Organic Chemistry Basics Result Question 1
1. $25 \mathrm{~g}$ of an unknown hydrocarbon upon burning produces $88 \mathrm{~g}$ of $\mathrm{CO} _2$ and $9 \mathrm{~g}$ of $\mathrm{H} _2 \mathrm{O}$. This unknown hydrocarbon contains
(a) $20 \mathrm{~g}$ of carbon and $5 \mathrm{~g}$ of hydrogen
(b) $22 \mathrm{~g}$ of carbon and $3 \mathrm{~g}$ of hydrogen
(c) $24 \mathrm{~g}$ of carbon and $1 \mathrm{~g}$ of hydrogen
(d) $18 \mathrm{~g}$ of carbon and $7 \mathrm{~g}$ of hydrogen
(2019 Main, 12 April II)
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Answer:
Correct Answer: 1. ( c )
Solution:
- Hydrocarbon containing $\mathrm{C}$ and $\mathrm{H}$ upon burning produces $\mathrm{CO} _2$ and water vapour respectively. The equation is represented as
$\mathrm{C}_x \mathrm{H}_y +(x+y / 4) \mathrm{O}_2 \longrightarrow x \mathrm{CO}_2+(y / 2) \mathrm{H}_2 \mathrm{O} $
$\text { Mass of carbon } =\frac{12}{44} \times \text { mass of } \mathrm{CO}_2 $
$ =\frac{12}{44} \times 88 \mathrm{~g}=24 \mathrm{~g} $
$\text { Mass of hydrogen } =\frac{2}{18} \times \text { mass of } \mathrm{H}_2 \mathrm{O} $
$ =\frac{2}{18} \times 9=1 \mathrm{~g}$
So, the unknown hydrocarbon contains $24 \mathrm{~g}$ of carbon and $1 \mathrm{~g}$ of hydrogen.
BETA
