JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Practical and Analytical Chemistry

2.0 g of a bromo hydrocarbon ( X ) was subjected to Carius analysis, gave 3.36 g of AgBr . The percentage of carbon in the compound ( X ) is

26.7%

. Total number of carbon atoms in the empirical formula for compound ( X ) is ______ .
(Given molar mass in

g mol^{- 1}

H: 1, C: 12, Br: 80, Ag : 108)

Answer: 5

Solution

From the Carius result, moles of AgBr

= \frac{3.36}{108 + 80} = \frac{3.36}{188} = 0.01787 mol

, which equals the moles of bromine in X, so the bromine mass is

0.01787 \times 80 = 1.429 g

. The carbon mass is

\frac{26.7}{100} \times 2.0 = 0.534 g

, giving

\frac{0.534}{12} = 0.0445 mol

C. The hydrogen mass is

2.0 - 1.429 - 0.534 = 0.037 g

, i.e.

0.037 mol

H. The mole ratio

C : H : Br = 0.0445 : 0.037 : 0.01787 \approx 2.5 : 2 : 1

, which on clearing fractions is

5 : 4 : 2

. The empirical formula is

C X_{5} H X_{4} Br X_{2}

, so the number of carbon atoms is 5.

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Integer answer