# Graham's Law

### Written by tutor Sharon Y.

Graham’s Law is a law that is applied very often in chemistry and physics. This law was formulated in 1846 by Thomas Graham, a physical chemist, and applies only to gases. It states that the **rate of effusion of a gas is inversely proportional to the square root of the molar mass of its particles**, as shown in the following equation:

First, let’s define some of the words I just mentioned:

• **Gas** – not to be confused with the other three states of matter (solid, liquid, and plasma), a gas has no definite shape or volume, and its particles move in a random motion.

• **Effusion** – not to be confused with diffusion, effusion is the process by which particles flow through a very small hole or opening. This is different from diffusion, which is the process by which particles flow through a larger opening, where multiple particles may flow through simultaneously.

• **Inversely proportional** – when two variables are inversely proportional to each other, it means that as one of the variables increases, the other variable will decrease, and vice versa.

• **Molar mass** – molar mass is also known as the atomic weight of a given substance. It is most commonly seen with units of grams/mol, or g/mol.

Now, let’s get back to the law.

Imagine you are riding a bicycle, and your friend is driving their pickup truck. You arrive at your house, and you each proceed to park your vehicles inside your garage. Who will have an easier time getting into the garage?

If you guessed that the bike would have an easier time getting into the garage, you’re exactly right. Because the bike is smaller than the pickup truck, it can fit through the garage door more easily, and more quickly, than the pickup truck.

The same logic goes for Graham’s Law. The smaller a gas molecule is, and therefore the lower its molar mass, the more quickly or more easily it will effuse through a small hole. Notice how this shows the inversely proportional relationship? As the molar mass of a gas decreases, its rate of effusion increases.

Let’s use the equation in an example to compare the rate of effusion of two common gases, Hydrogen and Oxygen:

* Remember, Hydrogen and Oxygen are both diatomic gases, meaning that in their most preferred state, they are found in groups of two.

* The molar mass of Oxygen is 16 g/mol, and the molar mass of Hydrogen is 1 g/mol. However, since these gases are diatomic, what we input into the equation is double each of the molar masses.

So, what does this mean? It means that the rate at which hydrogen gas will effuse through a small hole will be 4 times as fast as the rate at which oxygen gas will effuse through that same small hole. If you think about this answer as it relates to the bike and pickup truck analogy, it makes perfect sense. The smaller and lighter gas, hydrogen, will effuse through a small opening faster than the larger and heavier gas, oxygen.

Always remember that this equation is comparing two gases that are effusing through a small hole **under the same conditions**. Since gases change their behavior as a result of changing temperature or pressure, this equation is only accurate when you are comparing two gases that are under the same conditions.

This equation can also be used to determine the molar mass of an unknown gas. Imagine that we have two gases, A and B. We know that gas A is Nitrogen, with a molar mass of 28 g/mol. However, we have an unknown gas, which we will call gas B. We also know that gas A effuses 1.069 times as fast as gas B. Can you determine what gas B is?

If we look back at our periodic table, we do not see any gases with a molar mass around 32 g/mol. However, we know that Oxygen gas is most commonly seen as a diatomic gas with a molar mass of 32 g/mol. Therefore, we have determined that gas B is actually **oxygen**.

## Graham's Law Practice Quiz

When under the same pressure and temperature, at what rate does carbon dioxide (CO_{2}) effuse from a small opening compared to the rate of effusion of methane gas (CH_{4})?

**A.**CO

_{2}effuses 0.354 times slower than CH

_{4}

**B.**CO

_{2}effuses 1.658 times faster than CH

_{4}

**C.**CO

_{2}effuses 0.603 times slower than CH

_{4}

**D.**CO

_{2}effuses 0.25 times slower than CH

_{4}

**C**.

You are comparing the rate of effusion of CO_{2} to the rate of effusion of CH_{4}. This means you are taking the square root of the molar mass of CH_{4} (16 g/mol) divided by the molar mass of CO_{2} (44 g/mol).

A sample of nitrous oxide (N_{2}O) effuses at a rate of 0.655 cm/sec. A sample of toluene effuses at a rate of 0.453 cm/sec. What is the molecular weight of toluene?

**A.**44 g/mol

**B.**92 g/mol

**C.**1.446 g/mol

**D.**64 g/mol

**B**.

Using the periodic table, you can determine the molar mass of N2O. You now have the rate of N2O, the rate of toluene, and the molar mass of N2O. This leaves you with only one unknown variable, the molar mass of toluene. Just plug in your known values into the Graham’s Law equation, and solve for your unknown.