Question

1 gm of organic compound gave 1.32g $\text{CO}_2$, 0.53g of same compound gave 0.75g AgBr. If molecular formula of compound is $\text{C}_x\text{H}_y\text{Br}_z$ then calculate percentage of hydrogen in the given compound.

• A. 2%
• B. 3%
• C. 5%
• D. 4%

Hint:

Always check the sample mass used for each specific estimation analysis. It differs here (1g for C, 0.53g for Br).

Answer 1

The Correct Option is D

Percentage of Carbon:
Mass of C in 1.32g $\text{CO}_2$ = $\frac{12}{44} \times 1.32 = 0.36 \text{ g}$.
$% \text{C} = \frac{0.36}{1.0} \times 100 = 36%$.
Percentage of Bromine:
Mass of Br in 0.75g $\text{AgBr}$ (Molar Mass 188) = $\frac{80}{188} \times 0.75$.
Calculation: $0.75 \times \frac{80}{188} \approx 0.3191 \text{ g}$.
Wait, use exact fraction: $\frac{80}{188} = \frac{20}{47}$. $0.75 = 3/4$.
Mass Br $= \frac{20}{47} \times \frac{3}{4} = \frac{15}{47} \approx 0.3191 \text{ g}$.
Sample mass for Br estimation is 0.53g.
$% \text{Br} = \frac{0.3191}{0.53} \times 100 \approx 60.2%$.
Percentage of Hydrogen:
$% \text{H} = 100 - (% \text{C} + % \text{Br})$.
$% \text{H} = 100 - (36 + 60.2) = 100 - 96.2 = 3.8%$.
Rounding to the nearest integer gives $4%$.