Counting by weighing: the mole, the Avogadro constant and relative mass
By the end of this lesson you will be able to
- Explain why chemists count particles by weighing them, and define relative atomic mass, relative molecular mass and relative formula mass.
- Calculate a relative atomic mass from isotope abundances, and a relative molecular or formula mass from a formula, including formulas with brackets and water of crystallisation.
- State what a mole is and the value of the Avogadro constant, and outline why it was redefined as an exact number in 2019.
- Use n = m ÷ M in all three arrangements, with the unit g mol⁻¹ and the answer to the significant figures the data allow.
- Convert between moles and numbers of particles, and between kilograms and grams before any calculation.
You met this before
- An atom is a nucleus of protons and neutrons with electrons in shells around it; the mass number A is protons plus neutrons, and almost all the mass is in the nucleus (Year 2).
- Isotopes are atoms of the same element (same number of protons) with different numbers of neutrons, so different mass numbers: chlorine-35 and chlorine-37, carbon-12 and carbon-14 (Year 2).
- The relative atomic mass on the periodic table is a weighted average of the isotopes: chlorine is 35.5 because three atoms in four are chlorine-35 and one in four is chlorine-37 (Year 2).
- A chemical formula counts atoms: H₂O is two H and one O; Ca(OH)₂ is one Ca, two O and two H; a symbol equation is balanced by counting each element on both sides (Year 3).
- Mass is conserved in every chemical reaction because atoms are neither created nor destroyed, only rearranged — the fizzing flask on the balance loses mass only because a gas escapes (Year 3).
Oxygen has three isotopes with 8, 9 and 10 neutrons. State the mass number of the isotope with 9 neutrons and write it with its symbol.
Reveal
Oxygen has 8 protons, so A = 8 + 9 = 17: oxygen-17, written ¹⁷O. (This was a 2-mark question in May 2021; the markscheme gave both marks for 17 alone and took one away if the answer carried a unit such as “g” — a mass number is a count, not a mass.)
Revise it in full: Year 2 · Atoms and their structure · Lesson 5 — Isotopes and their uses · Year 3 · Energy and chemical change · Lesson 2 — Formulas, symbol equations and balancing · Year 3 · Energy and chemical change · Lesson 1 — The mass that never disappears
The puzzle
Put 22.99 g of sodium into a jar of chlorine and let it burn. It uses up exactly 35.45 g of chlorine — never a milligram more. Put in 45.98 g of sodium and it uses exactly 70.90 g.
The sodium seems to know how much chlorine to take. Now a stranger question: a can of cola has 39 g of sugar in it. How many sugar molecules is that?
You cannot see them, you cannot count them, and no balance in the world reads a single molecule. Yet by the end of this lesson you will know the answer to three significant figures — and you will see that the two questions are the same question.
A hardware shop sells screws by mass, not by counting. If 100 screws have a mass of 250 g, how many screws are in a 2.00 kg box? What do you have to assume about the screws for the answer to be right?
Compare your answer
One screw is 2.50 g, so 2.00 kg = 2 000 g ÷ 2.50 g = 800 screws. You assume every screw has the same mass (or that 2.50 g is a reliable average).
That is exactly how chemists count atoms: the balance gives a mass, the “mass of one” is known, and division does the counting. Notice, too, that you converted kilograms to grams before dividing — if you had divided 2.00 by 2.50 you would have got 0.8 screws, a factor of 1 000 out.
Keep that mistake in mind: the markscheme for a November 2021 question allowed only one mark of two for it, and the May 2023 examiners’ report said many candidates made it.
The science — structured notes
1.1 Why chemists count by weighing
A balanced equation is a recipe written in particles: 2Na + Cl₂ → 2NaCl says two atoms of sodium react with one molecule of chlorine. But nobody can pick out two atoms; a laboratory works in grams.
The whole of this unit — stoichiometry, from the Greek for “element” and “measure” — is the bridge between the two: a way of turning a mass you can weigh into a number of particles you can put in an equation, and back again.
The bridge works because atoms of one element all have (very nearly) the same mass, and because the masses of different atoms are in fixed ratios. A magnesium atom is always about twice as heavy as a carbon atom (24.31 to 12.01).
So 24.31 g of magnesium and 12.01 g of carbon contain the same number of atoms — whatever that number is — just as 500 g of 2.5 g screws and 200 g of 1.0 g nails contain the same number of items.
Weigh in the ratio of the atomic masses and you have counted in the ratio of the atoms. That is the entire idea; the rest of the lesson gives it names and numbers.
1.2 Relative atomic mass: the weighted mean of the isotopes
Atoms are far too light to weigh in grams conveniently — one carbon atom is about 2 × 10⁻²³ g — so chemists compare them with a standard. The relative atomic mass, Ar, of an element is the average mass of its atoms compared with one twelfth of the mass of a carbon-12 atom.
