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anonymous
 4 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?
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[C_ n H_{2n+2}\] 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94 u have a mistake molar mass of carbon is 12 u have written 14

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ohh Sorry. Thanks for notifying me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94 r u there??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Umm, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Co2 = 0.059 H2o= 0.071

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Very 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0H2o = 0.71 Co2 = 0.29

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just find out their wholeno. 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@open_study1 I haven't checked out the calculations.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@siddhantsharan sorry the hydrocarbon is C6H14

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw: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 / 44Equation 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@siddhantsharan i will solve it plz tell me whether u got the answer as C6H14

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I didnt solve it. Just do the calculations.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@apoorvk wat u say i think finding % will be more easy...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0follow both processes, and see for yourself! :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@apoorvk but how to find % with given info @siddhantsharan do u have idea

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I dont know. Just dont message me tens of links to your questions. 1 link is fine.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then do u have otherway plz show it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@apoorvk not getting help

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Aqua666 shall show you the way now... .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0give me a minute to figure out how to explain this in english...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually.. 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..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh..no aproovk show by % method

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, that sums it... don't follow my procedure then. though i wonder why it's be inefficient..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0**it's inefficient. Hmm, now I know.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0DO what @Aqua666 and @siddhantsharan say in this regard.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@apoorvk % is more easier i think plz help me to find after finding the % we can find emperical

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That 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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