What is the partial pressure of oxygen in the oxygen and helium mixture?

Question

A mixture of oxygen and helium is 92.3% by mass oxygen. It is collected at atmospheric pressure (735 Torr). What is the partial pressure of oxygen in this mixture?

a) 299 Torr
b) 412 Torr
c) 688 Torr
d) 333 Torr
e) 446 Torr

(an explanation would be very helpful, especially what the purpose of the 92.3% would help indicate or just any help at all.)

Answer 1

First molecular masses: O2: 32, He: 4.

A mixture of oxygen and helium is 92.3% by mass oxygen. Thus is 7.7% helium by mass.
Now to find the mole percentage:
O2: {92.3%/32} / {92.3%/32 + 7.7%/4} = 60%
He: {7.7%/4} / {92.3%/32 + 7.7%/4} = 40%
partial pressure of oxygen in this mixture: 60%735 Torr = 441 Torr.

Answer 2

I'm not sure if this is correct, but from my supposedly not-rusty, but feels rusty memory of pchem, use P(O2) = P*(O2) X P*(O2) = 0.980 atm = pressure of pure oxygen (for this problem) X = 0.923 = mole fraction of oxygen P(O2) = partial pressure of O2 P(O2) = (0.980 atm)(0.923) P(O2) = 0.90454 atm round to significant figures 0.90 atm I hope my answer is right, and my rounding in significant figures. =) If it's not correct, then I hope you find the answer to your question soon! EDIT: kudos to chemteam. They're correct in solving for mole fraction. =)