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open_study1 Group Title

An unknown hydrocarbon weighing 0.858g on complete combustion gives 2.63g of CO2 and 1.28g H2O . THe molecular weight of the hydro carbon is?

  • 2 years ago
  • 2 years ago

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  1. open_study1 Group Title
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    @Ishaan94 plz help

    • 2 years ago
  2. Ishaan94 Group Title
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    \[C_ n H_{2n+2}\] CO2 mass 12+32=44. H2O = 18 No. of moles of H2O 1.28/18 No. of moles of CO2 2.63/44 Mass of Hydrocarbon must be n*14 + 2n+2 = 16n + 2. No. of moles = .858/(16n+2). Now equate them.

    • 2 years ago
  3. open_study1 Group Title
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    @Ishaan94 u have a mistake molar mass of carbon is 12 u have written 14

    • 2 years ago
  4. Ishaan94 Group Title
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    Ohh Sorry. Thanks for notifying me.

    • 2 years ago
  5. open_study1 Group Title
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    n=???

    • 2 years ago
  6. open_study1 Group Title
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    @Ishaan94 r u there??

    • 2 years ago
  7. apoorvk Group Title
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    Umm, the hydrocarbon may not necessarily be an alkane, @Ishaan94 . The crucial thing here is the ratio of moles of CO2 and H2O produced in the reaction. Find that out.

    • 2 years ago
  8. siddhantsharan Group Title
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    Is the answer C5H12?

    • 2 years ago
  9. open_study1 Group Title
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    Co2 = 0.059 H2o= 0.071

    • 2 years ago
  10. apoorvk Group Title
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    Very good. Now if 0.071 is the number of moles of H2O, then the no. of atoms of H in there is twice, that is 0.142. SO, 0.142 moles of H per 0.059 moles of C. What is their ratio?

    • 2 years ago
  11. open_study1 Group Title
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    H2o = 0.71 Co2 = 0.29

    • 2 years ago
  12. apoorvk Group Title
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    Just find out their whole-no. ratios, that 'll give you atleast an empirical formula of the compound. And in this particular case, the molecular formula will be the same as the empirical one, can you guess why?

    • 2 years ago
  13. apoorvk Group Title
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    @open_study1 I haven't checked out the calculations.

    • 2 years ago
  14. open_study1 Group Title
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    @siddhantsharan sorry the hydrocarbon is C6H14

    • 2 years ago
  15. open_study1 Group Title
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    @apoorvk

    • 2 years ago
  16. siddhantsharan Group Title
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    |dw:1338950362943:dw| Now Moles of CxHy = 0.858/ 12x + y Moles of CO2 = 2.63/44 Moles of H20 = 1.28/18 Now 1 mole of cxHy produces x moles of CO2. => 0.858/(12x + y) moles produce x(0.858)/(12x + y) moles of CO2 = 2.63 / 44----Equation 1. Similarly 1 mole of cxHy produces y/2 moles of H20 => 0.858(y/2)/(12x + y) = 1.28/18 ------Equation 2. Soolve the two equations.

    • 2 years ago
  17. open_study1 Group Title
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    @siddhantsharan i will solve it plz tell me whether u got the answer as C6H14

    • 2 years ago
  18. siddhantsharan Group Title
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    I didnt solve it. Just do the calculations.

    • 2 years ago
  19. open_study1 Group Title
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    @apoorvk wat u say i think finding % will be more easy...

    • 2 years ago
  20. apoorvk Group Title
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    follow both processes, and see for yourself! :P

    • 2 years ago
  21. open_study1 Group Title
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    @apoorvk but how to find % with given info @siddhantsharan do u have idea

    • 2 years ago
  22. siddhantsharan Group Title
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    I dont know. Just dont message me tens of links to your questions. 1 link is fine.

    • 2 years ago
  23. Aqua666 Group Title
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    Are you guys still working on the molecular weight of the hydrocarbon? If so than I think you guys are making it way to complicated.. escpecially if the combustion reaction was hydrocarbon + O2 -> Co2 and H2O...

    • 2 years ago
  24. open_study1 Group Title
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    then do u have otherway plz show it

    • 2 years ago
  25. open_study1 Group Title
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    @apoorvk not getting help

    • 2 years ago
  26. apoorvk Group Title
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    @Aqua666 shall show you the way now... -.-

    • 2 years ago
  27. Aqua666 Group Title
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    give me a minute to figure out how to explain this in english...

    • 2 years ago
  28. open_study1 Group Title
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    ok waiting

    • 2 years ago
  29. Aqua666 Group Title
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    actually.. I calculated it pretty much the same as sidd.. I just didn't write it down, which makes it seem a hell of a lot easier.. but I can't explain it much better that him either.. not in writing.. and he already showed you the way..

    • 2 years ago
  30. open_study1 Group Title
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    oh..no aproovk show by % method

    • 2 years ago
  31. apoorvk Group Title
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    well, that sums it... don't follow my procedure then. though i wonder why it's be inefficient..

    • 2 years ago
  32. apoorvk Group Title
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    **it's inefficient. Hmm, now I know.

    • 2 years ago
  33. apoorvk Group Title
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    DO what @Aqua666 and @siddhantsharan say in this regard.

    • 2 years ago
  34. Aqua666 Group Title
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    I never said it was inefficient.. It really isn't.. you just don't need %, if you feel comfortable with your method than you should use it.. I just didn't quite understand what the hell you were doing.. However, now that I was trying to write it down I see why your method might be easier.. especially if you're trying to explain it to someone else...

    • 2 years ago
  35. open_study1 Group Title
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    @apoorvk % is more easier i think plz help me to find after finding the % we can find emperical

    • 2 years ago
  36. apoorvk Group Title
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    That ratio thingy is pretty unhelpful actually, since it will only help you find the empirical formula, and will become pretty cumbersome in cases of species with more than one atom. May help in this particular case, but calculations are killers, especially since you won't have a calculator with during the exam. Stick to the equation method, 'easy' is not always right. -.-

    • 2 years ago
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