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Combustion Analysis Numerical – JEE Main PYQs | Some Basic Concepts of Chemistry

0.5 g of an organic compound on combustion gave 1.46 g of CO₂ and 0.9 g of H₂O. The percentage of carbon in the compound is ______. (Nearest integer)

Given: Molar mass (in g mol⁻¹) C = 12, H = 1, O = 16

Step 1: Calculate moles of CO₂ formed

Moles of CO₂ = 1.46 / 44 = 0.03318 moles

Step 2: Calculate moles of carbon present

1 mole of CO₂ contains 1 mole of carbon.
Moles of carbon = 0.03318 moles

Step 3: Calculate mass of carbon

Mass of carbon = 0.03318 × 12 = 0.398 g

Step 4: Calculate percentage of carbon

Percentage of carbon = (0.398 / 0.5) × 100
= 79.6%

Step 5: Nearest integer value

Percentage of carbon ≈ 80%

✅ Final Answer: 80%

In Dumas' method for estimation of nitrogen, 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K and 715 mm Hg pressure

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Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.