Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
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In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
S1.4.1 Amount of substance, n
Amount of substance, n
The physical quantity that counts the number of specified elementary entities in a sample, expressed in moles (mol). It is not a mass or a volume: 1 mol of hydrogen molecules and 1 mol of carbon dioxide molecules contain the same number of molecules but have different masses. The symbol is n.
Mole (mol)
The SI unit of amount of substance. One mole contains exactly 6.02214076 × 10²³ elementary entities, the number fixed by the Avogadro constant; in IB calculations this is used as 6.02 × 10²³. The entity must always be specified: 1 mol of H₂O molecules contains 1 mol of oxygen atoms, 2 mol of hydrogen atoms and 3 mol of atoms in total.
Avogadro constant, N_A
The number of elementary entities per mole of substance, N_A = 6.02 × 10²³ mol⁻¹ (exactly 6.02214076 × 10²³ mol⁻¹). Its unit is mol⁻¹, so it converts an amount into a number of entities: N = n × N_A. For entities within a formula unit, multiply by the number of those entities in each unit: 0.500 mol H₂O contains 0.500 × 3 × 6.02 × 10²³ = 9.03 × 10²³ atoms.
Elementary entity
The particle being counted when an amount of substance is stated: an atom, a molecule, an ion, an electron, any other particle, or a specified group of particles such as the formula unit NaCl. The same sample has different amounts of different entities: 0.20 mol of CaCl₂ contains 0.20 mol of Ca²⁺ ions, 0.40 mol of Cl⁻ ions and 0.60 mol of ions in total.
Students often think The mole is a quantity of mass: 'amount of substance' means how much mass there is, so a value in mol can be read as the same value in g (and one mole of anything has a mass of 12 g). In fact No. The mole is the unit of amount of substance, a count of entities. The mass of one mole depends on the substance: 1 mol of H₂O has a mass of 18.02 g, 1 mol of NaCl 58.44 g.
Students often think One mole of any substance contains 6.02 × 10²³ atoms, whatever the substance and whatever its particles are. In fact Only when the specified entities are atoms. One mole contains 6.02 × 10²³ of whichever entity is specified: 1 mol of CH₄ contains 6.02 × 10²³ molecules and 5 × 6.02 × 10²³ = 3.01 × 10²⁴ atoms.
S1.4.2 Relative atomic mass, A_r
Relative atomic mass, A_r
The weighted mean mass of an atom of an element, taking account of the natural abundance of its isotopes, relative to one twelfth of the mass of an atom of carbon-12. On this scale a ¹²C atom has a relative mass of exactly 12. A_r is a ratio of masses and has no units. Values to two decimal places are used in calculations, e.g. A_r(Cl) = 35.45, A_r(C) = 12.01 (natural carbon contains some carbon-13).
Relative formula mass, M_r
The sum of the relative atomic masses of all the atoms in a formula, relative to one twelfth of the mass of a carbon-12 atom; called relative molecular mass for a molecular substance. It has no units. Every atom counts, including those multiplied by a subscript after a bracket or by the number of water molecules in a hydrated formula: M_r((NH₄)₂SO₄) = 2(14.01 + 4 × 1.01) + 32.07 + 4 × 16.00 = 132.17.
Students often think M_r and molar mass are the same quantity, so M_r carries the unit g mol⁻¹ (or molar mass has no unit). In fact No. M_r is a ratio of masses and has no unit. Molar mass M is mass per mole, with unit g mol⁻¹. They have the same numerical value: M_r(CO₂) = 44.01, M(CO₂) = 44.01 g mol⁻¹.
Students often think M_r is the mass of one molecule (or formula unit) in grams, so one CO₂ molecule has a mass of 44.01 g. In fact No. 44.01 g is the mass of one mole of CO₂ molecules. One molecule has a mass of 44.01 g ÷ 6.02 × 10²³ = 7.31 × 10⁻²³ g. M_r(CO₂) = 44.01 is a pure number.
S1.4.3 Molar mass, M
Molar mass, M
The mass per mole of a substance, M = m/n, with units g mol⁻¹. For practical purposes it has the same numerical value as M_r: M_r(CH₄) = 16.05 and M(CH₄) = 16.05 g mol⁻¹. Its value in g mol⁻¹ equals the mass in grams of 6.02 × 10²³ of the specified entities, not of one entity: one CH₄ molecule has a mass of 16.05 g ÷ 6.02 × 10²³ = 2.67 × 10⁻²³ g.
Relationships between mass, amount and number of particles
Amount of substance n (mol) = mass m (g) ÷ molar mass M (g mol⁻¹), and number of entities N = n × N_A. The units check the arrangement: g ÷ g mol⁻¹ = mol, and mol × mol⁻¹ = a pure number. The molar mass must be that of the entity specified: chlorine gas is Cl₂, M = 70.90 g mol⁻¹.
