Law of conservation of mass (Lavoisier) — proof, examples: Worked examples

Before any symbol appears, one plain-language anchor:

The scenario matrix

Every problem on this topic is one of these cells. We will hit each one.

Cell	What varies	The trap / skill	Example
A. Closed, find product	all reactants known	just add	Ex 1
B. Closed, find missing reactant	one reactant unknown	subtract	Ex 2
C. Open, gas leaves	product mass looks too small	add back the escaped gas	Ex 3
D. Open, gas enters	product mass looks too big	subtract the absorbed gas	Ex 4
E. Air composition twist	only part of the air reacts	use only the reacting fraction	Ex 5
F. Degenerate: zero reaction	limiting case	mass trivially unchanged	Ex 6
G. Word problem (sealed jar)	forecast then verify	"invisible" mass counts	Ex 7
H. Nuclear twist	law appears to fail	$E = m c^{2}$ mass defect	Ex 8

Two axes run through this table, and you should always ask them first:

Cell A — Closed system, find the product

Worked example Example 1 (Cell A) — Add the reactants

$4.0 g$ of hydrogen gas reacts completely in a sealed bomb with $32.0 g$ of oxygen gas to form water. What mass of water is produced?

Forecast: Before reading on, guess — will the water be more, less, or equal to $36.0 g$ ? (Trust the LEGO picture.)

Steps

Check the box: "sealed bomb" ⇒ closed, nothing escapes. Why this step? Only in a closed box may we write $M_{R} = M_{P}$ directly.
Reactant pile: $M_{R} = m_{H_{2}} + m_{O_{2}} = 4.0 + 32.0 = 36.0 g$ . Why this step? Every reactant atom must end up somewhere in the product; summing their masses counts all those atoms.
Product pile: the only product is water, so $M_{P} = M_{R} = 36.0 g$ . Why this step? Conservation forces $M_{P} = M_{R}$ ; there is nothing else for the mass to be.

Verify: Reactant total $= 36.0 g$ ; product total $= 36.0 g$ . Equal ✓. Units: grams throughout ✓. Forecast was "equal to $36.0$ " ✓.

Cell B — Closed system, find a missing reactant

Worked example Example 2 (Cell B) — Subtract to find what was consumed

In a sealed chamber, $24.0 g$ of carbon burns to give exactly $88.0 g$ of carbon dioxide. What mass of oxygen was consumed?

Forecast: The product is heavier than the carbon alone. Where did the extra weight come from — magic, or oxygen?

Steps

Closed box ⇒ $M_{R} = M_{P}$ . Reactants are carbon and oxygen; product is $CO_{2}$ only. $m_{C} + m_{O_{2}} = m_{CO_{2}}$ Why this step? Writing every pile explicitly stops us from forgetting a reactant.
Solve for the unknown by rearranging: $m_{O_{2}} = m_{CO_{2}} - m_{C} = 88.0 - 24.0 = 64.0 g$ Why this step? Subtraction is the conservation law rearranged — the "extra" mass in the product must have entered as oxygen.

Verify: $24.0 + 64.0 = 88.0 g$ ✓. The extra $64.0 g$ is oxygen, not magic — forecast confirmed.

Cell C — Open system, gas leaves

Here mass looks destroyed. See the picture: the balance drops, but only because an invisible passenger walked out.

Figure — Law of conservation of mass (Lavoisier) — proof, examples

Worked example Example 3 (Cell C) — Add back the escaped gas

$50.0 g$ of calcium carbonate ( $CaCO_{3}$ ) is heated in an open crucible: $CaCO_{3} \to CaO + CO_{2} ↑$ The solid left behind weighs $28.0 g$ . Was mass destroyed? If not, what mass of gas escaped?

Forecast: The dish is $22.0 g$ lighter. Guess: is that mass gone forever, or hiding in the air of the room?

Steps

Check the box: "open crucible" and the $↑$ arrow ⇒ matter can leave. The law does not apply to the crucible, only to the universe. Why this step? This is the whole trap of cell C — the law never broke; the box did.
In the universe, $M_{R} = M_{P}$ : $m_{CaCO_{3}} = m_{CaO} + m_{CO_{2}}$ Why this step? We restore closure by counting the escaped gas as a product too.
Solve for the escaped gas: $m_{CO_{2}} = 50.0 - 28.0 = 22.0 g$ Why this step? The missing mass of the dish is precisely the mass that flew off as $CO_{2}$ .

Verify: $28.0 (CaO) + 22.0 (CO_{2}) = 50.0 g$ = original ✓. Mass not destroyed — it left the box. Forecast confirmed.

Cell D — Open system, gas enters

The mirror image of Cell C: now the product looks too heavy, because a passenger walked in.

Worked example Example 4 (Cell D) — Subtract the absorbed gas

A strip of magnesium of mass $12.0 g$ is burned in open air. It reacts with oxygen from the surrounding air: $2 Mg + O_{2} \to 2 MgO$ The white ash left weighs $20.0 g$ — more than the metal. What mass of oxygen was pulled in from the air?

Forecast: The ash is heavier than the metal you started with. Does that break conservation of mass? (Watch the arrow direction.)

Steps

Check the box: open, but this time gas comes in (oxygen from the air is a hidden reactant). Why this step? In Cell C the surprise was a leaving product; here it is an entering reactant — same law, opposite sign.
Count all reactants (including the invisible one): $m_{Mg} + m_{O_{2}} = m_{MgO}$ Why this step? Air-oxygen is easy to forget precisely because you didn't pour it in.
Solve for absorbed oxygen: $m_{O_{2}} = m_{MgO} - m_{Mg} = 20.0 - 12.0 = 8.0 g$ Why this step? The gain in the solid's mass equals the mass of oxygen it grabbed from the air.

