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Combustion Train Analysis

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    When 2.66 grams of a compound containing only carbon, hydrogen, and oxygen is burned completely, 4.50 grams of CO2 and 1.10 grams of H2O are produced. What is the empirical formula of the compound?
    Answer Options:
    C5H6O4
    C5H6O5
    C5H12O5
    C4H6O3
    C4H8O3
    C3H4O3
    C4H4O5
    C5H8O4
    C4H4O3

    2. Relevant equations
    Dimensional Analysis


    3. The attempt at a solution
    4.5gCO2*(1molCO2/44gCO2)*(1molCO2/1molC)*(12gC/1molC)=1.23gC
    1.1gH20*(1molH2)/18gH2O*(1molH2O/1molH)*(1gH/1molH)=.122gH
    2.66g-1.23g-.122g=1.308gO

    1.23gC*(1molC/12gC)=.1023molC/.0819mol=1C
    .122gH*(1molH/1gH)=.122molH/.0819mol=1H
    1.31gO*(1molO/16gO)=.0819molO/.0819mol=1O
    Empirical Formula:CHO<--not an available answer
     
  2. jcsd
  3. Oct 28, 2007 #2
    how many mol of H in H2O?
     
  4. Oct 28, 2007 #3
    Sorry that was a typo, in my calculation I used 2 mol of H in H2O.
     
  5. Oct 28, 2007 #4
    you can't round down to 1. Keep the decimals and multiply all mole numbers by the same factor until you get whole numbers
     
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