Constant Composition: Basic Laws of Chemistry - Chemistry (Grade 10)

Compounds Keep a Mass Ratio

The law of constant composition states that every pure compound contains its constituent elements in a fixed mass ratio. It is also called the law of definite proportions or Proust's law.

OpenStax Chemistry: Atoms First 2e explains that Joseph-Louis Proust showed samples of a pure compound contain the same elements in the same mass proportion in openstax.org. Britannica summarizes the law as the statement that a chemical compound has fixed element proportions by mass in britannica.com.

The phrase pure compound matters. Sugar water can be made sweeter or more dilute, but pure water is still composed of hydrogen and oxygen in the same pattern. The IUPAC Gold Book uses the idea of constant composition when defining a chemical substance in goldbook.iupac.org.

Water Keeps the Same Ratio

The formula for water is $\mathrm{H_2O}$ . That means each unit of water contains $2$ hydrogen atoms and $1$ oxygen atom. Using the simple relative atomic masses $A_r\mathrm{(H)} = 1$ and $A_r\mathrm{(O)} = 16$ , the hydrogen-to-oxygen mass ratio in water is:

\begin{aligned} m_{\mathrm{H}} : m_{\mathrm{O}} &= (2 \times 1) : (1 \times 16) \\ &= 2 : 16 \\ &= 1 : 8 \end{aligned}

So even when different amounts of water form, the mass ratio of hydrogen to oxygen that actually reacts stays $1:8$ . PubChem lists water as $\mathrm{H_2O}$ with a molecular weight of about $18.015\ \mathrm{g/mol}$ in its pubchem.ncbi.nlm.nih.gov.

Extra Reactant Stays Extra

Change the starting masses in the model. The leftover substance changes. The hydrogen-to-oxygen ratio that enters water does not.

Water Ratio Gate

Watch water form only from the reactant portion that satisfies the mass ratio

1:8

All reactants form water because the hydrogen and oxygen masses already match the $1:8$ ratio.

Reacting ratio: $2:16 = 1:8$ , the water mass ratio.
Left after reaction: No excess substance remains.

This law is often misread as "all starting material must be used up." Not quite. If the starting masses do not match the compound's ratio, only the matching portion reacts and the rest remains as excess reactant.

Testing Calcium Oxide Data

Now read the data like a researcher. Calcium oxide has the formula $\mathrm{CaO}$ . PubChem lists that formula and a molecular weight of $56.08\ \mathrm{g/mol}$ in its pubchem.ncbi.nlm.nih.gov.

Suppose two experiments give the following data.

Trial	Mass of $\mathrm{CaO}$	Mass of $\mathrm{O}$	Mass of $\mathrm{Ca}$	Ratio $\mathrm{Ca:O}$
$1$	$2.8\ \mathrm{g}$	$0.8\ \mathrm{g}$	$2.0\ \mathrm{g}$	$2.0:0.8 = 2.5:1$
$2$	$3.5\ \mathrm{g}$	$1.0\ \mathrm{g}$	$2.5\ \mathrm{g}$	$2.5:1.0 = 2.5:1$

The calcium mass is the compound mass minus the oxygen mass.

\begin{aligned} m_{\mathrm{Ca,1}} &= 2.8 - 0.8 = 2.0\ \mathrm{g} \\ m_{\mathrm{Ca,2}} &= 3.5 - 1.0 = 2.5\ \mathrm{g} \end{aligned}

The two experiments form different masses of compound, but the calcium-to-oxygen mass ratio is the same, $2.5:1$ . That is evidence that the samples fit the law of constant composition.

Calculating Mass from a Formula

For iron(III) oxide, the formula is $\mathrm{Fe_2O_3}$ . PubChem's ferric oxide data also lists $\mathrm{Fe_2O_3}$ in its pubchem.ncbi.nlm.nih.gov.

Using $A_r\mathrm{(Fe)} = 56$ and $A_r\mathrm{(O)} = 16$ , the mass ratio of iron to oxygen is:

\begin{aligned} m_{\mathrm{Fe}} : m_{\mathrm{O}} &= (2 \times 56) : (3 \times 16) \\ &= 112 : 48 \\ &= 7 : 3 \end{aligned}

If the mass of $\mathrm{Fe_2O_3}$ is $64\ \mathrm{g}$ , the total ratio parts are $7 + 3 = 10$ .

\begin{aligned} m_{\mathrm{Fe}} &= \frac{7}{10} \times 64 = 44.8\ \mathrm{g} \\ m_{\mathrm{O}} &= \frac{3}{10} \times 64 = 19.2\ \mathrm{g} \end{aligned}

Therefore, $64\ \mathrm{g}$ of iron(III) oxide contains $44.8\ \mathrm{g}$ of iron and $19.2\ \mathrm{g}$ of oxygen bonded together.

Once one compound has a fixed ratio, the next question is what happens when the same two elements can form more than one compound. That is where the law of multiple proportions begins.