Law of definite proportions (Proust): Worked examples

The scenario matrix

Every problem this topic can throw is one of these cells. The worked examples below are labelled with the cell(s) they hit, and together they cover all of them.

Cell	Case class	What makes it tricky
C1	Scale up a known ratio	Just multiply — the "easy" cell
C2	Scale down / find the smaller mass	Divide instead of multiply
C3	Zero / degenerate input	What if a mass is $0$ ?
C4	Verify "same compound?"	Compare two ratios exactly
C5	Reject an impostor (mixture / bad data)	Ratios disagree → not the same compound
C6	Limiting reactant + leftover	Excess does not shift composition
C7	Total-mass split	Given only the total, find each part
C8	Percentage composition link	Ratio → percent by mass
C9	Real-world word problem	Strip the story, find the ratio
C10	Exam twist (three elements)	Ratio has three numbers, same logic

Worked Examples

Worked example Example 3 — Zero / degenerate input

(Cell C3) A student claims they made "water" using $5.0$ g hydrogen and $0$ g oxygen. Is this water?

Forecast: Can a compound contain none of one of its elements?

Compute the ratio. $r = \frac{m _{O}}{m _{H}} = \frac{0}{5.0} = 0$ . Why this step? We test the claim against the law's fixed value $r = 8$ .
Compare to the required constant. $0 \neq = 8$ . Why this step? Definite proportions demands exactly $8$ ; any other value (including $0$ ) is disqualifying.
Interpret the degenerate case. A ratio of $0$ means the "oxygen bucket" on our balance is empty — there are no O atoms at all. With no oxygen, no water molecule can form. It is pure hydrogen, not a compound of H and O.

Answer: Not water — it is just hydrogen gas. The zero input breaks the fixed ratio.

Verify: For genuine water $m_{O} = 8 \times 5.0 = 40$ g would be required; $0 \neq = 40$ ✓ the claim fails.

Worked example Example 6 — Limiting reactant + leftover

(Cell C6) · figure You mix $4.0$ g hydrogen with $16.0$ g oxygen to make water ( $O : H = 8 : 1$ ). What mass of water forms, and what is left over?

Forecast: With $8 : 1$ needed, does the hydrogen or the oxygen run out first?

Find hydrogen's oxygen demand. $4.0$ g H needs $8 \times 4.0 = 32.0$ g O. Only $16.0$ g O is present. Why this step? The fixed ratio tells us the exact demand — compare demand to supply.
Oxygen is the limiter. So oxygen sets how much reacts. $16.0$ g O consumes $16.0/8 = 2.0$ g H. Why this step? We convert the available O into the H it can pair with, via the same ratio.
Water formed = the two reacted masses. $16.0 + 2.0 = 18.0$ g water (mass is conserved — see Law of Conservation of Mass).
Leftover hydrogen. $4.0 - 2.0 = 2.0$ g H sits unreacted. Why this step? Excess reactant does not change the water's composition — the water is still exactly $8 : 1$ . Look at the red "excess H" bar in the figure: it stands outside the reacted block.

Answer: $18.0$ g water + $2.0$ g hydrogen left over.

Verify: Reacted O:H $= 16.0 : 2.0 = 8 : 1$ ✓; total mass in $= 4.0 + 16.0 = 20.0$ g $=$ $18.0$ water $+ 2.0$ excess ✓.

Worked example Example 9 — Real-world word problem

(Cell C9) A jeweller melts down two old brass buttons to check they are the same alloy... no — trick! Brass is a mixture. Instead: a factory makes carbon dioxide. Batch 1 burns $6.0$ g carbon and captures $22.0$ g of $CO_{2}$ . Batch 2 must produce $44.0$ g of $CO_{2}$ . How much carbon must batch 2 burn?

Forecast: Double the product ⇒ double the carbon?

Strip the story to a ratio. In batch 1, carbon $= 6.0$ g and the product $CO_{2} = 22.0$ g, so the fixed ratio $\frac{m _{C}}{m _{C O_{2}}} = \frac{6.0}{22.0}$ . Why this step? The law fixes carbon-per-gram-of-product for this compound; that is our lever.
Scale to the new product mass. $m_{C} = \frac{6.0}{22.0} \times 44.0 = 12.0$ g. Why this step? Same compound ⇒ same ratio; multiply by the new product mass.

Answer: $12.0$ g carbon.

Verify: $12.0/44.0 = 0.2727 = 6.0/22.0$ ✓; and $44.0 = 2 \times 22.0$ , carbon also doubled ( $12.0 = 2 \times 6.0$ ) ✓, matching the forecast.

Worked example Example 10 — Exam twist: three elements

(Cell C10) · figure Sulfuric acid $H_{2} SO_{4}$ always contains H : S : O in the fixed mass ratio $2 : 32 : 64$ . A sample contains $8.0$ g of sulfur. Find the masses of hydrogen and oxygen, and the total mass.

Forecast: Three numbers now, but the logic is identical to two — anchor on the element you know.

Anchor on sulfur. The ratio part for S is $32$ , matching $8.0$ g. So one "ratio unit" $= 8.0/32 = 0.25$ g. Why this step? Every element shares the same fixed recipe; find the value of one ratio-unit once, reuse it.
Hydrogen. $2$ units $\times 0.25 = 0.50$ g.
Oxygen. $64$ units $\times 0.25 = 16.0$ g. Why this step? Multiply each element's ratio number by the shared unit — look at the three stacked bars in the figure, all scaled by the same red unit.
Total mass. $0.50 + 8.0 + 16.0 = 24.5$ g.

Answer: $0.50$ g H, $16.0$ g O, total $24.5$ g.

Verify: Ratio check $0.50 : 8.0 : 16.0 = 2 : 32 : 64$ (divide by $0.25$ ) ✓; total $= (2 + 32 + 64) \times 0.25 = 98 \times 0.25 = 24.5$ g ✓.

Active Recall

Recall Which cell does each cue belong to?

"Given carbon, find product mass" ::: Scale up/down a known ratio (C1/C2/C9). "0 g of one element — still a compound?" ::: Degenerate case C3 — no, ratio $\neq =$ constant. "Two samples, same ratio?" ::: Verification C4 — equal ratios ⇒ same compound. "Ratios differ — what law is knocking?" ::: C5 → Law of Multiple Proportions. "Excess reactant left over" ::: Limiting-reactant cell C6 — composition unchanged. "Split a total into element masses" ::: Parts method, cell C7. "Ratio to percent by mass" ::: Cell C8 — divide each part by total $\times 100$ .

Recall Numeric drills

$7.0$ g H → mass of O in water? ::: $56$ g. $27.0$ g water splits into? ::: $24.0$ g O + $3.0$ g H. Percent O in water? ::: $88.89%$ . $8.0$ g S in $H_{2} SO_{4}$ → total mass? ::: $24.5$ g.

Connections

Law of definite proportions (Proust) — the parent law these examples exercise.
Law of Multiple Proportions — where Cell C5 leads when ratios legitimately differ.
Law of Conservation of Mass — used in C6 to sum reacted masses.
Percentage Composition — Cell C8 is its direct entry point.
The Mole Concept — the natural next step once mass ratios are second nature.
Empirical and Molecular Formula — fixed ratios like C10's decode into formulas.
Dalton's Atomic Theory — the atom-counting that makes every ratio here fixed.