On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g and 0.307 g . The percentages of hydrogen and oxygen in the given organic compound respectively are: [2025]

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Solution

Q.
On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g $H_{2} O$ and 0.307 g ${CO}_{2}$ . The percentages of hydrogen and oxygen in the given organic compound respectively are: [2025]

1 53.41, 39.6

2 6.72, 53.41

3 7.55, 43.85

4 6.72, 39.87

Ans.
(2)

Moles of $H_{2} O$ $(n_{H_{2} O}) =$ $\frac{Given mass of H_{2} O}{Molar mass of H_{2} O}$ $= \frac{0.127}{18} mol$

Moles of hydrogen in organic compound $(n_{H})$ = moles of hydrogen in water $= 2 \times n_{H_{2} O} = 2 \times \frac{0.127}{18} mol$

Mass of hydrogen in organic compound $(W_{H})$ = $n_{H} \times$ molar mass of H = $2 \times \frac{0.127}{18} \times 1$ gm

Percentage of hydrogen in organic compound = $\frac{W_{H}}{W_{organic compound}} \times 100$

$= 2 \times \frac{0.127}{18 \times 0.210} \times 1 \times 100 = 6.72 %$

Moles of ${CO}_{2}$ $(n_{C O_{2}}) = \frac{Given mass of {CO}_{2}}{Molar mass of {CO}_{2}} = \frac{0.307}{44} mol$

Moles of carbon in organic compound $(n_{C})$ = moles of carbon in ${CO}_{2}$ $= \frac{0.307}{44} mol$

Mass of carbon in organic compound $(W_{C})$ $= n_{C} \times$ molar mass of C $= \frac{0.307}{44} \times 12$ gm

Percentage of hydrogen in organic compound $= \frac{W_{C}}{W_{organic compound}} \times 100$

$= \frac{0.307}{44 \times 0.210} \times 12 \times 100 = 39.87 %$

$% of O = 100 - (% of C + % of H)$

$= 100 - (39.87 + 6.72) = 53.41 %$

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