Graham's Law of Effusion
Problems 1-10

Probs 11-25

Ten Examples

Examples and Problems only

Return to KMT & Gas Laws Menu


Problem #1: If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster?

Solution:

1) Set rates and get atomic weights:

rate1 = He = x
rate2 = Ar = 1

The atomic weight of He = 4.00
The atomic weight of Ar = 39.95

2) Graham's Law is:

r1 / r2 = MM2 / MM1

3) Substituting, we have:

x / 1 = 39.95 / 4.00

x = 3.16

Helium escapes faster than Ar. It does so at 3.16 times the rate of the argon.


Problem #2: What is the molecular weight of a gas which diffuses 1/50 as fast as hydrogen?

Solution:

1) Set rates and molecular weights:

rate1 = unknown gas = 1
rate2 = H2 = 50

Remember, 1/50th of 50 is 1.

The molecular weight of H2 = 2.016
The molecular weight of the other gas = x.

2) By Graham's Law (see the solution to problem #1), we have:

1 / 50 = 2.016 / x

0.0004 = 2.016 / x <--- remember to square both sides

x = 5040 g/mol


Problem #3: Two porous containers are filled with hydrogen and neon respectively. Under identical conditions, 2/3 of the hydrogen escapes in 6 hours. How long will it take for half the neon to escape?

Solution:

1) Set rates and weights:

rate1 = H2 = x
rate2 = Ne = 1

The molecular weight of H2 = 2.016
The atomic weight of Ne = 20.18

2) By Graham's Law:

x / 1 = 20.18 / 2.016

x = 3.164

3) Since the H2 escapes 3.164 times as fast as Ne, this calculation determines the amount of Ne leaving in 6 hours:

0.667 / 3.164 = 0.211

4) Calculate the time needed for half the Ne to escape, knowing that 0.211 escapes in 6 hours:

0.211 / 6 = 0.50 / x

x = 14.2 hours


Problem #4: If the density of hydrogen is 0.090 g/L and its rate of effusion is 5.93 times that of chlorine, what is the density of chlorine?

Solution:

1) Set rates and weights:

rate1 = H2 = 5.93
rate2 = Cl2 = 1

The molecular weight of H2 = 2.016
The molecular weight of Cl2 = x.

2) By Graham's Law:

5.93 / 1 = x / 2.016

35.1649 = x / 2.016

x = 70.89 g/mol

3) Determine gas density using the molar volume:

70.89 g / 22.414 L = 3.163 g/L

Problem #5: How much faster does hydrogen escape through a porous container than sulfur dioxide?

Solution:

1) Rates & weights:

rate1 = H2 = x
rate2 = SO2 = 1

The molecular weight of H2 = 2.016
The molecular weight of SO2 = 64.06

2) State Graham's Law:

r1 / r2 = MM2 / MM1

3) Substituting:

x / 1 = 64.06 / 2.016

x = 5.637

Hydrogen gas effuses 5.637 times as fast as SO2 gas does.


Problem #6: Compare the rate of diffusion of carbon dioxide (CO2) & ozone (O3) at the same temperature.

Solution:

1) Weights first (there's a reason):

The molecular weight of CO2 = 44.009
The molecular weight of O3 = 47.997

2) Set rate of the heavier:

O3 rate = 1
assign it to be r2 (which puts a 1 in the denominator)

The CO2 rate will be assigned the unknown 'x.' Since it is the lighter

3) Graham's Law:

x / 1 = 47.997 / 44.009 <--- don't forget to take the square root after dividing

x = 1.044

CO2 effuses 1.044 times as fast as O3


Problem #7: 2.278 x 10¯4 mol of an unidentified gaseous substance effuses through a tiny hole in 95.70 s. Under identical conditions, 1.738 x 10¯4 mol of argon gas takes 81.60 s to effuse. What is the molar mass of the unidentified substance?

Solution:

1) The first thing we need to do is compute the rate of effusion for each gas:

unknown gas: 2.278 x 10¯4 mol / 95.70 s = 2.380 x 10¯6 mol/s
argon: 1.738 x 10¯4 mol / 81.60 s = 2.123 x 10¯6 mol/s

2) Now, we are ready to use Graham's Law. Please note: (a) I will drop the 10¯6 from each rate and (b) we know the molar mass of argon from reference sources. Let argon be r1:

2.123 / 2.380 = x / 39.948

3) Square both sides and solve for x:

0.7957 = x / 39.948

x = 31.79 g/mol (to 4 significant figures)


Problem #8: A compound composed of carbon, hydrogen, and chlorine diffuses through a pinhole 0.411 times as fast as neon. Select the correct molecular formula for the compound:

(a) CHCl3
(b) CH2Cl2
(c) C2H2Cl2
(d) C2H3Cl

Solution:

1) Set rates:

Let r1 = 0.411; this means r2 (the rate of effusion for Ne) equals 1.

2) Inserting values into Graham's Law yields:

0.411 / 1 = 20.18 /x

20.18 is the atomic weight of Ne.

3) Squaring both sides gives:

0.16892 = 20.18 / x

4) Solving for x yields:

x = 119.46 g/mol

5) Examining the formulas for the possible answers, we see that answer choice (a) (CHCl3) gives a molecular weight of about 119.5.


Problem #9: Which pair of gases contains one which effuses at twice the rate of the other in the pair?

(a) He and Ne
(b) Ne and CO2
(c) He and CH4
(d) CO2 and HCl
(e) CH4 and HCl

Solution:

1) We can solve this problem by solving a simplified problem:

Set rate1 = 2
Set rate2 = 1

2) We now have a gas (rate1) effusing twice as fast as another gas (rate2). We now want to know how much heavier the slower gas is.

