Master IB Chemistry S1.4: Counting particles, mass. The mole with notes created by examiners and strictly aligned with the syllabus.
S1.4.1 The mole and Avogadro constant
S1.4.2 Relative atomic mass and relative formula mass
S1.4.3 Molar mass and mass–amount–particle calculations
S1.4.4 Empirical and molecular formulas
Atoms, ions and molecules are much too small to count one at a time in the laboratory. Yet reactions still take place particle by particle. The mole gives chemists the link they need: it connects the invisible particle scale with masses and volumes that can actually be measured.
A mole is an SI unit of amount of substance that contains exactly 6.02214076 × 10²³ specified elementary entities. The unit symbol is mol. An amount of substance is a physical quantity that measures how many specified elementary entities are present, using the mole as its unit.
An elementary entity is the specified particle, or specified group of particles, being counted. It may be an atom, molecule, ion, electron, formula unit, or a stated group such as pairs of ions. Always state what entities your mole refers to: 1 mol of oxygen atoms is not the same thing as 1 mol of oxygen molecules.
Samples of different substances that are one mole each contain the same number of specified entities, but they do not have the same mass. That is why 1 mol of aluminium, 1 mol of water molecules and 1 mol of sodium chloride formula units look like very different quantities on the bench.

The Avogadro constant is the physical constant equal to the number of elementary entities per mole. Its symbol is Nₐ and its unit is mol⁻¹. The value is given in the data booklet.
The particle–mole relationship is:
N = nNₐ, where N is the number of specified elementary entities (unit 1, a count), n is the amount of substance (mol) and Nₐ is the Avogadro constant (mol⁻¹).
To convert from moles to particles, multiply by Nₐ. To convert from particles to moles, divide by Nₐ. It’s like converting between kilograms and grams; the conversion factor is just much larger.
If 0.250 mol of H₂O molecules is present, there are 0.250 mol of water molecules. Each water molecule contains three atoms, so the same sample contains 0.500 mol of hydrogen atoms and 0.250 mol of oxygen atoms, or 0.750 mol of atoms in total. Read the wording carefully: atoms, molecules, ions and electrons are different things to count.
Atomic masses are extremely small, so chemists use a relative scale instead. Relative atomic mass is the weighted mean mass of the atoms of an element compared with one-twelfth of the mass of an atom of carbon-12. Its symbol is Aᵣ. It has no unit, since it’s a ratio.
Relative formula mass is the mass of one formula unit of a substance compared with one-twelfth of the mass of an atom of carbon-12. Its symbol is Mᵣ, and it has no unit either. “Formula” is the useful word here, because it can refer to molecular substances, ionic compounds and giant covalent substances.
A formula unit is the simplest whole-number collection of particles shown by the formula of a substance. In an ionic compound such as MgCl₂, the formula unit contains one Mg²⁺ ion and two Cl⁻ ions, even though the solid forms a giant lattice rather than separate molecules.
To find Mᵣ, add the Aᵣ values for every atom in the formula. In calculations, use the relative atomic masses from the data booklet to two decimal places, as required by the guide.
For example, for Ca(NO₃)₂:
Mᵣ = Aᵣ(Ca) + 2 × Aᵣ(N) + 6 × Aᵣ(O)
Pay attention to brackets. The subscript 2 outside (NO₃) doubles everything inside the bracket. Hydrates follow the same idea: in CuSO₄·5H₂O, the 5 applies to the entire H₂O group.
Because relative masses are ratios, don’t write g or g mol⁻¹ after Aᵣ or Mᵣ. Use units for molar mass in the next section.
As you go down a group in the periodic table, atoms generally have larger relative atomic masses because they contain more protons and neutrons. Some properties can be linked partly to this increase, including density trends and the mass contribution to melting or boiling behaviour. Don’t give mass too much credit, though: reactivity, metallic character and ion formation are usually explained more directly by electronic structure, nuclear charge and atomic radius.
Molar mass means the mass of one mole of a substance. Its symbol is M, and in this topic the usual chemistry unit is g mol⁻¹. For IB calculations, its numerical value is the same as Aᵣ for atoms and Mᵣ for formulas, to the precision needed.
For example, if Mᵣ(H₂O) = 18.02, then M(H₂O) = 18.02 g mol⁻¹. The first value has no units because it is a relative ratio. The second gives mass per mole.
