D. Quantities in Chemical Reactions · Quantities in Chemical ReactionsThe Law of Definite Proportions and Percentage Composition
D3.1 D2.2 D2.1 D1.1 A2.2About 70 minutes
Learning goals- I can state and explain the law of definite proportions.
- I can calculate the percentage composition of a compound from its formula.
- I can calculate percentage composition from experimental mass data.
- I can use percentage composition to find the mass of an element in a sample and to identify a compound.
Words to know
law of definite proportions
A pure compound always contains the same elements in the same proportions by mass, no matter how much there is or where it came from. Also called the law of constant composition.
percentage composition
The percentage by mass of each element in a compound.
mass percent
(mass of part ÷ mass of whole) × 100%.
law of multiple proportions
When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a small whole-number ratio.
pure substance / mixture
A pure substance has a fixed composition; a mixture can have any proportions of its parts.
Water is water
Meltwater from a snowbank in Sudbury, a drop from Lake Superior and water made in a chemistry lab are all the same compound. If you could break each sample down into its elements, you would find the same thing every time: about 11.2% hydrogen and 88.8% oxygen by mass. Not 12%, not 10%. Always the same.
That is the law of definite proportions, and it is one of the reasons chemists can write a single formula, H₂O, for every sample of water.
The law of definite proportions
A pure compound always contains the same elements combined in the same proportions by mass.
The French chemist Joseph Proust argued for this law around 1800, using careful analyses of compounds such as copper carbonate. Some chemists at the time disagreed, because some of their samples were really mixtures. Proust's careful work won, and the law helped John Dalton build his atomic theory.
Why is it true? A compound is made of identical particles, each with a fixed number of each kind of atom. Every water molecule has 2 H and 1 O, so every sample, big or small, has the same ratio of masses. A mixture, such as salt water or air, can have any proportions, so it does not follow this law.
Percentage composition from a formula
The percentage composition of a compound lists the percent by mass of each element. To find it from a formula, use one mole of the compound:
% element = (number of atoms × molar mass of element) ÷ (molar mass of compound) × 100%
The percentages must add to 100% (or very close, because of rounding).
Worked exampleExample 1: Percentage composition of water
Find the percentage composition of water, H₂O.
- M(H₂O) = 2(1.01) + 16.00 = 18.02 g/mol.
- % H = (2 × 1.01) ÷ 18.02 × 100% = 2.02 ÷ 18.02 × 100% = 11.21%
- % O = 16.00 ÷ 18.02 × 100% = 88.79%
- Check: 11.21% + 88.79% = 100.00%.
Water is 11.21% H and 88.79% O by mass.
A 100% bar for glucose. More than half of the mass of sugar is oxygen, even though only a quarter of its atoms (6 of 24) are oxygen atoms; oxygen atoms are heavy.Percentage composition from experimental data
In the lab you cannot read a formula off a powder. Instead you measure masses: for example, you heat a known mass of metal so it combines with oxygen, then weigh the product. Then:
% element = (mass of element ÷ mass of compound) × 100%
Comparing an experimental percentage composition with the values calculated from possible formulas lets you identify the compound or check its purity. You will do this yourself in the next lesson's investigation.
Worked exampleExample 2: Which copper oxide?
Copper forms two oxides, CuO and Cu₂O. A 2.50 g sample of a copper oxide is found to contain 2.00 g of copper. Which oxide is it?
- Experimental: % Cu = 2.00 ÷ 2.50 × 100% = 80.0%; % O = 20.0%.
- CuO: M = 63.55 + 16.00 = 79.55 g/mol; % Cu = 63.55 ÷ 79.55 × 100% = 79.89%.
- Cu₂O: M = 2(63.55) + 16.00 = 143.10 g/mol; % Cu = 127.10 ÷ 143.10 × 100% = 88.82%.
- 80.0% is very close to 79.89% and far from 88.82%.
The sample is copper(II) oxide, CuO.
Worked exampleExample 3: Mass of an element in a sample
Rust is mainly iron(III) oxide, Fe₂O₃. What mass of iron is in 250 g of pure Fe₂O₃?
- M(Fe₂O₃) = 2(55.85) + 3(16.00) = 159.70 g/mol.
- % Fe = 111.70 ÷ 159.70 × 100% = 69.94%.
- Mass of Fe = 0.6994 × 250 g = 174.9 g.
