About the Graham's Law Calculator
This Graham's law calculator compares how fast two gases effuse through a tiny opening (or, approximately, diffuse through each other). Pick two gases from the list or enter custom molar masses to get the rate ratio and the time ratio; enter a measured rate ratio to identify an unknown gas by its molar mass; or enter one gas's rate to predict the other's.
It is meant for general chemistry students working gas-law and kinetic-molecular-theory problems, and is a quick way to see why helium balloons deflate faster than air-filled ones, why the HCl/NH₃ smoke ring forms nearer the HCl end of a tube, or why uranium enrichment by gaseous diffusion needed thousands of stages.
Graham's law assumes both gases are at the same temperature and pressure and that molecules pass through the opening individually (true effusion). Diffusion in real mixtures follows the trend only approximately because molecules also collide with each other.
How to use the graham's law calculator
- 1Choose whether you want a rate ratio, an unknown molar mass or a second rate.
- 2Select the gases from the list, or choose Custom and type a molar mass.
- 3For an unknown gas, enter the measured rate ratio (rate of gas 1 ÷ rate of unknown).
- 4Read the result, the rate and time ratios, and the comparison chart.
Formula and method
Graham's law of effusion states that the rate at which a gas escapes through a pinhole is inversely proportional to the square root of its molar mass. It follows from kinetic molecular theory: at the same temperature all gases have the same average kinetic energy ½mv², so lighter molecules move faster, with average speed proportional to 1/√M.
The time needed for a fixed amount to effuse is inversely proportional to the rate, so the time ratio is the reciprocal of the rate ratio. Rearranging gives the molar mass of an unknown gas from a measured rate or time ratio against a known reference gas. Both gases must be at the same temperature and pressure.
- r₁, r₂
- Effusion rates of gas 1 and gas 2
- M₁, M₂
- Molar masses (g/mol)
- t₁, t₂
- Times for equal amounts of each gas to effuse
Worked examples
Helium versus nitrogen
r(He) ÷ r(N₂) = √(28.014 ÷ 4.0026) = √6.999 = 2.646. Helium escapes about 2.6 times faster, which is why a helium balloon goes flat much sooner than one filled with air.
Identify an unknown gas
Oxygen effuses 1.415 times faster than the unknown gas. M₂ = 31.998 × 1.415² = 31.998 × 2.002 = 64.07 g/mol, which matches sulfur dioxide (SO₂, 64.06 g/mol).
Rate of HCl given the rate of NH₃
r(NH₃) ÷ r(HCl) = √(36.46 ÷ 17.031) = 1.463, so if ammonia diffuses 10 cm in a given time, HCl covers about 6.83 cm. That is why the white NH₄Cl ring forms closer to the HCl end of the tube.
Uranium hexafluoride isotopes
²³⁵UF₆ (349.03 g/mol) effuses faster than ²³⁸UF₆ (352.04 g/mol) by a factor of √(352.041 ÷ 349.034) = 1.0043 — a tiny enrichment per stage.
Frequently asked questions
What is Graham's law?+
Graham's law says the rate of effusion of a gas is inversely proportional to the square root of its molar mass: r₁/r₂ = √(M₂/M₁). A gas four times heavier effuses half as fast.
What is the difference between effusion and diffusion?+
Effusion is gas escaping through a hole smaller than the mean free path of its molecules, so molecules leave one at a time. Diffusion is the mixing of gases by random motion, where collisions matter. Graham’s law is exact for effusion and only approximate for diffusion.
How do you find molar mass using Graham’s law?+
Measure the rate (or time) of the unknown gas relative to a known gas under the same conditions. Then M_unknown = M_known × (r_known ÷ r_unknown)², or equivalently M_known × (t_unknown ÷ t_known)².
Does temperature affect Graham’s law?+
The law compares two gases at the same temperature, so temperature cancels out of the ratio. Raising the temperature speeds up both gases by the same factor (proportional to √T), leaving their ratio unchanged.
Why does a helium balloon deflate faster than an air balloon?+
Helium (4.00 g/mol) is about seven times lighter than air (about 29 g/mol), so it effuses through the tiny pores in latex roughly √7 ≈ 2.7 times faster.