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URL: https://nakafa.com/en/subjects/chemistry/basic-chemistry-laws/multiple-proportions-law
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Learn how to read simple whole-number mass ratios when the same two elements form more than one compound.

---

## Two Elements Can Make Several Compounds

The law of multiple proportions applies when **the same two elements** form **more than one compound**. If the mass of one element is kept fixed, the masses of the other element that combine with it form a simple whole-number ratio.

OpenStax Chemistry: Atoms First 2e explains this law as a fixed mass of one element reacting with masses of another element in a small whole-number ratio 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 same mass pattern in [law of multiple proportions](https://www.britannica.com/science/law-of-multiple-proportions).

So this law does not compare any two substances at random. The compounds must be different, but their elements must be the same.

## Keep One Ratio Base Fixed

Suppose two compounds are both made from elements $$\mathrm{A}$$ and $$\mathrm{B}$$. To read Dalton's law, first hold the mass of $$\mathrm{A}$$ fixed, then compare the mass of $$\mathrm{B}$$ that combines with it.

Visible text: Suppose two compounds are both made from elements and . To read Dalton's law, first hold the mass of fixed, then compare the mass of that combines with it.

```math
\frac{m_{\mathrm{B\ in\ compound\ 2}}}{m_{\mathrm{B\ in\ compound\ 1}}} = \text{simple whole-number ratio}
```

The fixed mass is the important move. If both element masses change at once, the pattern can look messy. Once one element is put on the same basis, the multiple ratio becomes visible.

## Compare Related Compounds

Choose a compound pair below. The model shows the element held fixed and the element whose mass changes.

Component: MultipleProportionsLab
Props:
- title: Ratio Pairing
- description: Compare two compounds made from the same elements. One element mass is
held fixed, then the other is read as a simple ratio.
- labels: {
chooseMode: "Choose compound pair",
comparisonView: "Law of multiple proportions model",
fixedLabel: "Mass held fixed",
changingLabel: "Changing ratio",
modes: {
"water-peroxide": {
tab: "Water",
tabLabel: "Water and Peroxide",
helperCaption: (
<>
With the same mass of $$\mathrm{H}$$,{" "}
$$\mathrm{H_2O_2}$$ carries twice the mass of{" "}
$$\mathrm{O}$$ as $$\mathrm{H_2O}$$.
</>
),
fixed: (
<>
The hydrogen mass is fixed at $$2\ \mathrm{g}$$.
</>
),
changing: (
<>
The oxygen masses are $$16:32 = 1:2$$.
</>
),
},
"carbon-oxides": {
tab: "Carbon Oxides",
tabLabel: "Carbon Oxides",
helperCaption: (
<>
With the same mass of $$\mathrm{C}$$,{" "}
$$\mathrm{CO_2}$$ carries twice the mass of{" "}
$$\mathrm{O}$$ as $$\mathrm{CO}$$.
</>
),
fixed: (
<>
The carbon mass is fixed at $$12\ \mathrm{g}$$.
</>
),
changing: (
<>
The oxygen masses are $$16:32 = 1:2$$.
</>
),
},
"nitrogen-oxides": {
tab: "Nitrogen Oxides",
tabLabel: "Nitrogen Oxides",
helperCaption: (
<>
With the same mass of $$\mathrm{O}$$,{" "}
$$2\mathrm{N_2O}$$ carries four times the mass of{" "}
$$\mathrm{N}$$ as $$\mathrm{NO_2}$$.
</>
),
fixed: (
<>
The oxygen mass is fixed at $$32\ \mathrm{g}$$.
</>
),
changing: (
<>
The nitro ... [truncated; 1246 chars]
  Visible text: {
chooseMode: "Choose compound pair",
comparisonView: "Law of multiple proportions model",
fixedLabel: "Mass held fixed",
changingLabel: "Changing ratio",
modes: {
"water-peroxide": {
tab: "Water",
tabLabel: "Water and Peroxide",
helperCaption: (
<>
With the same mass of ,{" "}
 carries twice the mass of{" "}
 as .
</>
),
fixed: (
<>
The hydrogen mass is fixed at .
</>
),
changing: (
<>
The oxygen masses are .
</>
),
},
"carbon-oxides": {
tab: "Carbon Oxides",
tabLabel: "Carbon Oxides",
helperCaption: (
<>
With the same mass of ,{" "}
 carries twice the mass of{" "}
 as .
</>
),
fixed: (
<>
The carbon mass is fixed at .
</>
),
changing: (
<>
The oxygen masses are .
</>
),
},
"nitrogen-oxides": {
tab: "Nitrogen Oxides",
tabLabel: "Nitrogen Oxides",
helperCaption: (
<>
With the same mass of ,{" "}
 carries four times the mass of{" "}
 as .
</>
),
fixed: (
<>
The oxygen mass is fixed at .
</>
),
changing: (
<>
The nitro ... [truncated; 1246 chars]

In each pair above, one element is used as the comparison base. Once that base is the same, the element whose count changes is read as a simple mass ratio.

