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Homework Help: A gaseous mixture contains ethane and propane problem

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A 9.780-g gaseous mixture contains ethane (C2H6) and propane
    (C3H8). Complete combustion to form carbon dioxide
    and water requires 1.120 mole of oxygen gas. Calculate the
    mass percent of ethane in the original mixture.

    2. Relevant equations

    3. The attempt at a solution
    7/2 mol O2 burns 1 mole ethane, then x mole O2 burns 2x/7 mole ethane (A)
    5 mole O2 burns 1 mole propane, then (1.12-x) mole O2 burns (1.12-x)/5 mole propane (B)
    Multiplying A and B with ethane's and propane's molar masses and adding them will give 9,78g.
    So solving that for x and then recalculating A for ethane and finally then dividing by 9.78 should give mass percent of ethane.
    But, this solution disagrees with solution manual, mine is %29,14, textbook's %38.
    I do understand the solution(there's a solution besides an answer) of textbook but I do not understand why mine is incorrect!
  2. jcsd
  3. Apr 11, 2015 #2


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    Staff: Mentor

    Show your math, description is just hard to follow.

    Generally speaking it is an example of a problem in which the final result is extremely sensitive to errors (so called ill-conditioned set of equations). My approach to the solution called for solving two equations. Depending on molar masses used I got very different results:



    (click on "approximate form" to see the decimal result).

    Note the plot - it contains two lines, each representing one of the equations. They are almost parallel, which means shifting one of them by a minute amount (which is what I did by using more precise molar mass) results in a huge difference in the final result.

    Actually it is not surprising the lines are parallel, when "normalized" these equations look like

    3.07x + 4.50y = 1
    3.13x + 4.46y = 1

    Do you see how similar they are?
    Last edited: Apr 11, 2015
  4. Apr 11, 2015 #3
    I use my blackboard for solutions so I had to take a picture of all, hope you will read my writing :)

    Attached Files:

  5. Apr 11, 2015 #4


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    Staff: Mentor

    I still don't understand why you expect me to spend time guessing what your x is, instead of explaining it so that I don't have to waste time.

    It is number of moles of oxygen that reacted with ethane, right?

    Your equation looks OK, what I wrote still holds. Put more precise molar masses into the equation in the upper right corner and you will end with substantially different result.
  6. Apr 11, 2015 #5
    Sorry, I thought it was pretty obvious :)
    Yes, it is the number of moles of O2.

    Thank you for the help.
  7. Apr 11, 2015 #6


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    Staff: Mentor

    It is not quite a natural choice of a single unknown. I would go for amount of ethane or amount of propane (after all, that's what you want to find). Not that the choice of unknown invalidates your approach, it just makes it more difficult to analyze.
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