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.

Question

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?

Answer 1

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

Right?

Answer 2

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

Answer 3

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.

Answer 4

I get the same answer as before = 0.271atm

According to the program, the right answer is 0.391atm

Answer 5

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
X_SO2 * (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.