Graham's law of effusion - diffusion (rate ∝ 1 - √M): Worked examples

Everything here rests on the one formula the parent built from scratch: $\frac{r _{1}}{r _{2}} = \frac{M _{2}}{M _{1}}$ where $r$ = rate (amount of gas moving per unit time) and $M$ = molar mass (grams per mole). The swap — mass 2 sits above mass 1 — comes from rate being inverse to $M$ . If any of that feels shaky, reread the parent and Root Mean Square Speed first.

The scenario matrix

Every Graham's-law problem is one of these shapes. The trick is always: turn whatever you're given into a rate ratio, then apply the formula.

#	Case class	What's given → what's asked	Hit by
A	Direct rate compare	two masses → speed/rate ratio	Ex 1
B	Time inversion	rate is $V / t$ , same $V$ → time from rate	Ex 2
C	Unknown mass	a rate ratio → solve for $M$	Ex 3
D	Distance / diffusion tube	tube length → where gases meet	Ex 4 (figure)
E	Density form	densities instead of masses	Ex 5
F	Degenerate: equal masses	$M_{1} = M_{2}$ → ratio must be exactly 1	Ex 6
G	Limiting / tiny mass difference	isotopes, ratio ≈ 1 → real separation factor	Ex 7
H	Mixture / compound	find $M$ of a molecule, then identify it	Ex 8
I	Exam twist: pressure-drop timing	pressure falls as gas leaks → ratio of $Δ t$	Ex 9

We cover cells A–I below. Each example says which cell it belongs to.

Worked Examples

Worked example Example 1 — Cell A: direct rate compare (He vs CH₄)

How many times faster does helium ( $M = 4$ ) effuse than methane ( $M = 16$ )?

Forecast: He is 4× lighter. Do you expect it 4× faster, or less? (Remember the square root!)

Write the ratio with the other gas's mass on top: $\frac{r _{H e}}{r _{C H_{4}}} = \frac{M _{C H_{4}}}{M _{H e}} = \frac{16}{4}$ Why this step? Rate $\propto 1/ M$ , so the lighter gas (He) must come out faster — its ratio must be $> 1$ . Putting the heavier mass (16) on top guarantees that.
Simplify: $\frac{16}{4} = 4 = 2$ Why this step? The square root turns "4× lighter" into "2× faster" — this is why the answer is 2, not 4.

Verify: He is lighter, so it should be faster ⇒ ratio $> 1$ . ✓ And 4× lighter → $4 = 2$ × faster, exactly as the parent's H₂/O₂ example (16× lighter → 4×). Pattern holds.

Worked example Example 2 — Cell B: time inversion (Ar vs Ne)

Equal volumes of argon ( $M = 40$ ) and neon ( $M = 20$ ) effuse through the same hole. Neon finishes in 30 s. How long does argon take?

Forecast: Argon is heavier and slower — will its time be more or less than 30 s?

Rate ratio (lighter Ne is faster): $\frac{r _{N e}}{r _{A r}} = \frac{40}{20} = 2 \approx 1.414$ Why this step? We first find how the speeds compare before touching time.
Same volume $V$ means rate $= V / t$ , so for equal $V$ : $r \propto 1/ t$ , i.e. $t \propto 1/ r$ . Thus: $\frac{t _{A r}}{t _{N e}} = \frac{r _{N e}}{r _{A r}} = 2$ Why this step? Time is the inverse of rate at fixed volume. The slower gas takes proportionally longer — so we flip the rate ratio to get the time ratio.
Solve: $t_{A r} = 2 \times 30 s \approx 42.4 s$

Verify: Argon is heavier ⇒ slower ⇒ longer time ⇒ $> 30$ s. ✓ Units: seconds. ✓

Worked example Example 4 — Cell D: diffusion tube geometry (H₂S vs SO₂)

A 100 cm tube: hydrogen sulfide $H_{2} S$ ( $M = 34$ ) enters the left end, sulfur dioxide $S O_{2}$ ( $M = 64$ ) enters the right end at the same moment. A yellow sulfur deposit forms where they meet. How far from the $H_{2} S$ end?

Forecast: $H_{2} S$ is lighter, so it travels farther. Will the meeting point be past the middle (>50 cm) or before it?

In the same time $t$ , distances are in the ratio of rates: $\frac{d _{H_{2} S}}{d _{S O_{2}}} = \frac{M _{S O_{2}}}{M _{H_{2} S}} = \frac{64}{34} = 1.882 \approx 1.372$ Why this step? Both gases diffuse for the same time before meeting, so distance $\propto$ speed $\propto$ rate. Look at the figure: the meeting point (green) is where the two coloured paths touch.
The two distances add to the tube length: $d_{H_{2} S} + d_{S O_{2}} = 100$ . Let $d_{S O_{2}} = x$ , so $d_{H_{2} S} = 1.372 x$ : $1.372 x + x = 100 ⟹ 2.372 x = 100 ⟹ x \approx 42.2 cm$ Why this step? The gases together cover the whole 100 cm; the ratio tells us how to split it.
Distance from the $H_{2} S$ end: $d_{H_{2} S} = 100 - 42.2 \approx 57.8 cm$

Verify: $57.8 > 50$ , so the lighter $H_{2} S$ went past the middle. ✓ The deposit sits closer to the heavier $S O_{2}$ end — exactly as the parent's NH₃/HCl example predicts.

