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D3.1 · Explain the law of definite proportions

Learn to explain the law of definite proportions through clear examples and targeted practice.

Ontario Grade 11 Chemistry

Quantities in Chemical Reactions

Why a pure compound always contains its elements in the same mass ratio

A pure sample of water has the same chemical identity whether it comes from rain, a lake, or a bottle. Its source and sample size can change, but its composition does not. The law of definite proportions describes this steady composition: a particular compound always contains the same elements in the same fixed proportions by mass. To understand the law, first review what a compound and a mass ratio are.

What you will learn

State the law of definite proportions in your own words.
Explain how the fixed composition of a compound relates to its particles and formula.
Use a simple mass ratio to compare or predict the composition of a compound.

1. Prerequisites: compounds and mass ratios

An element is a pure substance made of one kind of atom. A compound is a pure substance whose particles contain atoms of two or more different elements joined in a fixed arrangement. Water is a compound made from hydrogen and oxygen. The chemical formula for water is

\mathrm{H_2O}

. The small 2 means each water molecule has two hydrogen atoms; oxygen has no small number, so there is one oxygen atom.

A ratio compares amounts. A mass ratio compares the masses of two substances or elements. For example, if a sample contains 2 g of one element for every 16 g of another, their mass ratio is 2:16. Dividing both numbers by 2 gives the equivalent, simpler ratio 1:8. The units cancel when the same mass unit is used for both amounts.

Keep atom counts and mass amounts distinct. A formula tells the number of each kind of atom in one particle of a compound. It does not say that the elements have equal masses. Atoms of different elements have different masses, so a fixed atom count leads to a particular mass ratio.

\mathrm{H_2O}

A compound contains more than one element in a fixed composition.
A chemical formula shows the kinds and counts of atoms in a particle.
A mass ratio compares the masses of elements, not just their atom counts.

2. Observable pattern and particle model

Imagine comparing pure samples of the same compound from different sources. A larger sample contains more total material, but the relative amounts of its elements stay the same. If one sample contains twice as much of the compound as another, it contains twice as much of each element. The ratio between the element masses remains unchanged.

At the particle level, every particle of a particular compound has the same formula. Each particle therefore has the same number of atoms of each element. When many identical particles are present, increasing the number of particles increases the amounts of all their atoms together. It does not change the elements' mass ratio.

This particle model explains why the composition is fixed. The formula specifies the atom counts, and the masses of those atoms set the corresponding mass proportions. A different proportion would not describe the same pure compound. This lesson concerns pure compounds; a mixture can contain different substances in varying amounts.

Changing the sample size changes the total mass, not the composition of the compound.
The same compound has the same formula and fixed element proportions.
A mixture is not a single pure compound and may have variable composition.

3. The law and how to use it

The law of definite proportions states that a given pure compound always contains the same elements combined in the same fixed proportions by mass, regardless of the sample's source or size. The proportions are also called the compound's mass composition. This is a statement about a compound, not a rule that every substance has the same proportions.

For a simple formula-based example, carbon dioxide is represented by

\mathrm{CO_2}

. Each molecule contains one carbon atom and two oxygen atoms. The two oxygen atoms together have a mass about 32 times the mass of one carbon atom, using course-level atomic masses of about 16 for oxygen and 12 for carbon. Thus, the oxygen-to-carbon mass ratio is about 32:12, or 8:3. The atom ratio is 2:1, but the mass ratio is not 2:1.

To use the law in a comparison, identify the two element masses in each sample and compare their ratios in the same order. Reduce each ratio by dividing both masses by the same value, or compare the resulting mass per unit mass. A matching ratio supports the conclusion that the samples have the same fixed composition. If the given ratio is different, do not claim that both represent the same pure compound as described.

When predicting a missing mass, use the fixed ratio as a proportion. Keep units with the known masses while calculating. Round the final result to a sensible number of significant digits based on the information provided. Significant digits are the digits that communicate the precision of a measured or stated value.

m_{\mathrm{O}}:m_{\mathrm{C}}=32:12=8:3

The law applies to a given pure compound and compares its elements by mass.
A formula's subscripts give atom counts; atomic masses are needed to find mass proportions.
Compare ratios in the same order and use consistent units.

4. Reading the claim carefully

The law does not say that every sample has the same total mass. It says that the relative masses of the elements in the same pure compound stay fixed. A small and a large sample can have different total masses while keeping the same ratio.

