Graham's Law Calculator

How it is calculated

The tool computes rate A ÷ rate B = √(MB ÷ MA), where M_A and M_B 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.

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.