Dalton's atomic theory — postulates and limitations: Worked examples

Before anything else, two words must be earned.

The scenario matrix

Every problem this topic can throw at you falls into one of these cells (a "cell" is just a labelled row of the table — we tag each worked example with the cell it demonstrates, e.g. C3 means row C3). Each worked example below is tagged with the cell it covers.

Cell	Case class	What makes it tricky	Covered by
C1	Multiple proportions, ratio simplifies cleanly	the "textbook" $1 : 2$	Ex 1
C2	Multiple proportions, THREE compounds at once	must fix the same element across all three	Ex 2
C3	Ratio that looks ugly ( $\frac{7}{4}$ ) but is still whole-number	don't panic — $7 : 4$ is small integers	Ex 3
C4	Definite proportions with an odd sample size	non-integer grams, ratio must still hold	Ex 4
C5	Conservation of mass with a leftover reactant	not all mass becomes product	Ex 5
C6	Conservation with an escaping gas (open system trap)	mass "vanishes" — apparent violation	Ex 6
C7	Degenerate / zero input	one element mass $= 0$ → no compound	Ex 7
C8	Real-world word problem	pull the numbers out of prose	Ex 8
C9	Exam twist — data given as percentages	convert $%$ to grams first	Ex 9

The single picture behind every multiple-proportions example is below: fix one element's mass, then count how much of the other element rides along. We return to it explicitly in Ex 1 and Ex 2.

Figure — Dalton's atomic theory — postulates and limitations

Read the figure like this: the amber blocks are the fixed element (kept the same mass in both compounds), and the cyan blocks are the varying element we count. Compound 1 carries one cyan block, Compound 2 carries two — so the cyan totals stand in the ratio $16 : 32 = 1 : 2$ . Every example below is just this picture with different numbers.

The examples

Worked example Ex 1 — Multiple proportions, clean ratio (cell C1)

Sulphur forms two oxides. In the first, $32$ g S combines with $32$ g O. In the second, $32$ g S combines with $48$ g O. Show they obey the Law of Multiple Proportions.

Forecast: guess the simplified oxygen ratio before reading on. (It is not $1 : 2$ .)

Fix sulphur at $32$ g in both — this is the amber block in the figure above, held constant. Why this step? The law compares the other element for a fixed mass of one — so we hold S constant.
Oxygen masses are $32$ g and $48$ g → ratio $32 : 48$ . These are the cyan blocks we count. Why? These are the amounts of the varying element (O) per fixed S.
$g cd (32, 48) = 16$ , so divide both by $16$ : $32 : 48 = 2 : 3$ . Why? Simplify to lowest terms to expose the whole numbers. Result: $2 : 3$ — small whole numbers → law confirmed (this is $S O_{2}$ vs $S O_{3}$ ).

Verify: $\frac{32}{16} = 2$ , $\frac{48}{16} = 3$ , and $g cd (2, 3) = 1$ so it is fully reduced. Units: g÷g is dimensionless — correct, a ratio has no units.

Worked example Ex 2 — Three compounds at once (cell C2)

Nitrogen and oxygen form $N_{2} O$ , $NO$ , and $N O_{2}$ . For a fixed $28$ g of N, the oxygen masses found are $16$ g, $32$ g, $64$ g respectively. Verify multiple proportions across all three.

Forecast: three numbers now — what simple integer chain do they form?

Fix nitrogen at $28$ g. Why exactly $28$ ? One nitrogen atom has a relative atomic mass of about $14$ (this is the standard measured mass of a single N atom on the chemists' scale), so $28$ g is the mass of two nitrogen atoms' worth — a convenient fixed amount that lets whole molecules like $N_{2} O$ divide evenly. Why fix it at all? Same as before: hold one element constant to compare the other. (This is the fixed amber block, applied three times.)
Line up the oxygen masses: $16 : 32 : 64$ . Why? All measured against the same fixed $28$ g of N, so they are directly comparable.
$g cd (16, 32, 64) = 16$ , so divide every term by $16$ : $1 : 2 : 4$ . Why? Simplify the whole chain by the common factor. Result: $1 : 2 : 4$ — small whole numbers → law holds for three compounds simultaneously.

