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An organic compound contains 49.3% carbon, 6.84% hydrogen and its vapour density is 73. Molecular formula of compound is
- a)C₃H₅O₂
- b)C₆H₁₀O₄
- c)C₃H₁₀O₂
- d)C₄H₁₀O₂
Correct answer is option 'B'. Can you explain this answer?
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An organic compound contains 49.3% carbon, 6.84% hydrogen and its vapo...
Calculating the Molecular Formula
Given Data:
- % of carbon = 49.3%
- % of hydrogen = 6.84%
- Vapour density = 73
Step 1: Calculate the Empirical Formula
1. Assume 100g of the compound to make calculations easier.
2. Calculate the number of moles of carbon and hydrogen present in 100g of the compound.
- Moles of carbon = (49.3g / 12.01g/mol) = 4.10 mol
- Moles of hydrogen = (6.84g / 1.01g/mol) = 6.77 mol
3. Find the simplest whole-number ratio of carbon to hydrogen.
- Divide the moles of each element by the smallest number of moles.
- Ratio of C to H = 4.10 / 4.10 : 6.77 / 4.10 ≈ 1 : 1.65
- Multiply by a factor to get whole numbers. In this case, multiplying by 2 gives the ratio of 2 : 3.
4. The empirical formula is CH2.
Step 2: Calculate the Molecular Formula
1. Calculate the empirical formula mass.
- Mass of CH2 = 12.01 (C) + 2(1.01) (H) = 14.03 g/mol
2. Calculate the ratio of the vapour density to the empirical formula mass.
- Ratio = 73 / 14.03 ≈ 5.2
3. Multiply the subscripts in the empirical formula by the ratio obtained.
- Molecular formula = C(1)H(2) × 5.2 = C5H10
4. Therefore, the molecular formula of the compound is C5H10, which corresponds to option b) CHO.
Given Data:
- % of carbon = 49.3%
- % of hydrogen = 6.84%
- Vapour density = 73
Step 1: Calculate the Empirical Formula
1. Assume 100g of the compound to make calculations easier.
2. Calculate the number of moles of carbon and hydrogen present in 100g of the compound.
- Moles of carbon = (49.3g / 12.01g/mol) = 4.10 mol
- Moles of hydrogen = (6.84g / 1.01g/mol) = 6.77 mol
3. Find the simplest whole-number ratio of carbon to hydrogen.
- Divide the moles of each element by the smallest number of moles.
- Ratio of C to H = 4.10 / 4.10 : 6.77 / 4.10 ≈ 1 : 1.65
- Multiply by a factor to get whole numbers. In this case, multiplying by 2 gives the ratio of 2 : 3.
4. The empirical formula is CH2.
Step 2: Calculate the Molecular Formula
1. Calculate the empirical formula mass.
- Mass of CH2 = 12.01 (C) + 2(1.01) (H) = 14.03 g/mol
2. Calculate the ratio of the vapour density to the empirical formula mass.
- Ratio = 73 / 14.03 ≈ 5.2
3. Multiply the subscripts in the empirical formula by the ratio obtained.
- Molecular formula = C(1)H(2) × 5.2 = C5H10
4. Therefore, the molecular formula of the compound is C5H10, which corresponds to option b) CHO.
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An organic compound contains 49.3% carbon, 6.84% hydrogen and its vapour density is 73. Molecular formula of compound isa)CHOb)CHOc)CHOd)CHOCorrect answer is option 'B'. Can you explain this answer?
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An organic compound contains 49.3% carbon, 6.84% hydrogen and its vapour density is 73. Molecular formula of compound isa)CHOb)CHOc)CHOd)CHOCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An organic compound contains 49.3% carbon, 6.84% hydrogen and its vapour density is 73. Molecular formula of compound isa)CHOb)CHOc)CHOd)CHOCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An organic compound contains 49.3% carbon, 6.84% hydrogen and its vapour density is 73. Molecular formula of compound isa)CHOb)CHOc)CHOd)CHOCorrect answer is option 'B'. Can you explain this answer?.
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