Students often think The relationship between amount, mass and molar mass is used upside down, n = M/m; for example A_r is divided by the percentage by mass when finding an empirical formula. In fact No: n = m/M. Units check it: g ÷ g mol⁻¹ = mol, whereas M/m would have the unit mol⁻¹.
Students often think The 'Cl₂' in CaCl₂ is a chlorine molecule or a single Cl₂²⁻ ion, so each formula unit of CaCl₂ contains two ions. In fact No. Calcium chloride is an ionic lattice of Ca²⁺ and Cl⁻ ions in the ratio 1 : 2. Each formula unit comprises three separate ions: one Ca²⁺ and two Cl⁻.
S1.4.4 Empirical formula
Empirical formula
The formula giving the simplest whole-number ratio of the atoms of each element in a compound. It is found from percentage (or mass) composition by dividing the mass of each element by its A_r to obtain amounts in moles, dividing by the smallest amount, and, if the result is not a set of whole numbers, multiplying all values by the smallest integer that makes them whole (×2 for a ratio such as 1 : 1.50). Ethanoic acid, C₂H₄O₂, has the empirical formula CH₂O.
Molecular formula
The formula giving the actual number of atoms of each element in one molecule. It is a whole-number multiple of the empirical formula: molecular formula = (empirical formula)ₓ, where x = molar mass ÷ empirical formula mass. Butadiene has empirical formula C₂H₃ (empirical formula mass 27.05) and molar mass 54.10 g mol⁻¹, so x = 2 and the molecular formula is C₄H₆. A formula of either kind does not show which atoms are bonded to which.
Percentage composition by mass
The mass of each element in a compound as a percentage of the total mass: % of element = (number of atoms of the element in the formula × A_r) ÷ M_r × 100. It is a share of the mass, not of the atoms: in (NH₄)₂SO₄ nitrogen is 2 of 15 atoms but 21.20% of the mass. Compounds with the same empirical formula have the same percentage composition.
Students often think The percentage composition by mass gives the ratio of atoms directly, and the fraction of an element's atoms in a formula is its percentage by mass. In fact No. Mass percentages must be divided by the A_r values to give amounts in moles before a ratio of atoms is taken. Conversely, the fraction of atoms that belong to an element is not its percentage by mass: in (NH₄)₂SO₄ nitrogen is 2 of 15 atoms but 21.20% of the mass.
Students often think Any mole ratio can be rounded to the nearest whole number, so 1 : 1.50 becomes 1 : 1 or 1 : 2. In fact No. A value of 1.50 is not experimental scatter; it shows that the whole-number ratio is 2 : 3. Multiply every value by the smallest integer that makes all of them whole (×2 for .5, ×3 for .33 or .67).
S1.4.5 Molar concentration and square-bracket notation
Molar concentration and square-bracket notation
The amount of solute per unit volume of solution: C = n/V, usually in mol dm⁻³. Square brackets around a formula denote its molar concentration, e.g. [NaCl] = 0.50 mol dm⁻³. The volume is that of the solution, not of the solvent used, so a solution of known concentration is made by dissolving the solute and making the solution up to the required volume.
Mass concentration and conversion
Concentration may also be expressed as mass of solute per unit volume of solution, in g dm⁻³. The two forms are related by the molar mass of the solute: concentration in g dm⁻³ = [X] (mol dm⁻³) × M (g mol⁻¹), and [X] = concentration in g dm⁻³ ÷ M. Ignoring its reaction with water, a solution containing 4.00 g dm⁻³ of dissolved Cl₂ has [Cl₂] = 4.00 ÷ 70.90 = 0.0564 mol dm⁻³.
n = CV and volume units
The amount of solute in a volume V of solution of molar concentration C is n = CV. With C in mol dm⁻³, V must be in dm³: 1 dm³ = 1000 cm³, so 25.0 cm³ = 0.0250 dm³ and 25.0 cm³ of 0.100 mol dm⁻³ solution contains 2.50 × 10⁻³ mol of solute.
Students often think A concentration in g dm⁻³ is converted into mol dm⁻³ by multiplying it by the molar mass. In fact No. Divide by M: [X] = (concentration in g dm⁻³) ÷ M. Units: g dm⁻³ ÷ g mol⁻¹ = mol dm⁻³. Multiplying by M converts the other way, from mol dm⁻³ to g dm⁻³.
Students often think The volume can be put into n = CV in cm³ directly, whatever the unit of the concentration. In fact No. V must be in dm³ to match mol dm⁻³: 25.0 cm³ = 0.0250 dm³.
