Question

A 0.1647g of hydrocarbon sample is going through combustion process. It produces 0.5694g of CO2 and 0.0845g of H2O. Determine the percentage of these elements in this hydrocarbon.

Answer 1

CxHy + 3/2 O2------> xCO2 + Y/2H20
take out mole of CO2 as u are provided mass and molecularweight ok

Answer 2

hydrocarbon is CH4 i get that but now how to cal % ?

Answer 3

any idea?

Answer 4

hey i got C3H6

Answer 5

no how can be this possible i got x=1 and y=4 by that i cal hydrocarbon results CH4

Answer 6

u show me ur calculus and i ll show mine ok

Answer 7

K

Answer 8

CxHy + 3/2 O2------> xCO2 + Y/2H20

take out mole of CO2 as u are provided mass and molecular weight
x=0.5694/44
x= approx 1

let y mole of H2O
y=0.0845/18
y=4
mole og h20 =y/2=2

Answer 9

ur turn

Answer 10

I WAS DOING
Y/2= 0.004 => Y= 0.008
LOL

Answer 11

wat funny :P
anyways nw how to find percentage :?

Answer 12

I WILL NOW SOLVE % COMPOSITION WAIT!!

Answer 13

ok :) can i go now ? i have to do revision of hydrocarbon

Answer 14

H ---> 75%
C-25%
AM I RIGHT??

Answer 15

how u cal % ? frankly saying i forgot all things :( plz show me ur work

Answer 16

sry sry H-25%
C-75%

Answer 17

for H ,
4/16 x 100= 25 %
\
for C
12/16 x 100 = 75%

Answer 18

yup :)

Answer 19

hi5 we solve it :)

Answer 20

i alwz do silly mistake sry

Answer 21

even i too not a big prob its d symbool dat we are still learners:)

Answer 22

i have a different aprocah for this problem , shall i tell u

Answer 23

sure :)

Answer 24

1 mole of CO2----> 44 g
=> 1 g of CO2---> 1/44 mole
=> 0.5694 g of CO2--> 1/44 x 0.5694 =.01294

Answer 25

i knw this method :)

Answer 26

same thing for H20 == 0.004694 mole

Answer 27

:-) great

Answer 28

can i go now? plz dont mind but its urgnt for me to revise it as i dont knw the abc of alkene ... sorry

Answer 29

yea sure.... bye tnx for the help