chemistry

a 2.52-g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess O to yield 4.23 g of CO2 and 1.01 g of H2O as the only carbon and hydrogen containing products respectively.
Another sample of the same compound, of mass 4.14 g, yielded 2.11 g of SO3 as the only sulfur containing product.
A third sample, of mass 5.66 g, was burned under different conditions to yield 2.27 g of HNO3 as the only nitrogen containing product.


PLEASE HELP! THIS IS DUE VERY SOON AND I HAVE BEEN WORKING ON IT FOR HOURS BUT KEEP GETTING STUCK:(

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asked by Grace
  1. What is the question? The empirical formula?
    Convert 4.23g CO2 to g C and convert that to %C. Here is how you do that.
    4.23g CO2 x (atomic mass C/molar mass CO2) = 4.23 X 12/44 = about 1.15g C, then %C = 1.15/mass sample)*100 = about 45.8%C
    For g H it is 1.01g H2O x (2*1/18) = 0.112g and %H = (0.112/2.52)*100 = about 4.45%
    You should redo these to confirm and/or obtain with better accuracy.


    Do th same kind of thing for %S (in the different sample but % is %) and the same for %N. That leaves %O which is obtained by adding %C, %H, %N, %S and subtracting from 100.
    Now take a 100 g sample which gives you the same numbers as the % but without the % sign.
    Convert g each to mols by
    mol = g/atomic mass and I assume you can take it from here. Post your work if you get stuck.I didn't work it through to get an answer but this should be straight forward from here.

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