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Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can have
- a)43 g
- b)86 g
- c)129 g
- d)172 g
Correct answer is option 'A'. Can you explain this answer?
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Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂...
To determine the molecular mass of compound X, we need to analyze the products formed during its combustion.
Given:
Mass of compound X = 0.858 g
Mass of CO formed = 2.63 g
Mass of HO formed = 1.28 g
Step 1: Calculate the moles of CO formed
Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Moles of CO = Mass of CO formed / Molar mass of CO
Moles of CO = 2.63 g / 28.01 g/mol ≈ 0.094 mol
Step 2: Calculate the moles of HO formed
Molar mass of HO = 1.01 g/mol + 16.00 g/mol = 17.01 g/mol
Moles of HO = Mass of HO formed / Molar mass of HO
Moles of HO = 1.28 g / 17.01 g/mol ≈ 0.075 mol
Step 3: Determine the ratio of moles of CO to moles of HO
The ratio of moles of CO to moles of HO can be found by dividing the moles of CO by the moles of HO.
Ratio = Moles of CO / Moles of HO
Ratio = 0.094 mol / 0.075 mol ≈ 1.253
Step 4: Determine the empirical formula of compound X
The empirical formula represents the simplest whole number ratio of the elements present in the compound. To determine the empirical formula, we need to find the ratio of the elements present.
In compound X, the ratio of C to O can be determined from the moles of CO formed, and the ratio of H to O can be determined from the moles of HO formed.
Ratio of C to O = 1 (from CO)
Ratio of H to O = 2 (from HO)
Therefore, the empirical formula of compound X is C1H2O2.
Step 5: Determine the molecular formula of compound X
The molecular formula represents the actual number of atoms of each element in a compound. To determine the molecular formula, we need to know the molecular mass of compound X.
The molecular mass of compound X can be calculated by adding up the atomic masses of its constituent elements. The empirical formula C1H2O2 has a molecular mass of 29 g/mol.
Since the given options are in grams, we need to find the molecular mass in grams.
Molecular mass of X = 29 g/mol ≈ 29 g
Among the given options, the lowest molecular mass possible for compound X is 43 g (option A).
Given:
Mass of compound X = 0.858 g
Mass of CO formed = 2.63 g
Mass of HO formed = 1.28 g
Step 1: Calculate the moles of CO formed
Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Moles of CO = Mass of CO formed / Molar mass of CO
Moles of CO = 2.63 g / 28.01 g/mol ≈ 0.094 mol
Step 2: Calculate the moles of HO formed
Molar mass of HO = 1.01 g/mol + 16.00 g/mol = 17.01 g/mol
Moles of HO = Mass of HO formed / Molar mass of HO
Moles of HO = 1.28 g / 17.01 g/mol ≈ 0.075 mol
Step 3: Determine the ratio of moles of CO to moles of HO
The ratio of moles of CO to moles of HO can be found by dividing the moles of CO by the moles of HO.
Ratio = Moles of CO / Moles of HO
Ratio = 0.094 mol / 0.075 mol ≈ 1.253
Step 4: Determine the empirical formula of compound X
The empirical formula represents the simplest whole number ratio of the elements present in the compound. To determine the empirical formula, we need to find the ratio of the elements present.
In compound X, the ratio of C to O can be determined from the moles of CO formed, and the ratio of H to O can be determined from the moles of HO formed.
Ratio of C to O = 1 (from CO)
Ratio of H to O = 2 (from HO)
Therefore, the empirical formula of compound X is C1H2O2.
Step 5: Determine the molecular formula of compound X
The molecular formula represents the actual number of atoms of each element in a compound. To determine the molecular formula, we need to know the molecular mass of compound X.
The molecular mass of compound X can be calculated by adding up the atomic masses of its constituent elements. The empirical formula C1H2O2 has a molecular mass of 29 g/mol.
Since the given options are in grams, we need to find the molecular mass in grams.
Molecular mass of X = 29 g/mol ≈ 29 g
Among the given options, the lowest molecular mass possible for compound X is 43 g (option A).
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Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. Can you explain this answer?
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Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. 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 Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. Can you explain this answer?.
Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. 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 Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. Can you explain this answer?.
Solutions for Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of Complete combustion of 0.858 g of compound X gives 2.63 g of CO₂ and 1.28 g of H₂O. The lowest molecular mass X can havea)43 gb)86 gc)129 gd)172 gCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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