0.46 g of an organic compound was analysed. The increase in mass of CaCl2 U-tube was 0.54 g and potash bulb was 0.88 g. The percentage composition of the compound is

Question

0.46 g of an organic compound was analysed. The increase in mass of CaCl2 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 = 34.79 % (B) C = 50 %, H = 50 % (C) C=32.19 %, H = 18.01 %, O = 49.8 % (D) C = 72 %, H = 28 %

Tardigrade Admin · Accepted Answer

Mass of H2O = 0.54 g Mass of CO2 = 0.88 g % of C=(12/44)×(0.88/0.46)×100=52.17 % % of H =(2/18)×(0.54/0.46)×100=13.04 % % of O=100-(52.17+13.04)=34.79 %

Q. $0.46 g$ of an organic compound was analysed. The increase in mass of $C a C l_{2}$ $U$ -tube was $0.54 g$ and potash bulb was $0.88 g$ . The percentage composition of the compound is

$C = 52.17%, H = 13.04%, O = 34.79%$

$C = 50%, H = 50%$

$C = 32.19%, H = 18.01%, O = 49.8%$

$C = 72%, H = 28%$

Mass of $H_{2} O = 0.54 g$
Mass of $C O_{2} = 0.88 g$
$%$ 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 $O = 100 - (52.17 + 13.04) = 34.79%$

Q. 0.46g of an organic compound was analysed. The increase in mass of CaCl2​ U-tube was 0.54g and potash bulb was 0.88g. The percentage composition of the compound is

C=52.17%,H=13.04%,O=34.79%

C=50%,H=50%

C=32.19%,H=18.01%,O=49.8%

C=72%,H=28%