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Chemistry Combustion Reaction

  1. Mar 14, 2013 #1
    Burning a 3.230 g sample of the unknown compound in an excess of oxygen produces 4.7341 g of CO2 (carbon dioxide). The carbon initially in the unknown compound is completely converted to carbon in the CO2.

    If it is determined that the molecular weight of the unknown compound is 180.16 g/mol, what is the correct molecular formula?

    I don't see how I can do this without know the mass of H2O.
     
  2. jcsd
  3. Mar 14, 2013 #2

    Borek

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    Staff: Mentor

    Can you calculate mass of carbon in the compound sample?

    Note: while you are missing something obvious, even after taking it into account it is still not possible to solve the problem, as it turns out compound contains more hydrogen than carbon (in terms of mass - so the molar ratio is even worse).
     
  4. Mar 14, 2013 #3
    So there are 0.108 mols of C which is equal to 1.29g C. I still don't see how this helps since I do not know how much water was produced by this reaction.
     
  5. Mar 14, 2013 #4

    Borek

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    Staff: Mentor

    OK, I probably misunderstood your question. Not that I am sure I understand it now.

    My first idea was that if there was 3.230 g of the sample, and it contained 1.29 g of carbon, the remaining mass can be attributed to hydrogen - and then simply the answer is "no solution".

    Chances are your comment about water really means "not knowing amount of water I can't calculate amount of oxygen in the original sample". Sadly, question is ambiguous, as it doesn't say anything about the composition - could be the compound contains any other elements as well. But let's assume it is just CnHkOl only.

    We can easily calculate number of moles of the substance, so at least n in Cn is obvious. Then we know how much mass can be attributed to other elements. Then (with a brute force) we can check that there exist a reasonable CnHkOl formula that fits the data given - you just need such a pair of k, l that 16*l+k equals remaining part of the molar mass given.

    I retract my earlier comment about lack of solution.
     
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