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

Step 1 — What "mass" really is: a pile of tiny bricks, each with its own mass

WHAT. Before we can conserve mass, we must say what mass is made of. Picture every object as a collection of tiny balls. Each kind of ball (each element) has its own fixed mass.

WHY this idea and not "mass is just a number"? Because if mass is a smooth, featureless quantity, there is no reason it should be conserved — it could leak away like heat. But if mass is a count of unbreakable bricks each of fixed mass, then conservation becomes obvious: you can't lose a brick that can't be destroyed. We are choosing the picture that makes the law inevitable. This picture comes from Dalton's Atomic Theory.

PICTURE. Below, three element-bricks. The number written on each is its atomic mass $m_{i}$ — "the mass of one brick of element $i$ ." The subscript $i$ is just a name-tag: $i = H$ , $i = O$ , etc.

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

$m_{i}$ ::: the mass of a single atom of element labelled $i$ (a fixed number that never changes in chemistry).

Step 2 — Total mass = "count each kind of brick, times its mass, add up"

WHAT. To get the mass of a whole pile, we go element by element: multiply how many bricks of that element ( $N_{i}$ ) by the mass of each one ( $m_{i}$ ), then add across all the elements that actually appear in the pile.

WHY the $\sum$ (the "sum" symbol)? The stretched-E, $\sum_{i}$ , is shorthand for "add up the following for every value of $i$ ." Here $i$ runs over only the elements that are actually present in this reaction — not every element in the universe. (Elements that never appear just contribute $N_{i} = 0$ , so leaving them out changes nothing.) We use the symbol because a real substance has several elements and writing $+ \dots$ forever is clumsy. It is nothing more than "add them all."

PICTURE. Watch the tally board: each column is one element present, its height is $N_{i}$ , and we tag each column with its per-brick mass $m_{i}$ . The grand total at the bottom is $M$ .

$N_{i}$ ::: the number of atoms of element $i$ in this pile (a plain count: 0, 1, 2, 3, …).
$\sum_{i}$ ::: "for every element $i$ present, add the term after me."

Step 3 — A reaction is a rearrangement, not a birth or a death

WHAT. We now say what happens during a reaction, in brick language. Take the real reaction $2 H_{2} + O_{2} \to 2 H_{2} O$ . On the left we start with two H $_{2}$ molecules (that is 4 H-bricks) and one O $_{2}$ molecule (that is 2 O-bricks). On the right those very same $4$ H-bricks and $2$ O-bricks are re-clipped into two water molecules, each $H_{2} O$ using $2$ H and $1$ O.

WHY insist on "no brick added or removed"? This is the one physical assumption — Dalton's postulate — that does all the work. Everything else is arithmetic. If we did not assume it, mass could change. So we make the assumption explicit and then watch it force conservation. Notice the counts match perfectly: $4$ H and $2$ O on both sides — that is why we wrote 2H $_{2}$ and 2H $_{2}$ O, not one each.

PICTURE. Left: two H $_{2}$ molecules (4 H-bricks in two clipped pairs) and one O $_{2}$ molecule (2 O-bricks). Right: the same $4$ H-bricks and $2$ O-bricks, re-clipped into two water molecules. Trace any single brick from left to right — it is still there, just holding new hands.

Step 4 — Turn "no brick added or removed" into an equation

WHAT. We name the counts before the reaction $N_{i}^{before}$ and after $N_{i}^{after}$ , and Step 3 tells us they are equal — for each element separately, not just in total. (In our example: $N_{H} = 4$ and $N_{O} = 2$ , both before and after.)

WHY "for every element" and not just the grand total? Because carbon bricks can't turn into oxygen bricks. Their counts are locked individually. This stronger statement is exactly what Balancing Chemical Equations enforces when you make the little numbers match on both sides.

PICTURE. Two tally boards side by side — "before" and "after." Each coloured column has the same height in both boards. The dashed arrows show equal heights, element by element.

$N_{i}^{before}$ ::: count of element $i$ among the reactants.
$N_{i}^{after}$ ::: count of element $i$ among the products.

Step 5 — Substitute and watch the masses collapse into equality

WHAT. Mass of products $M_{P}$ uses the after counts; mass of reactants $M_{R}$ uses the before counts. Step 4 says those counts are identical term-by-term. The per-brick masses $m_{i}$ never changed (Step 1). So the two sums are the same sum written twice — hence equal.

WHY does this finish the job? Because $M_{R}$ and $M_{P}$ were the only two things we cared about. We have shown they are algebraically forced to match. No experiment needed — the picture guaranteed it.