Set MM1 = 1
Set MM2 = x

3) Our faster gas (rate1) is also our lighter gas (MM1). We know want to know the molar mass (MM2) of our heavier, slower (rate2) gas.

4) Notice how I set the lighter gas' mass equal to 1. I could have used any number, all I need to know is how many times larger the mass of the slower gas is.

5) Use Graham's Law:

2 / 1 = x / 1

x = 4

6) Our heavier gas is four times heaver than the lighter gas (remember that the lighter is effusing twice as fast as the heavier gas).

7) Answer the question:

We look for a pair of gases in which the heavier gas is four times as heavy as the lighter gas. We find the only choice which satisfies that criterion is answer choice (c).

8) To continue the 'twice as' theme, you could solve this problem, if you wish:

Oxygen weighs approximately twice as much as methane. Under the same conditions of temperature and pressure, how much faster does a sample of methane effuse than a sample of oxygen?

Problem #10: If a molecule of CH4 diffuses a distance of 0.530 m from a point source, calculate the distance (in meters) that a molecule of N2 would diffuse under the same conditions for the same period of time.

Solution:

1) Assume the gases each diffuse in one second, in order to create a rate.

Set rate1 = N2 = x
Set rate2 = CH4 = 0.530 m/s

The molecular weight of N2 = 28.014
The molecular weight of CH4 = 16.0426

2) Graham's Law:

r1/ r2 = MM2 / MM1

3) Substituting, we have:

x / 0.530 = 16.0426 / 28.014

x = 0.401 m/s


Bonus Problem #1: Calculate the density of a gas at STP, if a given volume of the gas effuses through an apparatus in 6.60 min and the same volume of nitrogen at the same temperature and pressure, effuses through this apparatus in 8.50 minutes.

Solution:

1) Let us assume 1.00 L of each gas effused. This allows us to calculate rates:

unknown ---> 1.00 L / 6.60 min = 0.1515152 L/min
N2 ---> 1.00 L / 8.50 min = 0.117647 L/min

It is a common mistake in this type of problem to use the times directly in the solution to the problem. This is a mistake that the ChemTem has made. Oops!

2) Next, we use Graham's Law:

r1/r2 = SQRT(MM2 / MM1)

I will call the unknown r2 because that will put its molar mass in the numerator (making for an easier solution sequence).

0.117647 / 0.1515152 = SQRT( x / 28.014)

0.881175 = x / 28.014

x = 24.685 g/mol

3) Since we know everything was at STP, the density is as follows:

a) using molar volume:
24.685 g/mol / 22.414 L/mol = 1.10 g/L

b) or, using PV = nRT:

(1.00 atm) (1.00 L) = (n) (0.08206 L atm / mol K) (273 K)

n = 0.044638 mol (this is the moles of gas in 1.00 L)

24.685 g/mol times 0.044638 mol = 1.10 g

Since the grams are in 1.00 L, the density is 1.10 g/L


Bonus Problem #2: At 25.0 °C and 380.0 mmHg, the density of sulfur dioxide is 1.31 g/L. The rate of effusion of sulfur dioxide through an orifice is 4.48 mL/s. (a) What is the density of a sample of gas that effuses through an identical orifice at the rate of 6.78 mL/s under the same conditions? (b) What is the molar mass of the gas?

Solution:

1) The molar mass of the gas (part (b) above) is a straight-forward use of Graham's Law:

r1 / r2 = SQRT[MM2 / MM1]

Let r2 and MM2 be the unknown gas.

4.48 / 6.78 = SQRT[x / 64.063]

0.436613 = x / 64.063

x = 28.0 g/mol (to three sig figs)

2) The solution to part (a) is a less straight-forward application of Graham's Law. Here is where we will wind up:

r1 / r2 = SQRT[d2 / d1]

where d is the density of each gas.

3) The path to get to the above relationship starts with the ideal gas law and the definition of molar mass (I will use MM to indicate molar mass):

PV = nRT

and

MM = mass / moles

4) Rearrange, substitute into ideal gas law, rearrange:

moles = mass / MM

PV = (mass / MM) RT

MM = (mass / V) (RT/P)

5) Since density = mass / V, we have this

MM = d (RT/P)

6) Since the equation just above is valid for all ideal gases, we can write it for two different gases, specifying the same T and P"

MM1 = d1 (RT/P)

MM2 = d2 (RT/P)

7) Dividing the 2nd by the first

(MM2 / MM1) = (d2 / d1) [(RT/P) / (RT/P)]

8) Since RT/P = RT/P, we have this:

(MM2 / MM1) = (d2 / d1)

9) State Graham's Law:

r1 / r2 = SQRT[MM2 / MM1]

10) Substitute d2 / d1 for MM2 / MM1 and solve:

r1 / r2 = SQRT[d2 / d1]

Let the unknown gas be d2.

4.48 / 6.78 = SQRT[x / 1.31]

0.436613 = x / 1.31

x = 0.572 g/L <--- the answer to (b)

11) Suppose you had solved (a) first. There is an alternate way to get the molar mass of the unknown. The relationship t be used is this:

(MM2 / MM1) = (d2 / d1)

12) I'll solve it symbolically, then substitute the values:

Let the unknown gas be MM2

MM2 = MM1 (d2 / d1)

MM2 = 64.063 (0.572 / 1.31)

MM2 = 28.0 g/mol (to three sig figs)


Probs 11-25

Ten Examples

Examples and Problems only

Return to KMT & Gas Laws Menu