Graham's Law of Effusion & Diffusion Calculator

Free Graham's Law of Effusion & Diffusion Calculator

This Graham's Law Calculator finds the effusion or diffusion rate of one gas, or the molar mass of one gas, from the other three quantities using Rate₁/Rate₂ = √(M₂/M₁). Choose which value you need, fill in the rest, and get the rate ratio, molar mass ratio, and time ratio alongside the answer — with the full step-by-step working and a labelled effusion-race diagram.

It is built for chemistry and physics students, lab write-ups, and anyone identifying an unknown gas from how quickly it moves. Four quick-fill presets — helium escaping a balloon faster than air gets in, identifying an unknown gas leak, comparing uranium hexafluoride isotopes, and a classic hydrogen-versus-oxygen race — show realistic numbers instantly.

What Is Graham's Law?

Graham's law of effusion (and diffusion) states that the rate at which a gas effuses or diffuses is inversely proportional to the square root of its molar mass. In other words, lighter gas molecules move faster on average than heavier ones at the same temperature, so they escape through a tiny opening (effusion) or spread through another gas (diffusion) more quickly. The law is named after Scottish chemist Thomas Graham, who studied gas effusion in the 1840s.

Effusion is the escape of a gas through a tiny hole into a vacuum or lower-pressure region; diffusion is the gradual mixing of one gas into another. Graham's law describes both processes with the same underlying formula, because both depend on the same average molecular speed.

Graham's Law Formula

The Graham's law formula is Rate₁/Rate₂ = √(M₂/M₁), where Rate₁ and Rate₂ are the effusion or diffusion rates of gas 1 and gas 2, and M₁ and M₂ are their molar masses. This calculator supports all four rearranged forms — Rate₁ = Rate₂√(M₂/M₁), Rate₂ = Rate₁√(M₁/M₂), M₁ = M₂(Rate₂/Rate₁)², and M₂ = M₁(Rate₁/Rate₂)² — covering the usual textbook question of comparing two known gases, and the reverse problem of identifying an unknown gas from a measured rate.

Because effusion time is inversely proportional to rate, the same relationship gives a time ratio: t₁/t₂ = √(M₁/M₂), shown automatically alongside every result.

Solving for an Unknown Rate

The most common Graham's law question: given the molar masses of two gases and the known rate of one, find the rate of the other. For example, helium (M = 4.003 g/mol) escapes from a balloon while air (average M ≈ 28.97 g/mol) leaks in. If helium's rate is set to 1, air's rate is 1 × √(4.003/28.97) ≈ 0.371 — air moves roughly a third as fast as helium, which is exactly why a helium balloon deflates and shrivels well before it goes fully flat with air.

Solving for an Unknown Molar Mass

A classic lab identification problem: measure how fast an unknown gas effuses compared to a known reference gas, and use that ratio to find the unknown's molar mass. For example, if hydrogen (M = 2.016 g/mol) effuses at a rate of 1 and an unknown gas effuses at a rate of 0.343 under the same conditions, then M(unknown) = 2.016 × (1/0.343)² ≈ 17.1 g/mol — close to the molar mass of ammonia (NH₃, 17.03 g/mol), a common real identification result.

Real-Life Examples of Graham's Law

Graham's law shows up anywhere gases of different masses move, escape, or mix.

• A helium party balloon deflating faster than an air-filled balloon of the same size, because helium molecules are much lighter than the average air molecule.
• Identifying an unknown gas in a lab by timing its effusion rate against a known reference gas and solving for molar mass.
• Uranium enrichment by gaseous diffusion, an early industrial method that separated uranium-235 from uranium-238 hexafluoride gas based on their tiny molar mass difference.
• Natural gas leak detection, where lighter components like methane diffuse and disperse faster than heavier hydrocarbon vapors.
• Scent and odor spreading through a room, which is diffusion in action — lighter, smaller scent molecules generally reach your nose sooner than heavier ones released at the same time.

Effusion vs Diffusion

Effusion and diffusion are related but distinct processes, and Graham's law applies to both with the same formula. Effusion is the passage of gas molecules through a small opening into an empty or lower-pressure space, one molecule at a time, without collisions in the opening itself. Diffusion is the more general spreading of one gas through another due to random molecular motion, and is slowed by collisions between the gas molecules along the way.

Because both processes depend on the same average molecular speed — which is set by temperature and molar mass — Graham's law predicts the relative rate of both equally well, even though real diffusion is typically slower and more complex than idealized effusion due to those extra molecular collisions.

Why Lighter Gases Move Faster

At a given temperature, all gas molecules — light or heavy — have the same average kinetic energy. Since kinetic energy is ½mv², a molecule with smaller mass m must have a higher average speed v to carry the same energy as a heavier molecule. That higher average speed is exactly why lighter gases effuse and diffuse faster, and it is the physical reason behind the square-root relationship in Graham's law.

Common Mistakes When Using Graham's Law

A handful of errors show up repeatedly in Graham's law problems.

• Forgetting the square root — Graham's law compares the square root of the molar mass ratio, not the ratio itself.
• Mixing up which gas is 1 and which is 2 — swapping them inverts the ratio and gives the reciprocal of the correct answer.
• Confusing a rate ratio with a time ratio — rate and time are inversely related, so the faster gas has the shorter effusion time.
• Applying Graham's law when temperature or pressure differs between the two gases being compared, which the formula assumes are equal.

How to Use This Calculator

Start by choosing which of the four quantities you need to find from the dropdown menu: rate of gas 1, rate of gas 2, molar mass of gas 1, or molar mass of gas 2. Then fill in the three values you already know, or click one of the quick-fill example buttons to load a realistic scenario automatically. The result, the rate and molar mass ratios, the time ratio, the effusion-race diagram, and the full step-by-step working all update immediately as you type.