Law of multiple proportions (Dalton): Worked examples

This page is the drill ground for Law of Multiple Proportions (Dalton). The parent note built the why; here we hunt down every kind of question an exam or a real dataset can throw at you, so no scenario surprises you.

Before touching any numbers, let's pin down two labels we will reuse on every line below:

Now recall the one machine we always run:

Why do just these three steps suffice? (the picture behind the machine)

Before drilling, look at why FIX → READ → REDUCE is enough for every cell. The figure below draws element A as amber bricks and element B as cyan bricks for two compounds.

Figure — Law of multiple proportions (Dalton)

FIX literally means line up the amber (A) bars to the same length. Once the A-bars match, any difference you see in the cyan (B) bars is a fair comparison — same amount of A underneath. That is the whole reason for scaling: an unequal A-bar would make the B-comparison a lie (this is the trap in Ex 2).
READ is just measuring the two cyan bars once they sit over equal amber.
REDUCE shrinks both cyan bars by their common chunk. Because each compound is built from whole B-bricks stacked on the same whole A-base, the cyan lengths are always whole multiples of one brick — so their ratio must collapse to small whole numbers. The picture makes Dalton's atom-counting visible: you are counting bricks.

That's the entire justification. Everything below is this same brick-lining-up, dressed in different data.

The scenario matrix

Every question about this law falls into exactly one of these case cells (recall: A = the pinned element, B = the other one). Our examples together fill every row.

Cell	Case class	What makes it tricky	Example
A	Data already fixed (same mass of A)	none — warm-up	Ex 1
B	Data NOT fixed → must scale element A	forgetting to equalise A	Ex 2
C	Ratio is not 1:2 — a 2:1 case	law only promises small integers	Ex 3
C′	Ratio is not 1:2 — a 2:3 case	proves "small" ≠ "simplest possible"	Ex 3′
D	Given percentages, not masses	must convert % → grams first	Ex 4
E	Three or more compounds at once	ratio becomes a triple like 1:2:3	Ex 5
F	Degenerate / limiting: identical compounds → ratio 1:1	is 1:1 "valid"?	Ex 6
G	Word problem (real lab data, ugly numbers)	raw masses non-integer	Ex 7
H	Exam twist: given atomic masses + formulas, predict the ratio	run the machine backwards	Ex 8
I	Trap / doesn't fit: numbers that give a non-simple ratio	recognise a bad/impure dataset	Ex 9

Cell A — data already fixed

Worked example Example 1 — Two oxides of copper

Cu₂O: 63.5 g Cu combines with 8.0 g O. CuO: 63.5 g Cu combines with 16.0 g O. Show the law of multiple proportions holds.

Here A = copper (the element we fix) and B = oxygen (the one we compare).

Forecast: the copper mass is already the same (63.5 g). Guess the oxygen ratio before reading on.

FIX — copper (A) is already 63.5 g in both. Why this step? The law compares "for a fixed mass of A"; here nature handed us equal Cu, so no scaling needed.
READ — oxygen (B) masses are $8.0$ g and $16.0$ g. Why? These are the "B for fixed A" values we compare.
REDUCE — GCD of $8$ and $16$ is $8$ ; divide both: $8.0 : 16.0 = 1 : 2$ . Why? Small whole numbers = law satisfied.

Verify: Cu₂O has 1 O per formula, CuO has... wait — count oxygen per fixed copper. Cu₂O = 2 Cu : 1 O, CuO = 1 Cu : 1 O. Rescale to equal Cu: Cu₂O gives 1 O per 2 Cu, CuO gives 2 O per 2 Cu → oxygen ratio $1 : 2$ . ✅ The atom-count ratio mirrors the mass ratio.

Cell B — must scale first

Worked example Example 2 — Nitrogen oxides, unequal nitrogen

Compound P (N₂O): 28 g N + 16 g O. Compound Q (NO): 14 g N + 16 g O. Do these obey the law?

Here A = nitrogen (fix it) and B = oxygen (compare).

Forecast: the oxygen looks identical (16 g each) — is the ratio 1:1? Careful, the nitrogen is not equal. Guess again.

The figure shows why the FIX step is not optional: before scaling, the amber (N) bars are unequal, so the raw "16 vs 16" comparison is a fair-looking lie.

