Law of Multiple Proportions

Carbon and oxygen do not form just one compound — they form two: carbon monoxide and carbon dioxide. Weigh the oxygen that combines with a fixed mass of carbon in each, and something remarkable falls out: the two oxygen masses stand in the ratio $2:1$ , a small whole number. This is not a coincidence. It happens every single time two elements form more than one compound, and it is exactly the pattern John Dalton needed to argue that matter is built from indivisible atoms combining in fixed integer ratios. The law of multiple proportions is the third of the classical laws of chemical combination in the CBSE Class 11 syllabus, following the law of conservation of mass and the law of definite proportions, and it leads straight into Dalton's atomic theory. This article restates the law precisely, derives the working ratio test used to verify it from mass data, and works through two worked examples — carbon's two oxides and nitrogen's two oxides — the same numbers used throughout the unit.

The core insight

When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are always in a ratio of small whole numbers — direct evidence that atoms combine in fixed integer counts.

The core idea

Statement. Proposed by John Dalton in 1803 as part of his atomic theory, the law says: when two elements A and B combine to form more than one compound, the different masses of A that combine with a fixed mass of B are in a ratio of small whole numbers.

Where it sits in the syllabus. Unit I, "Some Basic Concepts of Chemistry," lists three laws of chemical combination before introducing Dalton's atomic theory: conservation of mass (mass is neither created nor destroyed in a reaction), definite proportions (a given compound always has the same elements in the same fixed mass ratio), and multiple proportions (this law, which needs two compounds of the same pair of elements to test). Multiple proportions is the one that most directly foreshadows atoms, because a "ratio of small whole numbers" is exactly what discrete, countable particles would produce.

The test, in words. Given two compounds made from the same two elements X and Y: fix the mass of Y at the same value in both compounds (scale the raw data if it isn't already matched), then compare the two masses of X that go with that fixed mass of Y. If the law holds, that comparison reduces to a small whole-number ratio such as $1:2$ , $1:3$ , or $2:3$ .

\frac{m_{X,1}}{m_{X,2}}\bigg|_{\,m_Y \text{ fixed and equal}} = \text{small whole-number ratio} \tag{1}

where: $m_{X,1}$ and $m_{X,2}$ are the masses of element X combining with the same fixed mass $m_Y$ of element Y, in compound 1 and compound 2 respectively.

Figure

Bar chart comparing oxygen mass per fixed 3 gram carbon in carbon monoxide and carbon dioxide showing a 1 to 2 ratio

Derivation of the multiple-proportions ratio test

Derivation AThe multiple-proportions ratio test

The goal is to build, from two raw mass-composition data sets for the same pair of elements, a single number that either confirms or refutes the law of multiple proportions — the ratio test used in every textbook problem on this topic, including both worked examples below.

Step 1. Start with the raw compositions. Suppose compound 1 contains mass $a_1$ of element X combined with mass $b_1$ of element Y, and compound 2 contains mass $a_2$ of X combined with mass $b_2$ of Y. These four numbers come straight from experiment (percentage composition or combustion data), with no assumption yet about which ratio is "clean."

Step 2. Fix the mass of Y. Pick a convenient common mass of Y — often the smaller of $b_1, b_2$ , or a round number like $1\ \text{g}$ — and scale each compound's data proportionally so that both now carry that same mass of Y. If $b_1$ already equals $b_2$ , as in both worked examples below, this step needs no arithmetic at all; if not, multiply $a_1$ by $b_2/b_1$ or vice versa so the Y-masses match.

Step 3. Read off the corresponding masses of X once Y is equalised — call them $m_{X,1}$ and $m_{X,2}$ . These are directly comparable because they now sit on the same fixed mass of the other element.

Step 4. Compute the ratio $m_{X,1} : m_{X,2}$ and simplify it by dividing both terms by their common factor, or by multiplying through until both terms are integers if the raw ratio is a decimal (as in Example 2 below). If the simplified ratio is small — single-digit integers such as $1:2$ , $2:3$ , or $5:2$ — the law of multiple proportions is confirmed for this pair of compounds. Physical reading: a clean small integer ratio is exactly what results if compound 1 has, say, one atom of Y bonded to $n_1$ atoms of X and compound 2 has one atom of Y bonded to $n_2$ atoms of X — the mass ratio inherits the atom-count ratio $n_1 : n_2$ , because atoms come in whole units and nothing in between.