Worked example Example 7 — Cell G: limiting case, tiny mass difference (uranium isotopes)

In Isotope Separation (Uranium Hexafluoride), $^{235} U F_{6}$ ( $M = 349$ ) is separated from $^{238} U F_{6}$ ( $M = 352$ ). What is the ratio of effusion rates (the separation factor per stage)?

Forecast: The masses barely differ. Do you expect a big ratio, or something agonisingly close to 1?

Lighter isotope on... which gas is faster? The $^{235}$ one ( $M = 349$ ) is lighter, so: $\frac{r _{235}}{r _{238}} = \frac{352}{349} = 1.00859 \approx 1.0043$ Why this step? Even a 3-unit mass gap out of ~350 is tiny, so the ratio sits just above 1. This is the limiting behaviour of Graham's law: as $M_{1} \to M_{2}$ , the ratio $\to 1$ .
Interpret: each effusion stage enriches the lighter isotope by only $\approx 0.43%$ . Why this step? Because the factor is so close to 1, thousands of stages (a cascade) are needed — this is why isotope enrichment is so hard.

Verify: Ratio $> 1$ (lighter is faster) but only just. ✓ Contrast with Ex 1 (ratio 2): here mass difference is minute, so ratio hugs 1 — the correct limiting picture.

Worked example Example 8 — Cell H: identify a compound from its effusion rate

An unknown gaseous hydrocarbon effuses 0.707× as fast as methane, $C H_{4}$ ( $M = 16$ ). Which hydrocarbon is it?

Forecast: Slower than methane ⇒ heavier. $0.707 \approx 1/ 2$ — does that hint at a factor-of-2 mass?

Set up (unknown is slower, so its ratio is $< 1$ ): $\frac{r _{u nk}}{r _{C H_{4}}} = 0.707 = \frac{16}{M _{u nk}}$ Why this step? The known lighter gas $C H_{4}$ 's mass (16) on top gives a ratio $< 1$ for a heavier unknown — matching $0.707$ .
Square: $0.70 7^{2} = 0.500 = \frac{16}{M _{u nk}} ⟹ M_{u nk} = \frac{16}{0.5} = 32 g/mol$ Why this step? Squaring removes the root; $0.70 7^{2}$ is exactly $0.5$ , so the mass is exactly double methane's.
Identify: a hydrocarbon with $M = 32$ ? Carbon+hydrogen only: $C_{2} H_{?}$ gives $24 + 8 = 32$ ⇒ $C_{2} H_{8}$ is impossible (too many H). Actually $M = 32$ matches nothing sensible as a pure hydrocarbon — the closest real gas is... methanol $C H_{3} O H$ or hydrazine? This flags a trick: $M = 32$ is oxygen or methanol, not a hydrocarbon. Why this step? Always sanity-check the answer against chemistry. The maths gives 32; the "hydrocarbon" claim is a distractor. Correct reading: the gas has $M = 32$ g/mol.

Verify: $32 = 2 \times 16$ , and slower by $2$ means twice as heavy. ✓ Units g/mol. ✓

Worked example Example 9 — Cell I: exam twist, pressure-drop timing (O₂ then unknown)

A fixed container has a pinhole. It takes 40 s for the pressure of $O_{2}$ ( $M = 32$ ) to drop by a fixed amount. Refilled with an unknown gas to the same conditions, the same pressure drop takes 50 s. Find $M_{u nk}$ .

Forecast: The unknown takes longer ⇒ slower ⇒ heavier. Expect $M > 32$ .

Equal pressure drop = equal amount of gas leaving. Rate $=$ amount $/$ time, so for equal amounts, $r \propto 1/ t$ : $\frac{r _{O_{2}}}{r _{u nk}} = \frac{t _{u nk}}{t _{O_{2}}} = \frac{50}{40} = 1.25$ Why this step? We convert times into a rate ratio. The gas that takes longer has the smaller rate, so it goes in the denominator — hence $t$ ratio flips.
Now apply Graham's law: $\frac{r _{O_{2}}}{r _{u nk}} = \frac{M _{u nk}}{M _{O_{2}}} ⟹ 1.25 = \frac{M _{u nk}}{32}$ Why this step? $O_{2}$ is faster, so the unknown's (larger) mass sits on top, giving a ratio $> 1$ .
Square and solve: $1.2 5^{2} = 1.5625 = \frac{M _{u nk}}{32} ⟹ M_{u nk} = 32 \times 1.5625 = 50 g/mol$

Verify: $50 > 32$ , so heavier ⇒ slower ⇒ longer time (50 s). ✓ All three consistency arrows point the same way.

Recall

Recall Which quantity goes on top?

When you turn time into a rate ratio, which gas's time goes in the numerator? ::: The gas whose rate you want in the numerator — but time flips, so put the other gas's time on top: $r_{1} / r_{2} = t_{2} / t_{1}$ (equal amounts).

Recall The two-flip trap

A problem gives you times and asks for masses. How many "flips" happen? ::: Two — once from time→rate (flip), once inside the square root when solving for mass. Miss either and your answer is upside-down.

Recall Degenerate check

If two gases have equal molar mass, what is their rate ratio? ::: Exactly 1 — Graham's law only sees mass, so it cannot separate them.

Connections

Root Mean Square Speed — the $v_{r m s} = 3 R T / M$ that every ratio here rests on.
Kinetic Theory of Gases — why equal $T$ means equal average KE.
Maxwell-Boltzmann Distribution — the real spread of speeds behind the average.
Ideal Gas Equation — supplies $ρ \propto M$ used in Cell E.
Isotope Separation (Uranium Hexafluoride) — the real-world Cell G.