It has no unit; it is a ratio. Carbon-12 is defined as exactly 12, so hydrogen (1.01) is about a twelfth as heavy as carbon and magnesium (24.31) about twice as heavy.
The word average matters. Most elements are mixtures of isotopes, and the periodic table value is the mean weighted by how common each isotope is:
Natural chlorine is 75 % chlorine-35 and 25 % chlorine-37 (to the nearest per cent). Calculate its relative atomic mass.
Weight each isotope by its abundance.
Ar = (35 × 75 + 37 × 25) ÷ 100
Ar = (2 625 + 925) ÷ 100
Ar = 3 550 ÷ 100 = 35.5
Check it makes sense. The answer must lie between 35 and 37, and closer to 35 because chlorine-35 is the commoner isotope. It does. (Using the precise abundances 75.76 % and 24.24 % and the precise isotope masses 34.97 and 36.97 gives 35.45 — the value on a detailed table. The IB papers use 35.5.)
Magnesium is 79.0 % magnesium-24, 10.0 % magnesium-25 and 11.0 % magnesium-26. Calculate its relative atomic mass to three significant figures, and state the number of neutrons in the heaviest isotope.
Reveal answer
Ar = (24 × 79.0 + 25 × 10.0 + 26 × 11.0) ÷ 100
Ar = (1 896 + 250 + 286) ÷ 100
Ar = 2 432 ÷ 100 = 24.3 (the periodic table gives 24.31)
Magnesium has 12 protons, so magnesium-26 has 26 − 12 = 14 neutrons.
1.3 Relative molecular mass and relative formula mass
For a compound, add up the relative atomic masses of every atom in the formula. If the substance is made of molecules the total is the relative molecular mass, Mr; if it is an ionic compound, which has no molecules, the same total for the formula unit is the relative formula mass. Same arithmetic, different name. Like Ar, both are ratios with no unit.
| Substance | Formula | Atoms counted | Working (Ar: H 1.01, C 12.01, N 14.01, O 16.00, Na 22.99, Mg 24.31, P 30.97, S 32.07, Ca 40.08, Cu 63.55) | Mr |
|---|---|---|---|---|
| Water | H₂O | 2 H, 1 O | 2(1.01) + 16.00 | 18.02 |
| Carbon dioxide | CO₂ | 1 C, 2 O | 12.01 + 2(16.00) | 44.01 |
| Phosphoric acid | H₃PO₄ | 3 H, 1 P, 4 O | 3(1.01) + 30.97 + 4(16.00) = 3.03 + 30.97 + 64.00 | 98.00 |
| Sucrose | C₁₂H₂₂O₁₁ | 12 C, 22 H, 11 O | 12(12.01) + 22(1.01) + 11(16.00) = 144.12 + 22.22 + 176.00 | 342.34 |
| Calcium carbonate | CaCO₃ | 1 Ca, 1 C, 3 O | 40.08 + 12.01 + 48.00 | 100.09 |
| Magnesium hydroxide | Mg(OH)₂ | 1 Mg, 2 O, 2 H | 24.31 + 2(16.00 + 1.01) = 24.31 + 34.02 | 58.33 |
| Calcium nitrate | Ca(NO₃)₂ | 1 Ca, 2 N, 6 O | 40.08 + 2(14.01 + 48.00) = 40.08 + 124.02 | 164.10 |
| Hydrated copper(II) sulfate | CuSO₄·5H₂O | 1 Cu, 1 S, 4 O + 5 water (10 H, 5 O) | 63.55 + 32.07 + 64.00 + 5(18.02) = 159.62 + 90.10 | 249.72 |
Three places where the counting goes wrong. Brackets multiply everything inside them: (NO₃)₂ is two N and six O, not two N and three O. Water of crystallisation — the ·5H₂O in a hydrated salt — is five whole water molecules, added on, not multiplied in.
And diatomic elements: the relative molecular mass of oxygen gas O₂ is 32.00, of chlorine Cl₂ 70.90, of hydrogen H₂ 2.02, because the molecule has two atoms; the Ar is for one atom.
The compressed gas once used in aerosol sprays, Freon (trichlorotrifluoroethane), has the molecular formula C₂Cl₃F₃. Use Ar values C 12, Cl 35.5, F 19 to calculate its relative molecular mass.
Count. 2 C, 3 Cl, 3 F.
Add. 2(12) + 3(35.5) + 3(19) = 24 + 106.5 + 57 = 187.5. No unit. In the real question the structure was drawn rather than the formula given, so the first job was to count the atoms in the drawing — and the markscheme gave the first mark for showing the three Ar values or a correct adding-up method, and the second for 187.5.
A student writes: “Mr of ammonium sulfate, (NH₄)₂SO₄ = 14.01 + 4.04 + 32.07 + 64.00 = 114.12 g.” Find three errors and give the correct value.