S1.4.6 Avogadro's law
Avogadro's law
Equal volumes of all gases, measured at the same temperature and pressure, contain equal numbers of molecules. The law concerns molecules (or, for a monatomic gas such as argon, atoms), not atoms in general: equal volumes of H₂ and CO₂ contain equal numbers of molecules but different numbers of atoms, and different masses. It is obeyed by gases behaving ideally and approximately by real gases.
Reacting volumes of gases
Because volume is proportional to amount in moles for gases at the same temperature and pressure, the volumes of gaseous reactants and products are in the same ratio as their coefficients in the balanced equation. In 2SO₂(g) + O₂(g) → 2SO₃(g), 30.0 cm³ of SO₂ reacts with 15.0 cm³ of O₂ and forms 30.0 cm³ of SO₃. The same idea gives molar masses: equal volumes of two gases at the same T and p contain equal amounts, so their masses are in the ratio of their molar masses.
Students often think Equal volumes of gases at the same temperature and pressure contain equal numbers of atoms. In fact No. They contain equal numbers of molecules. 1 dm³ of H₂ and 1 dm³ of CO₂ contain the same number of molecules, but each CO₂ molecule has three atoms and each H₂ molecule two, so the CO₂ sample contains 1.5 times as many atoms.
Students often think Heavier or larger gas molecules take up more space, so a given volume holds fewer of them. In fact No. In a gas the molecules are far apart and their own volume is a tiny fraction of the gas volume. The volume depends on the number of molecules, the temperature and the pressure, not on the size or mass of the molecules.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement about one mole of a substance is correct?
Answer and reasoning
It contains 6.02 × 10²³ of whichever entity is specified. — One mole contains N_A = 6.02 × 10²³ elementary entities, and the entity must be named: 1 mol of H₂O is 6.02 × 10²³ molecules but 3 × 6.02 × 10²³ atoms.
It has a mass of 12 g, whichever substance it may be. — A student who thinks of the mole as a mass picks this. 12 g is the mass of one mole of carbon-12 atoms only; the mass of one mole is the molar mass, which differs between substances (18.02 g for H₂O, 58.44 g for NaCl).
It contains 6.02 × 10²³ atoms, whatever the substance. — A student who attaches the Avogadro constant to atoms picks this. One mole contains 6.02 × 10²³ of the specified entity; 1 mol of CH₄ molecules contains 5 × 6.02 × 10²³ atoms.
It occupies 22.7 dm³ at 273 K and 100 kPa, whatever its state. — A student who applies the gas molar volume to every substance picks this. 22.7 dm³ mol⁻¹ holds only for gases (behaving ideally) at 273 K and 100 kPa; one mole of liquid water occupies about 18 cm³.
2 A student writes: 'M_r of carbon dioxide, CO₂ = 44.01 g mol⁻¹'. Which comment on this statement is correct?
Answer and reasoning
It is correct: M_r and molar mass are one and the same quantity. — A student who treats M_r and molar mass as one quantity accepts the unit. They share the number 44.01, but M_r is a pure number and only the molar mass M has the unit g mol⁻¹.
The unit is wrong: M_r is the mass of one molecule in grams. — A student who thinks M_r is the mass of one molecule in grams picks this. 44.01 g is the mass of a mole of CO₂ molecules; one molecule has a mass of 7.31 × 10⁻²³ g. M_r itself has no unit.
The unit is wrong: M_r is measured in atomic mass units. — A student taught that relative masses are 'in atomic mass units' picks this. The mass of one CO₂ molecule is 44.01 u, but M_r is that mass divided by the atomic mass unit, so it has no unit.
The unit is wrong: M_r is a ratio of masses, with no unit. — M_r compares the mass of a CO₂ molecule with one twelfth of the mass of a ¹²C atom, so the units cancel: M_r(CO₂) = 44.01. The quantity with unit g mol⁻¹ is the molar mass, M(CO₂) = 44.01 g mol⁻¹.
3 The relative molecular mass of methane, CH₄, is 16.05. Which statement about the molar mass, M, of methane is correct?
Answer and reasoning
It is 16.05 g; one molecule of CH₄ has a mass of 16.05 g. — A student who thinks relative or molar mass is the mass of a single molecule picks this. One CH₄ molecule has a mass of 16.05 g ÷ 6.02 × 10²³ = 2.67 × 10⁻²³ g; 16.05 g is a mole of molecules.
It is 16.05, with no unit, as molar mass is the same as M_r. — A student who treats molar mass and M_r as one quantity transfers 'no unit' to M. M_r has no unit, but molar mass is a mass per mole with unit g mol⁻¹.
It is 16.05 g mol⁻¹; 1 mol of CH₄ molecules has mass 16.05 g. — Molar mass is mass per mole, in g mol⁻¹, and has the same number as M_r: 16.05 g is the mass of 1 mol (6.02 × 10²³) of CH₄ molecules, so M = 16.05 g mol⁻¹.
