When 3.517 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 11.59 grams of CO2 and 3.164 grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 40.06 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Find empirical formula and molecular formula.

4 answers

mols C = 11.59/44 = ?
mols H = 3.164*2/18 = ?

Now find the ratio of these two numbers with the smallest being 1.00. The easy way to do that is to divide the smaller number by itself and divide the other number by the same small number. Round to whole numbers UNLESS the numbers are larger than 1.2 or so. If that is the case then multiply each number by 2, 3, 4, 5 etc until they can be rounded to whole numbers. For example if the numbers are 1.0 C and 1.1 H you would round to C1H1 but if they are 1.0C and 1.25 H you would multiply by 2 to get 2.0 C and 2.5 H and that still can't be rounded to a whole number. Try 3 and you get 3C and 1.25*3 or 3.75 H. Still no good. Try 4 and you get 4.0C and 1.25*4 = 5.0 for H so the formula would be C4H5. That's the empircal formula (but of course that isn't the answer for this problem).
For the molecular formula do this.
Calculate the empirical formula of CxHy after you know what C and H are. Then empirical formula x ? = 40.06. Solve for ? and round to a whole number which I will call z. Then the molecular formula is (CxHy)z.
this completely confused me.... these are bonus problems we have not been taught for extra credit and i can't figure out the answers
for molecular i got 1 and 1

empirical i don't know what to do
I showed you exactly how to do it. Please read my response, step by step. You will see I follow what I told you to do. Here it is.
mols C = 11.59/44 = 0.263
mols H = 3.164*2/18 = 0.351
I suspect you did not do those calculations. I suspect you read the whole thing, said, Hey that's too long and complicated, and wrote me to work it for you. Here is the next step. I said to find the ratio of the numbers and the easy way to do that is to divide the smaller number by itself, then divide the other one by the same small number. Here is that step.
C = 0.263/0.263 = 1.00
H = 0.352/0.263 = 1.34

Now you round those to whole numbers EXCEPT the 1.34 is too far from a whole number. Note I said round UNLESS the number was more than 1.1 or so. So you must go through that multiplication thing.
C = 1.00 x 2 = 2.00
H = 1.34 x 2 = 2.68
Still can't round. Try 3
C = 1.00 x 3 = 3.00
H = 1.34 x 3 = 4.02.
That IS within rounding rules since the 4.01 is not larger than 4.1 so the formula is C3 and H4 or
C3H4. That's the empirical formula. That takes care of the first part of the problem.

The second part is to find the molecular formula. The mass of that empirical formula is (3*12) + (4*1) = 36 + 4 = 40.
Molecular formulas are whole number multiples of the empirical formula so your job is to find the whole number that is the multiple. You do it this way. The empirical formula mass x some number = molar mass.
40 + some number = 40.06. It should come as no surprise that the number is 40.06/40 = 1.001. Now you round that off to a whole number of 1.00. That tells you that the empirical formula and the molecular formula are the same so the empirical formula is C3H4 and the molecular formula is C3H4.

The ONLY thing I have done in this post that's different from my first response is that I've put in the numbers and calculated them. I honestly think that if you had followed the steps you would have seen how to do it too. You can't get intimidated by chemistry. It is no different from any other course EXCEPT you must be disciplined and think things through. In that first response I even put in the numbers for you to use and you didn't do the calculation to see if you could do it. I think you gave up far to easily. Thank you for using Jiskha. Don't get discouraged. You can do this if you put your mind to it and think things through. Good luck and post your questions here any time. We're here to help but we can be more effective if you tell us exactly what your problem is. I'm confused doesn't help much but I don't understand because ..... really does.