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Law of definite proportions (Proust)

If you analyze tap water from a faucet in Houston and meltwater from a glacier in Glacier National Park, Montana, will they have the same proportion of hydrogen and oxygen? The answer is yes, always, and that very simple fact is one of the most useful laws in chemistry. A compound like water has a fixed proportion, while a mixture, like salt dissolved in water or a trail mix, can have any proportion you like.

Proust's discovery

At the end of the 18th century, the French chemist Joseph Louis Proust analyzed the same compound obtained from very different sources (natural minerals, laboratory reactions...) and found something surprising: the proportion between the elements that made it up was always identical.

Three samples of water from different origins (rain, laboratory and glacier) have the same composition by mass: 1 g of hydrogen for every 8 g of oxygen. rainwater laboratory water glacier water 1 g H 8 g O 1 g H 8 g O 1 g H 8 g O in all three: 1 g of H for every 8 g of O
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Complete the sentence with the missing words: "When two or more elements combine to form a specific $\dots$, they always do so in the same proportion by $\dots$,…

Complete the sentence with the missing words: "When two or more elements combine to form a specific $\dots$, they always do so in the same proportion by $\dots$, regardless of the amount of $\dots$ or its origin."

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We analyze distilled water from a laboratory and collected, purified rainwater. In what ratio do hydrogen and oxygen combine in each? What does Proust's law say about…

We analyze distilled water from a laboratory and collected, purified rainwater. In what ratio do hydrogen and oxygen combine in each? What does Proust's law say about the origin of the samples?

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Calculating with a fixed proportion

If you know the proportion of a compound, you can calculate the mass of one element from the mass of the other, just as you would with a cooking recipe.

Try the widget below: it splits the 36 g of water from the example above into its $1:8$ parts, hydrogen and oxygen.

Notice that the more hydrogen you add, the more oxygen you need to keep exactly the same $1:8$ proportion.

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In water, hydrogen and oxygen combine in the ratio of $1\ \text{g}$ of hydrogen for every $8\ \text{g}$ of oxygen. How many grams of oxygen will combine with…

In water, hydrogen and oxygen combine in the ratio of $1\ \text{g}$ of hydrogen for every $8\ \text{g}$ of oxygen. How many grams of oxygen will combine with $4\ \text{g}$ of hydrogen?

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To form carbon dioxide, carbon and oxygen combine in the ratio of $3\ \text{g}$ of carbon for every $8\ \text{g}$ of oxygen. What mass of oxygen do you need to…

To form carbon dioxide, carbon and oxygen combine in the ratio of $3\ \text{g}$ of carbon for every $8\ \text{g}$ of oxygen. What mass of oxygen do you need to completely burn $9\ \text{g}$ of carbon?

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Calcium and oxygen combine in the ratio $5:2$ to form quicklime (calcium oxide). a) What mass of calcium combines with $16\ \text{g}$ of oxygen? b) What mass of…

Calcium and oxygen combine in the ratio $5:2$ to form quicklime (calcium oxide).

  1. a) What mass of calcium combines with $16\ \text{g}$ of oxygen?
  2. b) What mass of quicklime is obtained?
Solve
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If $1\ \text{g}$ of hydrogen reacts exactly with $8\ \text{g}$ of oxygen, what mass of water is formed? What other law of chemical reactions did you need besides…

If $1\ \text{g}$ of hydrogen reacts exactly with $8\ \text{g}$ of oxygen, what mass of water is formed? What other law of chemical reactions did you need besides Proust's law?

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In ammonia, nitrogen and hydrogen combine in the ratio $14:3$. a) What mass of hydrogen combines exactly with $28\ \text{g}$ of nitrogen? b) What mass of ammonia is…

In ammonia, nitrogen and hydrogen combine in the ratio $14:3$.

