## open_study1 3 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?

1. open_study1

@Ishaan94 plz help

2. Ishaan94

\[C_ n H_{2n+2}\] 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

@Ishaan94 u have a mistake molar mass of carbon is 12 u have written 14

4. Ishaan94

Ohh Sorry. Thanks for notifying me.

5. open_study1

n=???

6. open_study1

@Ishaan94 r u there??

7. apoorvk

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

9. open_study1

Co2 = 0.059 H2o= 0.071

10. apoorvk

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

H2o = 0.71 Co2 = 0.29

12. apoorvk

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

@open_study1 I haven't checked out the calculations.

14. open_study1

@siddhantsharan sorry the hydrocarbon is C6H14

15. open_study1

@apoorvk

16. siddhantsharan

|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

@siddhantsharan i will solve it plz tell me whether u got the answer as C6H14

18. siddhantsharan

I didnt solve it. Just do the calculations.

19. open_study1

@apoorvk wat u say i think finding % will be more easy...

20. apoorvk

follow both processes, and see for yourself! :P

21. open_study1

@apoorvk but how to find % with given info @siddhantsharan do u have idea

22. siddhantsharan

I dont know. Just dont message me tens of links to your questions. 1 link is fine.

23. Aqua666

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

then do u have otherway plz show it

25. open_study1

@apoorvk not getting help

26. apoorvk

@Aqua666 shall show you the way now... -.-

27. Aqua666

give me a minute to figure out how to explain this in english...

28. open_study1

ok waiting

29. Aqua666

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

oh..no aproovk show by % method

31. apoorvk

well, that sums it... don't follow my procedure then. though i wonder why it's be inefficient..

32. apoorvk

**it's inefficient. Hmm, now I know.

33. apoorvk

DO what @Aqua666 and @siddhantsharan say in this regard.

34. Aqua666

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

@apoorvk % is more easier i think plz help me to find after finding the % we can find emperical

36. apoorvk

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. -.-