What is Graham's Law of Effusion?

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass: r1/r2 = sqrt(M2/M1). Formulated by Scottish chemist Thomas Graham in 1848, it follows directly from the kinetic-molecular theory of gases — lighter molecules move faster at the same temperature, so they escape through a small orifice more quickly.

What is the difference between effusion and diffusion?

Effusion is the process of gas molecules passing through a tiny hole (or porous barrier) into a vacuum or lower-pressure region. Diffusion is the spreading of gas molecules through another gas. Graham's Law strictly describes effusion, but the same square-root relationship applies approximately to diffusion under similar conditions (equal temperature and pressure, ideal gases).

How do I find the molar mass of an unknown gas with this tool?

Select "Molar mass of Gas 1 (M1)" or "Molar mass of Gas 2 (M2)" from the Solve-for dropdown. Enter the molar mass of the known reference gas, then enter both measured effusion rates (in any consistent unit). The calculator rearranges to M1 = M2 * (r2/r1)^2 or M2 = M1 * (r1/r2)^2 and displays the result. This technique was historically used to identify radioactive uranium hexafluoride isotopologues during isotope separation.

How does effusion time relate to rate?

Effusion rate and effusion time are reciprocally related: if a gas effuses at twice the rate, it takes half the time to transfer the same number of molecules. The time form of Graham's Law is t1/t2 = sqrt(M1/M2) — note the molar masses are in the opposite order compared to the rate form. The note shown below the result box reminds you of this distinction.

Does Graham's Law work for all gases?

Graham's Law assumes ideal gas behaviour — molecules are point masses with no intermolecular attractions, and the only energy is translational kinetic energy. Real gases deviate at high pressure or low temperature. For the highly accurate isotope-separation calculations used in uranium enrichment, more refined models (e.g., including the van der Waals correction) are used, but for typical chemistry problems and educational purposes Graham's Law is very accurate at near-ambient conditions.

Why is H2 so much faster than other gases?

Hydrogen (H2, M = 2.016 g/mol) is by far the lightest diatomic gas. Compared with oxygen (O2, M = 32 g/mol), the rate ratio is sqrt(32/2.016) ≈ 3.98 — hydrogen effuses about four times faster. Compared with sulfur hexafluoride (SF6, M = 146 g/mol), the ratio rises to sqrt(146/2.016) ≈ 8.51. This is why hydrogen leaks spread so quickly and why hydrogen-filled pressure vessels require special attention to seal integrity.

Graham's Law of Effusion Calculator

Graham’s Law of Effusion is one of the most practically important relationships in gas-phase chemistry. It links the effusion rate of a gas directly to its molar mass through a simple square-root formula, and it underpins everything from laboratory gas-handling decisions to industrial isotope-separation processes.

The Formula

The core relationship is:

r₁ / r₂ = √(M₂ / M₁)

where r₁ and r₂ are the effusion rates of Gas 1 and Gas 2, and M₁ and M₂ are their molar masses (in any consistent unit, typically g/mol). A lighter gas (smaller M) always has a higher effusion rate at the same temperature and pressure, because its molecules have higher average speeds. The relationship is derived from the Maxwell-Boltzmann distribution: at temperature T, the root-mean-square speed of a gas is v_rms = sqrt(3RT/M), so the speed ratio (and therefore rate ratio) is sqrt(M2/M1).

The time form is the reciprocal: t₁ / t₂ = √(M₁ / M₂). If Gas 1 is lighter, it takes less time to effuse the same quantity — note the molar masses swap positions compared with the rate form.

How the Calculator Works

Select one of five quantities to solve for, then fill in the remaining knowns:

Rate ratio (r₁ / r₂) — the dimensionless multiplier comparing how fast Gas 1 moves relative to Gas 2. No rate measurements needed, just the two molar masses.
Rate₁ or Rate₂ — absolute effusion rate in mol/s, mol/min, L/s, L/min, or molecules/s, calculated from the known rate of the other gas and both molar masses.
M₁ or M₂ — identify an unknown gas by entering the known molar mass and both measured rates; the calculator rearranges to M_unknown = M_known × (r_known / r_unknown)².

Twenty common gas presets are built in (from H₂ at 2.016 g/mol to Xe at 131.29 g/mol and SF₆ at 146.06 g/mol), so you can explore comparisons instantly without looking up molar masses. The working box shows the full substitution so you can copy it directly into a report or homework answer.

Worked Example

Problem: H₂ and an unknown gas X are allowed to effuse through the same pinhole. Under identical conditions, H₂ effuses at 3.72 mol/s while gas X effuses at 0.932 mol/s. What is the molar mass of gas X?

Solution — solve for M₂ (unknown is Gas 2):

Rearranging Graham’s Law: M₂ = M₁ × (r₁ / r₂)²

M₁ = 2.016 g/mol (H₂)
r₁ = 3.72 mol/s, r₂ = 0.932 mol/s
M₂ = 2.016 × (3.72 / 0.932)² = 2.016 × (3.99)² = 2.016 × 15.94 ≈ 32.1 g/mol

This matches O₂ (31.999 g/mol) — the unknown gas is oxygen. Select “Molar mass of Gas 2” in the dropdown, set Gas 1 to H₂ (preset), enter r₁ = 3.72 and r₂ = 0.932, and the calculator confirms M₂ = 32.09 g/mol in one step.

Real-World Context

Graham’s Law has two landmark applications. First, isotope separation: during the Manhattan Project and subsequent uranium enrichment programs, gaseous uranium hexafluoride (UF₆) was pumped through thousands of porous membranes. The two isotopologues ²³⁵UF₆ (M = 349.03) and ²³⁸UF₆ (M = 352.04) have a rate ratio of only sqrt(352.04 / 349.03) ≈ 1.0043, so thousands of separation stages were needed to reach weapons-grade enrichment. Second, leak detection and safety: knowing that H₂ effuses roughly four times faster than O₂ helps engineers predict leak rates and ventilation requirements for hydrogen storage systems and fuel cells.