On complete combustion, 0.492 g of an organic compound gave 0.792 g of CO 2. The % of carbon in the organic compound is (Nearest integer)

Question

On complete combustion, 0.492 g of an organic compound gave 0.792 g of CO 2. The % of carbon in the organic compound is (Nearest integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

weight of C in 0.792 gm CO 2 =(12/44) × 0.792=0.216 % of C in compound =(0.216/0.492) × 100 =43.90 % =44

Q. On complete combustion, $0.492 g$ of an organic compound gave $0.792 g$ of $C O_{2}$ . The $%$ of carbon in the organic compound is ___ (Nearest integer)

weight of $C$ in $0.792 g m C O_{2}$
$= \frac{12}{44} \times 0.792 = 0.216$
$%$ of $C$ in compound $= \frac{0.216}{0.492} \times 100$
$= 43.90%$
$= 44$

Q. On complete combustion, 0.492g of an organic compound gave 0.792g of CO2​. The % of carbon in the organic compound is ___ (Nearest integer)

weight of C in 0.792gmCO2​ =4412​×0.792=0.216 % of C in compound =0.4920.216​×100 =43.90% =44

Q. On complete combustion, $0.492 g$ of an organic compound gave $0.792 g$ of $C O_{2}$ . The $%$ of carbon in the organic compound is ___ (Nearest integer)

weight of $C$ in $0.792 g m C O_{2}$
$= \frac{12}{44} \times 0.792 = 0.216$
$%$ of $C$ in compound $= \frac{0.216}{0.492} \times 100$
$= 43.90%$
$= 44$