It is 16.05 g mol⁻¹; 16.05 g of CH₄ contains 6.02 × 10²³ atoms. — A student who thinks a mole is always 6.02 × 10²³ atoms picks this. 16.05 g of methane contains 6.02 × 10²³ molecules, and each has 5 atoms, so it contains 3.01 × 10²⁴ atoms.
4 The empirical formula of ethanoic acid is CH₂O. What does this empirical formula tell you about ethanoic acid?
Answer and reasoning
Its atoms of C, H and O are in the ratio 1 : 2 : 1. — The empirical formula gives the simplest whole-number ratio of atoms: C₂H₄O₂ simplifies to 1 : 2 : 1, written CH₂O.
Each of its molecules has one C atom, two H atoms and one O atom. — A student who takes the empirical formula to be the molecular formula picks this. Each molecule of ethanoic acid contains 2 C, 4 H and 2 O atoms; CH₂O gives only their ratio.
The masses of its C, H and O are in the ratio 1 : 2 : 1. — A student who equates the ratio of atoms with the ratio of masses picks this. The atoms have different masses: the mass ratio C : H : O is 12.01 : 2.02 : 16.00, not 1 : 2 : 1.
Each of its C atoms is bonded to two H atoms and to one O atom. — A student who reads a formula as a map of bonding picks this. In CH₃COOH one carbon atom carries three H atoms and no O atom, and the other carries two O atoms and no H atom. Formulae give numbers of atoms, not which are joined.
5 What amount of sodium hydroxide is present in 25.0 cm³ of NaOH(aq) in which [NaOH] = 0.100 mol dm⁻³?
Answer and reasoning
0.0250 mol — A student who takes 1 dm³ as 100 cm³ converts 25.0 cm³ to 0.250 dm³: 0.100 × 0.250 = 0.0250 mol. 1 dm³ = (10 cm)³ = 1000 cm³, so 25.0 cm³ = 0.0250 dm³ and n = 0.00250 mol.
0.250 mol — A student who uses n = V/C calculates 0.0250 ÷ 0.100 = 0.250 mol. Concentration is amount per dm³, so n = C × V = 0.100 × 0.0250 = 0.00250 mol.
2.50 mol — A student who substitutes the volume in cm³ calculates 0.100 × 25.0 = 2.50 mol. The concentration is per dm³, so the volume must be in dm³: 0.0250 dm³, giving 0.00250 mol.
6 Two identical sealed flasks are at the same temperature and pressure. One is filled with methane, CH₄, and the other with chlorine, Cl₂. The methane flask contains 5.00 × 10²² atoms in total. How many atoms does the chlorine flask contain? (A_r: H 1.01, C 12.01, Cl 35.45)
Answer and reasoning
2.00 × 10²² — Equal volumes at the same T and p contain equal numbers of molecules. The CH₄ flask holds 5.00 × 10²² ÷ 5 = 1.00 × 10²² molecules, so the Cl₂ flask also holds 1.00 × 10²² molecules, which contain 2 × 1.00 × 10²² = 2.00 × 10²² atoms.
5.00 × 10²² — A student who thinks equal volumes hold equal numbers of atoms gives the same number of atoms as in the CH₄ flask. Avogadro's law counts molecules: each flask holds 1.00 × 10²² molecules, so the Cl₂ flask holds 2.00 × 10²² atoms.
1.25 × 10²³ — A student who applies the ratio of atoms per molecule the wrong way round calculates 5.00 × 10²² × 5/2 = 1.25 × 10²³. Each Cl₂ molecule has fewer atoms than each CH₄ molecule, so with equal numbers of molecules the Cl₂ flask holds fewer atoms: 5.00 × 10²² × 2/5 = 2.00 × 10²².
4.53 × 10²¹ — A student who thinks equal volumes have equal masses reasons that the Cl₂ flask holds fewer molecules, in the ratio 16.05 : 70.90: 1.00 × 10²² × 16.05 ÷ 70.90 = 2.26 × 10²¹ molecules, or 4.53 × 10²¹ atoms. Equal volumes contain equal numbers of molecules, not equal masses, so the answer is 2.00 × 10²² atoms.
Working By Avogadro's law the flasks contain equal numbers of molecules. N(CH₄ molecules) = 5.00 × 10²² ÷ 5 = 1.00 × 10²² = N(Cl₂ molecules). Each Cl₂ molecule has 2 atoms, so N(atoms) = 2 × 1.00 × 10²² = 2.00 × 10²². (The A_r values are not needed.)
7 How many atoms are there in 0.500 mol of water, H₂O? (N_A = 6.02 × 10²³ mol⁻¹)
Answer and reasoning
3.01 × 10²³ — A student who stops at n × N_A counts molecules: 0.500 × 6.02 × 10²³ = 3.01 × 10²³ H₂O molecules. The question asks for atoms, and each molecule contains 3, giving 9.03 × 10²³.
