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88 questions on this topic of Purification and Characterisation of Organic Compounds (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. An organic compound contains only carbon, hydrogen, and oxygen. On analysis, it gave 54.54% C and 9.09% H by mass. If its molecular weight is 88 g mol⁻¹, what is the molecular formula of the compound?

  1. C₅H₁₂O
  2. C₂H₄O
  3. C₃H₄O₃
  4. C₄H₈O₂
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Answer: (D) C₄H₈O₂

Mass % of O = 100 - (54.54 + 9.09) = 36.37%. Moles of C = 54.54 / 12 = 4.545; moles of H = 9.09 / 1 = 9.09; moles of O = 36.37 / 16 = 2.273. Mole ratio C : H : O = 2 : 4 : 1, giving empirical formula C₂H₄O (empirical mass = 44 g mol⁻¹). Since molar mass = 88 g mol⁻¹, n = 88 / 44 = 2. Molecular formula is C₄H₈O₂.

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2. A gaseous hydrocarbon contains 85.7% carbon by mass. If 112 mL of this gaseous hydrocarbon at STP weighs 0.21 g, what is its molecular formula? [Atomic masses: C = 12, H = 1]

  1. C₂H₄
  2. C₃H₆
  3. C₄H₈
  4. C₅H₁₀
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Answer: (B) C₃H₆

Ratio of moles: C = 85.7/12 = 7.14, H = 14.3/1 = 14.3 (Ratio 1:2), so empirical formula is CH₂ (formula mass = 14). Molar mass = (0.21 g / 112 mL) × 22400 mL = 42 g mol⁻¹. Value of n = 42 / 14 = 3. Thus, molecular formula is C₃H₆.

hard

3. A gaseous hydrocarbon contains 80% carbon by mass. The volume of oxygen gas at STP required for the complete combustion of 1.5 g of this hydrocarbon is:

  1. 2.24 L
  2. 5.60 L
  3. 4.48 L
  4. 3.92 L
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Answer: (D) 3.92 L

Empirical formula: C = 80 / 12 = 6.67; H = 20 / 1 = 20. Ratio C : H = 1 : 3, empirical formula CH₃. The only stable gaseous hydrocarbon with this ratio is ethane, C₂H₆ (molar mass = 30 g mol⁻¹). Combustion reaction: C₂H₆ + 3.5 O₂ → 2 CO₂ + 3 H₂O. Moles of C₂H₆ = 1.5 / 30 = 0.05 mol. Moles of O₂ required = 0.05 × 3.5 = 0.175 mol. Volume of O₂ at STP = 0.175 × 22.4 L = 3.92 L.

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4. In Kjeldahl's method, the ammonia evolved from 0.28 g of a nitrogenous organic compound was absorbed in 50 mL of 0.1 M H₂SO₄. The unreacted acid required 60 mL of 0.1 M NaOH for complete neutralization. The percentage of nitrogen in the compound is:

  1. 20%
  2. 14%
  3. 28%
  4. 35%
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Answer: (A) 20%

Initial millimoles of H₂SO₄ = 50 × 0.1 = 5.0 mmol (equivalent to 10.0 meq of H⁺). Millimoles of NaOH used = 60 × 0.1 = 6.0 mmol (6.0 meq of OH⁻). Millimoles of NH₃ neutralized = 10.0 - 6.0 = 4.0 mmol = 4.0 × 10⁻³ mol. Mass of nitrogen = 4.0 × 10⁻³ × 14 = 0.056 g. Percentage of N = (0.056 / 0.28) × 100 = 20%.

easy

5. An organic compound containing carbon, hydrogen, and oxygen has an empirical formula of C₂H₄O. If 0.10 mole of this compound weighs 8.8 g, its molecular formula is:

  1. C₄H₈O₂
  2. C₂H₄O
  3. C₆H₁₂O₃
  4. C₃H₆O₂
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Answer: (A) C₄H₈O₂

Molar mass of the compound = 8.8 g / 0.10 mol = 88 g mol⁻¹. The empirical formula mass of C₂H₄O = (2 × 12) + (4 × 1) + 16 = 44 g mol⁻¹. The ratio n = 88 / 44 = 2, so the molecular formula is (C₂H₄O)₂ = C₄H₈O₂.

hard

6. A monobasic organic acid containing C, H, and O has an empirical formula of CH₂O. A sample of 0.18 g of this acid requires 20.0 mL of 0.1 M NaOH for complete neutralization. What is the molecular formula of the acid?