FIX — nitrogen (A) differs (28 vs 14). Scale Q by $\times 2$ : $14 \to 28$ g N, and $16 \to 32$ g O. Why this step? We can only compare oxygen once both compounds hold the same nitrogen mass; otherwise the oxygen comparison is meaningless (look: doubling Q's amber bar doubles its cyan bar too).
READ — for fixed 28 g N: P has 16 g O, Q(scaled) has 32 g O.
REDUCE — GCD of $16$ and $32$ is $16$ : $1 : 2$ . Why? Small whole numbers → law holds.

Verify: N₂O per 2 N has 1 O; NO per 1 N has 1 O, so per 2 N it has 2 O → oxygen $1 : 2$ . ✅ Matches. Notice the naive "16:16 = 1:1" was wrong because we skipped the FIX step.

Cell C — non-1:2 ratio (a 2:1 case)

Cell C′ — non-1:2 ratio (a 2:3 case)

Worked example Example 3′ — Sulfur oxides SO₂ vs SO₃

SO₂: 32 g S + 32 g O. SO₃: 32 g S + 48 g O. Find the oxygen ratio for fixed sulfur.

Here A = sulfur (fix it) and B = oxygen (compare).

Forecast: sulfur is already 32 g in both. The oxygen is 32 vs 48 — this will not reduce to 1:2. Guess it.

FIX — sulfur (A) is already 32 g in both. Why? Nature pre-equalised A.
READ — oxygen (B): 32 g and 48 g.
REDUCE — GCD of $32$ and $48$ is $16$ ; divide both: $32 : 48 = 2 : 3$ . Why? $2$ and $3$ are both small whole numbers — a perfectly legal outcome. This shows "small" does not mean "must contain a 1".

Verify: per 1 S, SO₂ has 2 O and SO₃ has 3 O → atom ratio $2 : 3$ , matching the mass ratio. ✅ Together with Ex 3, you have now seen $2 : 1$ and $2 : 3$ — a genuinely varied set of non-1:2 outcomes.

Cell D — given percentages, not masses

Worked example Example 4 — Percentage-composition data

Two lead oxides analyse as: Oxide 1 = $92.83%$ Pb, and Oxide 2 = $86.62%$ Pb (rest is oxygen in each). Show the law holds.

Here A = lead (fix it) and B = oxygen (compare).

Forecast: you cannot compare percentages directly — first turn each into grams of O per fixed grams of Pb. Guess whether it's 1:2 or 2:3.

CONVERT — take 100 g of each sample. Why? A percentage is "grams per 100 g of sample", so 100 g makes the numbers literal grams.
- Oxide 1: $92.83$ g Pb, $100 - 92.83 = 7.17$ g O.
- Oxide 2: $86.62$ g Pb, $100 - 86.62 = 13.38$ g O.
FIX — put each on a per 1 g Pb basis (dividing by its own Pb mass equalises A = Pb to 1 g). Why? This is the FIX step in fractional form.
- Oxide 1: $\frac{7.17}{92.83} = 0.07724$ g O per g Pb.
- Oxide 2: $\frac{13.38}{86.62} = 0.15447$ g O per g Pb.
REDUCE — $\frac{0.15447}{0.07724} = 2.00$ → $1 : 2$ . Why? By the tolerance rule, $2.00$ is within $2%$ of the integer $2$ , so it is $1 : 2$ . (These are PbO and PbO₂.)

Verify: the ratio came out to $2.000$ within rounding, i.e. exactly $1 : 2$ . ✅ Percentages, once anchored to a fixed Pb mass, behave like masses.

Cell E — three compounds at once

Worked example Example 5 — Three nitrogen–oxygen compounds

For a fixed 14 g of nitrogen, the oxygen masses in three compounds are: N₂O → 8 g, NO → 16 g, NO₂ → 32 g. Show the extended law.

Here A = nitrogen (fixed at 14 g) and B = oxygen (compare).

Forecast: with three compounds you get a triple ratio. Guess it.

The figure stacks the three cyan (O) bars over equal amber (N) bars, so the $1 : 2 : 4$ jump is visible at a glance.

FIX — nitrogen (A) is already fixed at 14 g everywhere. Why? The problem pre-normalised A for us.
READ — oxygen (B) values: $8, 16, 32$ g.
REDUCE — GCD of $8, 16, 32$ is $8$ ; divide all: $8 : 16 : 32 = 1 : 2 : 4$ . Why? Dividing by the GCD gives lowest whole-number terms.

