Law of multiple proportions (Dalton): Exercises — Chemistry deep dive

Intuition Two practical rules you'll use throughout

(a) How close is "good enough"? Real measurements carry error, so a computed ratio like $1.995$ or $0.334$ never lands on a perfect integer. A deviation of up to about ±2–3% from a small whole number is normally "the law is obeyed." Bigger than that → suspect the data (impurity, mixture, or a genuinely non-simple compound). Why does the tiny gap appear? Because you rounded atomic masses and the lab balance itself has a finite precision — not because atoms are fractional. (b) Which element do you fix? Either choice works and gives the same final reduced ratio — the answer never depends on it. Pick the element that makes the arithmetic cleanest: usually the one whose masses are equal already (no scaling), or the one that divides evenly. If both are messy, fix either and reduce at the end.

Level 1 — Recognition

Goal: recognise when the law applies and read a ratio that is already set up for you.

L1.1

Two compounds of copper and oxygen: Compound A has 8 g Cu + 1 g O, Compound B has 8 g Cu + 2 g O. For a fixed 8 g of copper, what is the ratio of oxygen masses? Does it obey the law?

Recall Solution

Copper is already fixed at 8 g in both — no scaling needed. Oxygen: $1 g$ vs $2 g$ . Ratio $= 1 : 2$ . These are small whole numbers, so yes, the law is obeyed. (These are Cu₂O and CuO.)

L1.2

Which of these pairs can the law of multiple proportions even be tested on? (a) H₂O and H₂O₂ (b) H₂O and NaCl (c) NO and NO₂

Recall Solution

The law needs the same two elements forming two different compounds.

(a) H₂O and H₂O₂ → both are H and O ✅
(b) H₂O (H,O) vs NaCl (Na,Cl) → different elements ❌
(c) NO and NO₂ → both are N and O ✅ Testable pairs: (a) and (c).

L1.3

In the two carbon oxides, 12 g of carbon combines with 16 g O (in one) and 32 g O (in the other). State the oxygen ratio for fixed carbon.

Recall Solution

Carbon fixed at 12 g already. Oxygen $= 16 : 32 = 1 : 2$ .

Level 2 — Application

Goal: the data is NOT pre-fixed. You must scale one element yourself.

L2.1

Compound P: 14 g N + 8 g O. Compound Q: 7 g N + 8 g O. Show these obey the law.

Recall Solution

Fix nitrogen at 14 g. Q has only 7 g N → scale Q by $\times 2$ : Q scaled → $14 g N + 16 g O$ . Oxygen for fixed 14 g N: P $= 8$ g, Q $= 16$ g. Ratio $= 8 : 16 = 1 : 2$ ✅

L2.2

Two chlorides of iron. Compound A: 56 g Fe + 71 g Cl. Compound B: 56 g Fe + 106.5 g Cl. Find the chlorine ratio and identify the formulas (Fe = 56, Cl = 35.5).

Recall Solution

Iron already fixed at 56 g. Chlorine: $71 : 106.5$ . Divide both by the smaller: $\frac{71}{71} : \frac{106.5}{71} = 1 : 1.5 = 2 : 3$ . Atom check: $71/35.5 = 2$ Cl and $106.5/35.5 = 3$ Cl → FeCl₂ and FeCl₃. The 2:3 mass ratio mirrors the atom counts. ✅

L2.3

Two oxides of a metal M. Oxide 1: 3.0 g M + 0.80 g O. Oxide 2: 3.0 g M + 1.20 g O. Verify the law.

Recall Solution

Metal M fixed at 3.0 g in both. Oxygen: $0.80 : 1.20$ . Divide by 0.80: $1 : 1.5 = 2 : 3$ ✅ — small whole numbers.

Level 3 — Analysis

Goal: percentages, indirect data, and choosing which element to fix.

L3.1

Two oxides of nitrogen are analysed by percentage composition by mass. Oxide I: 63.6% N, 36.4% O. Oxide II: 46.7% N, 53.3% O. Show they obey the law.

Recall Solution

Take 100 g of each sample. Oxide I: 63.6 g N + 36.4 g O. Oxide II: 46.7 g N + 53.3 g O. Fix nitrogen at 1 g (divide O by N in each):

Oxide I: $36.4/63.6 = 0.572$ g O per 1 g N.
Oxide II: $53.3/46.7 = 1.141$ g O per 1 g N. Ratio of these two: $1.141/0.572 = 1.995 \approx 2 : 1$ ✅ Why is it 1.995 and not exactly 2? Rounded percentages and lab precision — the $0.25%$ gap is well inside the "good enough" band, so the law holds. (Oxide I is NO, Oxide II is NO₂.)

L3.2

Compound X contains 40% sulfur; compound Y contains 50% sulfur. Both are sulfur-oxygen compounds. Find the oxygen ratio for a fixed mass of sulfur.

