Constant Composition

Pure Compounds Have a Fixed 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.

Proust showed that samples of a pure compound contain the same elements in the same mass proportion. This fixed mass proportion is the central statement of the law of constant composition.

The law of constant composition applies to pure compounds. The composition of a mixture such as sugar water changes when the amount of sugar or water changes. Pure water always contains hydrogen and oxygen in the same mass ratio. A chemical substance has a constant composition and characteristic properties.

Water Keeps the Same Ratio

The formula for water is $\mathrm{H_2O}$ . That means each molecule 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}

Even when different amounts of water form, the mass ratio of hydrogen to oxygen that reacts stays $1:8$ with the rounded values used above. Water has the formula $\mathrm{H_2O}$ . The $2024$ abridged standard atomic weights are $A_r\mathrm{(H)}=1.0080$ and $A_r\mathrm{(O)}=15.999$ . Substituting those two atomic weights gives the following molar mass:

\begin{aligned} M(\mathrm{H_2O}) &=2(1.0080)+15.999 \\ &=18.015\ \mathrm{g\,mol^{-1}} \end{aligned}

Extra Reactant Stays Extra

Change the starting masses in the model. A different starting mixture may leave hydrogen or oxygen behind. The hydrogen and oxygen that form water always react in a $1:8$ mass ratio.

This model counts the mass of each element. The next equation shows how the molecules react. The $\mathrm{H}$ and $\mathrm{O}$ labels identify the element masses being counted. Elemental hydrogen and oxygen are represented in the balanced formation equation as $\mathrm{H_2}$ and $\mathrm{O_2}$ :

2\mathrm{H_2(g)}+\mathrm{O_2(g)}\rightarrow 2\mathrm{H_2O}

The Fixed Mass Ratio in Water

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.

If the starting masses differ from the compound's ratio, only the portion that satisfies the ratio reacts. The rest remains as excess reactant.

Testing Calcium Oxide Data

The next two experiments test the law of constant composition with calcium oxide. Calcium oxide has the formula unit $\mathrm{CaO}$ . The CIAAW values $A_r\mathrm{(Ca)}=40.078$ and $A_r\mathrm{(O)}=15.999$ give $M(\mathrm{CaO})=56.077\ \mathrm{g\,mol^{-1}}$ , which rounds to $56.08\ \mathrm{g\,mol^{-1}}$ .

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. Both have the same calcium-to-oxygen mass ratio, $2.5:1$ . This common ratio shows that the samples follow the law of constant composition.

Calculating Mass from a Formula

Iron(III) oxide has the formula unit $\mathrm{Fe_2O_3}$ .

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}

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

The law of multiple proportions describes cases in which the same two elements form more than one compound.