250 g of Fe₂O₃ contains about 175 g of iron. Mining companies use calculations like this to judge how much metal an ore can give.
Where you see percentage composition at home
A bag of lawn or garden fertilizer shows three numbers, such as 21-0-0. The first number is the percentage of nitrogen by mass. Pure ammonium sulfate, (NH₄)₂SO₄, is 21.2% nitrogen, which is why it is sold as 21-0-0. Gardeners and farmers use these percentages to calculate how much fertilizer to spread, so that plants get enough nitrogen without extra running off into lakes and rivers.
Just for funWhy did the compound never worry about gaining weight over the holidays?
The law of definite proportions says it always keeps the same figure.
Hands-on activitySnack-mix percentages (compound or mixture?)
You need: a kitchen scale, three dry snacks (for example cereal squares, raisins and pretzels), two bowls and a calculator. Check for allergies first.
- Make Batch A: weigh 60 g cereal, 30 g raisins and 10 g pretzels into a bowl. Calculate the mass percent of each (60%, 30%, 10%).
- Make Batch B, twice as big, with the same recipe (120 g, 60 g, 20 g). Calculate its percentages.
- Now scoop one small handful from Batch A without stirring. Sort and weigh each part, and calculate the percentages of the handful.
- Answer: Which batches behaved like a compound obeying the law of definite proportions? Which sample showed that a mixture can have any composition?
What to look for: Batches A and B have identical percentages even though their masses differ; that is like a compound. The handful almost certainly has different percentages, because a mixture is not fixed. Chemically, the analogy breaks down in one way: in a real compound the "recipe" is fixed by the formula and can never change, while you could choose any recipe for a snack mix.
Your turn
1Find the percentage composition of carbon dioxide, CO₂.
M = 44.01 g/mol. % C = 12.01 ÷ 44.01 × 100% = 27.29%; % O = 32.00 ÷ 44.01 × 100% = 72.71%.
2Find the percentage of nitrogen in ammonium nitrate, NH₄NO₃, a fertilizer.
M = 2(14.01) + 4(1.01) + 3(16.00) = 80.06 g/mol. % N = 28.02 ÷ 80.06 × 100% = 35.0%.
3A 4.03 g sample of magnesium oxide contains 2.43 g of magnesium. Find its percentage composition. Does it match the formula MgO?
% Mg = 2.43 ÷ 4.03 × 100% = 60.3%; % O = 39.7%. From the formula: % Mg = 24.31 ÷ 40.31 × 100% = 60.31%. Yes, it matches.
4What mass of copper is in 50.0 g of copper(II) sulfate pentahydrate, CuSO₄·5H₂O (M = 249.72 g/mol)?
% Cu = 63.55 ÷ 249.72 × 100% = 25.45%. Mass = 0.2545 × 50.0 = 12.7 g.
5Three nitrogen fertilizers are ammonium nitrate (NH₄NO₃), ammonium sulfate ((NH₄)₂SO₄) and urea (CO(NH₂)₂). Which has the highest percentage of nitrogen?
NH₄NO₃: 35.0%. (NH₄)₂SO₄: 28.02 ÷ 132.17 = 21.2%. Urea: M = 60.07, % N = 28.02 ÷ 60.07 = 46.6%. Urea has the most nitrogen per gram.
6Sample A of a carbon–oxygen compound has a mass of 2.80 g and contains 1.20 g of carbon. Sample B has a mass of 4.40 g and also contains 1.20 g of carbon. Are they the same compound? Explain.
% C in A = 1.20 ÷ 2.80 = 42.9%; in B = 1.20 ÷ 4.40 = 27.3%. The percentages differ, so by the law of definite proportions they are different compounds (they match CO, 42.88% C, and CO₂, 27.29% C).
More practice: warm-up
1State the law of definite proportions in your own words.
A pure compound always has the same elements in the same proportions by mass, whatever the sample's size or source.
2What is the percentage of hydrogen in water?
3What should the percentages in a percentage composition add up to?
100% (or very close, because of rounding).
4Find the percentage of sodium in sodium chloride, NaCl.
22.99 ÷ 58.44 × 100% = 39.34%
5Does air obey the law of definite proportions? Why or why not?
No. Air is a mixture; the amount of water vapour in it, for example, changes from day to day.
