Graham's Law Calculator
Compare gas effusion or diffusion rates from formulas or molar masses, then solve optional rate or time values.
What can you enter?
Use gas formulas or molar masses. The calculator can also use one known rate or one known time to solve the matching value for the other gas.
| Input type | Example | What it means |
|---|---|---|
| Gas formula | H2, O2, CO2, NH3 | The tool calculates molar mass from the formula. |
| Molar mass | 2.016 or 31.998 | Use direct g/mol values from a data table. |
| Known rate | rate1 = 20 mL/s | Solves the other gas rate using the rate ratio. |
| Known time | time2 = 30 s | Uses inverse rate relationship for equal gas amounts. |
Formula used
Worked example: hydrogen and oxygen
Input: gas 1 = H2 and gas 2 = O2.
- Molar mass of H2 is about 2.016 g/mol.
- Molar mass of O2 is about 31.998 g/mol.
- rate(H2) / rate(O2) = sqrt(31.998 / 2.016).
- The ratio is about 3.98, so hydrogen effuses about 4 times faster than oxygen.
- For the same amount of gas, hydrogen takes about one-fourth the time oxygen takes.
Built-in example data
| Example | Gas 1 | Gas 2 | Expected idea |
|---|---|---|---|
| Hydrogen vs oxygen | H2 | O2 | H2 is about 4 times faster. |
| Helium vs nitrogen | He | N2 | He is much faster; optional rate example included. |
| Ammonia vs carbon dioxide | NH3 | CO2 | NH3 is faster because it is lighter. |
| Sulfur dioxide vs methane | SO2 | CH4 | CH4 is faster; the preset solves a time comparison. |
| Uranium hexafluoride vs helium | UF6 | He | Heavy UF6 effuses far more slowly. |
Why this calculator is useful
Graham's law problems can be confusing because the lighter gas has the faster rate, but the square root uses the molar mass of the opposite gas in the numerator. This page keeps the molar masses, rate ratio, inverse time ratio, and conclusion visible together.
| Where it helps | Why it matters |
|---|---|
| Homework checks | Shows the exact substitution into the square-root formula. |
| Gas identification labs | Compares an unknown gas rate against a reference gas. |
| Diffusion demonstrations | Explains why lighter gases spread faster. |
| Exam revision | Separates rate ratios from time ratios. |
| Formula practice | Accepts chemical formulas and calculates molar masses automatically. |
Effusion vs diffusion
Effusion is gas escaping through a tiny opening into a lower-pressure space. Diffusion is gas spreading through another gas or through a space. Graham's law is most exact for effusion, but many classroom diffusion comparisons use the same inverse-square-root relationship as an approximation.
The key idea is kinetic: at the same temperature, gases have the same average kinetic energy. A lighter gas therefore has a higher average speed than a heavier gas.
Common mistakes to avoid
- Putting the molar masses in the wrong order for the requested rate ratio.
- Forgetting the square root and using M2 / M1 directly.
- Confusing rate ratio with time ratio. Time is inverse to rate for equal amounts.
- Using atomic mass for O instead of molecular molar mass for O2.
- Comparing gases under different temperature or pressure conditions without noting the assumption.
Rounding and result checking
Keep molar masses to at least three or four significant figures when the data are available, then round the final ratio to match your course rules. The lighter gas should always have the larger effusion or diffusion rate. If your result says the heavier gas is faster, recheck the order of M1 and M2.
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Graham's Law Calculator FAQs
What is Graham's Law of Effusion?
Given by Thomas Graham in 1848, this law states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass, under identical conditions of temperature and pressure. Lighter gases effuse faster because their molecules have higher average velocity (kinetic energy is the same for all gases at a given T, but lighter molecules must move faster to share the same energy). This law explained why hydrogen escapes from balloons faster than oxygen and led directly to isotope separation techniques. r1 / r2 = √(M2 / M1)
How to use Graham's Law of Effusion?
Apply the formula r1/r2 = √(M2/M1) where r is the rate of effusion and M is molar mass. The lighter gas always has the higher rate. Example: compare H2 (M = 2) and O2 (M = 32). rH2/rO2 = √(32/2) = √16 = 4. So hydrogen effuses 4 times faster than oxygen. To find an unknown molar mass, measure relative rates against a known gas and rearrange the formula.
How to solve Graham's Law of Effusion?
Step-by-step: (1) identify the two gases and their molar masses; (2) decide which is r1 and which is r2; (3) substitute into r1/r2 = √(M2/M1); (4) solve for the unknown. Take care with the inversion — it is the lighter gas (smaller M) that has the larger rate. If time of effusion (t) is given instead of rate, remember rate ∠1/time, so t1/t2 = √(M1/M2) (note: reversed!).
What is the equation for Graham's Law of Effusion?
The fundamental equation is: r1/r2 = √(M2/M1) = √(d2/d1), where r is the rate of effusion, M is the molar mass, and d is the gas density (since at fixed T and P, d ∠M). Equivalently, since rate is inversely proportional to time, t1/t2 = √(M1/M2). All forms come from the kinetic theory result vrms = √(3RT/M).
How do you calculate the molar mass of an unknown gas?
Compare the rate (or effusion time) of the unknown gas with that of a known gas under the same conditions. Use Munknown = Mknown × (rknown/runknown)2, or equivalently Munknown = Mknown × (tunknown/tknown)2. Example: an unknown gas effuses in 60 s while O2 takes 30 s in the same apparatus. Munknown = 32 × (60/30)2 = 32 × 4 = 128 g/mol.
How do you calculate effusion time instead of rate?
Since rate is inversely proportional to time of effusion (for equal volumes), t1/t2 = √(M1/M2) — note carefully this formula has the molar masses the same way round, opposite to the rate formula. Heavier gases take longer to effuse. Example: if H2 takes 10 s, the time for O2 = 10 × √(32/2) = 10 × 4 = 40 s. The relation is symmetric so you can work either way.
What is the difference between diffusion and effusion?
Both refer to the spreading of gas molecules, but they happen in different physical settings. Diffusion is the gradual mixing of gases that are already in the same container — molecules of one gas move randomly through another (e.g., perfume spreading in a room). Effusion is the escape of a gas through a tiny pinhole or small orifice into a vacuum or low-pressure region (e.g., helium leaking out of a balloon). Graham's law applies to both with the same √M dependence, although the actual rates differ.
Which gas effuses faster, Helium or Oxygen?
Helium effuses much faster. Helium's molar mass is 4 g/mol while oxygen's is 32 g/mol. By Graham's law, rHe/rO2 = √(32/4) = √8 ≈ 2.83. So helium effuses about 2.83 times faster than oxygen. That is why helium-filled balloons deflate noticeably within a day or two — the small, light helium atoms slip through the pores in rubber faster than the heavier oxygen and nitrogen molecules outside.
How does temperature affect the rate of effusion?
From kinetic theory, vrms = √(3RT/M), so the rate of effusion is proportional to √T. Higher temperature → higher rate of effusion. Doubling the absolute temperature increases the rate by a factor of √2 ≈ 1.41. However, when comparing two different gases at the same temperature, the temperature factor cancels out, leaving only the √M dependence. Temperature is therefore essential to specify in effusion experiments.