Verify: $12.0 (Mg) + 8.0 (O_{2}) = 20.0 (MgO)$ ✓. Heavier ash ≠ broken law — you just added oxygen. Forecast confirmed.

Cell E — Only part of the air reacts

Lavoisier's real experiment lives here: air is a mixture, and only its oxygen (~21% by mass in these idealised numbers) is consumed.

Worked example Example 5 (Cell E) — Air-composition twist

Mercury is heated in a sealed flask holding $10.0 g$ of air. Of that air, $2.3 g$ is oxygen. All the oxygen reacts: $2 Hg + O_{2} \to 2 HgO$ The mercury absorbs the $2.3 g$ of oxygen. (a) What is the total mass of the whole sealed flask + contents after heating, if it was $85.0 g$ before? (b) How much free gas remains inside afterwards?

Forecast: The metal got heavier by taking oxygen. Did the whole sealed flask get heavier too?

Steps

(a) Box is sealed ⇒ total mass of the flask is unchanged, no matter what rearranges inside. $M_{after} = M_{before} = 85.0 g$ Why this step? Inside a closed box, mass only moves between piles — the grand total is frozen.
(b) Only oxygen ( $2.3 g$ ) left the gas pile and joined the solid pile. Remaining free gas: $m_{gas left} = 10.0 - 2.3 = 7.7 g$ Why this step? This is the non-oxygen part of the air (mostly nitrogen), which does not react — it stays as gas.

Verify: Flask total still $85.0 g$ ✓ (nothing crossed the seal). Gas remaining $7.7 g$ + oxygen now locked in solid $2.3 g$ = $10.0 g$ original air ✓. This is exactly why Lavoisier saw no change in the flask's weight even though the metal gained mass.

Cell F — The degenerate case: no reaction

Always test the boring extreme — it confirms your method doesn't secretly assume a reaction happened.

Cell G — Word problem: forecast then verify

Worked example Example 7 (Cell G) — The sealed jar candle

A candle of wax mass $8.0 g$ sits on a scale inside a sealed jar full of air. It burns until it goes out. Someone claims: "the jar must weigh less now, because the wax is mostly gone."

Forecast: Decide before computing — will the sealed jar+contents weigh less, more, or the same?

Steps

Identify the box: sealed jar ⇒ closed. Nothing crosses the glass. Why this step? This single fact overrides the misleading visual of "the wax disappearing."
Where did the wax go? Its carbon and hydrogen atoms became $CO_{2}$ and water vapour — still inside the jar. Why this step? "Gone from sight" ≠ "gone from the box"; gases have mass too.
Apply the law to the whole sealed jar: $M_{jar, after} = M_{jar, before}$ The reading on the scale does not change. Why this step? Every atom of wax + every atom of consumed oxygen is still accounted for inside.

Verify: If jar+contents was, say, $500.0 g$ before, it reads $500.0 g$ after ✓. The $8.0 g$ of wax didn't vanish — it turned into an equal mass of trapped gas and vapour. Correct forecast: the same.

Cell H — The nuclear twist (where the law "fails")

Worked example Example 8 (Cell H) — Mass defect

In a nuclear reaction, $4.00000 g$ of a substance transforms and the products weigh $3.99920 g$ . A student panics: "Conservation of mass is broken!" (a) How much mass is 'missing'? (b) Is the classical chemistry law violated, and where did the mass go?

Forecast: In every chemistry example above, mass balanced exactly. Should you expect the same here?

Steps

(a) Compute the shortfall (the mass defect): $Δ m = 4.00000 - 3.99920 = 0.00080 g$ Why this step? We measure the discrepancy first before explaining it.
(b) This is a nuclear change: nuclei themselves are rebuilt, not just chemical bonds. The tiny lost mass became energy through Mass–Energy Equivalence (E=mc²): $E = Δ m c^{2}$ Why this step? $c$ (the speed of light) is enormous, so a microscopic mass loss releases huge energy — which is why nuclear, not chemical, reactions show it.
The classical Law of Conservation of Mass is a chemistry law. In chemistry, nuclei stay intact, so $Δ m$ is far too small to detect and $M_{R} = M_{P}$ holds to every decimal you can weigh. Why this step? Naming the domain of a law is as important as the law itself — cell H is exactly where the earlier cells' assumption (nuclei unchanged) is dropped.

Verify: $Δ m = 0.00080 g$ ✓, roughly $0.02%$ of the starting mass — invisible on a lab balance for any chemical reaction, but real for nuclear ones. So the law didn't "fail"; you left its jurisdiction. Forecast: no, mass–energy (not mass alone) is the fully general conserved quantity.

Recall Which cell am I in? (quick self-test)

Open dish, solid gets lighter after heating ::: Cell C — a gas left; add it back. Open dish, solid gets heavier after heating ::: Cell D — a gas entered; subtract it. Sealed jar, anything at all happens ::: Total flask mass unchanged (Cells A, E, G). Products weigh a whisker less than reactants, no gas involved ::: Cell H — nuclear mass defect, $E = m c^{2}$ . Only part of the air reacts ::: Cell E — use only the reacting oxygen fraction.

Connections

Balancing Chemical Equations — every "add/subtract" here is bookkeeping the balanced equation already guarantees.
Dalton's Atomic Theory — the indestructible-atom premise behind all eight cells.
Stoichiometry and the Mole — the next step: not just masses, but amounts of each substance.
Law of Definite Proportions — companion mass law; combines cleanly with Cell A/B arithmetic.
Mass–Energy Equivalence (E=mc²) — the domain boundary exposed in Cell H.