JEE Main 2026 — Apr 4, Shift 1 — Chemistry solution

JEE Main2026Apr 4, Shift 1Chemistry

$2.0$ 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 : $108$ )

Your answer

Solution

$n_{\mathrm{Br}} = \dfrac{3.36}{(108+80)}$

$r =$ no. of Br atoms in organic compound

$n_{\mathrm{organic\ compound}} = \dfrac{3.36}{(108+80)\times x} = \dfrac{2}{M}$

$M = \dfrac{2\times188\times x}{3.36}$

$M = \frac{1}{2} \times x$

$(x is integer)$

$W_C = \dfrac{112\times x\times26.7}{100} = 30\times x$

$n_C = \dfrac{30\times x}{12} = 2.5\times x$

take $x = 2, n_{C} = 5$

(Empirical formula is $C_{5} H_{4} B r$ )

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