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?
I'm big dumb, r2 = 0.033/1.73 = 0.019 was the correct r2 value. what I did was wrong for r2, thus causing me to mess up to final answer.
This is how I would do it.
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.
That 1.73 comes from r1/r2 and not r2/r1.
To obtain the value of 1.73 for the ratio (r1/r2), you need to consider the relationship between the rates of effusion and the time is taken for each gas to effuse through the same pinhole.
Let's break down the steps again:
Given:
r1 = 0.033 mol/min (rate of effusion for CO2)
r2 = 0.033 mol / 104 s (rate of effusion for the mystery gas)
To calculate the molar mass of the mystery gas (M2), we can use Graham's law of effusion:
M2 = M1 * (r1/r2)^2
In this formula, M1 represents the molar mass of CO2 (44.01 g/mol), and M2 represents the molar mass of the mystery gas that we want to find. The key is to manipulate the formula to solve for the ratio (r1/r2), which will help us find the value of 1.73.
Rearranging Graham's law formula:
(r1/r2)^2 = M2/M1
Now, substitute the known values:
(0.033 mol/min) / (0.033 mol / 104 s) = M2 / 44.01 g/mol
We can simplify the left-hand side of the equation to:
(0.033 mol/min) * (104 s / 0.033 mol) = M2 / 44.01 g/mol
This simplifies to:
3.64 min = M2 / 44.01 g/mol
To isolate M2, we multiply both sides by 44.01 g/mol:
3.64 min * 44.01 g/mol = M2
M2 ≈ 160.56 g/mol
So, the molar mass of the mystery gas is approximately 160.56 g/mol.
To determine the identity of the gas, you can refer to the periodic table. The closest monoatomic gas to this molar mass is xenon (Xe) with a molar mass of 131.29 g/mol. Thus, the identity of the mystery gas would be xenon (Xe).
It's important to note that the ratio (r1/r2) was not derived directly from the given data but was obtained by manipulating the formula. By following the steps above, you can find the value of 1.73 and arrive at the correct answer without prior knowledge.