1.00 × 10²³ — A student who applies the ratio of 3 atoms per molecule the wrong way round divides by 3: 3.01 × 10²³ ÷ 3 = 1.00 × 10²³. There are more atoms than molecules, so multiply: 3 × 3.01 × 10²³ = 9.03 × 10²³.
6.02 × 10²³ — A student who counts only the atoms with a written subscript takes each H₂O molecule to have 2 atoms: 0.500 × 2 × 6.02 × 10²³ = 6.02 × 10²³. The O has an unwritten subscript of 1, so each molecule has 3 atoms: 0.500 × 3 × 6.02 × 10²³ = 9.03 × 10²³.
Working N(molecules) = n × N_A = 0.500 mol × 6.02 × 10²³ mol⁻¹ = 3.01 × 10²³. Each H₂O molecule contains 2 H + 1 O = 3 atoms, so N(atoms) = 3 × 3.01 × 10²³ = 9.03 × 10²³.
8 The mass of one molecule of compound Z is 3.666 times the mass of one atom of carbon-12. What is the relative molecular mass, M_r, of Z?
Answer and reasoning
3.666 — A student who compares Z with a whole ¹²C atom treats the ratio as the relative mass. On the scale, a ¹²C atom is 12, not 1, so M_r = 3.666 × 12 = 43.99.
43.99 — Relative masses are measured against one twelfth of the mass of a ¹²C atom, on which scale ¹²C = 12 exactly. M_r(Z) = 3.666 × 12 = 43.99.
44.03 — A student who takes the reference to be natural carbon uses 12.01: 3.666 × 12.01 = 44.03. The stem compares Z with carbon-12, which is assigned exactly 12; 12.01 is the A_r of natural carbon, which contains some carbon-13.
3.273 — A student who applies the ratio upside down calculates 12 ÷ 3.666 = 3.273. Z is heavier than a ¹²C atom, so its relative mass must be larger than 12: 3.666 × 12 = 43.99.
Working On the relative mass scale one ¹²C atom = 12 exactly. M_r(Z) = (mass of Z molecule ÷ mass of ¹²C atom) × 12 = 3.666 × 12 = 43.992 ≈ 43.99.
9 How many ions are there in 5.55 g of calcium chloride, CaCl₂? (N_A = 6.02 × 10²³ mol⁻¹; A_r: Cl 35.45, Ca 40.08)
Answer and reasoning
9.03 × 10²² — M(CaCl₂) = 40.08 + 2 × 35.45 = 110.98 g mol⁻¹; n = 5.55 ÷ 110.98 = 0.0500 mol. Each formula unit contains 3 ions (one Ca²⁺, two Cl⁻): 0.0500 × 3 × 6.02 × 10²³ = 9.03 × 10²².
3.01 × 10²² — A student who stops at n × N_A counts formula units: 0.0500 × 6.02 × 10²³ = 3.01 × 10²². Each formula unit of CaCl₂ contains three ions, so there are 9.03 × 10²² ions.
6.02 × 10²² — A student who reads the Cl₂ in CaCl₂ as one unit counts two ions per formula unit: 0.0500 × 2 × 6.02 × 10²³ = 6.02 × 10²². CaCl₂ is a lattice of Ca²⁺ and separate Cl⁻ ions, three ions per formula unit.
3.61 × 10²⁵ — A student who inverts n = m/M calculates 110.98 ÷ 5.55 = 20.0 'mol', giving 20.0 × 3 × 6.02 × 10²³ = 3.61 × 10²⁵. The amount is mass ÷ molar mass: 5.55 ÷ 110.98 = 0.0500 mol.
Working M(CaCl₂) = 40.08 + 2(35.45) = 110.98 g mol⁻¹. n = m/M = 5.55 g ÷ 110.98 g mol⁻¹ = 0.0500 mol. Ions per formula unit = 1 Ca²⁺ + 2 Cl⁻ = 3. N(ions) = 0.0500 × 3 × 6.02 × 10²³ = 9.03 × 10²².
10 An oxide of iron contains 69.94% iron and 30.06% oxygen by mass. Which ratio of iron atoms to oxygen atoms gives its empirical formula? (A_r: O 16.00, Fe 55.85)
Answer and reasoning
Fe : O = 1 : 2 — A student who rounds the ratio 1 : 1.50 to the nearest whole number gets 1 : 2. A value of 1.50 is not scatter: it means the whole-number ratio is 2 : 3.
Fe : O = 3 : 2 — A student who inverts n = m/M divides A_r by the percentage: 55.85 ÷ 69.94 = 0.799 and 16.00 ÷ 30.06 = 0.532, a ratio of 1.50 : 1, i.e. 3 : 2. The amount is mass ÷ A_r, which gives 2 : 3.