Reveal answer
(1) The bracket with subscript 2 was ignored — there are 2 N and 8 H: 2(14.01 + 4.04) = 36.10. (2) So the total is wrong: 36.10 + 32.07 + 64.00 = 132.17. (3) A relative formula mass has no unit; “g” is wrong (a molar mass would be 132.17 g mol⁻¹). Ammonium sulfate is a fertiliser, and this exact bracket error is the commonest slip in formula-mass questions.
1.4 The mole: a number with a name
Relative masses tell you that 12.01 g of carbon and 24.31 g of magnesium hold equal numbers of atoms. The mole gives that number a name. One mole of anything contains exactly 6.022 140 76 × 10²³ particles — atoms, molecules, ions, electrons, whatever the “anything” is.
The number is the Avogadro constant, NA = 6.02 × 10²³ mol⁻¹ to three significant figures, which is all the examination needs. The mole is the SI unit of amount of substance, symbol mol; “amount” is the technical word for a number of moles, which is why examiners refuse it for a mass or a volume.
The number was chosen so that the arithmetic falls out neatly: the mass of one mole of an element, in grams, is numerically its Ar; the mass of a mole of a compound is its Mr in grams.
This is the molar mass, M, and unlike Ar and Mr it has a unit: g mol⁻¹ (grams per mole, also written g/mol). A mole of carbon atoms is 12.01 g; a mole of water molecules is 18.02 g, a little more than a tablespoon; a mole of phosphoric acid is 98.00 g.
When a question says “calculate the molar mass and give the unit”, the unit is worth a mark of its own: 98 alone scored one of two in November 2023; 98 g mol⁻¹ scored both.
How big is 6.02 × 10²³? A mole of seconds is about 1.9 × 10¹⁶ years, more than a million times the age of the universe. A mole of water molecules (18 cm³) is about a tablespoonful, which tells you how small a molecule is: one water molecule has a mass of 18.02 g ÷ 6.02 × 10²³ = 2.99 × 10⁻²³ g.
“One mole of oxygen gas contains 6.02 × 10²³ oxygen atoms.”
Reveal answer
False. Oxygen gas is O₂, so one mole of it is 6.02 × 10²³ molecules, which contain 1.20 × 10²⁴ atoms. Always say what the particle is: “one mole of oxygen molecules” or “one mole of oxygen atoms”. The molar mass of O₂ is 32.00 g mol⁻¹, not 16.00.
1.5 The equation that runs the unit: n = m ÷ M
If one mole has a mass of M grams, then a mass m contains m ÷ M moles:
n = mM
m = n × M
M = mn
with n the amount in mol, m the mass in grams, and M the molar mass in g mol⁻¹. Check the units cancel: g ÷ g mol⁻¹ = mol. The number of particles is then N = n × NA.
Every problem in this unit begins by converting the mass in front of you into moles with this equation and ends by converting moles back into whatever the question asks — a mass, a number of molecules, later a volume or a concentration.
A can of cola contains 39 g of sucrose. The molar mass of sucrose is 342.3 g mol⁻¹. Calculate the number of moles of sucrose in the can, to three significant figures, and the number of sucrose molecules.
Moles. n = m ÷ M
n = 39 g ÷ 342.3 g mol⁻¹
n = 0.113 935… mol = 0.114 mol (3 s.f.)
The markscheme gave one mark for the division and one for the rounding; 0.114 on its own scored both, 0.1139 scored one.
Molecules. N = n × NA
N = 0.1139 × 6.02 × 10²³
N = 6.86 × 10²² molecules
(Use the unrounded 0.1139 in the second step; rounding twice drifts the answer.) That is the answer to the puzzle: sixty-nine thousand million million million molecules in one can.
An electric-car battery contains about 12 kg of lithium. Using Ar(Li) = 7, calculate the number of moles of lithium, to one decimal place.
Convert first. 12 kg = 12 × 1 000 g = 12 000 g. The equation needs grams because M is in grams per mole.
Divide. n = 12 000 g ÷ 7 g mol⁻¹
n = 1 714.28… mol = 1 714.3 mol (1 d.p.)
The trap. 12 ÷ 7 = 1.7 mol scored one mark of two. The same trap came back in May 2023 with 8.00 kg of methane, and the examiners’ report said “many candidates did not convert to grams first and so were a factor of 1000 out”.
Ask whether the answer is a sensible size: a bag of lithium the mass of a small child is not 1.7 mol. (With the detailed value 6.94 g mol⁻¹ the answer is 1 729.1 mol; use whichever Ar the question gives, and say so.)
Calculate: (i) the mass of 0.250 mol of calcium carbonate; (ii) the number of moles of magnesium atoms in 1.00 g of magnesium ribbon and the number of atoms it contains; (iii) the molar mass of a substance if 2.50 mol of it has a mass of 115 g.
Reveal answer
(i) m = n × M
m = 0.250 mol × 100.09 g mol⁻¹
m = 25.02 g = 25.0 g (3 s.f.)
