Mole Concept Combustion Analysis Shortcut

 

❓ Question

An organic compound weighing:

$$500 \text{ mg}$$

produced:

$$220 \text{ mg of } CO_2$$

on complete combustion.

Find the percentage composition of carbon in the compound.

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🖼 Question Image

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✍️ Short Concept

In combustion problems:

👉 All carbon converts into:

$$CO_2$$

So first find:

$$\text{Mass of carbon in } CO_2$$

then calculate percentage.

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🔷 Step 1 — Carbon Fraction in $CO_2$ 💯

Molar mass:

$$CO_2 = 44$$

Carbon mass in one mole:

$$= 12$$

So carbon fraction:

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

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🔷 Step 2 — Carbon Present in Given $CO_2$

Given:

$$220 \text{ mg } CO_2$$

Carbon mass:

$$= 220 \times \frac{12}{44}$$
$$= 60 \text{ mg}$$

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🔷 Step 3 — Percentage of Carbon

Total compound mass:

$$500 \text{ mg}$$

Percentage carbon:

$$\frac{60}{500} \times 100$$
$$= 12\%$$

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✅ Final Answer

$$\boxed{12\%}$$

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⭐ Golden JEE Insight

Combustion analysis shortcut:

$$\text{Mass of C} = \text{Mass of } CO_2 \times \frac{12}{44}$$

Similarly:

$$\text{Mass of H} = \text{Mass of } H_2O \times \frac{2}{18}$$

These shortcuts directly save time in mole concept questions.