Laws of chemical combination: Lavoisier, Proust, Dalton

If you are in Year 2 of a technical school, it is early October and in integrated sciences (the chemistry part) you are reaching the laws of chemical combination by mass (in Italian leggi ponderali): Lavoisier, Proust, Dalton. They are usually the first “quantitative” block of chemistry, just before balancing reactions and the first moles. The test that covers them falls in five to eight weeks, so you can learn them calmly: they are a few ideas, but they have to be understood, not memorised as three sentences.

Here you will find the three laws with real numbers, Dalton’s model that explains them, the typical mistakes and three worked exercises. For balancing I have written a dedicated article; for the basic concepts of matter and measurement there is the one on Year 1 chemistry; for moles and proper stoichiometry, the Year 3 kind, mole and stoichiometry in Year 3. I do not repeat that content here: I stop at masses.

Why they are called “by mass”

The Italian word ponderale comes from pondus, weight: these are laws that speak of masses that combine. Before understanding what atoms were, in the late eighteenth and early nineteenth century, chemists had a balance and a precise idea: weigh everything, before and after the reaction. The three laws are what the balances showed.

Lavoisier’s law: mass is conserved

In a chemical reaction the total mass of the reactants equals the total mass of the products. It seems obvious, but it is not if you look at a burning candle: the wax “disappears”. In reality it becomes carbon dioxide and water vapour, gases that go away; if you collected them all and weighed them, the mass would come back.

An example with simple numbers: burning $12$ g of carbon with $32$ g of oxygen gives $44$ g of carbon dioxide.

Two points to know for a test. The law holds for a closed system: if the container is open and a gas escapes, the balance reads less, and this does not contradict Lavoisier. It also holds for metals that gain mass when they oxidise in air: iron that rusts weighs more, because it has taken up oxygen.

Proust’s law: definite proportions

A pure compound always has the same composition, whatever its origin and quantity. In water, for example, $2$ g of hydrogen combine with $16$ g of oxygen and form $18$ g of water. The ratio between the masses is always

So $9$ g of water contain $1$ g of hydrogen and $8$ g of oxygen, and $36$ g contain $4$ g of hydrogen and $32$ g of oxygen. The percentage composition by mass comes from here:

If one of the two reactants is in excess, the extra part does not react: Proust’s law does not say “you must use exactly those quantities”, it says that what reacts does so in that ratio.

Dalton’s law: multiple proportions

Some elements form more than one compound. Carbon with oxygen gives the monoxide (CO) and the dioxide (CO${}_2$). Let us fix the same mass of carbon, $12$ g, and see how much oxygen combines:

The masses of oxygen are in the ratio $16:32=1:2$, a ratio of small whole numbers. This is the law of multiple proportions: the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.

With the oxides of nitrogen the phenomenon is even more evident. Fixing $28$ g of nitrogen:

The masses of oxygen are $16:32:64$, that is $1:2:4$. If in your book you see a table with other compounds, the procedure is the same: fix the mass of one of the two elements, then compare the other.

Dalton’s atomic model: why the laws work

The laws describe what happens; Dalton proposed why. His model, in brief:

1. matter is made of indivisible atoms;
2. atoms of the same element are alike (same mass), those of different elements are different;
3. in a reaction atoms are rearranged, neither created nor destroyed;
4. compounds are formed by the combination of atoms in ratios of whole numbers.

Point 3 explains Lavoisier: if atoms recombine but do not disappear, the total mass does not change. Point 4 explains Proust and multiple proportions: if water is always two atoms of H with one atom of O, the ratio between the masses can only always be the same; and if carbon can bond to one atom of oxygen or to two, the masses of oxygen will be in the ratio $1:2$. Today we know that atoms are divisible and that isotopes exist, so the model is outdated, but it remains the right explanation for this chapter.

The bridge to the mole

Take water: the ratio $1:8$ is a ratio between masses, but it comes from a ratio between atoms, $2$ atoms of H for $1$ of O. An oxygen atom weighs about $16$ times a hydrogen atom, so $2\cdot 1:1\cdot 16=2:16=1:8$. When counting atoms one by one is impossible, you count them in gigantic groups: it is the mole, which you will meet in a few weeks. For now it is enough to know that all the laws of combination by mass are the “masses” side of a reasoning that later becomes “number of atoms”.

The mistakes I see most often

Adding instead of setting up a proportion. If $2$ g of hydrogen react with $16$ g of oxygen and you are asked how much oxygen is needed for $6$ g of hydrogen, you cannot “add $4$ to $16$”. The ratio stays, so you multiply everything by $3$: $48$ g of oxygen.

Confusing the ratios. In Proust you compare the masses within the same compound; in Dalton you compare the masses of the variable element in different compounds, keeping the other fixed. Mixing the two levels leads to meaningless ratios.

Forgetting the excess reactant. If you mix quantities not in the right ratio, part does not react: the final mass stays the same (Lavoisier), but not everything is product.

Thinking that mass changes when a gas “disappears”. If a mass decreases, ask yourself where the gas has gone.

Applying Dalton to a single compound. You need at least two compounds of the same two elements.

Three worked exercises

1. Lavoisier: the missing mass

Burning $12.0$ g of magnesium in air gives $20.0$ g of magnesium oxide. How much oxygen has combined?

By Lavoisier the mass of the reactants equals that of the products:

From here I also get Proust’s ratio: $12.0:8.0=3:2$. Every $3$ g of magnesium combine with $2$ g of oxygen.

2. Proust: excess reactant

With the Mg:O ratio = $3:2$ by mass, calculate what happens if $10.0$ g of magnesium are reacted with $10.0$ g of oxygen.

I calculate how much oxygen $10.0$ g of Mg needs:

The oxygen available ($10.0$ g) is more than needed, so the magnesium is used up and there is left over

The oxide formed is $10.0+6.67\approx 16.67$ g. Check with Lavoisier: $16.67+3.33=20.0$ g, the initial mass.

3. Dalton: multiple proportions

Two oxides of sulphur contain respectively $50.0\%$ and $40.0\%$ of sulphur by mass. Verify the law of multiple proportions.

I take $100$ g of each compound.

• First oxide: $50.0$ g of S and $50.0$ g of O. For $1$ g of S there is $1.00$ g of O.
• Second oxide: $40.0$ g of S and $60.0$ g of O. For $1$ g of S there is $60.0/40.0=1.50$ g of O.

The ratio between the masses of oxygen per fixed mass of sulphur is

that is $2:3$, a ratio of small whole numbers: the law is verified. They are SO${}_2$ and SO${}_3$.

How to prepare over the coming weeks

Redo the three exercises without looking, changing the numbers. Build by hand a table like Dalton’s for CO/CO${}_2$ and for NO/NO${}_2$. When in class you move on to balancing and moles, the stoichiometry and mole simulator helps you practise the next step: it is not dedicated to the laws of combination alone, so use it as a bridge. Other resources on the subject are on the sciences page.

How we work in lessons

I give 1-to-1 online science tutoring. On this topic I almost always start from an empty table that we fill in together with the masses: when you see the ratios come out with your own numbers, the three laws stop being sentences to repeat.

If you would like a hand

If the test is getting close and you want to see where you stand, you can book a free introductory call.