Carius Method Bromine Percentage Calculation

 

❓ Question

In Carius method of estimation of Br⁻:

$$1.53\text{ g}$$

of an organic compound gave:

$$1\text{ g AgBr}$$

The percentage of Br in the organic compound is:

Atomic masses:

$$Ag=108,\quad Br=80$$

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🖼 Question Image

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✍️ Short Explanation

This question is based on:

👉 Carius method
👉 Gravimetric analysis
👉 Mole concept using precipitate mass.

In Carius method:

$$Br^- \rightarrow AgBr$$

So bromine mass is calculated from AgBr mass.

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🔷 Step 1 — Molar Mass of AgBr 💯

$$M(AgBr)=108+80$$
$$=188$$

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🔷 Step 2 — Bromine Fraction in AgBr

In:

$$188\text{ g AgBr}$$

Mass of Br present:

$$80\text{ g}$$

So in:

$$1\text{ g AgBr}$$

Mass of Br:

$$=\frac{80}{188}$$
$$=0.4255\text{ g}$$

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🔷 Step 3 — Percentage of Bromine

Mass of organic compound:

$$1.53\text{ g}$$

So:

$$\%\text{ Br} = \frac{0.4255}{1.53}\times100$$
$$=27.81\%$$

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🔷 Step 4 — Final Matching

Correct option:

$$\boxed{27.81\%}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Direct AgBr mass ko Br mass maan lena

❌ Atomic masses add galat kar dena

❌ Percentage formula mein total compound mass bhool jaana

Remember:

$$\text{Use mass fraction of Br in AgBr first}$$

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✅ Final Answer

$$\boxed{27.81\%}$$

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📚 Related Topics

• Mole Concept Combustion Analysis Shortcut
• Find Percentage Dissociation Using i Factor
• First Order Growth and Second Order Decay Trick