## Water and Hydrogen Peroxide

Water has the formula $$\mathrm{H_2O}$$, while hydrogen peroxide has the formula $$\mathrm{H_2O_2}$$. PubChem records those formulas in its [water property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/water/property/MolecularFormula,MolecularWeight/JSON) and [hydrogen peroxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/hydrogen%20peroxide/property/MolecularFormula,MolecularWeight/JSON).

Visible text: Water has the formula , while hydrogen peroxide has the formula . PubChem records those formulas in its [water property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/water/property/MolecularFormula,MolecularWeight/JSON) and [hydrogen peroxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/hydrogen%20peroxide/property/MolecularFormula,MolecularWeight/JSON).

Using $$A_r\mathrm{(H)} = 1$$ and $$A_r\mathrm{(O)} = 16$$, the hydrogen mass in the two formulas can be made the same, $$2\ \mathrm{g}$$.

Visible text: Using and , the hydrogen mass in the two formulas can be made the same, .

| Compound | Mass of $$\mathrm{H}$$ | Mass of $$\mathrm{O}$$ |
| :------- | :--------------------------------------- | :--------------------------------------- |
| $$\mathrm{H_2O}$$ | $$2\ \mathrm{g}$$ | $$16\ \mathrm{g}$$ |
| $$\mathrm{H_2O_2}$$ | $$2\ \mathrm{g}$$ | $$32\ \mathrm{g}$$ |

Visible text: | Compound | Mass of | Mass of |
| :------- | :--------------------------------------- | :--------------------------------------- |
| | | |
| | | |

Because the hydrogen mass is already the same, compare the oxygen masses directly.

```math
m_{\mathrm{O\ in\ H_2O}} : m_{\mathrm{O\ in\ H_2O_2}} = 16:32 = 1:2
```

The ratio $$1:2$$ is a simple whole-number ratio. That means these data fit the law of multiple proportions.

Visible text: The ratio is a simple whole-number ratio. That means these data fit the law of multiple proportions.

## Nitrogen Oxides

Nitrogen and oxygen can form several compounds. Two examples are dinitrogen monoxide, or nitrous oxide, $$\mathrm{N_2O}$$, and nitrogen dioxide, $$\mathrm{NO_2}$$. PubChem records $$\mathrm{N_2O}$$ and $$\mathrm{NO_2}$$ in its [nitrous oxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/nitrous%20oxide/property/MolecularFormula,MolecularWeight/JSON) and [nitrogen dioxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/nitrogen%20dioxide/property/MolecularFormula,MolecularWeight/JSON).

Visible text: Nitrogen and oxygen can form several compounds. Two examples are dinitrogen monoxide, or nitrous oxide, , and nitrogen dioxide, . PubChem records and in its [nitrous oxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/nitrous%20oxide/property/MolecularFormula,MolecularWeight/JSON) and [nitrogen dioxide property record](https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/name/nitrogen%20dioxide/property/MolecularFormula,MolecularWeight/JSON).

To make the oxygen mass the same, compare $$2\mathrm{N_2O}$$ with $$\mathrm{NO_2}$$. Both contain $$32\ \mathrm{g}$$ of oxygen.

Visible text: To make the oxygen mass the same, compare with . Both contain of oxygen.

| Compared compound | Mass of $$\mathrm{O}$$ | Mass of $$\mathrm{N}$$ |
| :---------------- | :--------------------------------------- | :--------------------------------------- |
| $$2\mathrm{N_2O}$$ | $$32\ \mathrm{g}$$ | $$56\ \mathrm{g}$$ |
| $$\mathrm{NO_2}$$ | $$32\ \mathrm{g}$$ | $$14\ \mathrm{g}$$ |

Visible text: | Compared compound | Mass of | Mass of |
| :---------------- | :--------------------------------------- | :--------------------------------------- |
| | | |
| | | |

Now that the oxygen mass is the same, compare the nitrogen masses:

```math
m_{\mathrm{N\ in\ 2N_2O}} : m_{\mathrm{N\ in\ NO_2}} = 56:14 = 4:1
```

Multiplying $$\mathrm{N_2O}$$ to $$2\mathrm{N_2O}$$ is only a way to put oxygen on the same mass basis. The compound formula is still $$\mathrm{N_2O}$$, not a new compound.

Visible text: Multiplying to is only a way to put oxygen on the same mass basis. The compound formula is still , not a new compound.

## Mistakes to Avoid

- Do not compare total compound masses. Compare the mass of one element after the other element has been held fixed.
- Do not use this law for two substances made from different elements. It only applies to compound pairs made from the same elements.

The law of constant composition reads one compound. The law of multiple proportions reads several compounds made from the same element pair. That small shift is why Dalton's law became strong evidence that atoms combine in whole-number counts.