1The idea in one paragraph

A single atom weighs about 10⁻²³ g, so no balance can weigh one and no one can count them. Chemists get round this the way a bank gets round counting coins: by weighing them in a fixed batch. That batch is the mole, 6.02 × 10²³ particles, and it is chosen so that a mole of any substance has a mass in grams equal to its relative formula mass. 12 g of carbon-12 and 18.02 g of water each contain one mole of particles. Once you can turn a mass, a volume of solution or a volume of gas into moles, you can compare substances particle for particle, which is what an equation does. Figure 1 is the map for the whole subtopic: every conversion goes through the amount in moles.

2The mole and the Avogadro constant

The mole (mol) is the SI unit of amount of substance, symbol n. One mole contains exactly 6.02214076 × 10²³ elementary entities; the data booklet gives the Avogadro constant to three significant figures as 6.02 × 10²³ mol⁻¹. Its unit is per mole, because it is a number of particles per mole of them.

number of particles = amount (mol) × Avogadro constant, and so amount = number of particles ÷ Avogadro constant

An elementary entity is whatever you are counting: an atom, a molecule, an ion, an electron, or a named group such as a formula unit of NaCl. You must say which, because the answer changes. Take 0.250 mol of carbon dioxide.

The same care applies to ions. One mole of CaCl₂ contains one mole of Ca²⁺ and two moles of Cl⁻, three moles of ions in all, so 0.100 mol of CaCl₂ contains 0.300 × 6.02 × 10²³ = 1.81 × 10²³ ions.

"A mole of oxygen" is ambiguous: a mole of O atoms has a mass of 16.00 g, a mole of O₂ molecules 32.00 g. Write the formula every time.

3Relative masses on the carbon-12 scale

Atoms are compared on a relative scale on which one atom of carbon-12 has a mass of exactly 12. Relative atomic mass, Ar, is the weighted mean mass of an element's atoms on that scale (S1.2 showed how the weighting works). Relative formula mass, Mr, is the sum of the Ar values of all the atoms in the formula. Both are ratios of masses, so neither has units. Use the values in the data booklet, to two decimal places.

Brackets multiply everything inside them, and the water in a hydrated salt counts in full. "Relative formula mass" covers ionic compounds too, where there are no molecules to speak of; relative molecular mass is used for molecular substances, and both are written Mr.

4Molar mass: from grams to moles and back

The molar mass, M, is the mass of one mole of a substance, in g mol⁻¹. It has the same number as Mr, with a unit: Mr of water is 18.02, so M of water is 18.02 g mol⁻¹. That matching is the whole reason the mole was defined as it was.

n = m ÷ M amount (mol) = mass (g) ÷ molar mass (g mol⁻¹)

Three rearrangements, three worked lines.

Most questions chain two steps. How many hydrogen atoms are in 5.00 g of water?

Keep the unrounded value in your calculator between steps and round only the final answer, to the number of significant figures in the data (three here).

5Empirical and molecular formulas

The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in one molecule. Glucose has molecular formula C₆H₁₂O₆ and empirical formula CH₂O. Ionic compounds are only ever written as empirical formulas (NaCl, MgCl₂), because there is no discrete molecule to count.

Formula to percentage composition. Divide the mass of each element in one mole by the molar mass.

Percentage composition to empirical formula. Treat the percentages as grams in a 100 g sample, convert each to moles, and find the ratio. Figure 2 sets out the four steps for a compound that is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass.

Dividing by the smallest amount gives 1.00 : 1.99 : 1.00, which is 1 : 2 : 1 within the rounding of the data, so the empirical formula is CH₂O.

The guide flags approximation as a Nature of science point, and it is where marks go. Experimental percentages carry error, so a ratio of 1.99 or 2.02 means 2. But a ratio near 1.5, 1.33 or 1.25 is not rounding error: it means multiply every value by 2, 3 or 4. An oxide of iron that is 69.9% iron by mass gives:

Rounding 1.50 to 2 would give FeO₂, a compound that is not what the data describe.