Verify: $16/16 = 1$ , $32/16 = 2$ , $64/16 = 4$ ; all integers, $g cd (1, 2, 4) = 1$ . Sanity check: $N_{2} O$ has $1$ O per $2$ N, $NO$ has $1$ O per $1$ N (= $2$ per $2$ N), $N O_{2}$ has $2$ O per $1$ N (= $4$ per $2$ N) — matches $1 : 2 : 4$ . ✓

Worked example Ex 4 — Definite proportions, awkward sample (cell C4)

Water is always $1 : 8$ H:O by mass. A student weighs out a $22.5$ g sample of pure water. How many grams of H and O does it contain?

Forecast: $22.5$ is not a neat multiple of $9$ … or is it?

One "unit" of water = $1$ g H $+ 8$ g O $= 9$ g total. Why? The $1 : 8$ ratio means the smallest whole recipe weighs $1 + 8 = 9$ g.
Number of units $= 22.5/9 = 2.5$ . Why? Definite proportions is scale-free, so we just scale the recipe — fractional scaling is allowed for mass (we're not counting atoms here, only grams).
Hydrogen $= 2.5 \times 1 = 2.5$ g; Oxygen $= 2.5 \times 8 = 20$ g. Why? Multiply each part of the recipe by the scale factor. Result: $2.5$ g H and $20$ g O.

Verify: $2.5 + 20 = 22.5$ g ✓ (total conserved). Ratio $2.5 : 20 = 1 : 8$ ✓ (fingerprint unchanged despite the odd sample size).

Worked example Ex 5 — Conservation with a leftover reactant (cell C5)

$10$ g of hydrogen is mixed with $40$ g of oxygen. Water forms in the fixed ratio $1$ g H : $8$ g O. What mass of water forms, and what is left over?

Forecast: which reactant runs out first — H or O?

To burn all $10$ g H you would need $10 \times 8 = 80$ g O. Why? The $1 : 8$ recipe fixes how much O each gram of H demands.
Only $40$ g O is available $< 80$ g needed → oxygen is the limiting reactant. Why? Whichever runs out first caps the reaction.
$40$ g O consumes $40/8 = 5$ g H. Why? Invert the recipe: each $8$ g O needs $1$ g H.
Water formed $= 5 g H + 40 g O = 45$ g. Leftover H $= 10 - 5 = 5$ g. Why? Only the reacted atoms enter the product; the rest sits unused. Result: $45$ g water, $5$ g H unreacted.

Verify: Total before $= 10 + 40 = 50$ g. Total after $= 45 (water) + 5 (leftover H) = 50$ g ✓. Mass is still conserved — the leftover counts too. This is the classic trap: leftover reactant is not "lost" mass.

Worked example Ex 6 — The open-system trap (cell C6)

A candle ( $20$ g of wax) burns completely. Afterward only a tiny puddle weighing $0.1$ g remains. Did $19.9$ g of mass get destroyed?

Forecast: did Dalton's postulate 3 just break?

Wax + oxygen (from air) → carbon dioxide gas + water vapour + residue. Why? Burning is a chemical reaction; atoms rearrange, they don't vanish.
The $C O_{2}$ and water escape into the air as gas. Why? An open system lets product gas leave the balance — so the measured solid mass drops.
If we sealed the candle in a closed jar and weighed everything including the air, mass in $=$ mass out. Why? Conservation is about the whole system, not just what stays on the tray. Result: No violation. The "missing" $19.9$ g left as invisible gases. Postulate 3 holds.

Verify: conceptual — closed-system mass $(wax + O_{2} consumed) = (CO_{2} + H_{2} O + residue)$ . No numeric check needed; the lesson is always close the system before applying conservation.

Worked example Ex 7 — Degenerate / zero input (cell C7)

An analyst reports a "compound" of carbon and oxygen containing $12$ g C and $0$ g O. Is this a compound obeying the mass laws?

Forecast: what happens to a ratio when one term is zero?

Ratio C : O $= 12 : 0$ . Why? We attempt the usual comparison.
Division by zero is undefined — you cannot express "how many times the oxygen fits into the carbon" when there is no oxygen. Why? A ratio needs two non-zero piles to compare proportion. Result: With $0$ g O there is no compound of C and O at all — it is just pure carbon (an element). The mass laws describe combinations; a zero-mass partner is the degenerate case where no combination happened.

