The effusion rate calculator above applies Graham's law, which states that the rate at which a gas escapes through a small opening is inversely proportional to the square root of its molar mass. Give it two molar masses and it returns the rate ratio in both directions, the time the second gas takes given the time the first one took, and the root mean square molecular speed of each. Switch modes and it inverts the relation to give an unknown molar mass from a measured rate ratio, which is the classic laboratory use of the law.
Arb Digital publishes free calculators for the relationships that keep getting inverted the wrong way round. Graham's law is a leading example: the rate is inversely proportional to the square root of the mass, so the heavier gas is slower, and the ratio of rates equals the square root of the inverse ratio of masses. Half the errors on this topic are a reciprocal in the wrong place, and the page shows both directions at once so there is nothing to guess.
What This Effusion Rate Calculator Does
In its default mode it takes two molar masses, either from the preset list of common gases or typed in directly, and returns the ratio of their effusion rates. The supporting figures give the reciprocal ratio, so you never have to work out which way round you needed it, and the time comparison, which is how the measurement is usually made in practice: you time how long a fixed volume takes to escape, and a slower gas gives a longer time in exactly the same proportion.
The second mode is the analytical one. If you have measured that an unknown gas effuses at some fraction of the rate of a known reference, the molar mass of the unknown follows directly. That was a real method for characterising gases before mass spectrometry, and it remains a standard teaching problem because it demonstrates that a bulk kinetic measurement can reveal a molecular property.
The root mean square speeds are included because they are what the law is really about. Effusion rate is proportional to average molecular speed, and molecular speed is what depends on molar mass. Seeing that hydrogen molecules average nearly 2,000 metres per second at room temperature while sulfur hexafluoride manages about 225 makes the ratio concrete rather than abstract.
How to Use It
- Pick both gases from the preset lists, or choose Custom and type a molar mass for anything not listed.
- Read the rate ratio from the hero. A value above one means gas 1 is the faster of the two.
- Enter a measured time for gas 1 to see the time gas 2 would take under the same conditions.
- Switch to the second mode if you have a measured rate ratio and want the unknown molar mass instead.
- Set the temperature if you want the molecular speeds at a condition other than 298.15 K. The rate ratio itself does not depend on temperature.
The Formula and How It Is Calculated
Graham's law is written r₁ / r₂ = √(M₂ / M₁), where r is effusion rate and M is molar mass. Because rate and time are reciprocal for a fixed amount of gas, the same relation in time form is t₂ / t₁ = √(M₂ / M₁). Rearranged for an unknown mass, M₂ = M₁ × (r₁ / r₂)².
Working the default example: hydrogen at 2.016 g/mol against oxygen at 31.998 g/mol gives a mass ratio of 15.872, whose square root is 3.984. Hydrogen therefore effuses 3.984 times as fast as oxygen, and if a sample of hydrogen takes 10 seconds to escape, the same amount of oxygen takes 39.84 seconds. Working backwards, a measured ratio of 3.984 against hydrogen gives M₂ = 2.016 × 3.984² = 32.0 g/mol, correctly identifying oxygen.
The root mean square speed comes from kinetic theory as v = √(3RT/M), with M in kilograms per mole. At 298.15 K with R taken as the exact NIST CODATA molar gas constant of 8.314462618 J mol⁻¹ K⁻¹, hydrogen gives √(3 × 8.3145 × 298.15 / 0.002016) = 1,920 m/s. Molar masses for specific compounds can be confirmed against the NIST Chemistry WebBook.
Effusion Is Not Diffusion
The two words are used interchangeably in a great many textbooks and they should not be. Effusion is gas escaping through an opening small compared with the mean free path, so molecules pass through one at a time without colliding on the way. That is the situation Graham's law describes exactly, and it is why the rate depends on nothing but molecular speed and therefore on molar mass.
Diffusion is one gas spreading through another, and the molecules collide constantly. The molar mass dependence survives approximately, so Graham's law is often quoted for gaseous diffusion and gives roughly the right answer, but the process is genuinely more complicated and depends on the properties of both gases together rather than on the diffusing species alone. The classic ammonia and hydrogen chloride tube demonstration, where the white ring forms nearer the heavier gas, is a diffusion experiment that Graham's law predicts fairly well and not exactly.
For diffusion in a liquid the situation is different again, and molar mass is not the controlling variable at all. There the diffusion coefficient calculator applies the Stokes-Einstein relation, in which viscosity and particle radius set the answer. That is the boundary between the two pages: molar mass in a gas here, drag in a liquid there.
Why the Square Root Matters So Much
The square root flattens the effect enormously, and this has real consequences. Uranium isotope separation by gaseous diffusion works on uranium hexafluoride, where the two isotopic forms have molar masses of about 349 and 352 g/mol. The rate ratio is √(352/349) = 1.0043, so a single stage enriches the lighter isotope by less than half a percent. Achieving useful enrichment required cascades of thousands of stages, which is why those plants consumed such extraordinary quantities of electricity and why centrifuges eventually replaced them.
