Question

\(0.53\ \text{g}\) of an organic compound \(X\) when heated with excess concentrated nitric acid and then with silver nitrate gave \(0.75\ \text{g}\) of silver bromide precipitate. \(1.0\ \text{g}\) of \(X\) gave \(1.32\ \text{g}\) of \(\mathrm{CO_2}\) on combustion. Find the percentage of hydrogen in compound \(X\). (Nearest integer) [Given: Atomic masses (g mol\(^{-1}\)): H = 1, C = 12, Br = 80, Ag = 108, O = 16]

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
Elemental analysis of organic compounds involves estimating the mass of known oxides or salts formed.
Carbon is estimated via combustion to \(\text{CO}_2\) (Liebig's Method).
Bromine is estimated via precipitation as \(\text{AgBr}\) (Carius Method).
The percentage of Hydrogen is found by subtracting the percentages of all other elements from 100%.
Step 2: Key Formula or Approach:
1. \(% C = \frac{12}{44} \times \frac{\text{Mass of } \text{CO}_2}{\text{Mass of compound}} \times 100\).
2. \(% Br = \frac{\text{At. wt. of Br}}{\text{Mol. wt. of AgBr}} \times \frac{\text{Mass of AgBr}}{\text{Mass of compound}} \times 100\).
3. \(% H = 100 - (% C + % Br)\).
Step 3: Detailed Explanation:
1. Calculate Percentage of Carbon:
Mass of compound = \(1.0 \text{ g}\), Mass of \(\text{CO}_2 = 1.32 \text{ g}\).
\[ % C = \frac{12}{44} \times \frac{1.32}{1.0} \times 100 = \frac{12}{44} \times 132 = 12 \times 3 = 36% \]
2. Calculate Percentage of Bromine:
Mass of compound = \(0.53 \text{ g}\), Mass of \(\text{AgBr} = 0.75 \text{ g}\).
Molar mass of \(\text{AgBr} = 108 + 80 = 188 \text{ g/mol}\).
\[ % Br = \frac{80}{188} \times \frac{0.75}{0.53} \times 100 \]
\[ % Br = 0.4255 \times 1.415 \times 100 \approx 60.2% \]
3. Calculate Percentage of Hydrogen:
Assuming the compound only contains C, H, and Br:
\[ % H = 100 - (36 + 60.2) = 100 - 96.2 = 3.8% \]
Rounding to the nearest integer, we get 4.
Step 4: Final Answer:
The percentage of hydrogen in the compound is 4%.