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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...
Given:
Percentage of Carbon = 49.3%
Percentage of Hydrogen = 6.84%
Vapour density = 73
To find: Molecular formula of the compound
Solution:
Step 1: Calculate the empirical formula of the compound
Assume 100 g of the compound, then
Mass of Carbon = 49.3 g
Mass of Hydrogen = 6.84 g
Mass of Oxygen = (100 - 49.3 - 6.84) = 43.86 g
Number of moles of Carbon = 49.3/12 = 4.108
Number of moles of Hydrogen = 6.84/1 = 6.84
Number of moles of Oxygen = 43.86/16 = 2.741
Divide each mole value by the smallest mole value to get the simplest mole ratio.
Number of moles of Carbon = 4.108/2.741 = 1.5
Number of moles of Hydrogen = 6.84/2.741 = 2.5
Number of moles of Oxygen = 2.741/2.741 = 1
So, the empirical formula of the compound is CH2O
Step 2: Calculate the molecular formula of the compound
The molecular formula is a multiple of the empirical formula, so we need to find the value of 'n'.
The vapour density of the compound is given as 73.
Vapour Density = 2 × Molar mass of the compound / 22.4
Substituting the given values, we get
73 = 2 × Molar mass of the compound / 22.4
Molar mass of the compound = 73 × 22.4 / 2 = 816.8 g/mol
Now, we need to find the value of 'n' by dividing the molar mass by the empirical formula mass.
n = Molar mass of the compound / Empirical formula mass
n = 816.8 / (12 + 2 + 16) = 37.6
The molecular formula of the compound is (CH2O) × 37.6 = C37H74O37
Therefore, the correct option is (b) CHO.
Percentage of Carbon = 49.3%
Percentage of Hydrogen = 6.84%
Vapour density = 73
To find: Molecular formula of the compound
Solution:
Step 1: Calculate the empirical formula of the compound
Assume 100 g of the compound, then
Mass of Carbon = 49.3 g
Mass of Hydrogen = 6.84 g
Mass of Oxygen = (100 - 49.3 - 6.84) = 43.86 g
Number of moles of Carbon = 49.3/12 = 4.108
Number of moles of Hydrogen = 6.84/1 = 6.84
Number of moles of Oxygen = 43.86/16 = 2.741
Divide each mole value by the smallest mole value to get the simplest mole ratio.
Number of moles of Carbon = 4.108/2.741 = 1.5
Number of moles of Hydrogen = 6.84/2.741 = 2.5
Number of moles of Oxygen = 2.741/2.741 = 1
So, the empirical formula of the compound is CH2O
Step 2: Calculate the molecular formula of the compound
The molecular formula is a multiple of the empirical formula, so we need to find the value of 'n'.
The vapour density of the compound is given as 73.
Vapour Density = 2 × Molar mass of the compound / 22.4
Substituting the given values, we get
73 = 2 × Molar mass of the compound / 22.4
Molar mass of the compound = 73 × 22.4 / 2 = 816.8 g/mol
Now, we need to find the value of 'n' by dividing the molar mass by the empirical formula mass.
n = Molar mass of the compound / Empirical formula mass
n = 816.8 / (12 + 2 + 16) = 37.6
The molecular formula of the compound is (CH2O) × 37.6 = C37H74O37
Therefore, the correct option is (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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