Graham’s law calculator

What is Graham’s law?

Gases effuse at rates inversely proportional to the square roots of their molar masses: rate A ÷ rate B = √(M_B ÷ M_A). Lighter gases escape faster (OpenStax Chemistry 2e).

Example

Hydrogen (2 g/mol) against oxygen (32 g/mol): √(32 ÷ 2) = 4 — hydrogen effuses four times as fast.

How do I compare times?

Time is the inverse of rate. OpenStax: if xenon takes 243 s, neon takes 243 × √(20.2 ÷ 131.3) = 95.3 s.

Can I type a formula?

Yes — enter a formula such as H2, CO2 or UF6 and the calculator adds up the molar mass, or type the molar mass as a number.

Is effusion the same as diffusion?

Effusion is gas escaping through a tiny hole; diffusion is gases spreading into each other. Graham’s law is stated for effusion; lighter gases also tend to diffuse faster.

How do I find the molar mass of an unknown gas?

Measure how fast it effuses relative to a gas of known molar mass at the same temperature and pressure, then square the ratio: M(unknown) = M(known) ÷ (rate ratio)². In OpenStax Example 9.22 an unknown gas effuses 1.66 times faster than CO₂ (44.0 g/mol), so its molar mass is 44.0 ÷ 1.66² = 16.0 g/mol, consistent with methane. To confirm the ratio here, enter 16 as gas 1 and CO2 as gas 2.

Why does Graham’s law matter for uranium enrichment?

Uranium hexafluoride gas made with the lighter uranium-235 effuses slightly faster than the version made with uranium-238. OpenStax notes that one pass through a porous barrier enriches the gas by only about 0.4%, so many diffusers are chained in a cascade. Gas centrifuges have largely replaced diffusion plants because they use far less energy.

What conditions does the law need?

Both gases at the same temperature and pressure (OpenStax Chemistry 2e, section 9.4).

The law and what it compares

Effusion is the escape of gas molecules through a tiny opening, such as a pinhole, into a vacuum. For two gases A and B at the same temperature and pressure, Graham’s law says the ratio of their effusion rates equals the square root of the inverse ratio of their molar masses: rate A ÷ rate B = √(M_B ÷ M_A) (OpenStax Chemistry 2e, section 9.4). The lighter gas effuses faster. The result is a ratio, so it holds for any rate unit, and the time to effuse the same amount is the inverse of the rate. The calculator reads a formula such as H2 or UF6 and adds up its molar mass from IUPAC atomic weights.

Worked example

OpenStax Example 9.20 asks for the effusion rate of sulfur dioxide when nitrogen effuses at 79 mL/s under the same conditions. With M(N₂) = 28.0 g/mol and M(SO₂) = 64.1 g/mol, the rate of SO₂ relative to N₂ is √(28.0 ÷ 64.1) = 0.661, so SO₂ effuses at 79 × 0.661 = 52 mL/s. Type SO2 as gas 1 and N2 as gas 2 to get the ratio, then multiply by the known rate. A note on exact formulas: the first answer above rounds hydrogen and oxygen to 2 and 32 g/mol to get a ratio of 4, while the calculator, using the IUPAC atomic weights of H2 and O2, returns 3.98.

Common mistakes

The ratio is easy to turn upside down: the square root has the heavier mass on top for the rate of the lighter gas. If gas 1 is the lighter one the answer is greater than 1. A second mistake is entering atoms where molecules effuse: typing H instead of H2 (1.008 instead of 2.016 g/mol) overstates hydrogen’s lead over oxygen, 5.63 instead of 3.98. Third, the law gives only a ratio at one temperature and pressure, not an absolute rate, so you need one measured rate or time to get the other. Last, effusion through a pinhole is not the same as diffusion through another gas; OpenStax notes that the two rates are not equal, though it applies the same molar-mass relation to the ratio of diffusion rates in its uranium example.

Sources: OpenStax Chemistry 2e, section 9.4 “Effusion and Diffusion of Gases” (Graham’s law, uranium enrichment, Examples 9.20 to 9.22). Related: ideal gas law calculator, partial pressure calculator, combined gas law calculator and molar mass calculator.