Organic Compound Percentage of Hydrogen Trick

 

❓ Question

$1\text{ g}$ of an organic compound gives:

$$1.32\text{ g CO}_2$$

Also,

$$0.53\text{ g}$$

of the same compound gives:

$$0.75\text{ g AgBr}$$

If molecular formula of compound is:

$$C_xH_yBr_z$$

calculate percentage of hydrogen in the compound.

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Carbon estimation from $CO_2$
👉 Bromine estimation using Carius method
👉 Percentage composition.

Main idea:

First find:

$$\%C$$

and

$$\%Br$$

Then remaining percentage gives hydrogen.

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🔷 Step 1 — Percentage of Carbon 💯

Given:

$$1\text{ g compound} \rightarrow 1.32\text{ g }CO_2$$

Carbon present in:

$$CO_2$$

is:

$$\frac{12}{44}$$

So mass of carbon:

$$1.32\times\frac{12}{44}$$
$$=0.36\text{ g}$$

Hence:

$$\%C=36\%$$

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🔷 Step 2 — Bromine from AgBr

Given:

$$0.53\text{ g compound} \rightarrow 0.75\text{ g AgBr}$$

Molar mass:

$$AgBr=108+80=188$$

Fraction of bromine in AgBr:

$$\frac{80}{188}$$

Mass of bromine:

$$0.75\times\frac{80}{188}$$
$$=0.319\text{ g}$$

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🔷 Step 3 — Percentage of Bromine

Compound mass:

$$0.53\text{ g}$$

So:

$$\%Br= \frac{0.319}{0.53}\times100$$
$$\approx60.2\%$$

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🔷 Step 4 — Percentage of Hydrogen

Organic compound contains only:

$$C,H,Br$$

Therefore:

$$\%H = 100-(\%C+\%Br)$$
$$=100-(36+60.2)$$
$$=3.8\%$$

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🔷 Step 5 — Final Answer

$$\boxed{3.8\%}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ $CO_2$ mass ko direct carbon mass maan lena

❌ AgBr ka molar mass wrong lena

❌ Remaining percentage method bhool jaana

Remember:

$$\text{Mass of element} = \text{Compound mass}\times \frac{\text{Atomic mass}}{\text{Molar mass}}$$

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✅ Final Answer

$$\boxed{3.8\%}$$

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📚 Related Topics

• Silver Halide Formation in Organic Analysis
• Faraday Law and Nernst Equation Combined Concept
• Mole Fraction of CO2 After Combustion Calculation