In an experiment, a sample of CO2 gas was shown to effuse through a pinhole at a rate of 0.033 mol/min (where 1 min = 60 s). Using the same setup, 0.033 mol of a mystery gas effused through the same pinhole in 104 seconds.
a) What is the molar mass of the mystery gas?
b) If the mystery gas is monoatomic, what would the identity of the gas be (use a periodic table)?
I'm confused on this question, the answer was 1.32x10^2 g/mol and Xe respectively. I manipulated the Graham's Formula to get M2 = M1 * ((r1/r2)^2), my issue is with the r values. I know that the outcome for the r values should be 1.73, but I don't know how to obtain it. so how would you go about doing it, can you explain please.
What I did:
r1 = 0.033 mol/min
r2= (0.033 mol/min) * (104s * (1 min/60s)) = 0.033 * 1.73 = 0.0571
M1 = 44.01 g/mol
M2 = x
M2 = M1 * ((r1/r2)^2)
x = 44.01 * (0.033/0.0571)^2
x = 14.7 g/mol
I understand, that to get the 1.73 required for the answer, I must divide r2 by r1 to get it (1.73), but how would I manipulate the already manipulated Graham's Formula M2 = M1 * ((r1/r2)^2) to do such, on the first attempt without knowing the answer?
3 answers
rate 1 = CO2 = 0.033 moles/60second
rate 2 = unknown = x = 0.033 moles/104 seconds
(rate 1/rate 2) = sqrt (M2/M1) whee M1 and M2 or molar masses
(0.033/60)/(0.033/104) = sqrt (M2/44)
0.033/60 x (104/0.033) = sqrt (M2/44)
1.7333 = sqrt (M2/44)
3.004 = M2/44
M2 =3.004 x 44 = 132 = Xe
First you manipulated the equation to begin with. It looks correct to me but I never do that because it's just another chance for an error especially when we're in a hurry.
I think you can see how I obtained the 1.73 ratio. The other thing I didn't do is convert to seconds/min/ etc. Another chance for errors. I suspect that's why you didn't get 1.73 but I can check all of that math if you wish. You can do that do. Let me know if you want me to do anything in addition.