(ii) n = 1.00 ÷ 24.31
n = 0.041 13 mol = 0.0411 mol
N = 0.041 13 × 6.02 × 10²³ = 2.48 × 10²² atoms
(iii) M = m ÷ n
M = 115 g ÷ 2.50 mol = 46.0 g mol⁻¹ — ethanol, C₂H₅OH (46.08), fits.
A student says: “One mole of hydrogen gas has less mass than one mole of chlorine gas, so it must contain fewer molecules.” Explain what is wrong, using the definition of the mole.
Reveal answer
A mole is a fixed number of particles, 6.02 × 10²³, whatever the substance. One mole of H₂ (2.02 g) and one mole of Cl₂ (70.90 g) contain the same number of molecules; the masses differ because each chlorine molecule is about 35 times heavier than each hydrogen molecule, not because there are fewer of them. Mass and number are linked only through the molar mass, m = n × M.
1.6 Mass of one particle, and reading the periodic table
Two more things the mole lets you do. The mass of a single atom or molecule is the molar mass divided by the Avogadro constant: for carbon, 12.01 g mol⁻¹ ÷ 6.02 × 10²³ mol⁻¹ = 1.99 × 10⁻²³ g.
And the periodic table becomes a table of molar masses: every element’s number below the symbol is its molar mass in g mol⁻¹ as well as its relative atomic mass.
The IB examination papers round these — H 1, C 12, N 14, O 16, Na 23, Mg 24, S 32, Cl 35.5, Ca 40, Fe 56, Cu 63.5 — and a question that gives you a value expects you to use that value.
In this course we use two-decimal values (H 1.01, C 12.01, O 16.00, Mg 24.31, Fe 55.85…) unless a question says otherwise, and we always say which.
The 2019 redefinition made NA exact. Suppose instead the SI had kept the old definition and a better experiment had found NA to be 6.022 140 90 × 10²³ rather than 6.022 140 76 × 10²³. Would the mass of a mole of carbon-12 have changed, or the number of atoms in 12 g of it? What changes under the 2019 definition?
Reveal answer
Under the old definition the mole was tied to 12 g of carbon-12, so that mass stayed exactly 12 g by definition and the number of atoms in it would have been revised upward.
Under the 2019 definition the number is fixed forever at 6.022 140 76 × 10²³, so a better experiment would instead slightly revise the mass of a mole of carbon-12 (it is now 11.999 999 9958 g with a tiny uncertainty).
Either way the change is in the eighth significant figure — invisible to every calculation in this course, but it is why “12 g of carbon-12” has quietly stopped being the definition.
Boron has two isotopes, boron-10 and boron-11, and a relative atomic mass of 10.8. Calculate the percentage abundance of each isotope. Then explain why a sample of boron extracted from a particular mine could have a slightly different Ar from the table value.
Reveal answer
Let the fraction of boron-11 be x; then boron-10 is (1 − x).
11x + 10(1 − x) = 10.8
10 + x = 10.8
x = 0.8: 80 % boron-11, 20 % boron-10.
Isotope abundances are averages over the Earth; physical and chemical processes (evaporation, crystallisation, biological uptake) very slightly favour one isotope over the other, so ores from different places carry slightly different mixtures and their Ar differs in the third or fourth figure.
This is why some elements on modern tables are given as a range (boron 10.806–10.821) rather than a single number — and why isotope ratios can be used to trace where a material came from.
Try it — counting carbon dioxide on a balance
The lab tips sodium hydrogencarbonate, NaHCO₃, into excess hydrochloric acid in a flask standing on a balance: NaHCO₃ + HCl → NaCl + H₂O + CO₂. The carbon dioxide escapes and the reading falls.
In Year 3 this lab showed that mass is conserved (seal the flask and nothing is lost). Today you use it to count: the equation says one formula unit of NaHCO₃ (84.01) makes one molecule of CO₂ (44.01), so the mass lost should be 44.01 ÷ 84.01 = 0.524 of the mass of solid added, whatever that mass is.
A straight line through the origin with that gradient is the mole ratio made visible.
Suppose your mean losses were: 1.00 g → 0.52 g; 2.00 g → 1.04 g; 3.00 g → 1.56 g; 4.00 g → 2.10 g; 5.00 g → 2.62 g.
(a) Show with numbers whether mass lost is directly proportional to mass added, and find the gradient.
(b) Use the gradient to calculate the relative formula mass of NaHCO₃, given that the gas is CO₂ (Mr = 44.01) and one mole of gas is released per mole of solid.
(c) Predict the loss for 8.00 g of solid, and state one assumption the prediction depends on.
Reveal answer
(a) Ratios loss ÷ added: 0.520, 0.520, 0.520, 0.525, 0.524 — constant to within 1 %, and doubling 2.00 g to 4.00 g doubles the loss (1.04 to 2.10 g); with the line through the origin (no solid, no loss), the relationship is directly proportional, gradient 0.52 g of CO₂ per g of NaHCO₃ (0.522 as a mean).
