An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the sum of masses (in gm) of carbon dioxide and water produced when 0.20 gm of this compound is subjected to complete combustion. Report your answer as 10x, if x is the sum of masses in the nearest integer.

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Solution

Percentage of C $= \frac{12}{44} \times \frac{M a s s o f C O_{2} f o r m e d}{M a s s o f s u b s t a n c e t a k e n} \times 100$
$⟹ 69 = \frac{12}{44} \times \frac{M a s s o f C O_{2} f o r m e d}{0.2 g m} \times 100$
$∴$ Mass of $C O_{2}$ formed $= 69 \times 44 \times 0.2 12 \times 44 = 0.506 g m$
Percentage of H $= \frac{2}{18} \times \frac{M a s s o f H_{2} O F o r m e d}{M a s s o f s u b s t a n c e t a k e n} \times 100$
$⟹ 4.8 = \frac{2}{18} \times \frac{M a s s o f H_{2} O f o r m e d}{M a s s o f s u b s t a n c e t a k e n} \times 100$
$∴$ Mass of water formed $= \frac{4.8 \times 18 \times 0.2}{2 \times 100} = 0.0864 g m$
The sum of masses of $C O_{2}$ and water formed $= 0.506 + 0.0864 = 0.5924 g m \approx 0.6 g m$ .
$∴ 10 x = 6 g m$ .

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