posted by Luc .
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?
partial pressure is on the basis of mole composition, not mass.
Yes it is based on mole composition, not mass. But how do you figure out the moles?
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.
I get the same answer as before = 0.271atm
According to the program, the right answer is 0.391atm
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.