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URL: https://nakafa.com/en/subjects/chemistry/basic-chemistry-laws/constant-composition-law
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Learn why a pure compound always contains its elements in a fixed mass ratio, even when the starting masses are different.

---

## 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 [Early Ideas in Atomic Theory](https://openstax.org/books/chemistry-atoms-first-2e/pages/2-1-early-ideas-in-atomic-theory). Britannica summarizes the law as the statement that a chemical compound has fixed element proportions by mass in [law of definite proportions](https://www.britannica.com/science/law-of-definite-proportions).

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 [chemical substance](https://goldbook.iupac.org/terms/view/C01039/plain).

## 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:

Visible text: The formula for water is . That means each unit of water contains hydrogen atoms and oxygen atom. Using the simple relative atomic masses and , the hydrogen-to-oxygen mass ratio in water is:

```math
\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 [water property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/water/property/MolecularFormula,MolecularWeight/JSON).

Visible text: So even when different amounts of water form, the mass ratio of hydrogen to oxygen that actually reacts stays . PubChem lists water as with a molecular weight of about in its [water property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/water/property/MolecularFormula,MolecularWeight/JSON).

## 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.

Component: ConstantCompositionLab
Props:
- title: Water Ratio Gate
- description: Watch water form only from the reactant portion that satisfies the mass
ratio $$1:8$$.
  Visible text: Watch water form only from the reactant portion that satisfies the mass
ratio .
- labels: {
chooseMode: "Choose starting mass",
before: "Start",
after: "Formed",
reactionView: "Constant composition model for water formation",
ratioLabel: "Reacting ratio",
leftoverLabel: "Left after reaction",
modes: {
exact: {
tab: "Exact",
tabLabel: "Exact",
helperCaption: (
<>
All reactants form water because the hydrogen and oxygen masses
already match the $$1:8$$ ratio.
</>
),
readoutBefore: "2 g H + 16 g O",
readoutAfter: "18 g H₂O",
ratio: (
<>
$$2:16 = 1:8$$, the water mass ratio.
</>
),
leftover: <>No excess substance remains.</>,
},
"hydrogen-excess": {
tab: (
<>
Extra $$\mathrm{H}$$
</>
),
tabLabel: "Extra H",
helperCaption: (
<>
Water still forms from the $$1:8$$ ratio, and the
hydrogen outside that ratio remains extra.
</>
),
readoutBefore: "3 g H + 16 g O",
readoutAfter: "18 g H₂O + 1 g H",
ratio: (
<>
Only $$2:16 = 1:8$$ reacts.
</>
),
leftover: (
<>
$$1\ \mathrm{g}$$ of hydrogen does not become
water.
</>
),
},
"oxygen-excess": {
tab: (
<>
Extra $$\mathrm{O}$$
</>
),
tabLabel: "Extra O",
helperCaption: (
<>
Water still forms from the $$1:8$$ ratio, and the
oxygen outside that ratio remains extra.
</>
),
readoutBefore: "2 g H + 20 g O",
readoutAfter: "18 g H₂O + 4 g O",
ra ... [truncated; 1340 chars]
  Visible text: {
chooseMode: "Choose starting mass",
before: "Start",
after: "Formed",
reactionView: "Constant composition model for water formation",
ratioLabel: "Reacting ratio",
leftoverLabel: "Left after reaction",
modes: {
exact: {
tab: "Exact",
tabLabel: "Exact",
helperCaption: (
<>
All reactants form water because the hydrogen and oxygen masses
already match the ratio.
</>
),
readoutBefore: "2 g H + 16 g O",
readoutAfter: "18 g H₂O",
ratio: (
<>
, the water mass ratio.
</>
),
leftover: <>No excess substance remains.</>,
},
"hydrogen-excess": {
tab: (
<>
Extra 
</>
),
tabLabel: "Extra H",
helperCaption: (
<>
Water still forms from the ratio, and the
hydrogen outside that ratio remains extra.
</>
),
readoutBefore: "3 g H + 16 g O",
readoutAfter: "18 g H₂O + 1 g H",
ratio: (
<>
Only reacts.
</>
),
leftover: (
<>
 of hydrogen does not become
water.
</>
),
},
"oxygen-excess": {
tab: (
<>
Extra 
</>
),
tabLabel: "Extra O",
helperCaption: (
<>
Water still forms from the ratio, and the
oxygen outside that ratio remains extra.
</>
),
readoutBefore: "2 g H + 20 g O",
readoutAfter: "18 g H₂O + 4 g O",
ra ... [truncated; 1340 chars]

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 [calcium oxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/calcium%20oxide/property/MolecularFormula,MolecularWeight/JSON).

Visible text: Now read the data like a researcher. Calcium oxide has the formula . PubChem lists that formula and a molecular weight of in its [calcium oxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/calcium%20oxide/property/MolecularFormula,MolecularWeight/JSON).

Suppose two experiments give the following data.

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

Visible text: | Trial | Mass of | Mass of | Mass of | Ratio |
| :---- | :----------------------------------------- | :-------------------------------------- | :--------------------------------------- | :----------------------------------------- |
| | | | | |
| | | | | |

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

```math
\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.

Visible text: The two experiments form different masses of compound, but the calcium-to-oxygen mass ratio is the same, . 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 [ferric oxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/ferric%20oxide/property/MolecularFormula,MolecularWeight/JSON).

Visible text: For iron(III) oxide, the formula is . PubChem's ferric oxide data also lists in its [ferric oxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/ferric%20oxide/property/MolecularFormula,MolecularWeight/JSON).

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

Visible text: Using and , the mass ratio of iron to oxygen is:

```math
\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 $$7 + 3 = 10$$.

Visible text: If the mass of is , the total ratio parts are .

```math
\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.

Visible text: Therefore, of iron(III) oxide contains of iron and 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.