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Percent Composition

  1. Dec 31, 2008 #1
    1. The problem statement, all variables and given/known data
    A 5.00g sample of natural gas, containing methane, CH4, and ethylene, C2H4, was burned in excess oxygen, yielding 14.5g CO2 and some H2O as products. What percent of the sample was ethylene?

    2. Relevant equations
    Moles and percent composition

    3. The attempt at a solution
    I don't really understand how to correctly set up the problem. I know that ethylene is about 40% and methane is about 60% by guess-and-checking percents until the two gas more or less result 14.5g CO2 combined, from 5.00g sample, but this way is obviously inefficient. How is it correctly set up to result the correct and accurate answer?
  2. jcsd
  3. Dec 31, 2008 #2


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    Write the balanced reactions for each combustion - be certain to balance reactants and products correctly. Calculate the molecular weights for methane, ethane(ethylene), and carbon dioxide.

    You will have two unknown values; the mass of methane (call it x) and mass of ethylene (call it y). Using these as unknown variables, derive a formula for the resulting mass of carbon dioxide, do so for each gas compound reactant, not including the oxygen reactant (meaning just do this for the methane AND for the ethylene).

    Now, you know two things: the expression with x plus the expression with y must equal 14.5 grams, and x+y must equal 5 grams.

    If there were a way to construct a table into which I could put values and expressions, and if the typesetting system were easier to use, I would show you the table which I constructed on paper, but into this forum message. Try using the approach described after your quoted message.
  4. Dec 31, 2008 #3


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    Your two reactions should be:

    CH4 + O2 [tex]\rightarrow[/tex] CO2+ 2H2O

    CH2CH2 + 3O2 [tex]\rightarrow[/tex] 2CO2 + 2 waters
    Last edited: Dec 31, 2008
  5. Dec 31, 2008 #4
    CH4 + 2O2 [tex]\rightarrow[/tex] CO2+ 2H2O
    C2H4 + 3O2 [tex]\rightarrow[/tex] 2CO2 + 2H2O

    x = CH4
    y = C2H4

    x + y = 5.00 g

    x × (1 mol CH4/16.0426 g CH4) × (1 mol CO2 / 1 mol CH4) × (44.009 g CO2 / 1 mol CO2) [tex]\approx[/tex] 2.74326x

    y × (1 mol C2H4/28.0536 g C2H4) × (2 mol CO2 / 1 mol C2H4) × (44.009 g CO2 / 1 mol CO2) [tex]\approx[/tex] 3.137494y

    2.74326x + 3.137494y [tex]\approx[/tex] 14.5g CO2

    x + y = 5.00 g [tex]\rightarrow[/tex] y = 5.00 g - x

    2.74326x + 3.137494(5.00 g - x) [tex]\approx[/tex] 14.5g CO2
    2.74326x + 15.68747 g - 3.137494x [tex]\approx[/tex] 14.5g CO2
    1.18747 g [tex]\approx[/tex] 0.394234x
    3.01209434 [tex]\approx[/tex] x

    x + y = 5.00 g
    3.01209434 + y = 5.00 g
    y [tex]\approx[/tex] 1.988 g

    3.01209434 / 5.00 [tex]\approx[/tex] 0.602 [tex]\approx[/tex] 60.2% CH4 by weight.

    1.988 / 5.00 = 0.398 = 39.2% C2H4 by weight.

    This looks correct. Thank you.
    Last edited: Dec 31, 2008
  6. Dec 31, 2008 #5


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    312213 you found and corrected my error in the methane reaction: TWO oxygen molecules, not just one oxygen molecule. I trust that the rest of your work is good and will not need to actually check it. In any case, you would not need the actual coefficient of oxygen molecules from the reactants sides in the actual rest of the solution process - but maybe just to check that the reactions are being balanced.
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