Use this central relationship:
n = m / M, where m is mass (g in these chemistry calculations; 1 g = 10⁻³ kg) and M is molar mass (g mol⁻¹).
Rearrange it when you need to:
Keep units consistent. If M is in g mol⁻¹, use m in g. If the question gives kg or mg, convert the mass before putting it into the equation.
Most calculations here join two conversions:
mass ⇄ amount in mol ⇄ number of particles
Use n = m / M to go from mass to moles, then N = nNₐ to go from moles to particles. With compounds, check the formula carefully because one formula unit can contain more than one atom of a particular element. One mole of Al₂O₃ formula units contains 2 mol of aluminium ions and 3 mol of oxide ions.
A balanced chemical equation gives mole ratios, not mass ratios. Molar mass lets you change a measured mass of reactant into moles, apply the coefficients in the equation, then change moles of product back into mass. That’s the calculation pathway used to predict masses of products in stoichiometry.
An empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. A molecular formula gives the actual number of atoms of each element in one molecule.
For molecular substances, these two formulas may not match. C₆H₁₂O₆ has empirical formula CH₂O, since 6:12:6 simplifies to 1:2:1. For ionic compounds, the formula normally used is already an empirical formula because it shows the simplest ratio of ions in the lattice. Examples comparing molecular formulas with empirical formulas.
| Substance | Molecular formula | Atom ratio | Empirical formula | Outcome |
|---|---|---|---|---|
| Water | H₂O | 2:1 | H₂O | Already simplest |
| Carbon dioxide | CO₂ | 1:2 | CO₂ | Already simplest |
| Hydrogen peroxide | H₂O₂ | 2:2 → 1:1 | HO | Simplified |
| Glucose | C₆H₁₂O₆ | 6:12:6 → 1:2:1 | CH₂O | Simplified |
| Benzene | C₆H₆ | 6:6 → 1:1 | CH | Simplified |
Percentage composition by mass is the percentage of a compound’s mass contributed by each element. For an element X in a compound:
ωₓ = (mₓ / mcompound) × 100%, where ωₓ is the percentage by mass of element X (%), mₓ is the mass of element X in the sample (g), and mcompound is the mass of the compound sample (g).
To work out an empirical formula from percentages, follow this routine:
Rounding matters, but it has to be sensible. Ratios such as 1.00, 1.99 and 3.01 can be rounded to 1, 2 and 3 because small deviations are expected from measurement uncertainty. A ratio such as 1.33 should not be rounded to 1; it usually suggests multiplying all terms by 3 to give a whole-number ratio.
The molecular formula is a whole-number multiple of the empirical formula. Find the multiplier using:
k = M / Memp, where k is the formula multiplier (unit 1) and Memp is the molar mass of the empirical formula (g mol⁻¹).
Next, multiply every subscript in the empirical formula by k. If the data are reliable, k should be very close to a whole number.
You can also derive empirical formulas from mass changes in reactions. In a simple oxide experiment, a known mass of metal is heated in air until the mass becomes constant. The gain in mass is the mass of oxygen that combined with the metal. Convert the metal mass and oxygen mass into moles, then find the simplest ratio.

The repeated heat–cool–weigh cycle matters. Heating to constant mass gives evidence that the reaction is complete; otherwise the calculated formula may contain too little oxygen. A lid reduces loss of solid while still allowing oxygen to enter. Realistic improvements include using a more precise balance, heating for longer, controlling air access better, and avoiding loss of powder when lifting the lid.
Combustion analysis uses the same mole logic. In complete combustion of a compound containing carbon and hydrogen, all carbon atoms end up in CO₂ and all hydrogen atoms end up in H₂O. From the measured masses of CO₂ and H₂O, the moles of C and H in the original compound can be calculated.
A graph can show fixed composition too. If different groups heat different masses of magnesium, a plot of mass of magnesium oxide against mass of magnesium should be close to a straight line. Anomalies show experimental error, and the gradient reflects the constant mass ratio in the compound.

A solution is a homogeneous mixture where one or more solutes are spread uniformly through a solvent. A solute is a substance dissolved in a solution. A solvent is the component that dissolves the solute and usually sets the physical state of the solution.
An aqueous solution has water as the solvent. In school chemistry, most concentration calculations use aqueous solutions because they’re easy to prepare, transfer and mix accurately.