Worked example

Problem. Carbon and oxygen form two compounds. In carbon monoxide (CO), $3.0\ \text{g}$ of carbon combines with $4.0\ \text{g}$ of oxygen. In carbon dioxide ( $\text{CO}_2$ ), $3.0\ \text{g}$ of carbon combines with $8.0\ \text{g}$ of oxygen. Verify the law of multiple proportions, and separately check nitrogen's two oxides.

Step 1. The mass of carbon is already fixed at $3.0\ \text{g}$ in both compounds, so no scaling of Step 2 from the derivation is needed here — the data walks straight into the ratio step.

Step 2. Read off the oxygen masses at this fixed carbon mass: $4.0\ \text{g}$ in CO and $8.0\ \text{g}$ in $\text{CO}_2$ .

Step 3. Form the ratio: $\dfrac{\text{mass O in }\text{CO}_2}{\text{mass O in CO}} = \dfrac{8.0}{4.0} = 2$ , i.e. $2:1$ — already a small whole number, so the law is confirmed. This matches the chemical formulas directly: $\text{CO}_2$ carries twice as many oxygen atoms per carbon atom as CO.

Step 4. Now the second data set. Oxide A of nitrogen has $1.4\ \text{g}$ nitrogen with $1.6\ \text{g}$ oxygen; Oxide B has $1.4\ \text{g}$ nitrogen with $4.0\ \text{g}$ oxygen. Nitrogen is already fixed at $1.4\ \text{g}$ in both, so form $\dfrac{4.0}{1.6} = 2.5$ . This is not yet an integer ratio, so multiply both terms by $2$ : $2.5:1 \rightarrow 5:2$ . Answer: both pairs obey the law — CO and $\text{CO}_2$ give the ratio $2:1$ , and nitrogen's two oxides give the ratio $5:2$ , both small whole numbers.

Competitive stretch

If a third nitrogen oxide, Oxide C, carried $1.4\ \text{g}$ nitrogen with $2.4\ \text{g}$ oxygen, compare it to Oxide A ( $1.4\ \text{g}$ N, $1.6\ \text{g}$ O): the ratio is $2.4/1.6 = 1.5 = 3:2$ — again small whole numbers, and consistent with A, B, C being $\text{N}_2\text{O}_3$ -type, $\text{N}_2\text{O}_5$ -type and an intermediate oxide in oxygen content, though pinning exact formulas needs formula-mass reasoning beyond a mass-ratio check.

Common traps

Forgetting to fix the mass of one element before comparing the masses of the other — Step 2 of the derivation (equalising) must happen first, or the ratio comes out meaningless.
Confusing this law with the law of definite proportions — multiple proportions always compares two different compounds of the same two elements; definite proportions concerns just one compound's fixed composition.
Expecting a "nice" integer ratio on the first calculation — sometimes the raw ratio is a decimal (like $2.5$ in Example 2) and must be scaled up before the small whole numbers appear.
Applying the law to two substances that are not built from the same pair of elements — the comparison is invalid unless both compounds share exactly elements X and Y.

Ask the tutor: if your ratio in a multiple-proportions problem doesn't reduce to small whole numbers, ask the AI tutor to check whether you fixed the right element's mass before comparing.

How is multiple proportions different from definite proportions?

Definite proportions is about one compound: however you make pure water, it is always hydrogen and oxygen in the same $1:8$ mass ratio. Multiple proportions needs two compounds of the same two elements, like CO and $\text{CO}_2$ , and compares how the mass of one element scales between them for a fixed mass of the other. One law fixes a single compound's recipe; the other compares two recipes made from the same ingredients.

What if my ratio does not come out as a whole number at all, not even after scaling?

First recheck that you actually fixed the same element's mass in both compounds before comparing — mixing up which element is "fixed" and which is "compared" is the most common source of a messy ratio. If the fixed-element masses genuinely differ between the two data sets, scale one compound's numbers up or down until they match before taking the ratio of the other element, exactly as Step 2 of the derivation describes. Rounding in the original experimental data can also blur a ratio slightly, but it should still be recognisably close to a small integer.

Why does this law support Dalton's atomic theory?

Because a small whole-number mass ratio is precisely what you would predict if compounds are built from countable atoms combining in fixed proportions. If one nitrogen atom bonds to two oxygen atoms in one oxide and to five oxygen atoms in another, the oxygen mass ratio between those oxides (for the same amount of nitrogen) must land on $2:5$ — never $2.37:5$ or some other fraction — because you cannot have a fractional atom. Dalton used exactly this observed pattern, seen across many element pairs, as evidence that matter is discrete rather than continuous.