  1. a) What mass of hydrogen combines exactly with $28\ \text{g}$ of nitrogen?
  2. b) What mass of ammonia is formed?
Solve
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In a first experiment, $4\ \text{g}$ of copper combine exactly with $1\ \text{g}$ of oxygen. In a second experiment, we start with twice as much copper. a) What mass of…

In a first experiment, $4\ \text{g}$ of copper combine exactly with $1\ \text{g}$ of oxygen. In a second experiment, we start with twice as much copper.

  1. a) What mass of oxygen will combine with $8\ \text{g}$ of copper?
  2. b) What mass of oxide is obtained in each experiment?
  3. c) Does the combination ratio depend on the amount of sample used? Justify your answer.
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Splitting a total mass

Sometimes you know the total mass of the compound instead of the mass of one element. Then you split the total into parts that follow the fixed ratio. This works for synthesis (building the compound) and for analysis (breaking it down): the ratio is the same in both directions.

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We completely decompose $90\ \text{g}$ of water into its elements (ratio $1:8$ between hydrogen and oxygen). What mass of hydrogen and what mass of oxygen are obtained?

We completely decompose $90\ \text{g}$ of water into its elements (ratio $1:8$ between hydrogen and oxygen). What mass of hydrogen and what mass of oxygen are obtained?

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Magnesium and oxygen combine in the ratio $3:2$ when forming magnesium oxide. If we analyze $20\ \text{g}$ of magnesium oxide, what mass of magnesium and what mass of…

Magnesium and oxygen combine in the ratio $3:2$ when forming magnesium oxide. If we analyze $20\ \text{g}$ of magnesium oxide, what mass of magnesium and what mass of oxygen does it contain?

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In a first experiment, we form water by combining $2\ \text{g}$ of hydrogen with $16\ \text{g}$ of oxygen. In another experiment, we decompose $9\ \text{g}$ of water and…

In a first experiment, we form water by combining $2\ \text{g}$ of hydrogen with $16\ \text{g}$ of oxygen. In another experiment, we decompose $9\ \text{g}$ of water and obtain $1\ \text{g}$ of hydrogen and $8\ \text{g}$ of oxygen.

  1. a) Verify that both results comply with the law of definite proportions.
  2. b) Explain the difference between the two experiments (synthesis and analysis) and reason why the ratio is the same in both.
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Are they the same compound?

Proust's law also works the other way round: if you are given two samples with different masses of the same elements, you can check whether they are consistent with the same compound by calculating the proportion for each one.

Comparison of two samples of iron and oxygen with different masses: the quotient of the masses of Fe and O is 2.33 in both, so they are consistent with the same compound. masses of each sample quotient m(Fe) / m(O) sample A 14 g Fe + 6 g O sample B 21 g Fe + 9 g O 14 / 6 ≈ 2.33 21 / 9 ≈ 2.33 same proportion: consistent with the same compound (the masses differ, but the quotient is the same)
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One water sample contains $2\ \text{g}$ of hydrogen and $16\ \text{g}$ of oxygen; another sample contains $5\ \text{g}$ of hydrogen and $40\ \text{g}$ of oxygen. The…

One water sample contains $2\ \text{g}$ of hydrogen and $16\ \text{g}$ of oxygen; another sample contains $5\ \text{g}$ of hydrogen and $40\ \text{g}$ of oxygen. The figure compares the composition of the two samples.

hydrogen oxygen 16 g 2 g sample A 40 g 5 g sample B

Calculate the ratio between hydrogen and oxygen in each sample and explain whether they could be the same compound.

Solve
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One sample contains $3\ \text{g}$ of carbon and $8\ \text{g}$ of oxygen, and another contains $3\ \text{g}$ of carbon and $4\ \text{g}$ of oxygen. Could they be the same…

One sample contains $3\ \text{g}$ of carbon and $8\ \text{g}$ of oxygen, and another contains $3\ \text{g}$ of carbon and $4\ \text{g}$ of oxygen. Could they be the same compound? Justify your answer using Proust's law (the law of definite proportions).