Verify: $\frac{12}{0}$ is undefined; and pure carbon has definite proportions trivially ( $100%$ C). No violation — the law simply doesn't apply to a non-compound.

Worked example Ex 8 — Real-world word problem (cell C8)

A factory finds that $2.8$ kg of iron always reacts with $1.6$ kg of sulphur to make iron sulphide, with nothing left over. A batch uses $7.0$ kg of iron. How much sulphur is needed, and how much product forms?

Forecast: the ratio is hidden in prose — extract it first.

Read off the fixed ratio: Fe : S $= 2.8 : 1.6$ . Multiply both by $10$ to clear decimals ( $28 : 16$ ), then divide by $g cd (28, 16) = 4$ : $7 : 4$ . Why? Definite proportions guarantees this recipe is constant.
Scale factor for $7.0$ kg Fe: $7.0/2.8 = 2.5$ . Why? We are enlarging the same recipe by this factor.
Sulphur needed $= 1.6 \times 2.5 = 4.0$ kg. Why? Scale the S part by the same factor to keep the ratio fixed.
Product $= 7.0 + 4.0 = 11.0$ kg. Why? Conservation of mass: nothing left over means all reactant mass becomes product. Result: $4.0$ kg sulphur, $11.0$ kg iron sulphide.

Verify: ratio $7.0 : 4.0 = 1.75 = 2.8 : 1.6$ ✓. Mass: $7.0 + 4.0 = 11.0$ ✓.

Worked example Ex 9 — Exam twist: data as percentages (cell C9)

Compound X is $42.9%$ carbon and $57.1%$ oxygen by mass. Compound Y is $27.3%$ carbon and $72.7%$ oxygen. Do X and Y obey multiple proportions? (This is the classic CO vs CO₂ data — but given as percentages, the exam trap.)

Forecast: percentages can't be compared directly — what must you fix first?

Take $100$ g of each so percentages become grams. X: $42.9$ g C, $57.1$ g O. Y: $27.3$ g C, $72.7$ g O. Why? Percent = grams per $100$ g; converting removes the trap.
Fix carbon at $1$ g in each. In X: O per g C $= 57.1/42.9 = 1.331$ . In Y: O per g C $= 72.7/27.3 = 2.663$ . Why? Multiple proportions needs a fixed mass of the common element.
Ratio of oxygen-per-carbon: $2.663 : 1.331$ . Divide both by $1.331$ : $2.663/1.331 = 2.001 \approx 2$ , giving $2 : 1$ . Why? Simplify to hunt for small integers. Result: $2 : 1$ — small whole numbers → law obeyed (Y is $C O_{2}$ , X is $CO$ ).

How close is "close enough"? The measured value $2.001$ differs from the ideal integer $2$ by about $0.05%$ . Experimental mass percentages are typically only good to about $\pm 0.5%$ , so a deviation five to ten times smaller than the measurement error is well within rounding — we are entitled to read $2.001$ as the exact integer $2$ . If the ratio had come out as, say, $2.4$ , that would be too far from any small integer to excuse, and we would suspect impure samples or a wrong formula.

Verify: $57.1/42.9 \approx 1.3310$ ; $72.7/27.3 \approx 2.6630$ ; ratio $2.6630/1.3310 \approx 2.001$ , within $0.5%$ of the integer $2$ ✓.

Recall Recall checklist

(In the answers below, "C1"–"C9" refer to the labelled rows of the scenario matrix table above.) Which matrix row (cell) needs you to convert $%$ to grams first? ::: Row C9 (Ex 9) — take a $100$ g basis In Ex 5, why is oxygen the limiting reactant? ::: Only $40$ g O available but $80$ g needed to burn all H, so O runs out first What is the simplified oxygen ratio in Ex 2's three nitrogen oxides? ::: $1 : 2 : 4$ Why is $12 : 0$ not a valid compound ratio? ::: Division by zero is undefined; with no oxygen there is no compound, just pure carbon In Ex 4, how many "9 g water units" are in a 22.5 g sample? ::: $2.5$ units What does $g cd (a, b)$ mean? ::: The greatest common divisor — the largest whole number that divides both $a$ and $b$ exactly

Connections

Parent · Dalton's atomic theory
Laws of Chemical Combination
Mole Concept and Stoichiometry
Isotopes and Isobars
Modern Atomic Theory