Fe : O = 7 : 3 — A student who takes the percentages as the ratio of atoms gets 69.94 : 30.06 = 2.33 : 1, i.e. 7 : 3. Iron atoms are much heavier than oxygen atoms, so each percentage must first be divided by A_r.
Fe : O = 2 : 3 — n(Fe) = 69.94 ÷ 55.85 = 1.252; n(O) = 30.06 ÷ 16.00 = 1.879. Dividing by 1.252 gives 1 : 1.50, and multiplying by 2 gives 2 : 3, so the empirical formula is Fe₂O₃.
Working n(Fe) = 69.94 ÷ 55.85 = 1.252 mol; n(O) = 30.06 ÷ 16.00 = 1.879 mol (per 100 g). Divide by the smaller: Fe 1.00, O 1.879 ÷ 1.252 = 1.50. Multiply by 2: Fe : O = 2 : 3, so the empirical formula is Fe₂O₃.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
8 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 How many electrons are there in 0.200 mol of ammonia, NH₃? (N_A = 6.02 × 10²³ mol⁻¹; atomic numbers: H 1, N 7)
Answer and reasoning
1.20 × 10²³ — A student who stops at n × N_A counts molecules: 0.200 × 6.02 × 10²³ = 1.20 × 10²³ NH₃ molecules. Each molecule contains 10 electrons, so there are 1.20 × 10²⁴ electrons.
9.63 × 10²³ — A student who counts only the electrons in a Lewis formula uses 5 + 3 = 8 valence electrons: 0.200 × 8 × 6.02 × 10²³ = 9.63 × 10²³. Nitrogen also has two 1s electrons; the molecule has 10 electrons in total.
1.20 × 10²² — A student who applies the ratio of 10 electrons per molecule the wrong way round divides: 1.20 × 10²³ ÷ 10 = 1.20 × 10²². There are more electrons than molecules, so multiply by 10: 1.20 × 10²⁴.
2 What is the percentage by mass of nitrogen in ammonium sulfate, (NH₄)₂SO₄? (A_r: H 1.01, N 14.01, O 16.00, S 32.07)
Answer and reasoning
11.86% — A student who applies the 2 after the bracket only to the H atoms reads the formula as NH₈SO₄: M_r = 118.16 and % N = 14.01 ÷ 118.16 × 100 = 11.86%. The 2 multiplies the whole NH₄ group, giving 2 N atoms and M_r = 132.17.
13.33% — A student who treats the share of atoms as the share of mass counts 2 N atoms out of 15 atoms: 2 ÷ 15 × 100 = 13.33%. Each atom must be weighted by its A_r: 28.02 ÷ 132.17 × 100 = 21.20%.
21.20% — M_r = 2(14.01 + 4 × 1.01) + 32.07 + 4 × 16.00 = 132.17. Mass of N = 2 × 14.01 = 28.02. % N = 28.02 ÷ 132.17 × 100 = 21.20%.
26.90% — A student who divides the mass of nitrogen by the mass of the other elements calculates 28.02 ÷ (132.17 − 28.02) × 100 = 26.90%. A percentage by mass is a part of the whole: 28.02 ÷ 132.17 × 100 = 21.20%.
Working M_r((NH₄)₂SO₄) = 2 × (14.01 + 4 × 1.01) + 32.07 + 4 × 16.00 = 36.10 + 32.07 + 64.00 = 132.17. Mass of N per formula unit = 2 × 14.01 = 28.02. % N = 28.02 ÷ 132.17 × 100 = 21.20%.
3 A hydrocarbon contains 88.80% carbon and 11.20% hydrogen by mass. Its molar mass is 54.10 g mol⁻¹. What is its molecular formula? (A_r: H 1.01, C 12.01)
Answer and reasoning
C₂H₃ — A student who stops at the empirical formula picks C₂H₃. Its formula mass is 27.05, half the molar mass of 54.10 g mol⁻¹, so each molecule contains two C₂H₃ units: C₄H₆.
C₄H₆ — n(C) = 88.80 ÷ 12.01 = 7.394; n(H) = 11.20 ÷ 1.01 = 11.09; ratio 1 : 1.50, so the empirical formula is C₂H₃ (27.05). x = 54.10 ÷ 27.05 = 2, so the molecular formula is C₄H₆.
C₄H₄ — A student who rounds the ratio 1 : 1.50 down to 1 : 1 takes the empirical formula as CH (13.02); 54.10 ÷ 13.02 = 4.16 ≈ 4 gives C₄H₄. The ratio 1 : 1.50 must be doubled to 2 : 3, giving C₂H₃ and then C₄H₆. The misfit 4.16 is itself a warning.