The same square root explains why helium leaks out of a balloon so much faster than air but not dramatically faster: the mass ratio to nitrogen is about seven, so the rate ratio is only 2.6. And it explains why the effusion method could never distinguish gases of similar mass, such as nitrogen and carbon monoxide at 28.014 and 28.010 g/mol, which are effectively identical to this measurement.
Read the other way, the square root is what makes the unknown-mass mode usable. Because the mass enters as a square, a modest error in the measured rate ratio becomes a doubled percentage error in the molar mass. A rate ratio measured to one percent gives a molar mass good to about two percent, which is enough to distinguish oxygen from argon but not enough to distinguish isotopes.
Conditions the Law Assumes
Graham's law holds under a specific set of conditions, and departures from them are the usual reason an experiment does not reproduce the predicted ratio. The opening must be small compared with the mean free path of the molecules, so that flow through it is molecular rather than hydrodynamic. If the hole is large, the gas flows as a fluid, viscosity enters, and the mass dependence weakens toward the different scaling of bulk flow.
Both gases must also be at the same temperature and pressure. Temperature drops out of the ratio only because it is the same for both; comparing a hot gas with a cold one has no simple relation at all. Pressure must be equal for the same reason. And both must behave close to ideally, which is generally true at ordinary pressures but fails near condensation.
Where the underlying gas behaviour is what you need rather than the escape rate, the ideal gas law calculator relates pressure, volume, amount and temperature, while the Boyle's law calculator, Charles's law calculator, Gay-Lussac's law calculator and combined gas law calculator each hold one or more variables fixed. Graham's law sits alongside those as the kinetic-theory member of the family: the only one of the group that depends on what the gas is made of rather than only on how much of it there is.
Getting the Molar Mass Right
Every answer on this page is only as good as the two molar masses fed into it, and the most common error is using an atomic mass where a molecular one is needed. Oxygen gas is O₂ at 31.998 g/mol, not 16.00; nitrogen gas is N₂ at 28.014. Monatomic gases such as helium, neon and argon are the exception, and mixing the two conventions gives a rate ratio wrong by a factor of √2.
For a mixture rather than a pure gas, there is no single effusion rate, because the components effuse at different rates and the composition of what escapes differs from the composition of what remains. That is the basis of the separation effect. If you need an average molar mass for a mixture, weight by mole fraction, which the mole fraction calculator handles, and get individual molar masses from the molar mass calculator.
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Browse All Free Tools Suggest a ToolCommon Mistakes to Avoid
- Inverting the ratio — the rate ratio is the square root of the inverse mass ratio, so the heavier gas is the slower one.
- Forgetting the square root — a gas sixteen times heavier effuses four times slower, not sixteen times slower.
- Using atomic mass for a diatomic gas — oxygen gas is 32 g/mol, and using 16 changes the answer by a factor of about 1.41.
- Treating time as proportional to rate — they are reciprocal, so the faster gas gives the shorter time.
- Applying the law to a large opening — once flow is hydrodynamic rather than molecular, the mass dependence no longer holds.
Related Free Tools From Arb Digital
For the liquid-phase counterpart governed by viscosity rather than molar mass, use the diffusion coefficient calculator. The ideal gas law calculator handles the general state relation, with the Boyle's law calculator, Charles's law calculator and combined gas law calculator for the constrained cases. Get molar masses from the molar mass calculator and mixture composition from the mole fraction calculator. The full free online tools hub lists everything else.
Frequently Asked Questions
It states that the rate at which a gas effuses through a small opening is inversely proportional to the square root of its molar mass. For two gases, the ratio of rates equals the square root of the inverse ratio of their molar masses.
About 3.98 times. The molar masses are 2.016 and 31.998 g/mol, so the mass ratio is 15.87 and its square root is 3.984. If hydrogen takes 10 seconds, oxygen takes 39.84 seconds.
Effusion is escape through an opening smaller than the mean free path, so molecules pass without colliding. Diffusion is one gas spreading through another with constant collisions. Graham's law is exact for effusion and approximate for diffusion.
Multiply the known molar mass by the square of the rate ratio. If an unknown effuses at one third the rate of helium at 4.003 g/mol, its molar mass is 4.003 times nine, or about 36 g/mol.
Not if both gases are at the same temperature, because the temperature term cancels. It does affect the absolute rates and the molecular speeds, both of which rise with the square root of absolute temperature.
Because the square root flattens small mass differences. Uranium hexafluoride isotopes differ by three parts in 350, giving a single-stage rate ratio of about 1.0043, so thousands of stages are needed for useful enrichment.
The square root of the average of the squared molecular speeds, equal to the square root of 3RT divided by molar mass in kg/mol. Hydrogen averages about 1,920 metres per second at 298 K, and heavier gases proportionally less.
This calculator is provided for education and general reference. It computes a published physical law from values you supply and is not laboratory, handling or safety guidance for any gas; follow the procedures and risk assessments issued by your own institution.
