Law of multiple proportions (Dalton): Question bank

The one rule that governs every item below:

Symbols and the master formula (read this once before the traps)

Several traps below lean on one small piece of algebra. Let us build it from scratch, defining every letter with a picture before using it.

Definition What each symbol means

Picture the two elements as two colours of building brick (this is the LEGO idea from the parent note). Element A is one colour, element B is the other.

$m_{A}$ = the mass of one single atom of A (a fixed constant of nature). $m_{B}$ = the mass of one atom of B.
We look at two compounds. To keep them apart, we tag each quantity with a subscript — a tiny label written below and to the right:
- subscript 1 means "belongs to Compound 1", subscript 2 means "belongs to Compound 2".
$p$ = how many A-atoms sit in one formula unit ("the count of the first-colour bricks"). So $p_{1}$ = number of A-atoms in Compound 1, and $p_{2}$ = number of A-atoms in Compound 2.
$q$ = how many B-atoms sit in one formula unit. So $q_{1}$ = B-atoms in Compound 1, $q_{2}$ = B-atoms in Compound 2.
Every one of $p_{1}, p_{2}, q_{1}, q_{2}$ is a whole number — you cannot place half a brick.

Look at the bar-model below while you read the four steps.

Figure — Law of multiple proportions (Dalton)

Formula Building the master ratio, step by step

Step 1 — total mass of each element in a compound. In Compound 1 the A-bricks weigh $p_{1} m_{A}$ in total (count × mass-each), and the B-bricks weigh $q_{1} m_{B}$ . Same for Compound 2 with subscript 2.

Step 2 — "mass of B per fixed mass of A". Why divide by the A-mass? Because the law says "for a fixed mass of A". Dividing turns "how much B sits on top of this much A" into a single number that no longer depends on how big a lump we weighed: $(\frac{mass B}{mass A})_{1} = \frac{q _{1} m _{B}}{p _{1} m _{A}}, (\frac{mass B}{mass A})_{2} = \frac{q _{2} m _{B}}{p _{2} m _{A}}$

Step 3 — compare the two compounds by dividing one by the other. Why a ratio of ratios? Because the law's verdict is about how Compound 1's B-per-A stacks against Compound 2's: $\frac{( mass B per A ) _{1}}{( mass B per A ) _{2}} = \frac{q _{1} m _{B} / ( p _{1} m _{A} )}{q _{2} m _{B} / ( p _{2} m _{A} )}$

Step 4 — the atomic masses cancel. Both top and bottom carry one factor $m_{B}$ and one factor $m_{A}$ ; dividing kills them: $\frac{( mass B per A ) _{1}}{( mass B per A ) _{2}} = \frac{q _{1} p _{2}}{q _{2} p _{1}}$ Everything left is a whole number → the result is a ratio of integers → a small whole-number ratio. That $\frac{q _{1} p _{2}}{q _{2} p _{1}}$ is why the pattern in the traps below is guaranteed, and why you never need to know $m_{A}$ or $m_{B}$ .

True or false — justify

Raw combining masses like 5.14 g and 6.86 g disprove the law because they are not whole numbers

False. The law never constrains the raw masses; only the ratio of B after fixing A must reduce to small integers. Non-integer raw masses are completely normal.

If two compounds of the same elements give an oxygen ratio of 2 : 3, that violates the law

False. The law promises small whole numbers, not specifically 1 : 2. 2 : 3 (e.g. SO₂ vs SO₃) is a perfectly valid small-integer ratio.

A single compound like water can illustrate the law of multiple proportions

False. Multiple proportions needs two or more compounds of the same two elements. One compound is the domain of the Law of definite (constant) proportions instead.

The law applies to any pair of elements that combine

False. It only bites when a pair forms more than one compound. Two elements that make a single compound simply have nothing to compare.

If the mass ratio of B comes out as 7 : 3, the law is broken

False (as stated). 7 : 3 is still a ratio of small whole numbers — it is fully reduced and both parts are small integers, which is all the law requires.

A ratio that reduces to 1 : 1 satisfies the law

True. 1 : 1 is a ratio of small whole numbers. It appears when the second element's atom count is unchanged between compounds (rare, but legal).

The law of multiple proportions is independent of the law of conservation of mass

False in spirit. It rests on Law of conservation of mass (Lavoisier) and the idea that atoms are conserved, indivisible units — that indivisibility is exactly what forces integer counts.

Spot the error

"CO has 16 g O and CO₂ has 32 g O, so directly 16 : 32 = 1 : 2 — no need to check the carbon"

The step is only valid because carbon is already fixed at 12 g in both. The hidden assumption (equal C mass) is what makes the direct comparison legal; skip it when C differs and the answer is garbage.

