Graham's Law of Effusion
Ten Examples

Probs 1-10

Probs 11-25

Examples and Problems only

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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 = 12mv2, (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 / m 2 = 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.

Just above, I formatted Graham's Law using MathML. In a variety of places on the Internet, MathML cannot be used. Look for this:

r1 / r2 = SQRT(MM2 / MM1)
or this:
r12 / r22 = MM2 / MM1

as alternate ways to write Graham's Law.

There is also this way:

r1 / r2 = √(MM2 / MM1)

This method uses the square root symbol that is available through HTML code.

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). That means we are mostly looking at amounts that move per unit time, not how fast the individual particles are moving. For example, a rate unit might be mL/min, which is not a unit of speed. Another example would be moles per second.

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.

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) and quickly fixef the mistake and uploaded the new file.

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.

Example #6: If a gas effuses 4.25 times faster than iodine gas (I2), what is its molar mass?

A. 59.7 g/mol
B. 163 g/mol
C. 123 g/mol
D. 158 g/mol

Comment: this problem posed on Yahoo Answers has an interesting twist. The correct answer is not among the above choices. I think that the most reasonable scenario is that the person setting up the answers made a mistake in solving the problem.


1) Graham's Law:

r1 / r2 = MM2 / MM1

2) We will set the rate for I2 to be 1.00, which makes the rate for the unknown gas be 4.25. The molar mass of I2 is 253.809 and the molar mass of the unknown is x

1 / 4.25 = x / 253.809

I assigned the r2 and MM2 to the unknown gas so as to put x in the numerator.

3) Square both sides, then continue on to the value for x:

0.0553633 = x / 253.809

x = 14.0

4) I think that the most reasonable scenario is that the person who did the problem to get the answer key did not square the left-hand side of the equation. In other words, they only squared the right-hand side of the equation. So, they wound up with this:

1 / 4.25 = x / 253.809 <--- I left the 1/4.25 as a fraction rather than making it into a decimal value

And that leads to answer choice A.

Example #7: 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.


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

2) Now, we can 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.

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:

24.685 g/mol / 22.414 L/mol = 1.10 g/L

Example #8: If 0.0949 moles of NH3 effuses in 881 seconds, how many seconds would it take for the same number of moles of B2H6 to effuse?


1) Let us determine the rate at which ammonia effuses:

0.0949 mol / 881 s = 0.0001077185 mol/s

2) Write Graham's Law:

r1 / r2 = SQRT(MM2 / MM1)


r1 / r2 = MM2 / MM1

3) Substitute values into Graham's Law and solve:

r1 / 0.0001077185 mol/s = 17.0307 / 27.6694

r1 / 0.0001077185 mol/s = 0.7845423

r1 = 0.0000845097 mol/s

4) One last step:

0.0949 mol / 0.0000845097 mol/s = 1123 s

Example #9: The rate of diffusion of an unknown gas was determined to be 2.92 times greater than that of NH3. What is the approximate molar mass of the unknown gas?


1) Write Graham's Law:

r1 / r2 = SQRT(MM2 / MM1)

2) Enter appropriate values into the above equation:

1 / 2.92 = SQRT(x / 17.0307)

The sub-ones are for the ammonia and the sub-twos are the unknown gas.

3) Square both sides (remember example #6!):

0.1172828 = x / 17.0307

x = 2.00 g/mol

Although not asked for, the gas is hydrogen (H2).

Example #10: If a molecule of C2H6 diffuses a distance of 0.978 m from a point source, calculate the distance (m) that a molecule of CH4 would diffuse under the same conditions for the same period of time.


We can turn the 0.978 m into a rate by assuming some measure of time equal to 1, say 1 memtosecond. So, the C2H6 diffuses at the rate of 0.978 m / 1 memtosecond. When we do the calculation, the other rate will be our answer since it will be the distance traveled in 1 memtosecond also.

By the way, the unit of 1 memtosecond is completely fake. However, what we do know is that C2H6 does diffuse 0.978 m in some span of time and it is that span of time that we are calling 1 memtosecond. This is being done to turn a distance (0.978 m) into a rate (0.978 m / memtosecond).

r1 / r2 = MM2 / MM1

x / 0.978 = 30.07 / 16.043

x = 1.83 m

Bonus Example #1: The rate of effusion of an unknown gas at 480 K is 1.6 times the rate of effusion of SO2 gas at 300 K. Calculate the molecular weight of the unknown gas.


For a discussion about the equation that will be used to solve this problem, please go here.

1) We will use the second equation, but without the V1 and V2.

r1 / r2 = (T1M2) / (T2M1)

2) We set the rate of effusion for SO2 to be equal to 1. That means the rate of effusion for the unknown gas is 1.6. Let us use r2 for the SO2:

1.6 / 1 = (480 · 64.063) / (300 · x)

3) Square both sides:

2.56 = (480 · 64.063) / (300 · x)

4) Multiply and cross-multiply:

30750.24 = 768x

x = 40 g/mol

Bonus Example #2: Heavy water, D2O (molar mass = 20.0276 g mol¯1), can be separated from ordinary water, H2O (molar mass = 18.0152 g mol¯1), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion for H2O when compared to D2O.


1) Write Graham's Law:

r12 / r22 = MM2 / MM1

Not the usual ChemTeam writing of Graham's Law, but it still works.

2) The D2O is slower, so I'm going to assign it to r2 and give it a value of 1. That means that the lighter H2O rate will be in the numerator and be a value greater than one.

x2 / 12 = 20.0276 / 18.0152

x2 = 1.11170567

x = 1.05437454

3) H2O diffuses at a rate 1.05 times that of D2O.

Probs 1-10

Probs 11-25

Examples and Problems only

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