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A gas sample containing only SO2, PF3 and SF6 has the following mass percent composition: 30.4% SO2 and 28.4% PF3. Calculate the partial pressure (atm) of SO2 if the total pressure of the sample is 676.4 torr.

I tried taking the total pressure in atm and multiplying it with the percent SO2 given, and came up with 0.271atm; however, the homework program that I'm using said that the correct answer is 0.391atm. Can someone point me in the right direction with this question?

  • Chemistry -

    partial pressure is on the basis of mole composition, not mass.


  • Chemistry -

    Yes it is based on mole composition, not mass. But how do you figure out the moles?

  • Chemistry -

    30.4% SO2 = mole fraction 0.304.
    Atm total pressure = 676.4/760 = ??

    mole fraction SO2 x total P (in atm) = partial pressure SO2.
    mole fraction SO2 = 0.304.

  • Chemistry -

    I get the same answer as before = 0.271atm

    According to the program, the right answer is 0.391atm

  • Chemistry -

    Luc, I apologize for sending you astray. Here is the way you do it. I tried a shortcut that isn't legal. :-0
    You want to find the mole fraction of SO2, then
    XSO2 * (676.4/760) = partial pressure SO2.

    Take a 100 gram sample. That will give you
    30.4 g SO2.
    28.4 g PF3.
    100 - (30.4 + 28.4) = xxg SF6.

    Now convert all to moles.
    30.4/molar mass SO2 = moles SO2.
    28.4/molar mass PF3 = moles PF3.
    xx g SF6 molar mass SF6 = moles SF6.

    Add total moles.
    mole fraction SO2 = (moles SO2/total moles). Then use the above (X*total P = partial pressure SO2.
    This will give you 0.391 atm.

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