On complete combustion 0.30 g of an organic compound gave 0.20 g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is _____________. (Nearest integer)

C_xH_y + ( x + y 4 )O_2 x\,CO_2 + y 2H_2O Given organic compound C x H y = 0.3 gm Produced carbon dioxide (CO 2 ) = 0.2 gm Produced water (H 2 O) = 0.1 gm Moles of CO 2 = 0.2 44 Moles of C atom = 0.2 44 Mass of C atom = 0.2 44 12 = 0.0545 Moles of H<

On complete combustion 030 g of an organic compound gave 020: null (Chemistry)

{C_x}{H_y} + \left( {x + {y \over 4}} \right){O_2} \to x\,C{O_2} + {y \over 2}{H_2}O

Given organic compound C_xH_y= 0.3 gm

Produced carbon dioxide (CO₂) = 0.2 gm

Produced water (H₂O) = 0.1 gm

Moles of CO₂ =

{{0.2} \over {44}}

\therefore

Moles of C atom =

{{0.2} \over {44}}

\therefore

Mass of C atom =

{{0.2} \over {44}}

\times

12 = 0.0545

Moles of H₂O =

{{0.1} \over {18}}

\therefore

Moles of H atoms =

{{0.1} \over {18}}

\times

\therefore

Mass of H atoms =

{{0.1} \times2\over {18}}

\times

1 = 0.0111

\therefore

% of C atom =

{{0.0545} \over {0.3}}

\times

100 = 18%

Explanation

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