On complete combustion, 0.3 g of an organic compound gave 0.2 g of CO_2 and 0.1 g of H_2 O . The percentage composition of carbon and hydrogen in the compound, respectively is:

Correct answer: B. 18.18 \% and 3.70 \% Percentage of carbon = 12 × m_2 × 100 44 × m m= mass of organic compound =0.3 ~g m_2= mass of carbon dioxilde =0.2 ~g \% C=(12 × 0.2 × 100)/(44 × 0.3)=18.18 \% Percentage of hydrogen = 2 × m_1 × 100 18 × m m_1= mass of water =0.1 ~g \% H=(2 × 0.1 × 100)/(18 × 0.3)=3.70 \%

NEET 2024

Exam: NEET
Year: 2024
Subject: Chemistry
Chapter: General Organic Chemistry

NEET 2024 · Chemistry · General Organic Chemistry

On complete combustion, 0.3 g of an organic compound gave 0.2 g of...

On complete combustion, 0.3 g of an organic compound gave 0.2 g of

\mathrm{CO}_2

and 0.1 g of

\mathrm{H}_2 \mathrm{O}

. The percentage composition of carbon and hydrogen in the compound, respectively is:

Options

A $4.07 \%$ and $15.02 \%$
B $18.18 \%$ and $3.70 \%$
C $15.02 \%$ and $4.07 \%$
D $3.70 \%$ and $18.18 \%$

Correct answer

B. $18.18 \%$ and $3.70 \%$

Step-by-step solution

Percentage of carbon

=\frac{12 \times \mathrm{m}_2 \times 100}{44 \times \mathrm{m}}

\mathrm{m}=

mass of organic compound

=0.3 \mathrm{~g}

\mathrm{m}_2=

mass of carbon dioxilde

=0.2 \mathrm{~g}

\therefore \% C=\frac{12 \times 0.2 \times 100}{44 \times 0.3}=18.18 \%

Percentage of hydrogen

=\frac{2 \times \mathrm{m}_1 \times 100}{18 \times \mathrm{m}}

\mathrm{m}_1=

mass of water

=0.1 \mathrm{~g}

\therefore \% \mathrm{H}=\frac{2 \times 0.1 \times 100}{18 \times 0.3}=3.70 \%

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