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Graham's Law Calculator

Find how much faster a lighter gas effuses with Graham's law.

Result

3.9840
3.98 : 1

How it is calculated

The tool computes rate A ÷ rate B = √(MB ÷ MA), where MA and MB are the molar masses of gas A and gas B in g/mol. Only the ratio of the two masses matters, so any consistent mass unit gives the same answer, but g/mol is standard. Both masses must be greater than zero. The result is the ratio of rates for A relative to B, shown to four decimals, and also as a rounded "x : 1" display. A value above 1 means gas A effuses faster than gas B, which happens when A is the lighter gas. The law assumes ideal gases at the same temperature and pressure.

Worked example

Compare hydrogen (A, 2.016 g/mol) with oxygen (B, 31.998 g/mol). Divide the masses with B on top: 31.998 ÷ 2.016 = 15.872. Take the square root: √15.872 ≈ 3.984. The tool shows a ratio of 3.9840 and the display 3.98 : 1. Hydrogen effuses almost four times faster than oxygen.

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Common mistakes

  • Swapping A and B. The ratio is rate A over rate B, so putting the heavy gas in A gives a number below 1, the reciprocal of what you may expect.
  • Forgetting that it is a square root. A mass ratio of 16 gives a rate ratio of 4, not 16.
  • Using atomic masses for molecules, for example 16 for O2 instead of about 32, or 1 for H2 instead of about 2.
  • Applying it to gases at different temperatures. The law compares rates at the same temperature and pressure.
  • Using the rate ratio as a time ratio. Time to effuse is the inverse: the faster gas takes proportionally less time.

About this calculator

This calculator applies Graham's law of effusion: rate1 / rate2 = √(M2 / M1), where M1 and M2 are the molar masses of two gases. It shows that lighter gases effuse (or diffuse) through a small opening faster than heavier ones, with the rate ratio equal to the inverse square root of the molar mass ratio.

Chemistry students and instructors use it in classic effusion experiments (like comparing HCl and NH3 diffusion in a tube) and in problems involving isotope separation, where the same principle historically helped enrich uranium-235 from uranium-238 by exploiting their small molar mass difference as UF6 gas. Enter the two molar masses to get the rate ratio.

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