  1. CH₂O
  2. C₃H₆O₃
  3. C₂H₄O₂
  4. C₆H₁₂O₆
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Answer: (B) C₃H₆O₃

Moles of NaOH consumed = (20.0 × 0.1) / 1000 = 0.002 mol. Since the acid is monobasic, moles of acid = 0.002 mol. Molar mass of the acid = 0.18 g / 0.002 mol = 90 g mol⁻¹. The empirical formula mass of CH₂O is 30 g mol⁻¹. Hence, n = 90 / 30 = 3, which gives the molecular formula C₃H₆O₃ (e.g., lactic acid).

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7. In a Dumas estimation, 0.20 g of an organic compound gave 28.0 mL of dry nitrogen gas collected at 0 °C and 1.0 atm (STP). The percentage of nitrogen in the compound is:

  1. 14.0%
  2. 22.4%
  3. 17.5%
  4. 28.0%
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Answer: (C) 17.5%

At STP (0 °C, 1 atm), 22400 mL of dry N₂ corresponds to 28 g of nitrogen. Mass of N in 28.0 mL = (28 / 22400) × 28.0 = 0.035 g. Percentage of N = (0.035 / 0.20) × 100 = 17.5%.

easy

8. Given below are two statements: Statement I: Kjeldahl's method cannot be used to estimate nitrogen in nitro compounds, azo compounds, and pyridine because their nitrogen is not quantitatively converted into ammonium sulfate. Statement II: In Kjeldahl's method, copper sulfate acts as a catalytic agent while potassium sulfate raises the boiling point of concentrated sulfuric acid. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

Statement I is correct because the nitrogen present in nitro groups, azo groups, and heterocyclic rings (like pyridine) does not yield ammonium sulfate under Kjeldahl digestion conditions. Statement II is correct because K₂SO₄ raises the boiling point of H₂SO₄ to speed up digestion, and CuSO₄ acts as a catalyst.

easy

9. Given below are two statements: Statement I: During Kjeldahl's method, potassium sulfate is added to raise the boiling point of concentrated sulfuric acid. Statement II: Copper sulfate acts as a catalyst in the digestion step of Kjeldahl's method. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

In Kjeldahl's method, potassium sulfate (K₂SO₄) is added to elevate the boiling point of H₂SO₄ to ensure complete decomposition, while copper sulfate (CuSO₄) serves as a catalyst to accelerate the digestion process.

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10. Consider the following statements regarding quantitative analysis of organic compounds: Statement I: In Dumas method, nitrogen oxides formed during combustion are reduced back to dinitrogen gas by passing over a heated copper spiral. Statement II: In Kjeldahl's method, potassium sulfate acts as a catalyst while copper sulfate raises the boiling point of concentrated sulfuric acid. Which of the following is correct?

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct: nitrogen oxides produced during combustion are reduced to N₂ using a heated copper gauze/spiral. Statement II is incorrect: K₂SO₄ is added to raise the boiling point of H₂SO₄, whereas CuSO₄ (or Hg/selenium) acts as a catalyst.

easy

11. In a Carius halogen determination, 0.36 g of an organic compound gave 0.376 g of AgBr. What is the percentage of bromine in the compound? (Molar masses: AgBr = 188 g mol⁻¹, Br = 80 g mol⁻¹)

  1. 22.22%
  2. 44.44%
  3. 55.56%
  4. 33.33%
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Answer: (B) 44.44%

Percentage of bromine = [(80 / 188) × (0.376 / 0.36)] × 100 = (80 / 188) × 1.0444 × 100 = 44.44%.

easy

12. In the estimation of sulfur by the Carius method, 0.25 g of an organic compound gave 0.5825 g of barium sulfate (BaSO₄). What is the percentage of sulfur in the organic compound? (Molar mass of BaSO₄ = 233 g mol⁻¹, atomic mass of S = 32 u)

  1. 32.0%
  2. 16.0%
  3. 40.0%
  4. 24.0%
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Answer: (A) 32.0%

Mass of sulfur in BaSO₄ = (32 / 233) × 0.5825 g = 0.08 g. Percentage of sulfur = (0.08 / 0.25) × 100 = 32.0%.

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