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open_study1

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?

  • one year ago
  • one year ago

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  1. open_study1
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    @Ishaan94 plz help

    • one year ago
  2. Ishaan94
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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.

    • one year ago
  3. open_study1
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    @Ishaan94 u have a mistake molar mass of carbon is 12 u have written 14

    • one year ago
  4. Ishaan94
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    Ohh Sorry. Thanks for notifying me.

    • one year ago
  5. open_study1
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    n=???

    • one year ago
  6. open_study1
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    @Ishaan94 r u there??

    • one year ago
  7. apoorvk
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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.

    • one year ago
  8. siddhantsharan
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    Is the answer C5H12?

    • one year ago
  9. open_study1
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    Co2 = 0.059 H2o= 0.071

    • one year ago
  10. apoorvk
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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?

    • one year ago
  11. open_study1
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    H2o = 0.71 Co2 = 0.29

    • one year ago
  12. apoorvk
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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?

    • one year ago
  13. apoorvk
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    @open_study1 I haven't checked out the calculations.

    • one year ago
  14. open_study1
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    @siddhantsharan sorry the hydrocarbon is C6H14

    • one year ago
  15. open_study1
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    @apoorvk

    • one year ago
  16. siddhantsharan
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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.

    • one year ago
  17. open_study1
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    @siddhantsharan i will solve it plz tell me whether u got the answer as C6H14

    • one year ago
  18. siddhantsharan
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    I didnt solve it. Just do the calculations.

    • one year ago
  19. open_study1
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    @apoorvk wat u say i think finding % will be more easy...

    • one year ago
  20. apoorvk
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    follow both processes, and see for yourself! :P

    • one year ago
  21. open_study1
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    @apoorvk but how to find % with given info @siddhantsharan do u have idea

    • one year ago
  22. siddhantsharan
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    I dont know. Just dont message me tens of links to your questions. 1 link is fine.

    • one year ago
  23. Aqua666
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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...

    • one year ago
  24. open_study1
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    then do u have otherway plz show it

    • one year ago
  25. open_study1
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    @apoorvk not getting help

    • one year ago
  26. apoorvk
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    @Aqua666 shall show you the way now... -.-

    • one year ago
  27. Aqua666
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    give me a minute to figure out how to explain this in english...

    • one year ago
  28. open_study1
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    ok waiting

    • one year ago
  29. Aqua666
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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..

    • one year ago
  30. open_study1
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    oh..no aproovk show by % method

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

    • one year ago
  32. apoorvk
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    **it's inefficient. Hmm, now I know.

    • one year ago
  33. apoorvk
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    DO what @Aqua666 and @siddhantsharan say in this regard.

    • one year ago
  34. Aqua666
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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...

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

    • one year ago
  36. apoorvk
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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. -.-

    • one year ago
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