C₂H₆ — A student who multiplies only the atom next to the bracket in (C₂H₃)₂ gets C₂H₆. The multiplier applies to the whole empirical unit: (C₂H₃)₂ = C₄H₆.
Working Per 100 g: n(C) = 88.80 ÷ 12.01 = 7.394 mol; n(H) = 11.20 ÷ 1.01 = 11.09 mol. Ratio H : C = 11.09 ÷ 7.394 = 1.50, so C : H = 1 : 1.50 = 2 : 3 and the empirical formula is C₂H₃. Empirical formula mass = 2(12.01) + 3(1.01) = 27.05. x = 54.10 ÷ 27.05 = 2.00. Molecular formula = (C₂H₃)₂ = C₄H₆.
4 Which procedure produces a solution in which [NaCl] = 0.500 mol dm⁻³? (M(NaCl) = 58.44 g mol⁻¹)
Answer and reasoning
Add 0.125 mol of NaCl to 250 cm³ of water and stir until dissolved. — A student who divides by the volume of solvent calculates 0.125 ÷ 0.250 = 0.500. Concentration is per volume of solution; dissolving 7.31 g of NaCl increases the volume to about 252 cm³, so [NaCl] ≈ 0.496 mol dm⁻³, not 0.500 mol dm⁻³, and the exact volume is not known. The solution must be made up to a known volume.
Dissolve 2.92 g of NaCl and make up to 100 cm³ of solution. — [NaCl] = 0.500 mol dm⁻³ corresponds to 0.500 × 58.44 = 29.2 g dm⁻³; 100 cm³ of solution therefore needs 2.92 g (0.0500 mol), made up to the mark in a volumetric flask.
Dissolve 0.500 g of NaCl, then make the solution up to 1.00 dm³. — A student who reads mol as g puts 0.500 g in each dm³. That is 0.500 ÷ 58.44 = 0.00856 mol, so [NaCl] = 0.00856 mol dm⁻³. The unit mol dm⁻³ means 0.500 mol (29.2 g) per dm³ of solution.
Add 0.500 mol of NaCl to 250 cm³ of water and stir to dissolve it. — A student who reads [NaCl] as the amount of NaCl uses 0.500 mol, whatever the volume. The solution has a volume of about 260 cm³, so [NaCl] ≈ 1.9 mol dm⁻³; square brackets mean amount per dm³ of solution.
Working Correct procedure: n(NaCl) = 0.500 mol dm⁻³ × 0.100 dm³ = 0.0500 mol; m = nM = 0.0500 mol × 58.44 g mol⁻¹ = 2.92 g, made up to 100 cm³ of solution (equivalently, 0.500 mol dm⁻³ = 29.2 g dm⁻³). 0.500 g ÷ 58.44 g mol⁻¹ = 0.00856 mol, so 0.500 g in 1.00 dm³ gives 0.00856 mol dm⁻³. 0.500 mol in about 0.260 dm³ of solution gives about 1.9 mol dm⁻³. Adding 0.125 mol to 250 cm³ of water gives a solution whose volume is not 250 cm³, so its concentration is not 0.500 mol dm⁻³ (about 0.496 mol dm⁻³: the solution volume is about 252 cm³).
5 A solution of chlorine in water contains 4.00 g of dissolved chlorine, Cl₂, per dm³. Ignore any reaction of chlorine with water. What is [Cl₂]? (A_r: Cl 35.45)
Answer and reasoning
0.113 mol dm⁻³ — A student who uses A_r of chlorine as the molar mass of chlorine gas calculates 4.00 ÷ 35.45 = 0.113 mol dm⁻³. Dissolved chlorine is Cl₂, M = 70.90 g mol⁻¹, giving 0.0564 mol dm⁻³.
284 mol dm⁻³ — A student who multiplies by the molar mass calculates 4.00 × 70.90 = 284. Units show the error: g dm⁻³ × g mol⁻¹ is not mol dm⁻³. Divide: 4.00 ÷ 70.90 = 0.0564 mol dm⁻³.
0.0564 mol dm⁻³ — M(Cl₂) = 2 × 35.45 = 70.90 g mol⁻¹. [Cl₂] = 4.00 g dm⁻³ ÷ 70.90 g mol⁻¹ = 0.0564 mol dm⁻³.
4.00 mol dm⁻³ — A student who treats grams and moles as interchangeable changes only the unit. 4.00 g of Cl₂ is 4.00 ÷ 70.90 = 0.0564 mol, so [Cl₂] = 0.0564 mol dm⁻³.
Working M(Cl₂) = 2 × 35.45 = 70.90 g mol⁻¹. [Cl₂] = (4.00 g dm⁻³) ÷ (70.90 g mol⁻¹) = 0.05642 mol dm⁻³ ≈ 0.0564 mol dm⁻³.