"Compound P: 28 g N + 16 g O; Compound Q: 14 g N + 16 g O. Both have 16 g O, so the ratio is 1 : 1"

Error: they fixed the wrong element without realising it. If you fix oxygen, you must then compare nitrogen (28 vs 14 → 2 : 1), not declare 1 : 1. You cannot fix a mass and then take the ratio of that same fixed element.

"The masses of oxygen are 16 and 32, both divisible by 16, so the ratio is 16 : 32"

Not reduced. A "small whole-number ratio" must be in lowest terms: 16 : 32 = 1 : 2. Leaving it unreduced hides whether the numbers are actually small.

"Since atoms have fixed mass, the combining masses of oxygen must themselves be whole numbers"

The masses depend on how much sample you weighed and on the atomic masses — they are usually decimals. Fixed atomic mass forces integer ratios, not integer masses.

"NO and NO₂ have N : O ratios, so I compare 14 : 16 against 14 : 32 by comparing 14 to 14"

Comparing the equal N values tells you nothing; the whole point is to fix N and compare O: 16 vs 32 → 1 : 2. Comparing the fixed element to itself is the classic empty step.

Why questions

Why must the mass of one element be equalised before forming the ratio

Because "for a fixed mass of A" is the law's definition. If A's amount differs, more B could just mean more A, not a different formula — the ratio would be meaningless.

Why does the law count as evidence for atoms rather than just a data pattern

A neat 1 : 2 in mass only makes sense if oxygen is added in indivisible whole chunks — half-integer ratios never appear because you cannot add half an atom. The discreteness of the ratios mirrors the discreteness of matter.

Why can the ratio be 2 : 3 or 3 : 5 and not only 1 : 2

The master formula gives the ratio as

\frac{q _{1} p _{2}}{q _{2} p _{1}}

, built from the atom-counts

p

(A-atoms per unit) and

q

(B-atoms per unit), subscripted 1 and 2 for the two compounds. Any two integers there give a small-integer ratio; nature isn't restricted to consecutive integers, so 2 : 3, 3 : 5, etc. all occur.

Why is Law of definite (constant) proportions not enough to prove atoms exist

Definite proportions says one recipe is constant — consistent with atoms but also with a smooth "always-the-same mixture." Multiple proportions adds the jumps between recipes, and jumps demand countable units.

Why does fixing carbon (rather than oxygen) still give the correct answer for CO vs CO₂

Either element may be fixed; the law is symmetric. Fixing O instead gives the carbon ratio, which is also a small-integer ratio. You just must be consistent about which one you hold constant.

Why doesn't the law need to know the actual mass of a single atom

In the master formula the factors

m_{A}

and

m_{B}

appear once on top and once on the bottom, so they cancel (see the worked CO/CO₂ cancellation above) — only the integer atom-counts survive. Dalton could use the law long before anyone measured

m_{A}

m_{B}

Edge cases

If two elements form three compounds, how many mass ratios must be small integers

Every pairwise ratio (compound 1 vs 2, 1 vs 3, 2 vs 3) must reduce to small whole numbers — e.g. the oxides giving O ratios like 1 : 2 : 3 for a fixed metal mass.

What if one "compound" is actually a mixture of the other two

The law is only guaranteed for genuine pure compounds with fixed formulas. A variable mixture can give non-integer ratios — a sign it is not a single compound at all.

Can the law ever give a ratio like 1.5 : 1 for a valid pair of compounds

Only as an unreduced form: 1.5 : 1 = 3 : 2, which is small integers. A truly irreducible fractional ratio would signal a data/experimental error, since atom counts are integers.

Does the law apply if the two compounds share elements but also contain a third element

Not directly — it is stated for two elements. You would isolate the two elements of interest and fix one, but a third element complicates "mass for fixed mass of A" and usually needs the Law of reciprocal proportions (Richter) framework.

What happens to the ratio if you accidentally use different sample sizes and forget to normalise A

You get a ratio contaminated by the differing A amounts — it can come out as any ugly number. Concretely, take Compound P = 28 g N + 16 g O and Compound Q = 14 g N + 16 g O: comparing O directly gives a false 16 : 16 = 1 : 1. Fix N first (scale Q ×2 → 28 g N + 32 g O), and the true ratio 16 : 32 = 1 : 2 (these are N₂O and NO) reappears.

If two compounds have the same formula (say both CO), what ratio do you get

1 : 1 for the second element — trivially a small whole number. But it is not really "multiple" proportions, since there is only one distinct compound.

Connections

Law of multiple proportions (Dalton) — the parent law these traps drill.
Law of definite (constant) proportions — the one-compound sibling confused in several traps.
Law of conservation of mass (Lavoisier) — the conservation foundation.
Dalton's atomic theory — why integer ratios prove countable atoms.
Mole concept and Avogadro number — converts these mass ratios into atom counts.
Empirical and molecular formula — integer atom ratios applied.
Law of reciprocal proportions (Richter) — needed when a third element enters.