It also does not mean that different compounds made from the same elements must have the same composition. The claim is about one particular compound at a time. A formula identifies the compound being discussed, and its fixed atom counts give its fixed mass proportions.

When explaining the law, connect the statement to the particle model: the same compound has the same kind of particle, with the same atom counts. Then connect those counts to mass: the same atoms in those counts give the same element mass ratio. This is more complete than saying only that the formula stays the same.

Fixed proportions do not mean fixed sample size.
The comparison must concern the same compound.
A complete explanation links formula, particle composition, and mass ratio.

Worked example

Predicting oxygen mass in carbon dioxide

A hypothetical pure carbon dioxide sample contains 6.00 g of carbon. Using the approximate carbon dioxide mass ratio of 12 parts carbon to 32 parts oxygen, determine the oxygen mass. Treat the stated values as the data for this example.

Identify the fixed ratio
Carbon dioxide has the formula $\mathrm{CO_2}$ . One carbon atom has an approximate mass of 12 units, and two oxygen atoms together have an approximate mass of 32 units. The fixed oxygen-to-carbon mass ratio is therefore 32:12.
$m_{\mathrm{O}}:m_{\mathrm{C}}=32:12$
Set up the proportion
The sample has 6.00 g of carbon. Since the ratio is oxygen mass divided by carbon mass, set the unknown oxygen mass over 6.00 g equal to 32 over 12. Keeping the order consistent is important.
$\frac{m_{\mathrm{O}}}{6.00\ \mathrm{g}}=\frac{32}{12}$
Calculate and round
Multiply 6.00 g by 32 and divide by 12. The result is 16.0 g to three significant digits, matching the precision of the given carbon mass.
$m_{\mathrm{O}}=6.00\ \mathrm{g}\times\frac{32}{12}=16.0\ \mathrm{g}$

Answer: The sample contains 16.0 g of oxygen.

Check: The ratio is 16.0 g oxygen to 6.00 g carbon, which simplifies to 32:12. It matches the fixed carbon dioxide mass ratio.

Common mistakes and how to avoid them

Treating a formula's atom ratio as the mass ratio.

Correction: Use atomic masses as well as subscripts. In carbon dioxide, the atom ratio is 1 carbon to 2 oxygen atoms, but the mass ratio is about 12:32.

Thinking every sample of a compound must have the same total mass.

Correction: The total mass can vary. The law says the relative masses of the elements remain fixed.

Comparing element masses in a different order in the two samples.

Correction: Write both ratios in the same order, such as oxygen-to-carbon, before deciding whether they match.

Applying the law to a mixture as though it were one pure compound.

Correction: The law describes the fixed composition of a given pure compound. A mixture can contain its substances in varying amounts.

Lesson summary

The law of definite proportions says that a given pure compound always contains the same elements in the same fixed proportions by mass.
A compound's formula gives the atom counts in its particles. Those counts, together with the atoms' masses, explain the fixed mass ratio.
Sample size may change, but the element mass ratio in the same pure compound does not.
Compare ratios in the same order and keep units and appropriate significant digits in calculations.

Check your understanding

Question 1

Which statement best describes the law of definite proportions?

Every pure compound has the same total mass.
A given pure compound has the same element mass proportions in every sample.
Every sample that contains two elements has the same mass ratio.
The atom count and mass of each element must be equal.

Show answer and explanation

A given pure compound has the same element mass proportions in every sample.

The law concerns the fixed mass proportions in one particular pure compound. It does not require equal element masses or equal sample sizes.

Question 2

A compound has a fixed mass ratio of element A to element B of 3:5. If a sample has 6 g of A, how much B does it contain?

5 g
8 g
10 g
15 g

Show answer and explanation

10 g

The ratio 3:5 is doubled to match 6 g of A, so the mass of B is 10 g. The ratio remains 6:10, which reduces to 3:5.

Key terms

Element: A pure substance made of one kind of atom.
Compound: A pure substance whose particles contain atoms of different elements in a fixed composition.
Mass ratio: A comparison of the masses of two or more substances or elements.
Law of definite proportions: The rule that a given pure compound always contains the same elements in the same fixed proportions by mass.
Significant digits: The digits in a value that communicate its precision.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Chemistry (SCH3U), expectation D3.1. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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