Whether it comes from a river or from burning hydrogen in a lab, water always contains hydrogen and oxygen in a mass ratio of 1 : 8. The fact that the elements in a compound always combine in the same mass ratio is the law of definite proportions.
When copper is heated to form copper(II) oxide, copper and oxygen always combine in a mass ratio of 4 : 1, because the atoms join 1 : 1 and a copper atom is four times as heavy as an oxygen atom.
In the diagram, with oxygen fixed at 4 g, changing the amount of copper turns only the matching share into copper oxide, and whatever is in excess is left over.
That is why this law also lets you work out how much of a substance will remain.
Compounds Always Combine in the Same Ratio
🧩 A Recipe That Never Changes
①A water molecule is always two hydrogen atoms and one oxygen atom
②So wherever water comes from, its hydrogen-to-oxygen mass ratio is the same 1 : 8
③Proust first established this law by analyzing compounds from many sources
④In the diagram, oxygen is fixed at 4 g and only the amount of copper changes
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Copper and oxygen always combining 4 : 1 by mass to form copper oxide, with any excess left over
16 g
What Happens to the Excess?
🔍 Anything Beyond the Ratio Is Left Over
①16 g copper + 4 g oxygen → 20 g copper oxide; the ratio matches exactly and nothing is left
②24 g copper + 4 g oxygen → 20 g copper oxide, with 8 g copper left over
③8 g copper + 4 g oxygen → 10 g copper oxide, with 2 g oxygen left over
④The masses that react always add up to the mass of copper oxide formed (conservation of mass)
Summary
Law of definite proportions
The mass ratio of the elements in a compound is always constant
Example: in copper(II) oxide, CuO, copper : oxygen = 4 : 1 by mass
Composition of water
H₂O → hydrogen : oxygen = 1 : 8 (by mass)
1 g hydrogen reacting with 8 g oxygen gives 9 g water
🎯 Common Test Points
①Law of definite proportions: the mass ratio of the elements in a compound is always constant
②Water H₂O: hydrogen : oxygen = 1 : 8
③Copper(II) oxide CuO: copper : oxygen = 4 : 1
④A substance added beyond the ratio does not react and is left over
⑤A graph of copper oxide mass against copper mass is a straight line through the origin that levels off once the oxygen runs out
Examples and Unit Review
Example 1
In copper(II) oxide, copper and oxygen always combine 4 : 1 by mass. How much oxygen does 16 g of copper need to react completely, and how much copper oxide forms?
1
Since copper : oxygen = 4 : 1, the oxygen needed is 16 ÷ 4 = 4 g.
2
By conservation of mass, the copper oxide formed is 16 g + 4 g = 20 g.
▸ 4 g of oxygen is needed, and 20 g of copper oxide forms
With 32 g of copper you would need 8 g of oxygen and get 40 g of copper oxide; the ratio stays the same.
Example 2
24 g of copper reacts with 4 g of oxygen. How much copper oxide forms, and which substance is left over, and how much? (copper : oxygen = 4 : 1)
1
4 g of oxygen can react with 4 × 4 = 16 g of copper, so 16 + 4 = 20 g of copper oxide forms.
2
Only 16 g of the 24 g of copper reacts, so 24 − 16 = 8 g of copper is left over.
▸ 20 g of copper oxide forms, and 8 g of copper is left over
To find what is left over, first check which of the two substances falls short of the ratio.
Unit Test
Which statement about the law of definite proportions is correct?
①The mass ratio of elements in a compound changes with how it is made
②In water (H₂O) the mass ratio of hydrogen to oxygen is 1 : 8
③A substance added beyond the fixed ratio still reacts completely
④The law of definite proportions holds for mixtures too
⑤In copper oxide the mass ratio of copper to oxygen is 1 : 4
▸ ② In water (H₂O) the mass ratio of hydrogen to oxygen is 1 : 8
1
The mass ratio of elements in a compound is always constant (① wrong), and a substance beyond the fixed ratio cannot react and is left over (③ wrong).
2
The law holds only for compounds, while a mixture has no fixed ratio (④ wrong), and in copper oxide Cu : O = 4 : 1 (⑤ wrong); in water H : O = 1 : 8 (② correct).