Graham's Law
Problems 1-10

Probs 11-25

Seven Examples

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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?


Set rate1 = He = x
Set rate2 = Ar = 1

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

Graham's Law is:

r1 over r2 = √MM2 over √MM1

Substituting, we have:

x / 1 = √(39.95 / 4.00)

x = 3.16 times as fast.

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


Set rate1 = other gas = 1
Set rate2 = H2 = 50

The molecular weight of H2 = 2.02
The molecular weight of the other gas = x.

By Graham's Law (see the answer to question #1), we have:

1 / 50 = √(2.02 / x)
x = 5050 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?


Set rate1 = H2 = x
Set rate2 = Ne = 1

The molecular weight of H2 = 2.02
The molecular weight of Ne = 20.18

By Graham's Law:

x / 1 = √(20.18 / 2.02)
x = 3.16

Since the H2 escapes 3.16 times as fast as Ne, this calculation determines the amount of Ne leaving in 6 hours:

0.67 / 3.16 = 0.211

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 diffusion is 5.93 times that of chlorine, what is the density of chlorine?


Set rate1 = H2 = 5.93
Set rate2 = Cl2 = 1

The molecular weight of H2 = 2.02
The molecular weight of Cl2 = x.

By Graham's Law:

5.93 / 1 = √(x / 2.02)
x = 71.03 g/mol

Determine gas density using the molar volume:

71.03 g / 22.414 L = 3.169 g/L

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


Set rate1 = H2 = x
Set rate2 = SO2 = 1

The molecular weight of H2 = 2.02
The molecular weight of SO2 = 64.06

By Graham's Law:

x / 1 = √(64.06 / 2.02)
x = 5.63 times as fast

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


The molecular weight of CO2 = 44.0
The molecular weight of O3 = 48.0

Do two things:

set O2 rate = 1 (since it is the heavier)
assign it to be r2 (since r2 is in the denominator)

Graham's Law:

r1 over r2 = √[(molec. wt2 over molec. wt1)]


x over 1 = √(48 over 44)
x = 1.04

CO2 diffuses 1.04 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?


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

Now, we are ready to use Graham's Law. Please note: (1) I will drop the 10¯6 from each rate and (2) we know the molar mass of argon from reference sources. Let argon be r1:

2.123 / 2.380 = √(x / 39.948)

Square both sides and solve for x:

0.7957 = x / 39.948
x = 31.786 g/mol (31.79 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


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

Inserting values into Graham's Law yields:

0/411 / 1 = √(20.18 /x )

the 20.18 is the atomic weight of Ne.

Squaring both sides gives:

0.16892 = 20.18 / x

Solving for x yields:

x = 119.46 g/mol

Examining the formulas for the possible answers, we see that answer 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


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

Set rate1 = 2
Set rate2 = 1

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

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.

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.

2) Use Graham's Law:

2/1 = √(x/1)

x = 4

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

3) 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 c.

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.


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 = 28.0
The molecular weight of CH4 = 16.0

Graham's Law is:

r1 over r2 = √MM2 over √MM1

Substituting, we have:

x / 0.530 = √(16.0 / 28.0)

x = 0.400 m/s

Probs 11-25

Seven Examples

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