Using Graham's Law

Graham's Law:

allows us to calculate molar masses compared to the rate at which two gases effuse or diffuse.


Here is a sample problem:
Hydrogen gas diffuses 4.67 times as fast as another gas. What is the molar mass of the second gas?

To solve this, we'll make the unknown gas "a" and hydrogen gas "b". We'll also need to remember that the molar mass of hydrogen is 2.02 g/mol (since hydrogen comes as H₂, not H).

For this problem then, Graham's Law looks like this:

Rearranging this equation to solve for the MM of "a" give us...

Substituting in, give us:

Squaring both sides give us: MM_a = 44.09 g/mol.

However, with a small change, we can very different problems.

From physics, we know that rate = distance/time. If we substitute that into the equation, we get:

This format allows us to solve problems like:

If it takes helium 38.3 seconds to cross a 100.0 meter room, how long will it take oxygen to cross a 400.0 meter room?

For this problem, we'll let oxygen be gas "'a" and helium be gas "b". We also need to remember that oxygen gas is O₂, not O.

First, we'll substitute in our gases:

Algebra ends up being the most problematic part of Graham's Law problems, so let's deal with that first. Dividing by fractions is ugly, so let's remember that dividing by a fraction is the same as multiplying by its reciprical.

That gives:

Next, we'll solve for time O₂:

Then, we can substitute our numbers and solve: