The law of constant proportions

Is water always the same?

Water from a river, from a well and from the rain is made of hydrogen and oxygen. Is the ratio of the two the same? A French chemist, Joseph Proust, studied this question by weighing, about the year 1800.

He took many samples of the same compound, from different places and made in different ways, and weighed the elements in each. The result was the same every time.

The law

He found that a compound always has the same elements, in the same ratio by mass. This is the law of constant proportions (also called the law of definite proportions).

In a chemical substance, the elements are always present in definite proportions by mass.

Water

In every sample of pure water, hydrogen and oxygen are in the ratio $1 : 8$ by mass.

• $9\ \text{g}$ of water gives $1\ \text{g}$ of hydrogen and $8\ \text{g}$ of oxygen.
• $36\ \text{g}$ of water gives $4\ \text{g}$ of hydrogen and $32\ \text{g}$ of oxygen.

It does not matter where the water came from, or how it was made.

The ratio $1 : 8$ comes from the formula $\text{H}_2\text{O}$. Two hydrogen atoms have a mass of $2 \times 1\ \text{u} = 2\ \text{u}$. One oxygen atom has a mass of $16\ \text{u}$. So the ratio is $2 : 16 = 1 : 8$. In general, the ratio by mass is found from the formula and the atomic masses.

Other compounds

You should find that every mass of magnesium gives an oxide $\dfrac{5}{3}$ times as heavy: the points never leave the line.

Using the law

The ratio lets us find the mass of each element in any mass of the compound. The steps are: add the parts of the ratio, find what fraction each part is, and multiply by the mass.

Example 1. How much hydrogen is there in $45\ \text{g}$ of water? The ratio is $1 : 8$, so the total is $9$ parts. Hydrogen is $\dfrac{1}{9}$ of the mass: $\dfrac{1}{9} \times 45\ \text{g} = 5\ \text{g}$. Oxygen is $\dfrac{8}{9} \times 45\ \text{g} = 40\ \text{g}$.

Example 2. How much carbon is there in $22\ \text{g}$ of carbon dioxide? The ratio $3 : 8$ has $11$ parts. Carbon is $\dfrac{3}{11} \times 22\ \text{g} = 6\ \text{g}$, and oxygen is $\dfrac{8}{11} \times 22\ \text{g} = 16\ \text{g}$.

Example 3. How much oxygen is needed to burn $12\ \text{g}$ of carbon completely to carbon dioxide? The ratio $\text{C} : \text{O} = 3 : 8$, so $12\ \text{g}$ of carbon needs $12\ \text{g} \times \dfrac{8}{3} = 32\ \text{g}$ of oxygen.

What the law tells us

The law says that a compound cannot have just any composition. If we mix $5\ \text{g}$ of hydrogen with $8\ \text{g}$ of oxygen, only $1\ \text{g}$ of the hydrogen joins with the $8\ \text{g}$ of oxygen, to make $9\ \text{g}$ of water. The other $4\ \text{g}$ of hydrogen is left over.

This law is what separates a compound from a mixture. A mixture can have its parts in any ratio. We can mix sand and salt in any amounts, but we cannot make water with any amounts of hydrogen and oxygen.

The law is also an early proof for the existence of atoms. If a compound is made of atoms that join in a fixed pattern (two hydrogen atoms for each oxygen atom), then the ratio of the masses has to be fixed. John Dalton used this idea in his atomic theory, which the next lessons describe.

Practise with new numbers

Try it in Python

This program checks the ratio in four samples of water.

Common mistakes

• Mixing up the ratio by mass with the ratio of atoms. In water, the ratio of atoms is $2 : 1$ (H : O), and the ratio by mass is $1 : 8$.
• Forgetting to add the parts before taking a fraction. For $1 : 8$, the total is $9$.
• Thinking that the law is true for mixtures. It is true only for pure compounds.
• Forgetting the left-over. If the elements are not in the right ratio, the extra part of one element does not join.

Remember these

• Law of constant proportions: the ratio by mass of the elements in a pure compound is fixed.
• Water: $\text{H} : \text{O} = 1 : 8$. Ammonia: $\text{N} : \text{H} = 14 : 3$. Carbon dioxide: $\text{C} : \text{O} = 3 : 8$.
• This law helps us tell a compound from a mixture.
• To find the mass of an element in a compound, use its fraction of the ratio.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How much hydrogen and how much oxygen are there in $99\ \text{g}$ of water?

2. $6\ \text{g}$ of carbon burns in plenty of oxygen. What mass of carbon dioxide forms? In it, $\text{C} : \text{O} = 3 : 8$.

3. A sample said to be water has $3\ \text{g}$ of hydrogen and $12\ \text{g}$ of oxygen. Can it be pure water?

4. $7\ \text{g}$ of hydrogen is mixed with $40\ \text{g}$ of oxygen and sparked. How much water forms, and what is left over?