question

A 25g of an unknown hydrocarbon upon burning produces $88{\text{ g}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ and $9{\text{ g}}$ of ${{\text{H}}_{\text{2}}}{\text{O}}$. This unknown hydrocarbon contains:
(A) $18{\text{ g}}$of carbon and $7{\text{ g}}$ of hydrogen
(B) $20{\text{ g}}$of carbon and $5{\text{ g}}$ of hydrogen
(C) $22{\text{ g}}$of carbon and $3{\text{ g}}$ of hydrogen
(D) $24{\text{ g}}$of carbon and $1{\text{ g}}$ of hydrogen

Answer 1

Hint: The products obtained when a hydrocarbon is burned are water and carbon dioxide. We have to calculate the amount of carbon and hydrogen in the products. To solve this calculate the amount of carbon and hydrogen in the amount of carbon dioxide and water produced.

Complete step by step solution:
We have to calculate the amount of carbon in $88{\text{ g}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ as follows:

From the molecular formula of ${\text{C}}{{\text{O}}_{\text{2}}}$, we can see that one mole of carbon dioxide contains one mole of carbon. Thus,
$1{\text{ mol C}}{{\text{O}}_{\text{2}}} = 1{\text{ mol C}}$
Thus, using the molar masses of carbon and carbon dioxide,
$44{\text{ g C}}{{\text{O}}_{\text{2}}} = 12{\text{ g C}}$
Thus,
Amount of carbon in $88{\text{ g}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ $ = 88{\text{ g C}}{{\text{O}}_2} \times \dfrac{{12{\text{ g C}}}}{{88{\text{ g C}}{{\text{O}}_2}}}$
Amount of carbon in $88{\text{ g}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ $ = 24{\text{ g C}}$
Thus, the amount of carbon in $88{\text{ g}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ is $24{\text{ g C}}$.
We have to calculate the amount of hydrogen in $9{\text{ g}}$ of ${{\text{H}}_{\text{2}}}{\text{O}}$ as follows:
From the molecular formula of ${{\text{H}}_{\text{2}}}{\text{O}}$, we can see that one mole of water contains two moles of hydrogen. Thus,
$1{\text{ mol }}{{\text{H}}_{\text{2}}}{\text{O}} = 2{\text{ mol H}}$
Thus, using the molar masses of water and hydrogen,
$18{\text{ g C}}{{\text{O}}_{\text{2}}} = 2{\text{ g C}}$
Thus,
Amount of hydrogen in $9{\text{ g}}$ of ${{\text{H}}_{\text{2}}}{\text{O}}$ $ = 9{\text{ g }}{{\text{H}}_2}{\text{O}} \times \dfrac{{2{\text{ g H}}}}{{18{\text{ g }}{{\text{H}}_2}{\text{O}}}}$
Amount of hydrogen in $9{\text{ g}}$ of ${{\text{H}}_{\text{2}}}{\text{O}}$ $ = 1{\text{ g H}}$
Thus, the amount of hydrogen in $9{\text{ g}}$ of ${{\text{H}}_{\text{2}}}{\text{O}}$ is $1{\text{ g H}}$.
Thus, $25{\text{ g}}$ of an unknown hydrocarbon upon burning produces $88{\text{ g}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ and $9{\text{ g}}$ of ${{\text{H}}_{\text{2}}}{\text{O}}$. This unknown hydrocarbon contains $24{\text{ g}}$of carbon and $1{\text{ g}}$ of hydrogen.

Thus, the correct option is (D) $24{\text{ g}}$of carbon and $1{\text{ g}}$ of hydrogen.

Note: We know that hydrocarbons are the compounds that contain carbon and hydrogen only. When hydrocarbons are burned, the hydrocarbons decompose and produce carbon dioxide and water. Thus, the products of hydrocarbon combustion are always carbon dioxide and water.