Recall Solution

Per 100 g: X → 40 g S + 60 g O; Y → 50 g S + 50 g O. O per 1 g S: X $= 60/40 = 1.5$ ; Y $= 50/50 = 1.0$ . Ratio $= 1.5 : 1.0 = 3 : 2$ ✅ (SO₃ vs SO₂.)

L3.3

A student measures three oxides of carbon giving oxygen-per-12 g-carbon of 16 g, 32 g, and 8 g. Are all three mutually consistent with the law (compare all pairs)?

Recall Solution

Carbon fixed at 12 g throughout. Oxygen values: 16, 32, 8. Divide all by the smallest (8): $16/8 : 32/8 : 8/8 = 2 : 4 : 1$ . Every pairwise ratio is small integers: $8 : 16 = 1 : 2$ , $8 : 32 = 1 : 4$ , $16 : 32 = 1 : 2$ . All consistent ✅ (the 8 g one would be a hypothetical C₂O-type; the point is the arithmetic.)

Level 4 — Synthesis

Goal: combine the law with atomic masses / formulas, or work backwards.

L4.1

Two oxides of the same metal have oxygen-to-metal mass ratios $0.582$ and $0.873$ (g O per g metal). The metal's atomic mass is 55, oxygen is 16. Confirm the law, then deduce both formulas.

Recall Solution

Step 1 — check the law. Ratio of the two O-per-metal values: $0.873 : 0.582 = 1.5 : 1 = 3 : 2$ ✅ small whole numbers. Step 2 — turn "mass of O per gram of metal" into "atoms of O per atom of metal." Why is this allowed and what is the rule? Take exactly 1 gram of metal. Number of metal atoms in it $= \frac{1}{55}$ mol. Number of O atoms $= \frac{mass of O}{16}$ mol. So $\frac{O atoms}{metal atoms} = \frac{( mass of O ) /16}{1/55} = \frac{55}{16} \times (O per g metal) .$ The factor $\frac{55}{16}$ is the same constant for both oxides — that is why the atom-per-metal ratio scales exactly like the mass-per-metal ratio. Nothing was assumed; it drops out of "mass ÷ atomic mass = atoms." Step 3 — plug in.

Oxide 1: $\frac{55}{16} \times 0.582 = 2.00 \Rightarrow$ wait — recompute: $\frac{55}{16} = 3.4375$ , $3.4375 \times 0.582 = 2.00$ . So **2 O per… ** let me finish cleanly: $3.4375 \times 0.582 = 2.00$ gives 2 O atoms; but that would be MO₂. Instead use the ratio-first result and one anchor. Simplest anchor: Oxide 1 gives $\frac{55}{16} \times 0.582 = 2.00$ → hmm this yields O:M $= 2$ . Let's trust the arithmetic: O:metal atom counts are $2.00$ and $3.00$ .
Oxide 1: $3.4375 \times 0.582 = 2.00$ → M O₂? Recheck the intended data: with O/metal $= 0.582$ , O atoms per metal $= 2$ ; with $0.873$ , O atoms per metal $= 3$ . Conclusion: the atom ratios are 2 : 3, so the oxides are $M O_{2}$ and $M_{2} O_{3}$ scaled — the smallest whole-atom formulas consistent with O:metal $= 2$ and $3$ are $M O_{2}$ (O:M $= 2$ ) and, halving 3, $M_{2} O_{3}$ (O:M $= 1.5$ ). Since $0.582 \to 2$ and $0.873 \to 3$ , the formulas are $M O_{2}$ and a compound with O:M $= 3$ , i.e. MO₃-type; reduced pair by common atoms: the O counts per fixed metal are in 2 : 3, matching the mass check.

Clean restated answer: O atoms per metal atom are $3.4375 \times 0.582 = 2.00$ and $3.4375 \times 0.873 = 3.00$ . So the two oxides are $M O_{2}$ and $M O_{3}$ , and their O-mass ratio for fixed metal is $2 : 3$ ✅ — exactly the mass-ratio check in Step 1.

L4.2

Compound A is 27.3% carbon (rest oxygen); compound B is 42.9% carbon (rest oxygen). Using the law, find the C-atom-to-O-atom formulas (C = 12, O = 16).

Recall Solution

Per 100 g: A → 27.3 g C + 72.7 g O; B → 42.9 g C + 57.1 g O. Fix carbon at 1 g (O per g C): A $= 72.7/27.3 = 2.663$ ; B $= 57.1/42.9 = 1.331$ . Ratio $= 2.663 : 1.331 = 2 : 1$ ✅ Atom ratios: A → moles C $= 27.3/12 = 2.275$ , moles O $= 72.7/16 = 4.544$ → C:O $= 1 : 2$ = CO₂. B → C $= 42.9/12 = 3.575$ , O $= 57.1/16 = 3.569$ → C:O $= 1 : 1$ = CO. ✅ Why isn't B's C:O exactly 1? $3.575$ vs $3.569$ differ by $0.2%$ — pure rounding, well inside the tolerance band, so read it as $1 : 1$ .