More practice: on your own
1Find the percentage composition of glucose, C₆H₁₂O₆.
M = 180.18 g/mol. % C = 72.06 ÷ 180.18 = 39.99%; % H = 12.12 ÷ 180.18 = 6.73%; % O = 96.00 ÷ 180.18 = 53.28%.
2Iron ores contain FeO, Fe₃O₄ or Fe₂O₃. Which oxide has the highest percentage of iron?
FeO: 55.85 ÷ 71.85 = 77.73%. Fe₃O₄: 167.55 ÷ 231.55 = 72.36%. Fe₂O₃: 69.94%. FeO has the highest percentage of iron (one O for every Fe).
3A 10.00 g sample of an iron oxide contains 7.00 g of iron. Which of the three oxides in the last question is it?
% Fe = 7.00 ÷ 10.00 × 100% = 70.0%. That matches Fe₂O₃ (69.94%).
4What mass of nitrogen is in a 20.0 kg bag of pure urea, CO(NH₂)₂?
% N = 46.65%. Mass N = 0.4665 × 20.0 kg = 9.33 kg.
5Two students analyse table salt: one uses a 2 g sample from a salt mine near Goderich, Ontario, the other a 50 g sample of sea salt that has been purified. Predict what they will find for the percentage of sodium, and explain.
Both should find about 39.3% sodium, because pure NaCl always has the same composition (law of definite proportions). The source and the sample size do not matter, as long as the samples are pure.
Challenge
1Nitrogen dioxide, NO₂, and dinitrogen tetroxide, N₂O₄, have exactly the same percentage composition. Show this and explain why it does not break the law of definite proportions.
NO₂: % N = 14.01 ÷ 46.01 = 30.45%. N₂O₄: % N = 28.02 ÷ 92.02 = 30.45%. Both have N : O = 1 : 2. The law says each pure compound has a fixed composition; it does not say two compounds cannot share one. Percentage composition alone cannot tell these two apart; you also need the molar mass (you will see this with molecular formulas).
2What percentage of the mass of blue copper(II) sulfate pentahydrate crystals, CuSO₄·5H₂O, is water? If you heat 10.0 g of crystals until all the water is driven off, what mass should remain?
% water = 5(18.02) ÷ 249.72 × 100% = 90.10 ÷ 249.72 = 36.08%. Mass remaining = 10.0 × (1 − 0.3608) = 6.39 g of anhydrous CuSO₄.
3In one copper oxide, 1.000 g of copper is combined with 0.2518 g of oxygen. In another, 1.000 g of copper is combined with 0.1259 g of oxygen. Find the ratio of the oxygen masses and suggest formulas for the two oxides.
0.2518 ÷ 0.1259 = 2.000, a small whole-number ratio (the law of multiple proportions). The first oxide has twice as much oxygen per copper: CuO and Cu₂O. Check: in CuO, O ÷ Cu = 16.00 ÷ 63.55 = 0.2518.
Quiz
1. Which statement is the law of definite proportions?
2. What is the percentage of oxygen in carbon dioxide, CO₂?
3. A 5.00 g sample of a compound contains 1.50 g of sulfur. What is the percentage of sulfur?
4. Which of these would NOT be expected to obey the law of definite proportions?
5. What mass of iron is in 100. g of Fe₂O₃ (69.94% Fe)?
6. A fertilizer bag is labelled 34-0-0. What does the 34 mean?
Big idea: A pure compound has a fixed composition by mass (law of definite proportions, D3.1). Percentage composition can be calculated from a formula or from lab data (D2.2), and comparing the two identifies a compound. The fertilizer example links to D1.1 (chemical quantities at home), and Proust's work is an example of a scientist's contribution (A2.2).
How to help: Look at a fertilizer bag, a multivitamin label or a food label together and talk about what percentages by mass mean. The snack-mix activity helps students see the difference between a compound (fixed) and a mixture (variable).
Common difficulties:
- Forgetting to multiply by the subscript (in Fe₂O₃ the iron part is 2 × 55.85).
- Confusing "percent of atoms" with "percent by mass". In CO₂, two-thirds of the atoms are oxygen, but oxygen is 72.7% of the mass.
- Dividing by the wrong total: always divide by the mass of the whole compound.
- Thinking the law means no two compounds can share a percentage composition. NO₂ and N₂O₄ do; the law is about each pure compound being consistent.