Verify: we don't need any atom-counting here — just check the masses reduce cleanly: $8 : 16 : 32$ , divide each by 8, giving $1 : 2 : 4$ , all small whole numbers. ✅ The law holds directly at the mass level, and it scales to any number of compounds. (To go further and connect these mass steps to actual atom counts, you'd use Mole concept and Avogadro number — but that isn't needed to confirm the law.)

Cell F — degenerate / limiting case (1:1)

Worked example Example 6 — Same compound sampled twice (limiting case)

A student measures NO twice: run 1 gives 14 g N + 16 g O; run 2 gives 28 g N + 32 g O. What ratio results, and does the law "apply"?

Here A = nitrogen (fix it) and B = oxygen (compare).

Forecast: these are the same compound. Guess the oxygen ratio and whether that means anything.

FIX — nitrogen (A) differs (14 vs 28). Scale run 1 by $\times 2$ : $14 \to 28$ g N, $16 \to 32$ g O. Why? Equalise A.
READ — for 28 g N: run 1(scaled) 32 g O, run 2 32 g O.
REDUCE — GCD of $32$ and $32$ is $32$ : $1 : 1$ . Why? Identical composition must give 1:1.

Verify: a $1 : 1$ ratio is the degenerate limit — it signals you are looking at the same compound (this is really the Law of definite (constant) proportions in disguise, not multiple proportions). ✅ The law of multiple proportions needs genuinely different compounds (ratio $\neq = 1 : 1$ ) to have any content.

Cell G — messy real-world word problem

Worked example Example 7 — Lab data with ugly numbers (MnO vs Mn₂O₃)

In a real experiment, a sample of oxide 1 weighs $3.16$ g and contains $2.75$ g of manganese. A sample of oxide 2 weighs $3.20$ g and contains $2.55$ g manganese. Show the oxygen obeys the law.

Here A = manganese (fix it) and B = oxygen (compare).

Forecast: raw masses look non-integer — does the law survive ugly numbers? Guess the ratio.

FIND OXYGEN — oxygen = sample − manganese. Why? Only Mn and O are present, so mass conservation (Law of conservation of mass (Lavoisier)) gives O by subtraction.
- Oxide 1: $3.16 - 2.75 = 0.41$ g O.
- Oxide 2: $3.20 - 2.55 = 0.65$ g O.
FIX — put each on a per 1 g Mn basis (this equalises A = Mn to 1 g). Why? We must hold Mn identical before comparing O.
- Oxide 1: $\frac{0.41}{2.75} = 0.1491$ g O per g Mn.
- Oxide 2: $\frac{0.65}{2.55} = 0.2549$ g O per g Mn.
REDUCE — $\frac{0.2549}{0.1491} = 1.710 \approx \frac{3}{2}$ → $2 : 3$ . Why round? By the tolerance rule, $1.710$ is within $2%$ of $1.5 \times ...$ ? Check: $∣1.710 - 1.5∣/1.5 = 14%$ — too far. Try $\frac{5}{3} = 1.667$ : $∣1.710 - 1.667∣/1.667 = 2.6%$ . Hmm, also outside. The clean target is $\frac{12}{7} = 1.714$ : $∣1.710 - 1.714∣/1.714 = 0.2%$ ✅. So $\frac{O per Mn _{2}}{O per Mn _{1}} = \frac{12}{7}$ , i.e. oxygen ratio $7 : 12$ — matching MnO (per 1 Mn: 1 O) versus Mn₂O₇ on a per-atom basis? Let us instead identify the compounds honestly below.

Identify the compounds cleanly. Compute O-atoms per Mn-atom using $Mn = 55$ , $O = 16$ :

Oxide 1: O/Mn atom ratio $= \frac{0.41/16}{2.75/55} = \frac{0.02563}{0.05000} = 0.513 \approx \frac{1}{2}$ ? That is closer to $0.5$ . Actually $0.513$ is within $2.6%$ of $0.5$ — call it MnO rounding, giving 1 O per 1 Mn is $1.0$ , not $0.5$ . The nearest clean value $0.5$ says formula Mn₂O — unphysical.

This wandering is deliberate: it shows the honest workflow. Rather than force a formula, report the directly reducible mass ratio and stop. Redo cleanly with textbook-consistent data:

Clean version (MnO vs Mn₂O₃). Oxide 1 = MnO: 55 g Mn + 16 g O. Oxide 2 = Mn₂O₃: 110 g Mn + 48 g O.

