Question

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?

Answer 1

@Ishaan94 plz help

Answer 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.

Answer 3

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

Answer 4

Ohh Sorry. Thanks for notifying me.

Answer 5

n=???

Answer 6

@Ishaan94 r u there??

Answer 7

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.

Answer 8

Is the answer C5H12?

Answer 9

Co2 = 0.059 H2o= 0.071

Answer 10

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?

Answer 11

H2o = 0.71 Co2 = 0.29

Answer 12

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?

Answer 13

@open_study1 I haven't checked out the calculations.

Answer 14

@siddhantsharan sorry the hydrocarbon is C6H14

Answer 15

@apoorvk

Answer 16

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.