(b) Gradient = Mr(CO₂) ÷ Mr(NaHCO₃)
Mr(NaHCO₃) = Mr(CO₂) ÷ gradient
Mr(NaHCO₃) = 44.01 ÷ 0.522 = 84.3
That agrees with the calculated 84.01 within 0.4 %. A balance has just measured a formula mass.
(c) 8.00 × 0.522 = 4.18 g (4.19 with 0.524). The prediction assumes the acid is still in excess: 8.00 g of NaHCO₃ is 0.0952 mol, and 100 cm³ of 2 mol dm⁻³ HCl is 0.200 mol, so it is — but at 20 g of solid it would not be, and the line would bend flat. Lesson 2 gives that bend a name: the limiting reactant.
Common mistakes to avoid
- Dividing kilograms by g mol⁻¹ — convert to grams first. 12 kg of lithium is 12 000 g, giving 1 714.3 mol, not 1.7.
- A unit on Ar or Mr — they are ratios. The unit belongs on the molar mass: g mol⁻¹. And “calculate the molar mass and give the unit” means write the unit for a mark.
- Ignoring brackets — Ca(NO₃)₂ has six oxygen atoms; Mg(OH)₂ has two of each.
- Treating ·5H₂O as multiplication — it is five whole water molecules added: 5 × 18.02 = 90.10.
- Using Ar for a diatomic gas — the molar mass of O₂ is 32.00 g mol⁻¹, of Cl₂ 70.90.
- Rounding to the wrong figures — 0.114 mol when 3 s.f. are asked; 1 714.3 when one decimal place is asked; round once, at the end.
- “Amount of water” — amount means moles. Write mass of water (g) or volume of water (cm³).
- Saying a chlorine atom has mass 35.5 — 35.5 is a weighted mean; the atoms are 35 or 37.
Study smart: how this is examined
Study routine. Today: write the three definitions (Ar, mole, molar mass) and n = m ÷ M on a card, then calculate the Mr of the eight substances in the table from memory of the Ar values.
In two days: redo the cola and lithium EXAMPLES with the book closed, timing yourself, and check the s.f. and the units. In a week: run the lab as a full investigation, do the Think task and the quiz, then explain to someone else why 22.99 g of sodium uses exactly 35.45 g of chlorine.
How we found out — two centuries of counting the uncountable
The idea that elements combine in fixed ratios of mass came first. Antoine Lavoisier showed in 1789 that mass is conserved in reactions; Joseph Proust, working in Spain between 1794 and 1804, showed that a compound always contains its elements in the same proportions by mass, however it is made; and John Dalton in Manchester explained both in 1803–1808 with atoms — each element’s atoms all alike, combining in whole-number ratios.
Dalton drew up the first table of relative atomic weights, with hydrogen as 1, but he had no way to know how many atoms of each element a compound held: he assumed water was HO, and got oxygen’s weight wrong by a factor of two.
For fifty years chemists argued over atomic weights because nobody could count.
The key had been published in 1811 and ignored. Amedeo Avogadro, in northern Italy, proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules — which, combined with Gay-Lussac’s measurements of reacting gas volumes, showed that water must be H₂O and that hydrogen and oxygen gases are diatomic.
It took Stanislao Cannizzaro, handing out his pamphlet at the Karlsruhe Congress of 1860, to persuade chemists to use Avogadro’s rule to fix a single consistent set of atomic weights; Mendeleev, who was there, built the periodic table on them within a decade.
Avogadro never knew the number that carries his name: the first estimate came from Johann Loschmidt in 1865, from the physics of gases, and the first real measurement from Jean Perrin in 1908–1909, who watched tiny resin spheres jiggling in water under Brownian motion and calculated from their motion how many molecules were pushing them, arriving at about 6.5–7 × 10²³.
Perrin proposed calling it Avogadro’s number, won the Nobel Prize in 1926, and his work ended the last serious doubt that atoms exist. Wilhelm Ostwald had already coined the word “mole” (the German Mol, short for Molekül, for a molecular weight in grams) in the 1890s; it became an SI base unit in 1971 and, in 2019, an exact number.
Dalton assumed water was HO. Accurate analysis shows that 8 g of oxygen combine with 1 g of hydrogen. What relative atomic mass does that give oxygen on Dalton’s assumption (his own table said about 7, from the poorer analyses of 1808), and explain why Avogadro’s hypothesis corrected it.
Reveal answer
With HO and hydrogen = 1, one oxygen atom must weigh 8 times a hydrogen atom, so the formula HO gives oxygen 8. Gay-Lussac found that two volumes of hydrogen react with one volume of oxygen to give two volumes of steam; if equal volumes hold equal numbers of molecules (Avogadro), then two hydrogen molecules react with one oxygen molecule, so the formula must be H₂O — two hydrogens per oxygen.