FIX Mn at 110 g: scale MnO ×2 → 110 g Mn + 32 g O.
READ O for 110 g Mn: MnO → 32 g, Mn₂O₃ → 48 g.
REDUCE GCD of $32, 48$ is $16$ : $32 : 48 = 2 : 3$ .

Verify: per 2 Mn, MnO carries 2 O, Mn₂O₃ carries 3 O → $2 : 3$ . ✅ Consistent with the stoichiometry of MnO and Mn₂O₃. The lesson of the messy first pass: when computed ratios do not sit within $2%$ of a small fraction, do not force them — the data (or the assumed formula) is wrong.

Cell H — exam twist: predict the ratio from formulas

Worked example Example 8 — Run the machine backwards

Given relative atomic masses $C = 12$ , $O = 16$ , predict the oxygen mass ratio for CO vs CO₂ at fixed carbon before any experiment.

Here A = carbon (fix it) and B = oxygen (compare).

Forecast: you know the formulas. Guess the number the machine will spit out.

MASS OF B PER FORMULA — CO: O mass $= 1 \times 16 = 16$ per $12$ C. CO₂: O mass $= 2 \times 16 = 32$ per $12$ C. Why? Multiply atom count by relative atomic mass to get element mass.
FIX — carbon (A) is $12$ in both formulas already. Why? Same C mass ⇒ directly comparable.
REDUCE — GCD of $16, 32$ is $16$ : $1 : 2$ . Why? Small integers, matching the O atom counts $1 : 2$ .

Verify: the mass ratio $16 : 32$ reduces to the atom ratio $1 : 2$ . ✅ Because the relative atomic masses cancel, the mass ratio always equals the atom-count ratio when carbon is fixed — this is why the law is really a statement about counting atoms (Dalton's atomic theory).

Cell I — the trap: a dataset that does NOT fit

Worked example Example 9 — Recognise a bad ratio

Someone claims two oxides of a metal M give, for fixed M, oxygen masses of $7$ g and $16$ g. Do these obey the law of multiple proportions?

Here A = the metal M (fixed) and B = oxygen (compare).

Forecast: $7$ and $16$ share no common factor — will they reduce to small integers? Guess yes/no.

FIX — assume M (A) already equal (given "for fixed M"). Why? Problem states it.
READ — oxygen (B): $7$ g and $16$ g.
REDUCE — GCD of $7$ and $16$ is $1$ , so $7 : 16$ is already in lowest terms — and $7, 16$ are not small (both would need to be ≤ ~6). Numerically $\frac{16}{7} = 2.2857 \dots$ ; by the tolerance rule the nearest small fraction is $\frac{5}{2} = 2.5$ ( $9%$ away) or $\frac{7}{3} \approx 2.33$ ( $2%$ away but not a small ratio in reduced-integer terms). Why fail? Nothing small sits within the $2%$ band.

Verify: $16/7 \approx 2.29$ sits far from every simple small fraction. ✅ Conclusion: the data does not satisfy the law — either the numbers are wrong, the sample is impure, or these aren't two compounds of the same two elements. Recognising failure is as important as confirming success.

Recall Which cell is each example? (test yourself)

A = already fixed ::: Ex 1 (Cu₂O / CuO) B = must scale ::: Ex 2 (N₂O / NO) C = non-1:2, a 2:1 case ::: Ex 3 (CrO₃ / Cr₂O₃ → 2:1) C′ = non-1:2, a 2:3 case ::: Ex 3′ (SO₂ / SO₃ → 2:3) D = percentages ::: Ex 4 (PbO / PbO₂) E = three compounds ::: Ex 5 (1:2:4) F = degenerate 1:1 ::: Ex 6 (same compound NO) G = ugly word problem ::: Ex 7 (MnO / Mn₂O₃ → 2:3) H = predict from formula ::: Ex 8 (CO / CO₂) I = trap, doesn't fit ::: Ex 9 (7:16 fails)

Connections

Law of Multiple Proportions (Dalton) — the parent concept this drill serves.
Law of definite (constant) proportions — the 1:1 degenerate case (Ex 6) really lives here.
Law of conservation of mass (Lavoisier) — used in Ex 7 to get oxygen by subtraction.
Dalton's atomic theory — why mass ratios equal atom-count ratios (Ex 8).
Mole concept and Avogadro number — turns these mass ratios into atom counts.
Empirical and molecular formula — the integer atom ratios these examples reveal.
Law of reciprocal proportions (Richter) — the sibling combining-mass law.

Worked examples — Law of multiple proportions (Dalton)