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4gram of hydrocarbon on complete combustion gave 12 • 571 gram of CO2 and 5 •143 gram of water. What is the empirical formula of the hydrocarbon
- a)CH
- b)CH2
- c)CH3
- d)C2H5
Correct answer is option 'B'. Can you explain this answer?
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4gram of hydrocarbon on complete combustion gave 12 • 571 gram of...
Given Data:
- 4g of hydrocarbon on complete combustion gave 12.571g of CO2 and 5.143g of water
Calculating Moles of CO2 and H2O:
- Moles of CO2 = 12.571g / 44g/mol = 0.285 mol
- Moles of H2O = 5.143g / 18g/mol = 0.285 mol
Calculating Moles of Carbon and Hydrogen:
- Moles of Carbon in CO2 = 0.285 mol
- Moles of Hydrogen in H2O = 0.285 mol
- Moles of Hydrogen in hydrocarbon = 0.285 mol - 0.285 mol = 0 mol
- Moles of Carbon in hydrocarbon = 0.285 mol
Calculating Empirical Formula:
- Ratio of Carbon to Hydrogen = 0.285 mol : 0 mol = 1 : 0
- The empirical formula is CH2
Therefore, the empirical formula of the hydrocarbon is CH2.
- 4g of hydrocarbon on complete combustion gave 12.571g of CO2 and 5.143g of water
Calculating Moles of CO2 and H2O:
- Moles of CO2 = 12.571g / 44g/mol = 0.285 mol
- Moles of H2O = 5.143g / 18g/mol = 0.285 mol
Calculating Moles of Carbon and Hydrogen:
- Moles of Carbon in CO2 = 0.285 mol
- Moles of Hydrogen in H2O = 0.285 mol
- Moles of Hydrogen in hydrocarbon = 0.285 mol - 0.285 mol = 0 mol
- Moles of Carbon in hydrocarbon = 0.285 mol
Calculating Empirical Formula:
- Ratio of Carbon to Hydrogen = 0.285 mol : 0 mol = 1 : 0
- The empirical formula is CH2
Therefore, the empirical formula of the hydrocarbon is CH2.
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4gram of hydrocarbon on complete combustion gave 12 • 571 gram of CO2 and 5 •143 gram of water. What is the empirical formula of the hydrocarbona)CHb)CH2c)CH3d)C2H5Correct answer is option 'B'. Can you explain this answer?
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4gram of hydrocarbon on complete combustion gave 12 • 571 gram of CO2 and 5 •143 gram of water. What is the empirical formula of the hydrocarbona)CHb)CH2c)CH3d)C2H5Correct 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 4gram of hydrocarbon on complete combustion gave 12 • 571 gram of CO2 and 5 •143 gram of water. What is the empirical formula of the hydrocarbona)CHb)CH2c)CH3d)C2H5Correct 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 4gram of hydrocarbon on complete combustion gave 12 • 571 gram of CO2 and 5 •143 gram of water. What is the empirical formula of the hydrocarbona)CHb)CH2c)CH3d)C2H5Correct answer is option 'B'. Can you explain this answer?.
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