Solve
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A laboratory analyzes two samples of an iron oxide. Sample A contains $14\ \text{g}$ of iron and $6\ \text{g}$ of oxygen; sample B contains $21\ \text{g}$ of iron and…

A laboratory analyzes two samples of an iron oxide. Sample A contains $14\ \text{g}$ of iron and $6\ \text{g}$ of oxygen; sample B contains $21\ \text{g}$ of iron and $9\ \text{g}$ of oxygen. Calculate the ratio for each and decide if they are the same compound.

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Real measurements and impure samples

In a real lab, measurements are never perfect, so the ratio you calculate is a little off. A small difference is measurement error. A big difference may mean the sample is not pure.

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A class measures the composition of water and obtains $10.2\ \text{g}$ of hydrogen and $79.8\ \text{g}$ of oxygen. a) Calculate the oxygen/hydrogen ratio (to one decimal…

A class measures the composition of water and obtains $10.2\ \text{g}$ of hydrogen and $79.8\ \text{g}$ of oxygen.

  1. a) Calculate the oxygen/hydrogen ratio (to one decimal place) and compare it with the theoretical ratio $8:1$.
  2. b) Should we conclude that Proust's law fails? Reason your answer.
  3. c) Point out two possible causes for the slight discrepancy with the theoretical value.
Solve
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A laboratory receives $250\ \text{g}$ of a white mineral that the supplier claims is pure calcium oxide (calcium:oxygen ratio $5:2$). Analysis reveals $175\ \text{g}$ of…

A laboratory receives $250\ \text{g}$ of a white mineral that the supplier claims is pure calcium oxide (calcium:oxygen ratio $5:2$). Analysis reveals $175\ \text{g}$ of calcium and $75\ \text{g}$ of oxygen.

  1. a) Check whether the mineral could be pure calcium oxide.
  2. b) What mass of oxygen should accompany $175\ \text{g}$ of calcium if it were?
  3. c) Propose an explanation for the result of the analysis.
Solve

Two elements, more than one compound

Sometimes two elements combine in more than one fixed proportion, and each proportion gives rise to a different compound, with different properties.

Atomic diagram of two different oxides of copper: one with one copper atom for each oxygen atom (8 g Cu to 2 g O) and one with two copper atoms for each oxygen atom (8 g Cu to 1 g O). The formulas are shown only as particle models. CuO 1 Cu : 1 O 8 g Cu : 2 g O CuCuO 2 Cu : 1 O 8 g Cu : 1 g O
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Copper can form two different oxides. In one sample, $8\ \text{g}$ of copper combine with $2\ \text{g}$ of oxygen; in another sample, $8\ \text{g}$ of copper combine…

Copper can form two different oxides. In one sample, $8\ \text{g}$ of copper combine with $2\ \text{g}$ of oxygen; in another sample, $8\ \text{g}$ of copper combine with $1\ \text{g}$ of oxygen. The figure compares the composition of the two samples.

sample 1 8 g 2 g sample 2 8 g 1 g copper oxygen

Is it the same oxide? How do you know?

Solve
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We analyze three samples of "iron oxide". Sample A contains $14\ \text{g}$ of iron and $6\ \text{g}$ of oxygen; sample B, $14\ \text{g}$ of iron and $4\ \text{g}$ of…

We analyze three samples of "iron oxide". Sample A contains $14\ \text{g}$ of iron and $6\ \text{g}$ of oxygen; sample B, $14\ \text{g}$ of iron and $4\ \text{g}$ of oxygen; sample C, $21\ \text{g}$ of iron and $9\ \text{g}$ of oxygen.

