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Question
JEE Main
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 $\text{g~mol}^{-1}$ $\text{H}: 1, \text{C}: 12, \text{Br} : 80, \text{Ag} : 108$)

Bromo Hydrocarbon Empirical Formula Calculation using Carius Analysis

This solution details the steps to determine the empirical formula of a bromo hydrocarbon (X) using Carius analysis data and the given percentage of carbon.

Calculating Bromine Content

First, we calculate the amount of Bromine (Br) present in the sample.

  • Molar mass of Silver Bromide (AgBr) = Molar mass of Ag + Molar mass of Br = $108 \text{ g~mol}^{-1} + 80 \text{ g~mol}^{-1} = 188 \text{ g~mol}^{-1}$.
  • Moles of AgBr formed = $\frac{\text{Mass of AgBr}}{\text{Molar mass of AgBr}} = \frac{3.36 \text{ g}}{188 \text{ g~mol}^{-1}}$.
  • From Carius analysis, moles of Br in the compound = moles of AgBr. Moles of Br = $\frac{3.36}{188}$ mol.
  • Mass of Br in the compound = Moles of Br $\times$ Molar mass of Br = $\left( \frac{3.36}{188} \text{ mol} \right) \times (80 \text{ g~mol}^{-1}) = \frac{268.8}{188} \text{ g}$.
  • Percentage of Br in compound (X) = $\frac{\text{Mass of Br}}{\text{Mass of compound (X)}} \times 100 = \frac{\frac{268.8}{188} \text{ g}}{2.0 \text{ g}} \times 100$.
  • Percentage of Br = $\frac{268.8 \times 100}{188 \times 2.0} \% = \frac{26880}{376} \% \approx 71.49\%$.

Determining Atomic Ratios

We use the percentages of Carbon (C) and Bromine (Br) to find the simplest molar ratio. Since it's a hydrocarbon, Hydrogen (H) is also present.

Given:

  • Percentage of Carbon (C) = 26.7%
  • Calculated Percentage of Bromine (Br) $\approx$ 71.49%

The percentage of Hydrogen (H) is:

Percentage of H = $100\% - (\% \text{C} + \% \text{Br}) = 100\% - (26.7\% + 71.49\%) = 100\% - 98.19\% = 1.81\%$.

To find the empirical formula, consider a 100 g sample:

Atomic Ratios Calculation
Element Percentage (%) Mass (g) Moles (Mass/Molar Mass) Ratio (Moles / Smallest Moles) Final Simplest Ratio
C 26.7 26.7 $\frac{26.7}{12} \approx 2.225$ $\frac{2.225}{0.8936} \approx 2.49$ 5
H 1.81 1.81 $\frac{1.81}{1} \approx 1.811$ $\frac{1.811}{0.8936} \approx 2.02$ 4
Br 71.49 71.49 $\frac{71.49}{80} \approx 0.8936$ $\frac{0.8936}{0.8936} = 1$ 2

The smallest molar amount is approximately 0.8936 (for Br). Dividing all molar amounts by this value gives the ratio C : H : Br $\approx 2.49 : 2.02 : 1$.

Empirical Formula Derivation

To obtain whole numbers, we multiply the ratio by 2:

C : H : Br $\approx (2.49 \times 2) : (2.02 \times 2) : (1 \times 2) \approx 4.98 : 4.04 : 2$.

This simplifies to the whole number ratio $5 : 4 : 2$.

Therefore, the empirical formula for compound (X) is $C_5H_4Br_2$.

The total number of carbon atoms in the empirical formula is 5.

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