Law of definite proportions (Proust): Visual walkthrough

This page unpacks the boxed result from Law of definite proportions (Proust): $\frac{m _{A}}{m _{B}} = \frac{n _{A} m _{A}^{atom}}{n _{B} m _{B}^{atom}} = constant$ We are going to earn every letter in that formula, one picture at a time.

Step 1 — Two kinds of "atom", and the marbles that stand for them

WHAT. Before any formula, let's agree what an atom is for us: a tiny ball that cannot be cut. We draw two flavours — a small red ball we call element $A$ and a bigger blue ball we call element $B$ . (For water, think $A =$ hydrogen, $B =$ oxygen.)

WHY. The whole law rests on one fact from Dalton's Atomic Theory: atoms come in whole pieces. You can't have $2.3$ of a red ball. If we never let a ball split, then any recipe using them is forced onto whole numbers — and that is exactly what "fixed ratio" will turn out to mean.

PICTURE. Look at the two marbles below. The red one has a fixed weight; the blue one has a different fixed weight. We haven't combined anything yet — we're just meeting the actors.

Figure — Law of definite proportions (Proust)

Step 2 — One molecule is a fixed snap-together count

WHAT. Now we snap balls together into one molecule — the smallest unit of the compound. Water's molecule is always $2$ red $+ 1$ blue. We call these counts $n_{A}$ (how many red) and $n_{B}$ (how many blue).

WHY. This is the heart of everything. A molecule isn't a random handful — it's a specific blueprint. The counts $n_{A}$ and $n_{B}$ are fixed small whole numbers baked into the compound itself. Change them and you have a different compound (that's the sister law, Step 8).

PICTURE. Below, one molecule: two red balls and one blue ball locked in place. The little labels $n_{A} = 2$ and $n_{B} = 1$ sit right on the cluster so you can count them.

Step 3 — Weigh the red team and the blue team in one molecule

WHAT. We total the weight of just the red balls, then just the blue balls, inside a single molecule.

WHY. The law talks about mass, not about counts. To get from "how many balls" to "how much it weighs", we multiply each count by the weight of one ball. This is the bridge from Step 2's counting to a measurable weight on a scale.

PICTURE. The molecule is split into a red pile and a blue pile. Each pile's weight is written as a multiplication, term-by-term.

$weight of all red in one molecule m_{A}^{mol} = how many red n_{A} \times weight of one red m_{A}^{atom}$

$weight of all blue in one molecule m_{B}^{mol} = how many blue n_{B} \times weight of one blue m_{B}^{atom}$

Step 4 — The ratio: comparing the two piles

WHAT. We divide the red-pile weight by the blue-pile weight for that one molecule.

WHY. A ratio answers "how many times heavier is one pile than the other?" — a single number that captures the proportion. We divide (not subtract) because we want a scale-free comparison: " $8$ times as heavy" stays true whether we measure in grams or kilograms.

PICTURE. A balance-style diagram: the red pile on one side, blue on the other, with the division written underneath and each symbol labelled.

$\frac{m _{A}^{mol}}{m _{B}^{mol}} = \frac{n _{A} m _{A}^{atom}}{n _{B} m _{B}^{atom}}$

Reading it aloud: (count of red)(weight of one red) over (count of blue)(weight of one blue). Every one of these four symbols is a fixed constant of nature — nothing here can wobble.

Step 5 — Scale up to a whole bucket: the $N$ cancels

WHAT. A real sample isn't one molecule — it's a huge number of identical molecules. Call that number $N$ . We weigh all the red and all the blue in the bucket.

WHY. We must prove the ratio doesn't depend on how big your sample is — a teardrop of water and an ocean of water should give the same $8 : 1$ . Watch $N$ appear on top and bottom, then vanish.

PICTURE. Many copies of the Step-2 molecule fill a bucket. The total red weight and total blue weight are stacked, and the ratio is formed.

Total red weight in the bucket, and total blue weight: $M_{A} = N \cdot n_{A} m_{A}^{atom}, M_{B} = N \cdot n_{B} m_{B}^{atom}$

where

$N$ ::: the number of molecules in your sample (could be a trillion).
$M_{A}$ ::: total weight of element $A$ in the whole sample.
$M_{B}$ ::: total weight of element $B$ in the whole sample.

Now divide: $\frac{M _{A}}{M _{B}} = \frac{N n _{A} m _{A}^{atom}}{N n _{B} m _{B}^{atom}} = \frac{n _{A} m _{A}^{atom}}{n _{B} m _{B}^{atom}}$

Step 6 — Plug in real water and see the 8:1

WHAT. Let's make it concrete. Water: $A =$ hydrogen ( $m^{atom} = 1$ ), $B =$ oxygen ( $m^{atom} = 16$ ), with $n_{A} = 2$ , $n_{B} = 1$ .