  1. a) Calculate the iron:oxygen ratio for each sample.
  2. b) Which of them are the same compound?
  3. c) Explain why iron can give rise to more than one different oxide.
Solve
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Carbon and oxygen can form two different gases. In sample A, there are $6\ \text{g}$ of carbon and $8\ \text{g}$ of oxygen; in sample B, there are $6\ \text{g}$ of…

Carbon and oxygen can form two different gases. In sample A, there are $6\ \text{g}$ of carbon and $8\ \text{g}$ of oxygen; in sample B, there are $6\ \text{g}$ of carbon and $16\ \text{g}$ of oxygen. The figure shows the model of each molecule.

sample A C O 6 g of C 8 g of O sample B O C O 6 g of C 16 g of O
  1. a) Calculate the carbon:oxygen ratio for each sample.
  2. b) Are they the same compound?
  3. c) Given that one of the gases is carbon monoxide (highly toxic) and the other is carbon dioxide, explain why Proust's law (the law of definite proportions) allows us to distinguish between them even though they are made of the same elements.
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When there is reactant left over

If you mix two elements without respecting their exact proportion, not everything reacts: the one in excess is left over without combining.

Bars to scale: 10 g of hydrogen compared with 64 g of oxygen, which is only enough for 8 g of hydrogen; the remaining 2 g of hydrogen are left over without reacting. Scale: 8 px = 1 g of H, and the O bar is drawn as the H it can use (64 g O ≙ 8 g H) Hydrogen (10 g) 8 g react 2 g of H left over without reacting Oxygen (64 g) 64 g O equivalent to only 8 g of H Reacts Left over (excess)
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In a sealed container, we place $10\ \text{g}$ of hydrogen and $64\ \text{g}$ of oxygen and cause them to react (the combination ratio is $1:8$). The figure shows the…

In a sealed container, we place $10\ \text{g}$ of hydrogen and $64\ \text{g}$ of oxygen and cause them to react (the combination ratio is $1:8$). The figure shows the container before the reaction.

hydrogen 10 g oxygen 64 g
  1. a) Does all the hydrogen react?
  2. b) What mass of water is formed?
  3. c) Is there any gas left over? What mass?
Solve
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Will $6\ \text{g}$ of carbon combine completely with $16\ \text{g}$ of oxygen to form carbon dioxide, with nothing left over? What mass of carbon dioxide will form?…

Will $6\ \text{g}$ of carbon combine completely with $16\ \text{g}$ of oxygen to form carbon dioxide, with nothing left over? What mass of carbon dioxide will form? (Remember: the ratio is $3\ \text{g}$ of carbon for every $8\ \text{g}$ of oxygen.)

Solve
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We burn $12\ \text{g}$ of carbon with $40\ \text{g}$ of oxygen (the combination ratio is $3:8$). a) How much oxygen is consumed? b) What mass of carbon dioxide is…

We burn $12\ \text{g}$ of carbon with $40\ \text{g}$ of oxygen (the combination ratio is $3:8$).

  1. a) How much oxygen is consumed?
  2. b) What mass of carbon dioxide is formed?
  3. c) What mass of oxygen remains unreacted?
Solve
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We mix $16\ \text{g}$ of hydrogen with $100\ \text{g}$ of oxygen and trigger the reaction (ratio $1:8$). a) What mass of each gas reacts? b) What mass of water is…

We mix $16\ \text{g}$ of hydrogen with $100\ \text{g}$ of oxygen and trigger the reaction (ratio $1:8$).

  1. a) What mass of each gas reacts?
  2. b) What mass of water is formed?
  3. c) What mass of hydrogen is left over?
Solve
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A student claims: "If I add twice as much oxygen, the water that forms will contain twice as much oxygen." To test this, we combine $4\ \text{g}$ of hydrogen with…

A student claims: "If I add twice as much oxygen, the water that forms will contain twice as much oxygen." To test this, we combine $4\ \text{g}$ of hydrogen with $64\ \text{g}$ of oxygen (twice as much as necessary).

  1. a) Calculate the mass of water formed and the mass of oxygen left over.
  2. b) Use Proust's law (the law of definite proportions) to explain why the student is mistaken.
Solve

Summary

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