Concentrated and dilute are handy words, but they’re too vague for calculations. A numerical concentration is clearer: it tells another chemist exactly how much solute is present in a stated volume of solution.
Molar concentration is the amount of solute per unit volume of solution. The guide uses this relationship:
n = CV, where C is molar concentration (mol dm⁻³ in this topic; 1 mol dm⁻³ = 1000 mol m⁻³) and V is volume of solution (dm³ in this topic; 1 dm³ = 10⁻³ m³).
You will also see the same relationship written as C = n / V. The volume means the final volume of the solution, not just the volume of water added.
Square brackets show molar concentration. For example, [Cl⁻] means the molar concentration of chloride ions, and [NaOH] = 0.200 mol dm⁻³ means the concentration of sodium hydroxide is 0.200 mol dm⁻³. The brackets refer to a particular solute or ion, not to the whole solution.
Concentrations can also be given in g dm⁻³. Mass concentration is the mass of solute per unit volume of solution.
ρ = msolute / V, where ρ is mass concentration (g dm⁻³ in this topic) and msolute is the mass of solute (g).
Molar concentration and mass concentration are connected by molar mass:
ρ = CM
and therefore:
C = ρ / M
So grams per dm³ can be changed into moles per dm³ by using grams per mole.
A standard solution is a solution with an accurately known concentration. To make one from a solid, weigh the solute, dissolve it in a small volume of deionized water, transfer it quantitatively into a volumetric flask, rinse the beaker and funnel into the flask, then make up to the calibration mark and mix thoroughly.
Glassware affects uncertainty. Use a volumetric flask when you need an accurate fixed final volume. Use a volumetric pipette to transfer one accurate fixed volume. A measuring cylinder is faster but less precise, so it is not the best choice for preparing a high-quality standard solution.

A serial dilution is a sequence of dilutions where each new solution is prepared from the previous one. It helps when you need several lower concentrations from one stock solution, especially for calibration curves. The dilution idea is conservation of solute: adding solvent changes volume and concentration, but not the amount of solute transferred.
For a dilution:
CinitialVinitial = CfinalVfinal, where Cinitial is the concentration before dilution (mol dm⁻³), Vinitial is the volume transferred before dilution (dm³), Cfinal is the concentration after dilution (mol dm⁻³), and Vfinal is the final volume after dilution (dm³).
A calibration curve is a graph linking a measured signal from known standard solutions to their concentrations. For a coloured solution, that signal is often absorbance from a colorimeter or spectrophotometer.
To use one, prepare standards covering the likely concentration range, measure their absorbance, plot absorbance against concentration, and draw a best-fit line or curve. Then measure the absorbance of the unknown and read its concentration from the calibration curve. If the unknown lies outside the reliable range, dilute it so it falls within the range, then account for the dilution.

Good calibration relies on sensible glassware and careful technique: clean cuvettes, consistent wavelength, standards made accurately, and concentrations that cover the unknown rather than clustering in one small region.
Avogadro’s law says that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. Because the number of particles is proportional to the amount in moles, the gas volume is proportional to the amount in moles too, provided temperature and pressure stay fixed.
For two gases under the same conditions:
n₁ / n₂ = V₁ / V₂, where n₁ and n₂ are the amounts of gas 1 and gas 2 (mol), and V₁ and V₂ are their gas volumes (dm³ in typical calculations; 1 dm³ = 10⁻³ m³).
That is why reacting gas volumes can be worked straight from the coefficients in a balanced equation, as long as all the gas volumes are measured at the same temperature and pressure.

Begin with the balanced equation. Its coefficients give the mole ratio. For gases under the same conditions, they give the volume ratio as well.
For example, if an equation shows 2A(g) reacting with 3B(g), then 2 volumes of A react with 3 volumes of B under the same conditions. So 20 cm³ of A would require 30 cm³ of B. You don’t need molar mass unless a mass is involved.
Watch the states. Avogadro’s law applies to gases, so you cannot use a gas volume ratio to find the volume of a liquid product or a solid reactant.
Avogadro’s law is exact for ideal gases. A real gas has particles with finite volume and may experience intermolecular forces, so it can deviate from ideal behaviour. These deviations are greatest at high pressure, where particle volume becomes significant, and at low temperature, where attractions between particles matter more. Gases with stronger intermolecular forces or larger particles tend to deviate more under comparable conditions.