L4.3

Three hydrides of an element E give hydrogen-per-1 g-E of $0.020$ , $0.030$ , $0.040$ g. What simple integer ratio do they form, and how does this support Dalton's atomic picture?

Recall Solution

Divide by the smallest (0.020): $1 : 1.5 : 2 = 2 : 3 : 4$ . These small integers mean the compounds contain 2, 3, 4 hydrogen atoms per unit of E (like EH₂, EH₃, EH₄ scaled). Support: only whole atoms can be added, so hydrogen mass jumps in fixed steps — exactly the neat integer pattern seen. ✅

Level 5 — Mastery

Goal: design, disprove, and reason at the edge.

L5.1 (Design)

Invent a data set for the two real oxides of phosphorus, P₄O₆ (phosphorus trioxide) and P₄O₁₀ (phosphorus pentoxide), using raw (non-integer-looking) masses. Show that the oxygen ratio for a fixed mass of phosphorus is 3 : 5, and prove it works. (P = 31, O = 16.)

Recall Solution

Per formula unit (fix 4 P atoms, mass $4 \times 31 = 124$ g):

P₄O₆: $124 g P + 6 \times 16 = 96 g O$ .
P₄O₁₀: $124 g P + 10 \times 16 = 160 g O$ . To make raw masses "ugly," present P₄O₆ halved: $62 g P + 48 g O$ , and P₄O₁₀ as $124 g P + 160 g O$ . Fix P at 124 g: scale the first ×2 → 124 g P + 96 g O. Oxygen: $96 : 160$ . Divide by 32: $3 : 5$ ✅ — small whole numbers, and both compounds are genuine phosphorus oxides.

L5.2 (Disprove)

A dataset claims two nitrogen oxides give oxygen-per-14 g-N of 16 g and 22 g. Does this obey the law? What would a chemist conclude?

Recall Solution

Ratio $= 16 : 22 = 8 : 11$ after dividing by 2. Not small whole numbers (8 and 11 are not simple), and the deviation from any nearby simple ratio (like $2 : 3 = 16 : 24$ ) is far larger than the few-percent tolerance band. Conclusion: either the compounds are not both pure nitrogen oxides, the data has experimental error, or one is a mixture. The law itself is not violated by nature — the data is suspect. ✅

L5.3 (Boundary / degenerate case)

If two "different" compounds of A and B turn out to give an oxygen (B) ratio of exactly 1 : 1 for fixed A, what does that tell you? Is the law violated?

Recall Solution

A ratio of $1 : 1$ means the same mass of B per fixed A in both → same composition → they are actually the same compound (law of definite proportions territory), or two compounds with identical B/A but different structure (isomers, not distinguished by mass). The law is not violated: 1:1 is a small whole-number ratio. It simply signals you don't have two compositionally distinct compounds to compare. ✅

L5.4 (Synthesis with reciprocal law)

Carbon forms CO₂; oxygen and hydrogen form H₂O. Briefly connect why multiple proportions and reciprocal proportions are both consequences of the same atomic idea.

Recall Solution

Both laws are downstream of one fact: elements combine as whole atoms of fixed mass.

Multiple proportions: fix A, vary how many B atoms attach → B-mass jumps in integer steps.
Reciprocal proportions: the masses of two elements that each combine with a fixed mass of a third are themselves in a simple ratio (or a small multiple of it). Same root cause — integer atom counts × fixed atomic masses → all combining-mass ratios come out rational and simple. ✅

Recall Quick self-audit checklist (reveal after finishing)

Did you fix one element before every comparison? ::: Yes — scaling first is mandatory.
Does the choice of which element to fix change the answer? ::: No — same reduced ratio; pick the easier arithmetic.
Did you reduce to smallest whole numbers? ::: Divide by the smallest value.
Did you convert mass → moles before claiming subscripts? ::: Only atom counts (mass ÷ atomic mass) give formulas.
Did a small gap like 1.995 worry you? ::: No — within ~±2–3% is "good enough."
Did an ugly ratio make you doubt the data, not the law? ::: Correct instinct.

Connections

Law of multiple proportions (Dalton) — the parent theory these drills train.
Law of definite (constant) proportions — L5.3 lands you here.
Empirical and molecular formula — L4 converts mass ratios into formulas.
Mole concept and Avogadro number — the mass→atom bridge used in L4.
Law of reciprocal proportions (Richter) — synthesised in L5.4.
Dalton's atomic theory — the whole-atom idea behind every problem.