$M_{R} = M_{P}$

PICTURE. The two total bars — reactant total and product total — rise to exactly the same line. The balance beam stays level.

$M_{R}$ ::: total mass of all reactants $= \sum_{i} N_{i}^{before} m_{i}$ .
$M_{P}$ ::: total mass of all products $= \sum_{i} N_{i}^{after} m_{i}$ .

Step 6 — The degenerate case: nothing reacts (or a physical change)

WHAT. What if the "reaction" is trivial — ice melting to water, or no change at all? Then every brick sits still: $N_{i}^{before} = N_{i}^{after}$ holds for the boring reason that nothing moved.

WHY show this? A good law must survive the empty case. If melting ice changed mass, our derivation would be wrong. It doesn't: the sum in Step 2 is untouched because both the counts and the masses are untouched. The bonds between water molecules rearrange, but atom counts don't even flicker.

PICTURE. Same bricks, same tally, before = after — the balance beam does not even wobble.

Step 7 — The "leaky" case: an open system (why mass seems to vanish)

WHAT. Our proof assumed the same bricks stay in the box (closed system). If the box is open and gas leaves, then the bricks we weigh are fewer than the bricks that exist. The law about the universe is fine; the reading on our scale is not.

WHY a separate step? Because this is where 90% of exam mistakes live. The equation $M_{R} = M_{P}$ is only true when you count all bricks — including the ones that flew away. Draw the dashed boundary of the container and ask: did any brick cross it?

PICTURE. Heating $CaCO_{3}$ in an open dish. $10.0 g$ of solid becomes $5.6 g$ of solid $CaO$ — but $4.4 g$ of $CO_{2}$ escaped across the dashed boundary. Trap it and $5.6 + 4.4 = 10.0 g$ : nothing lost.

$CaO, stays 5.6 + CO_{2}, escapes 4.4 = original 10.0 g$

Step 8 — The true limit: nuclear reactions break a brick

WHAT. Everything above rested on "no brick is destroyed." Nuclear reactions destroy that premise: they crack the nucleus. Then a sliver of mass $Δ m$ leaves as pure energy.

WHY does chemistry still get away with it? In chemistry only the connections between bricks change (bond energy), and the mass equivalent of that energy is far too small to measure — a billionth of a gram. So the classical law is exact for practical chemistry and only bends at the nuclear scale.

PICTURE. A nucleus splits; the two pieces together weigh slightly less than the original, and the missing sliver flies off as energy. This is the one place the balance beam finally tips.

$Δ m$ ::: the tiny "mass defect" — mass that became energy; zero (undetectable) in chemistry, real in nuclear reactions.

The one-picture summary

Everything above, compressed: bricks in → rearranged → bricks out; counts locked (Step 4) force the totals to match (Step 5); the only escapes are an open lid (Step 7, mass still conserved for the universe) or a cracked nucleus (Step 8, the genuine limit).

Recall Feynman: tell the whole walkthrough to a 12-year-old

Everything is made of tiny LEGO bricks, and each colour of brick has its own mass that never changes. To find how heavy something is, you just count each colour of brick and add up the masses. When a "reaction" happens, all you do is unclip and re-clip the bricks into new shapes — you never make a new brick and never throw one away. So of course the total mass before equals the total mass after: it's the same bricks! If your scale ever reads less, it's because some bricks turned into invisible gas and floated out of the open box — trap them and the mass comes right back. The only time you can truly lose a sliver of mass is if you actually smash a brick apart (a nuclear reaction), and even then it didn't vanish — it turned into energy. Atoms just dance; they never leave.

Recall Quick self-test

A sealed flask holds reactants weighing $50.0 g$ . After a violent reaction inside, what does the sealed flask weigh? ::: Exactly $50.0 g$ — no brick crossed the sealed boundary (Step 5, Step 7). $8.00 g$ of methane burns in a sealed jar with $32.00 g$ of oxygen. Total mass of products? ::: $40.00 g$ (Step 5: $M_{P} = M_{R}$ ). Which step's assumption fails in a nuclear reaction? ::: Step 3 — "bricks (nuclei) are unbreakable."

Connections

Dalton's Atomic Theory — supplies the unbreakable-brick premise used in Step 3.
Balancing Chemical Equations — the practical form of Step 4 (equal counts per element).
Law of Definite Proportions — the companion mass law from the same brick picture.
Stoichiometry and the Mole — turns the conserved counts into weighable amounts.
Mass–Energy Equivalence (E=mc²) — the Step 8 limit where the classical law bends.