Question

$0.46g$ of an organic compound was analysed. The increases in mass of $CaCl_{2}$ U – tube was $0.54$g and potash bulb was $0.88$ g. the percentage composition of the compound is.

• A. C = $52.17\%$, H = $13.04\%,$ O = $34.7\%$
• B. C = 50%, H = 50%
• C. C = $32.19\%$, H = $18.01\%,$ O = $49.8\%$
• D. C = 72%, H = 28%

Answer 1

Answer: (A)

C = $52.17\%$, H = $13.04\%,$ O = $34.7\%$

Answer 2

: mass of $H_{2}O = 0.54g$

Mass of $CO_{2} = 0.88g$

% of $C = \frac{12}{44} \times \frac{0.88}{0.46} \times 100 = 52.17\%$

% of $H = \frac{2}{18} \times \frac{0.54}{0.46} \times 100 = 13.04\%$

% of $O = 100 - (52.17 + 13.04) = 34.79\%$