Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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
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

This Question is Closed

open_study1Best ResponseYou've already chosen the best response.0
@Ishaan94 plz help
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.2
\[C_ n H_{2n+2}\] 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

open_study1Best ResponseYou've already chosen the best response.0
@Ishaan94 u have a mistake molar mass of carbon is 12 u have written 14
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.2
Ohh Sorry. Thanks for notifying me.
 one year ago

open_study1Best ResponseYou've already chosen the best response.0
@Ishaan94 r u there??
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
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

siddhantsharanBest ResponseYou've already chosen the best response.0
Is the answer C5H12?
 one year ago

open_study1Best ResponseYou've already chosen the best response.0
Co2 = 0.059 H2o= 0.071
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
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

open_study1Best ResponseYou've already chosen the best response.0
H2o = 0.71 Co2 = 0.29
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
Just 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?
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
@open_study1 I haven't checked out the calculations.
 one year ago

open_study1Best ResponseYou've already chosen the best response.0
@siddhantsharan sorry the hydrocarbon is C6H14
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
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 / 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.
 one year ago

open_study1Best ResponseYou've already chosen the best response.0
@siddhantsharan i will solve it plz tell me whether u got the answer as C6H14
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
I didnt solve it. Just do the calculations.
 one year ago

open_study1Best ResponseYou've already chosen the best response.0
@apoorvk wat u say i think finding % will be more easy...
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
follow both processes, and see for yourself! :P
 one year ago

open_study1Best ResponseYou've already chosen the best response.0
@apoorvk but how to find % with given info @siddhantsharan do u have idea
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
I dont know. Just dont message me tens of links to your questions. 1 link is fine.
 one year ago

Aqua666Best ResponseYou've already chosen the best response.0
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

open_study1Best ResponseYou've already chosen the best response.0
then do u have otherway plz show it
 one year ago

open_study1Best ResponseYou've already chosen the best response.0
@apoorvk not getting help
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
@Aqua666 shall show you the way now... .
 one year ago

Aqua666Best ResponseYou've already chosen the best response.0
give me a minute to figure out how to explain this in english...
 one year ago

Aqua666Best ResponseYou've already chosen the best response.0
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

open_study1Best ResponseYou've already chosen the best response.0
oh..no aproovk show by % method
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
well, that sums it... don't follow my procedure then. though i wonder why it's be inefficient..
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
**it's inefficient. Hmm, now I know.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
DO what @Aqua666 and @siddhantsharan say in this regard.
 one year ago

Aqua666Best ResponseYou've already chosen the best response.0
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

open_study1Best 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
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
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
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.