⚗️ Empirical Formula from Combustion Data – Chemistry JEE | Doubtify JEE

🧪 Question:

The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6:1. If one molecule of the above compound contains half as much oxygen as required to burn one molecule of compound CxHy completely to CO₂ and H₂O, then what is the empirical formula of the compound?

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📌 Why This Question is Important:

This question combines two highly testable concepts in JEE Chemistry – combustion analysis and empirical formula derivation. You’re not only working with mass percent ratio, but also applying oxygen balancing logic in a combustion reaction. Such questions are frequently seen in organic chemistry chapters, particularly under Basic Principles of Organic Chemistry and Stoichiometry.

Mastering this helps you crack more complex analysis-based questions in JEE Mains and even Advanced.

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💡 Concept Used:

To find the empirical formula, we first convert the mass percent ratio of C and H into moles, assuming 100 g of substance. Then we apply a stoichiometric approach to determine the amount of oxygen atoms needed for complete combustion and use the condition given:

Oxygen atoms in compound = ½ × Oxygen required for combustion

This balances the internal oxygen with the external demand of combustion, helping us derive the correct formula.

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✅ Step-by-Step Solution:

1. Mass percent ratio is given as C : H = 6 : 1
   → In 100g compound:

   • Carbon = 85.7 g

   • Hydrogen = 14.3 g

2. Convert to moles:

   • Moles of C = 85.7 / 12 ≈ 7.14

   • Moles of H = 14.3 / 1 = 14.3

3. Assume empirical formula is CₓHᵧO𝓏, now consider CxHy burning completely:

   • Combustion of C and H requires:
     C → CO₂ ⇒ 1 mol needs 1 mol O₂
     H → H₂O ⇒ 1 mol H needs 0.5 mol O₂

   • Total O needed = x + y/4

4. But compound contains half this amount internally:
   → z = ½(x + y/4)

5. Now, try fitting x = 4, y = 8
   → z = ½(4 + 8/4) = ½(4 + 2) = 3

✅ So, empirical formula is C₄H₈O₃

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🧠 Solution (Image):

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