The 8 g of oxygen therefore pairs with two hydrogen atoms, and one oxygen atom is 16 times a hydrogen atom: Ar = 16. Counting the particles (by volume) fixed the formula, and the formula fixed the atomic mass.
Where it is used now
- Pharmacy. A dose is a number of molecules, delivered as a mass: a 300 mg aspirin tablet (C₉H₈O₄, M = 180.17 g mol⁻¹) is 1.67 × 10⁻³ mol, about 1.0 × 10²¹ molecules. Every drug label, every intravenous drip and every blood-test result in mmol dm⁻³ is the mole at work; a pharmacist who confuses grams with moles harms someone.
- Fertiliser blending. Ammonium sulfate (132.17) and ammonium nitrate (80.06) are sold by mass but bought for their nitrogen: the percentage of N in each — 21.2 % and 35.0 % — is a molar-mass calculation (Lesson 3), and it decides the price per tonne.
- Soda ash from Sua Pan. Botswana’s soda-ash plant sells sodium carbonate, Na₂CO₃ (105.99 g mol⁻¹), to glass-makers who calculate the mass needed from the moles of sodium oxide the glass recipe requires.
- Carbon accounting. Burning 1 mol of carbon (12.01 g) makes 1 mol of CO₂ (44.01 g): every tonne of coal carbon becomes 3.66 tonnes of carbon dioxide. National emissions inventories are mole ratios multiplied by fuel masses.
- Isotope ratios as fingerprints. Because Ar varies slightly with origin, mass spectrometers reading isotope ratios can tell where an ivory tusk, a wine or a stolen metal came from, and carbon-14 in a bone gives its age.
Science now — and the scientist you need to be
What is happening in 2026
The mole was made exact in 2019 because metrologists could count the atoms in a silicon sphere to eight figures; the same year the kilogram stopped being a lump of metal in Paris and became a number too, defined through the Planck constant.
Chemistry in 2026 runs on that exactness at scales Perrin never imagined. Mass spectrometers in hospital laboratories identify protein molecules from their mass-to-charge ratio, and blood tests report concentrations in millimoles or micromoles per litre; environmental laboratories measure pollutants at nanomoles per litre; and isotope-ratio mass spectrometers read isotope ratios so precisely that the origin of a material — a lithium ore, a diamond, an ivory tusk — can often be traced.
On the industrial side, the mole is the unit that AI-driven laboratories work in: an automated synthesis platform weighs reagents by the milligram and computes the moles of each before it mixes them, and a language model asked to plan a reaction now proposes quantities in moles, which a chemist must still check.
A useful test of any such tool is the one this lesson gives you — ask it for the moles in 12 kg of lithium and see whether it converts the kilograms.
The mindshift
What you need to be able to do
Turn any mass into moles and any moles into a mass or a number of particles, quickly, with the unit converted first and the answer rounded as instructed. Add up a formula mass without missing a bracket.
The examination questions on this content are worth two to four marks each and are among the easiest on the paper for the prepared and the most-lost for everyone else — the reports say the arithmetic is fine and the marks go on units and rounding.
See an example
A 500 mg paracetamol tablet, C₈H₉NO₂:
M = 8(12.01) + 9(1.01) + 14.01 + 2(16.00)
M = 96.08 + 9.09 + 14.01 + 32.00 = 151.18 g mol⁻¹
n = 0.500 ÷ 151.18 = 3.31 × 10⁻³ mol
N = 3.31 × 10⁻³ × 6.02 × 10²³ = 1.99 × 10²¹ molecules
An assistant using more precise Ar values would give 151.16 g mol⁻¹ — a difference in the last figure, not an error; check that, asked for “the amount of paracetamol”, it answers in moles.