6 Sulfur dioxide reacts with oxygen: 2SO₂(g) + O₂(g) → 2SO₃(g). What volume of oxygen reacts completely with 30.0 cm³ of sulfur dioxide? All volumes are measured at the same temperature and pressure.
Answer and reasoning
60.0 cm³ — A student who uses the mole ratio upside down calculates 30.0 × 2/1 = 60.0 cm³. Only one O₂ is needed for every two SO₂, so V(O₂) = 30.0 × 1/2 = 15.0 cm³.
30.0 cm³ — A student who thinks gases react in equal volumes gives 30.0 cm³. The equation's coefficients fix the ratio: 2 volumes of SO₂ react with 1 volume of O₂.
10.0 cm³ — A student who thinks equal volumes hold equal numbers of atoms sets the volumes in the ratio of the atoms in the equation: 2SO₂ contains 6 atoms and O₂ contains 2, so V(O₂) = 30.0 × 2/6 = 10.0 cm³. Avogadro's law counts molecules: the volumes are in the ratio of the coefficients, 2 : 1, so V(O₂) = 30.0 × 1/2 = 15.0 cm³.
15.0 cm³ — At the same T and p, gas volumes are in the ratio of the coefficients: V(O₂) = 30.0 cm³ × 1/2 = 15.0 cm³.
Working At constant T and p, V ∝ n. From the equation, n(O₂) : n(SO₂) = 1 : 2, so V(O₂) = 30.0 cm³ × 1/2 = 15.0 cm³.
7 At the same temperature and pressure, a flask holds 0.640 g of oxygen, O₂, when filled with oxygen and 1.28 g of gas X when filled with X. What is the molar mass of X? (A_r: O 16.00)
Answer and reasoning
32.0 g mol⁻¹ — A student who uses A_r of oxygen as the molar mass of oxygen gas gets n = 0.640 ÷ 16.00 = 0.0400 mol and M(X) = 1.28 ÷ 0.0400 = 32.0 g mol⁻¹. Oxygen gas is O₂, M = 32.00 g mol⁻¹.
16.0 g mol⁻¹ — A student who applies the mass ratio the wrong way round calculates 32.00 × 0.640 ÷ 1.28 = 16.0 g mol⁻¹. X gives the heavier sample with the same number of molecules, so M(X) = 32.00 × 1.28 ÷ 0.640 = 64.0 g mol⁻¹.
64.0 g mol⁻¹ — Equal volumes at the same T and p contain equal amounts: n = 0.640 g ÷ 32.00 g mol⁻¹ = 0.0200 mol. M(X) = 1.28 g ÷ 0.0200 mol = 64.0 g mol⁻¹.
2.00 g mol⁻¹ — A student who treats a mass in grams as an amount in moles takes n(X) = n(O₂) = 0.640 mol and calculates 1.28 ÷ 0.640 = 2.00 g mol⁻¹. 0.640 g of O₂ is 0.640 ÷ 32.00 = 0.0200 mol, so M(X) = 1.28 ÷ 0.0200 = 64.0 g mol⁻¹.
Working Same flask, same T and p, so by Avogadro's law n(X) = n(O₂) = 0.640 g ÷ 32.00 g mol⁻¹ = 0.0200 mol. M(X) = 1.28 g ÷ 0.0200 mol = 64.0 g mol⁻¹.
8 What is the relative formula mass, M_r, of hydrated calcium nitrate, Ca(NO₃)₂·4H₂O? (A_r: H 1.01, N 14.01, O 16.00, Ca 40.08)
Answer and reasoning
164.10 — A student who leaves out the water of crystallisation calculates 40.08 + 2(14.01 + 3 × 16.00) = 164.10. The four H₂O molecules are part of the formula unit: add 4 × 18.02 = 72.08 to give 236.18.
236.18 — Add every atom, including the four water molecules: 40.08 + 2(14.01 + 3 × 16.00) + 4(2 × 1.01 + 16.00) = 40.08 + 124.02 + 72.08 = 236.18. M_r has no unit.
222.17 — A student who lets the 2 after the bracket multiply only the O₃ counts 1 N and 6 O: 40.08 + 14.01 + 6 × 16.00 + 72.08 = 222.17. The 2 multiplies the whole NO₃ group, 2 N and 6 O: M_r = 236.18.
182.12 — A student who treats the 4 like a coefficient in an equation counts one water molecule: 164.10 + 18.02 = 182.12. The 4 is part of the formula: each formula unit contains four H₂O, so add 4 × 18.02 = 72.08 to give 236.18.
Working M_r = 40.08 + 2(14.01 + 3 × 16.00) + 4(2 × 1.01 + 16.00) = 40.08 + 124.02 + 72.08 = 236.18. M_r has no unit.
That was your twenty minutes. Real practice on S1.4 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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