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open_study1

  • 2 years ago

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?

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  1. open_study1
    • 2 years ago
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    @Ishaan94 plz help

  2. Ishaan94
    • 2 years ago
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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.

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

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

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

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

  7. apoorvk
    • 2 years ago
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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.

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

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

  10. apoorvk
    • 2 years ago
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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?

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

  12. apoorvk
    • 2 years ago
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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?

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

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

  15. open_study1
    • 2 years ago
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    @apoorvk

  16. siddhantsharan
    • 2 years ago
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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.

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

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

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

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

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

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

  23. Aqua666
    • 2 years ago
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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...

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

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

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

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

  28. open_study1
    • 2 years ago
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    ok waiting

  29. Aqua666
    • 2 years ago
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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..

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

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

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

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

  34. Aqua666
    • 2 years ago
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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...

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

  36. apoorvk
    • 2 years ago
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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. -.-

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