WHY. A formula you can't test is just decoration. Substituting real numbers turns the abstract ratio into the famous $8 : 1$ from the parent note — closing the loop.

PICTURE. The water molecule with weights written on each ball, and the arithmetic marching to $8$ .

$\frac{M _{O}}{M _{H}} = \frac{n _{O} m _{O}^{atom}}{n _{H} m _{H}^{atom}} = \frac{1 \times 16}{2 \times 1} = \frac{16}{2} = 8$

So oxygen-to-hydrogen is $8 : 1$ by mass — exactly what the parent note promised, now derived rather than stated.

Step 7 — The degenerate case: adding excess doesn't move the ratio

WHAT. What if we pour in extra blue balls that find no red partner? Do the finished molecules change?

WHY. Beginners fear that "more oxygen ⇒ oxygen-richer water." We must show the leftover simply sits outside, untouched — the formed molecules keep their $8 : 1$ . This is the edge case where the law's word compound (not mixture) does the heavy lifting.

PICTURE. Finished water molecules on the left (all still $2 : 1$ red:blue), a pile of lonely, unbonded blue balls on the right marked "excess — not part of the compound."

The leftover blue is a separate substance (mass is still conserved, just not combined). Inside every water molecule the ratio is frozen at $8 : 1$ .

Step 8 — The other edge case: different blueprint ⇒ different (but still fixed) ratio

WHAT. Take the same two elements but a different snap-count: carbon monoxide $CO$ ( $n_{O} = 1$ ) versus carbon dioxide $CO_{2}$ ( $n_{O} = 2$ ).

WHY. We must show the law doesn't say "same elements ⇒ same ratio." It says "each compound has its own fixed ratio." Changing the blueprint gives a new constant — this is the boundary between this law and the Law of Multiple Proportions.

PICTURE. Two molecules side by side, each with its own arithmetic; both ratios are constants, just different numbers.

$CO: \frac{M _{O}}{M _{C}} = \frac{1 \times 16}{1 \times 12} = \frac{16}{12} = 1.33 \dots$ $CO_{2} : \frac{M _{O}}{M _{C}} = \frac{2 \times 16}{1 \times 12} = \frac{32}{12} = 2.67 \dots$

Each ratio is rock-solid within its own compound. They differ only because $n_{O}$ changed. No contradiction — different blueprints, different constants.

The one-picture summary

WHAT. One figure that runs the whole chain: balls → fixed counts → weigh each pile → divide → the $N$ cancels → a constant number.

$fixed n_{A}, n_{B} \times m^{atom} pile weights \div \frac{M _{A}}{M _{B}} = \frac{n _{A} m _{A}^{atom}}{n _{B} m _{B}^{atom}} = constant$

Recall Feynman: the whole walkthrough in plain words

Picture red and blue marbles that never break. Every "water toy" is glued from exactly $2$ reds and $1$ blue — that's the blueprint. Weigh the reds and weigh the blues in one toy: reds count times the weight of a red, blues count times the weight of a blue. Divide those two weights and you get one number — for water, $8$ . Now build a million toys: you've multiplied both weights by a million, but when you divide, the million on top and the million on bottom rub each other out, so you're back to $8$ . That's why a raindrop and an ocean give the same $8 : 1$ . Pour in spare blue marbles with no red partner? They just sit in a corner — the glued toys are still $2 : 1$ . And if you glue a different number of marbles, you've made a different toy with its own fixed number. Fixed blueprint in, fixed ratio out. That's the whole law.

Connections

Dalton's Atomic Theory — supplies the unbreakable balls and fixed counts (Steps 1–2).
Law of Conservation of Mass — the excess in Step 7 is conserved, just uncombined.
Law of Multiple Proportions — Step 8's different-blueprint case belongs to this sister law.
The Mole Concept — the number $N$ we cancelled in Step 5 is what the mole counts.
Empirical and Molecular Formula — the fixed ratio is exactly what a formula records.
Percentage Composition — the constant ratio turns into a constant percent by mass.

Active Recall

Why does the sample-size

N

not affect the mass ratio?

Because every molecule is identical,

N

multiplies both total masses equally and cancels in the division.

In one molecule, how do you get the weight of element

A

Multiply the count

n_{A}

by the weight of one

A

ball,

m_{A}^{atom}

For water, plug in the numbers for O:H by mass.

\frac{1 \times 16}{2 \times 1} = 8

, so

8 : 1

Does excess oxygen change water's composition?

No — unbonded oxygen is a spectator; formed water stays

8 : 1

CO gives

16/12

and CO

_{2}

gives

32/12

. Does this break the law?

No — each compound has its own fixed ratio; different blueprints give different constants.