Graham's Law

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Discovered by Thomas Graham of Scotland sometime in the 1830's (the ChemTeam is not sure exactly when).

To discuss this law, please consider samples of two different gases at the same Kelvin temperature.

Since temperature is proportional to the kinetic energy of the gas molecules, the kinetic energy (KE) of the two gas samples is also the same.

In equation form, we can write this:

KE1 = KE2

Since KE = (1/2) mv2, (m = mass and v = velocity) we can write the following equation:

m1v12 = m2v22

Note that the value of one-half cancels.

The equation above can be rearranged algebraically into the following:

√(m1 / m2) = v2 / v1

You may wish to assure yourself of the correctness of this rearrangement (as well as the one just below).

Graham's Law is often stated as follows:

r1 / r2 = √(MM2 / MM1)

where MM means the molar mass of the substance in question. Often, in these types of problems, you will be called upon to determine the molar mass of an unknown gas.

Note that I have substituted r for v. In Graham's Law, we will look at the rate of effusion (movement of gas through a small pinhole into a vacuum) more often than we will look at a speed (like a root mean square speed). For example, a rate unit might be mL/min, which is not a unit of speed.

Last point: often the rate of effusion of one gas is given relative to the rate of effusion of the other gas. That allows you to set the rate of effusion for one of the gases to a numerical value of 1. This is employed often, look for it.

Go to problems 1 - 10

Go to problems 11 - 25

Example #1: 8.278 x 10¯4 mol of an unidentified gaseous substance effuses through a tiny hole in 86.9 s Under identical conditions, 1.740 x 10¯4 mol of argon gas takes 81.3 s to effuse.

a) What is the molar mass of the unidentified substance (in g/mol)?
b) What is the molecular formula of the substance?
c) Under identical conditions, how many moles of ethene (C2H4) gas would effuse in 91.0 s?


1) Calculate the rates of effusion:

unknown ⇒ 8.278 x 10¯4 mol / 86.9 s = 9.525892 x 10¯6 mol/s
argon ⇒ 1.740 x 10¯4 / 81.3 s = 2.140221 x 10¯6 mol/s

Note that these are not speeds. A speed is an amount of distance covered in a unit amount of time. The above is a rate, a number of moles of gas effuse through a pinhole in a unit amount of time.

2) Use Graham's Law:

r1 / r2 = √(MM2 / MM1)

Assign the unknown molar mass to be MM2. I will cancel the exponent on the rates, since they are both 10¯6.

2.140221 / 9.525892 = √(MM2 / 39.948)

3) Solve:

0.05047844 = MM2 / 39.948

MM2 = 2.0165 g/mol (the answer to part a)

The gas is hydrogen, H2 (the answer to part b; no other gas weighs 2).

4) The solution to part c:

the molecular weight of ethene is 28.0536 g/mol

let us use the data from argon

let the rate for ethene be r1

r1 / r2 = √(MM2 / MM1)

x / 2.140221 x 10¯6 = √(39.948 / 28.0536 )

x / 2.140221 x 10¯6 = 1.19331

x = 2.55395 x 10¯6 mol/s

2.55395 x 10¯6 mol/s times 91.0 s = 2.324 x 10¯4 mol

Example #2: It takes 354 seconds for 1.00 mL of Xe to effuse through a small hole. Under the same conditions, how long will it take for 1.00 mL of nitrogen to effuse?


1) Write the rates for each gas:

xenon ⇒ 1.00 mL / 354 s
nitrogen ⇒ 1.00 mL / x

A common temptation is to use the times directly, as in Xe rate = 354 sec and N2 rate = x. However, please remember that rates have time in the denominator.

I'm going to leave the rates as fractions when entering them into Graham's Law. Since I knew what I was going to do when I wrote this solution, I'm going to assign nitrogen to r2.

2) Graham's Law:

r1 / r2 = √(MM2 / MM1)

[(1/354) / (1/x)] = √(28.014 / 131.293)

Watch what happens to the left-hand side.

x / 354 = 0.46192

x = 163.5 s

Note that this is a reasonable answer. Since the nitrogen is lighter than xenon, it takes less time for 1.00 mL of nitrogen to effuse. If you had used the times in the numerator of the rate (as opposed to being in the denominator, where they are supposed to be), you would have calculated an answer much larger than 354 s.

Note that the common mistake described can be made by anybody! The ChemTeam made the above mistake on one of the problems in one of the linked files and did not catch it until a student emailed with a question about the mistaken solution. The ChemTeam was thankful (and embarrassed).

Example #3: What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 4.2 mol/min?


1) Let us set the molar masses:

gas A = 1
gas B = 2

I'm going to assign gas B to r1 (and MM1, of course) since we want to know the effusion rate for the gas that is twice the molar mass of the other. (Personal note: I like putting the unknown into the numerator. It seems to fit better with my brain.)

Notice how I simply use 1 and 2 for the molar masses. We only know that one is twice the other, we do not know the actual values.

2) Use Graham's Law:

r1 / r2 = √(MM2 / MM1)

x / 4.2 = √(1 / 2)

x / 4.2 = 0.70710678

x = 2.97 mol/min

to two significant figures, the answer is 3.0 mol/min

Example #4: It takes 110. seconds for a sample of carbon dioxide to effuse through a porous plug and 275 seconds for the same volume of an unknown gas to effuse under the same conditions. What is the molar mass of the unknown gas (in g/mol)?


1) Set the rates, assuming 1.00 mole of each gas effuses:

r1 = 1.00 mol / 110. s = 0.00909091 mol/s
r2 = 1.00 mol / 275 s = 0.003636364 mol/s

Pease note that I did not use the times directly in the problem. I used them to create the rates. In step 5 below, I used the two fractions directly as opposed to what I did just above.

2) Set the molar masses:

MM1 = 44.009 g/mol
MM2 = x

3) Set up and substitute into Graham's Law:

r1 / r2 = √(MM2 / MM1)

0.00909091 / 0.003636364 = √(x / 44.009)

4) Square both sides:

6.25 = x / 44.009

x = 275 g/mol

5) Another way:

[(1/110) / (1/275)] = √(x / 44.009)

275 / 110 = √(x / 44.009)

2.5 = √(x / 44.009)

6.25 = x / 44.009

x = 275 g/mol

Example #5: What is the molar mass of a compound that takes 2.65 times as long to effuse through a porous plug as it did for the same amount of XeF2 at the same temperature and pressure?

Solution: Comment: you have to be careful reading the problem, in order to keep the gases with the correct subscript. Note how I assign a rate of 1 to the XeF2. I do this because the unknown gas takes 2.65 times longer than the XeF2 to effuse.

1) Set the rates:

r1 = 2.65
r2 = 1

2) Set the molar masses:

MM1 = x
MM2 = 169.286 g/mol

3) Set up and substitute into Graham's Law:

r1 / r2 = √(MM2 / MM1)

2.65 / 1 = √(x / 169.286)

4) Square both sides:

7.0225 = x / 169.286

x = 1188.8 g/mol

Three significant figures seems best, so 1190 g/mol.

Go to problems 1 - 10

Go to problems 11 - 25

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