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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?
Most Upvoted Answer
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.
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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?
Question Description
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