Empirical to molecular formula. The molecular formula is a whole-number multiple of the empirical one. Divide the molar mass by the empirical formula mass to find the multiple.

From combustion data. The guide's linking question points to a common experiment: burn a known mass of a compound of carbon and hydrogen in excess oxygen and weigh the carbon dioxide and water produced. All the carbon ends up in the CO₂ and all the hydrogen in the H₂O, so n(C) = n(CO₂) and n(H) = 2 × n(H₂O). From those two amounts you find the ratio exactly as above. Try it Q3 does one.

6Molar concentration

A solution's molar concentration is the amount of solute per unit volume of solution:

n = CV amount (mol) = concentration (mol dm⁻³) × volume (dm³)

The volume must be in dm³. Burettes, pipettes and flasks are marked in cm³, and 1 dm³ = 1000 cm³, so divide cm³ by 1000. Concentration is written with square brackets: [NaOH] = 0.250 mol dm⁻³ means "the concentration of sodium hydroxide is 0.250 mol dm⁻³". The guide requires that notation.

Mass concentration is in g dm⁻³. Convert between the two with the molar mass:

Worked example. 2.50 g of sodium hydroxide (M = 40.00 g mol⁻¹) is dissolved and made up to 250.0 cm³ of solution. Calculate [NaOH] in both units.

Ions in solution. A formula unit may release more than one of an ion. In 0.150 mol dm⁻³ MgCl₂, [Mg²⁺] = 0.150 mol dm⁻³ but [Cl⁻] = 2 × 0.150 = 0.300 mol dm⁻³.

Making a standard solution. A standard solution is one whose concentration is known accurately. Figure 3 shows the steps. The glassware is chosen for accuracy: a balance reading to at least two decimal places, and a volumetric flask, which holds one exact volume when filled to its mark.

To make 250.0 cm³ of 0.100 mol dm⁻³ copper(II) sulfate from the blue hydrated crystals, you need 0.100 × 0.2500 = 0.0250 mol, which is 0.0250 × 249.72 = 6.24 g of CuSO₄·5H₂O. Using the anhydrous molar mass by mistake gives the wrong mass, and the wrong concentration.

Dilution. Adding water changes the volume but not the amount of solute, so c₁V₁ = c₂V₂. Pipetting 25.0 cm³ of 0.250 mol dm⁻³ NaOH into a 250.0 cm³ volumetric flask and making up to the mark gives 0.250 × 25.0 ÷ 250.0 = 0.0250 mol dm⁻³. (The volumes here can stay in cm³, because the same unit is on both sides.) A serial dilution, Figure 4, repeats a fixed dilution step to make a set of standards, each a known fraction of the last. That set is what you need for a calibration curve: measure a property such as absorbance for each standard, plot it against concentration, and read the concentration of an unknown off the line. Try it Q5 is a calibration question.

7Avogadro's law and gas volumes

Avogadro's law: equal volumes of all gases, measured at the same temperature and pressure, contain equal numbers of molecules.

Figure 5 shows why this is surprising and why it works. A molecule of carbon dioxide is far bigger and heavier than one of hydrogen, but in a gas the molecules are so far apart that their own size hardly matters: the volume is almost all empty space. So the volume of a gas depends on how many molecules there are, not on what they are. (S1.5 explains the limits of that assumption.)

The useful consequence: for gases at the same temperature and pressure, the ratio of volumes is the ratio of moles, so the coefficients of a balanced equation give volume ratios directly. Figure 6 shows it for the synthesis of ammonia.

Worked example. 20 cm³ of propane is burned in 150 cm³ of oxygen. All volumes are measured at the same room temperature and pressure, at which water is a liquid. Find the volume and composition of the gas at the end.

Only gases follow the volume ratio. A liquid or a solid in the equation takes up almost no volume compared with a gas, so leave it out of the volume count. To turn a gas volume into moles directly you need the molar volume at a stated temperature and pressure, which S1.5 covers.