[justlearning]

Complete combustion of 19.48 g of a particular hydrocarbon yielded 62.93 g CO 2 and 20.61 g of H 2 O. What is the empirical formula of this hydrocarbon? (Enter your answer in the spaces below.)
Molar masses (in g mol -1 )
H, 1.00794    C, 12.011    O, 15.9994


the formula is c_h_. you have to find the blanks..
i tried working with this problem and it just left me with wrong answers.
i was hoping on help as to how to solve it..
this is what i tried

62.93g/molar mass c02 = 1 and     20.61g h20/molarmassh20= 1 , 19.48/molarmass02 = .6087
then i took the lowest number and divided each thing by it and my answer was wrong..

i also tried answreing the question without using any oxygen in the problem

so far my answers have been c1h4, c2h2, c1h1 and they all are wrong
PLEASE *delete me*!

[Mitch]

Start by writing us a general chemical equation.

[justlearning]

h20 +c02 + 02 -----> c_h_

the thing is i am trying to find out the C_H_

[mike]

You are burning your hydrocarbon in oxygen, rewrite the equation.

Also write down the molecular weights(mass) of CO2, H2O, O2.

[justlearning]

hmm..i did all of that above...

thats all i can think of  though

[mike]

No, your equation is wrong  

It is your hydrocarbon CH + O2 ---> etc

also if you have worked out the molecular weights then how did you get 1, 1 and 0.6087 as your answers? these look wrong too, sorry!

[AWK]

CxHy + zO2 =
hint
each carbon atom needs 1 molecule of O2
for 4 hydrogen atoms you need 1 molecule of O2