Two organic compounds A and B, both containing only C and H yield, on analysis, the same percentage composition by weight: C = (12/13) × 100% and H = (1/13) × 100% A decolourises Br2 water but B does not. Identify A and B.

Question

Two organic compounds A and B, both containing only C and H yield, on analysis, the same percentage composition by weight: C = (12/13) × 100% and H = (1/13) × 100% A decolourises Br2 water but B does not. Identify A and B. (A) A = C2H2, B = C6H6 (B) A = C6H6, B = C2H2 (C) A = C2H4, B = C2H6 (D) A = C2H2, B = C3H8

Tardigrade Admin · Accepted Answer

C: H = (12×100/13×12): (1×100/13×1) = 1: 1 Empirical formula = CH Since A decolourises bromine water, but B does not therefore, A = C2H2 and B = C6H6.

Q. Two organic compounds A and B, both containing only C and H yield, on analysis, the same percentage composition by weight:
$C = (12/13) \times 100$ and $H = (1/13) \times 100$
A decolourises $B r_{2}$ water but B does not. Identify A and B.

$A = C_{2} H_{2}, B = C_{6} H_{6}$

$A = C_{6} H_{6}, B = C_{2} H_{2}$

$A = C_{2} H_{4}, B = C_{2} H_{6}$

$A = C_{2} H_{2}, B = C_{3} H_{8}$

$C : H = \frac{12 \times 100}{13 \times 12} : \frac{1 \times 100}{13 \times 1} = 1 : 1$
Empirical formula $= C H$
Since A decolourises bromine water, but B does not therefore, $A = C_{2} H_{2}$ and $B = C_{6} H_{6}$ .

Q. Two organic compounds A and B, both containing only C and H yield, on analysis, the same percentage composition by weight: C=(12/13)×100 and H=(1/13)×100 A decolourises Br2​ water but B does not. Identify A and B.

A=C2​H2​,B=C6​H6​

A=C6​H6​,B=C2​H2​

A=C2​H4​,B=C2​H6​

A=C2​H2​,B=C3​H8​