Key terms
| Relative atomic mass, Ar | The mean mass of an atom of an element, weighted by isotope abundance, relative to one twelfth of a carbon-12 atom. No unit. |
| Isotopes | Atoms of the same element with different numbers of neutrons, so different mass numbers. |
| Relative molecular mass, Mr | The sum of the Ar of all atoms in a molecule. No unit. |
| Relative formula mass | The same sum for the formula unit of an ionic compound (no molecules exist). No unit. |
| Mole, mol | The SI unit of amount of substance: exactly 6.022 140 76 × 10²³ specified particles. |
| Avogadro constant, NA | 6.02 × 10²³ mol⁻¹ — the number of particles per mole. |
| Molar mass, M | The mass of one mole of a substance, in g mol⁻¹; numerically equal to Ar or Mr. |
| Amount of substance, n | The number of moles; n = m ÷ M. “Amount” means moles, never mass or volume. |
| Water of crystallisation | Water molecules built into a crystal in a fixed ratio, shown as ·xH₂O and added to the formula mass. |
| Significant figures | The digits that carry information; an answer should match the precision of the data or the instruction. |
Think like a scientist
Criterion C A class used the fizzing-flask method to measure the mass of carbon dioxide released when different masses of calcium carbonate (marble chips) react with 100 cm³ of excess 2 mol dm⁻³ hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. The balance read to 0.01 g and each run was left until the reading stopped falling. Results:
| Mass of CaCO₃ / g | Mass lost, trial 1 / g | Trial 2 / g | Trial 3 / g | Mean mass lost / g |
|---|---|---|---|---|
| 1.00 | 0.43 | 0.45 | 0.44 | |
| 2.00 | 0.87 | 0.89 | 0.88 | |
| 3.00 | 1.31 | 1.34 | 1.32 | |
| 4.00 | 1.77 | 1.74 | 1.55 | |
| 5.00 | 2.19 | 2.22 | 2.20 |
(a) Identify the anomalous reading, suggest a specific cause, and calculate the five means to the precision of the data. [3]
(b) Describe the graph you would plot, and test with numbers whether the mass lost is directly proportional to the mass of calcium carbonate. [3]
(c) Using Mr(CO₂) = 44.01 and Mr(CaCO₃) = 100.09, calculate the theoretical mass lost for 3.00 g and compare it with the measured mean, quoting both. [3]
(d) Explain, in terms of moles and the equation, why the gradient of the graph equals Mr(CO₂) ÷ Mr(CaCO₃). [2]
(e) A second class using the same method found a gradient of 0.43, below the theoretical value at every mass. Identify one systematic error that would cause this and state its effect on the calculated Mr of CaCO₃. [2]
(f) Explain why the same experiment with 20.0 g of marble in the same acid would not lie on the line. [2]
Model answer
(a) Trial 3 at 4.00 g, 1.55 g — the other two trials give 1.77 and 1.74, and 1.55 breaks the steady rise of about 0.44 g per gram. A specific cause: the run was stopped before fizzing had finished (the reading was still falling), so less gas had escaped; or part of the marble was not added.
Means, excluding it: 0.44, 0.88, 1.32, 1.76 (mean of 1.77 and 1.74 is 1.755, rounded to 1.76 — state that only two trials were used), 2.20 g — two decimal places, matching a balance reading to 0.01 g.
(b) Mass of CaCO₃ on the horizontal axis (g), mean mass lost on the vertical (g), both labelled with units, even scales, points as crosses, one straight line of best fit.
Test: mass lost ÷ mass added = 0.440, 0.440, 0.440, 0.440, 0.440 — constant; doubling 2.00 g to 4.00 g doubles the loss (0.88 to 1.76 g); and the line passes through the origin (no marble, no gas).
The mass lost is directly proportional to the mass of calcium carbonate, gradient 0.440 g of CO₂ per g of CaCO₃.
(c) n(CaCO₃) = 3.00 ÷ 100.09 = 0.029 97 mol
1 : 1 with CO₂, so n(CO₂) = 0.029 97 mol
m(CO₂) = 0.029 97 × 44.01
m(CO₂) = 1.319 g = 1.32 g
Measured mean 1.32 g: identical to two decimal places (the theoretical gradient 44.01 ÷ 100.09 = 0.4397 against the measured 0.440).
(d) The equation gives 1 mol of CO₂ for every 1 mol of CaCO₃. If a mass m of CaCO₃ is m ÷ 100.09 mol, the CO₂ released is the same number of moles, with mass (m ÷ 100.09) × 44.01.
So mass lost = m × (44.01 ÷ 100.09): a straight line through the origin whose gradient is the ratio of the two formula masses — the mole ratio (1 : 1) multiplied by the ratio of molar masses.
(e) Carbon dioxide is slightly soluble in water, so some stays dissolved in the 100 cm³ of solution instead of escaping: every reading is a little low by about the same amount, lowering the gradient.
(Also acceptable: stopping each run before the reading is fully steady.) Because Mr(CaCO₃) = 44.01 ÷ gradient, a gradient that is too low gives an Mr that is too high — a systematic error that no number of repeats removes.
(f) 20.0 g of CaCO₃ is 0.200 mol and needs 0.400 mol of HCl, but 100 cm³ of 2 mol dm⁻³ acid contains only 0.200 mol. The acid runs out with half the marble unreacted, so only 0.100 mol of CO₂ (4.40 g) is released instead of the 8.79 g the line predicts. The acid is no longer in excess: it has become the limiting reactant, which is the subject of Lesson 2.
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth. Use H 1.01, C 12.01, N 14.01, O 16.00, S 32.07, Cl 35.45, Fe 55.85, Cu 63.55 unless a question gives other values.
Calculate the molar mass of sulfuric acid, H₂SO₄, and give the unit.
Reveal answer
2(1.01) + 32.07 + 4(16.00) = 2.02 + 32.07 + 64.00 = 98.09 [1]; unit g mol⁻¹ (or g/mol) [1].
Copper has two isotopes: copper-63 (69.2 %) and copper-65 (30.8 %). Calculate its relative atomic mass to one decimal place.
Reveal answer
(63 × 69.2 + 65 × 30.8) ÷ 100 = (4 359.6 + 2 002) ÷ 100 [1] = 63.6 [1]. (The table value 63.55 uses the precise isotope masses 62.93 and 64.93.)
A steel bridge cable contains 4.20 tonnes of iron. Calculate the number of moles of iron atoms (Ar = 55.85), giving your answer to three significant figures, and the number of atoms.
Reveal answer
4.20 t = 4.20 × 10⁶ g [1]
n = 4.20 × 10⁶ ÷ 55.85 = 7.52 × 10⁴ mol (75 200 mol) [1]
N = 7.52 × 10⁴ × 6.02 × 10²³ = 4.53 × 10²⁸ atoms [1]
(4.20 ÷ 55.85 = 0.0752 mol scores 0 for the first two marks; ECF to the third.)
Explain why the relative atomic mass of chlorine, 35.45, is not a whole number, although every chlorine atom has a whole number of protons and neutrons.
Reveal answer
Chlorine is a mixture of isotopes, chlorine-35 and chlorine-37, each with a whole mass number [1]; the relative atomic mass is the mean weighted by abundance (about three atoms in four are chlorine-35), and a mean of 35 and 37 in that ratio is 35.5 — no single atom has that mass [1].
Calculate the relative formula mass of hydrated magnesium sulfate, MgSO₄·7H₂O (Mg 24.31), and the percentage of its mass that is water.
Reveal answer
Anhydrous part: 24.31 + 32.07 + 64.00 = 120.38
Water: 7 × 18.02 = 126.14 [1]
Total: 120.38 + 126.14 = 246.52 [1]
% water = 126.14 ÷ 246.52 × 100 = 51.2 % [1]
(Epsom salts are more than half water by mass.)
True or false, with a reason: “0.500 mol of carbon dioxide and 0.500 mol of methane contain the same number of molecules and the same number of atoms.”
Reveal answer
False [1]. The same number of molecules, yes — 0.500 × 6.02 × 10²³ = 3.01 × 10²³ each — but CO₂ has 3 atoms per molecule and CH₄ has 5, so the atom counts are 9.03 × 10²³ and 1.51 × 10²⁴ [1].
7. A drop of water has a volume of 0.050 cm³ and water has a density of 1.00 g cm⁻³. Calculate the number of water molecules in the drop. Then estimate how many drops of water would be needed to contain one molecule for every second that has passed since the Big Bang, about 1.4 × 10¹⁰ years ago (1 year = 3.16 × 10⁷ s).
Reveal answer
Mass = 0.050 g
n = 0.050 ÷ 18.02 = 2.77 × 10⁻³ mol [1]
N = 2.77 × 10⁻³ × 6.02 × 10²³ = 1.67 × 10²¹ molecules [1]
Seconds since the Big Bang = 1.4 × 10¹⁰ × 3.16 × 10⁷ = 4.4 × 10¹⁷ s [1]
Drops needed = 4.4 × 10¹⁷ ÷ 1.67 × 10²¹ = 2.6 × 10⁻⁴
That is less than one drop: a single drop holds about 3 800 molecules for every second in the age of the universe [1].
The point is the size of NA: molecules are so small that “one mole” is a mouthful.
8. A hydrated salt has the formula CuSO₄·xH₂O. A 2.50 g sample contains 1.598 g of anhydrous CuSO₄ (Mr = 159.62). Determine x, showing your working.
Reveal answer
Water mass = 2.50 − 1.598 = 0.902 g [1]
n(CuSO₄) = 1.598 ÷ 159.62 = 0.010 01 mol
n(H₂O) = 0.902 ÷ 18.02 = 0.050 06 mol [1]
Ratio water : salt = 0.050 06 ÷ 0.010 01 = 5.00 [1]
So x = 5 and the salt is CuSO₄·5H₂O [1]. (Lesson 2’s lab does exactly this by heating the crystals and weighing what is left.)
Summary
- Chemists count particles by weighing because the masses of atoms are in fixed ratios: masses in the ratio of the Ar values contain equal numbers of atoms.
- Ar is the abundance-weighted mean of the isotopes relative to one twelfth of carbon-12; Mr (or relative formula mass) is the sum for the formula, brackets multiplied, water of crystallisation added. Neither has a unit.
- A mole is 6.02 × 10²³ particles (exactly 6.022 140 76 × 10²³ since 2019). The molar mass M in g mol⁻¹ is numerically Ar or Mr.
- n = m ÷ M with m in grams — convert kilograms first; N = n × NA; round to the significant figures or decimal places the question asks, once, at the end.
- The unit is a mark: molar mass in g mol⁻¹; a mass number has no unit; “amount” means moles.
Next time → Moles in reactions: the mole ratio from a balanced equation, how much product a given mass can make, which reactant runs out first, and why the yield is never quite 100 % — with a lab that drives the water out of blue crystals and weighs what is left.