Two teaching hours at both levels, outcomes 1.1.1 to 1.1.3. There is no additional higher level content in Structure 1.1.
Guiding question: How can we model the particulate nature of matter?
Structure 1.1 · The particulate nature of matter
1Elements, compounds and mixtures 1.1.1 SL + HL Strong
A spoonful of salt, a glass of sea water and a lump of sodium look nothing alike, yet chemistry treats all three with one idea: everything is made of particles, and what matters is which particles are present and what holds them together. That idea sorts all matter into three categories, and the sorting is not bookkeeping. It predicts which properties a sample will have and how, if at all, it can be separated.
Elements are the primary constituents of matter, which cannot be chemically broken down into simpler substances.
Compounds consist of atoms of different elements chemically bonded together in a fixed ratio.
Mixtures contain more than one element or compound in no fixed ratio, which are not chemically bonded and so can be separated by physical methods.
The two phrases in bold carry the whole distinction. A compound has a fixed ratio because chemical bonds fix it: every molecule of water contains two hydrogen atoms and one oxygen atom, wherever the water came from. A mixture has no fixed ratio because nothing holds its components to one: a salt solution can be made weak or strong. Everything else follows from that one difference.
- Properties. A compound has its own properties, often completely unlike those of its elements. Sodium is a reactive metal and chlorine a toxic gas; sodium chloride is neither. In a mixture each component keeps its own properties: salt water tastes of salt and still contains water that boils.
- Separation. A compound can be broken down only by a chemical reaction. A mixture can be separated by physical methods that exploit a difference in a physical property of its components.
- Composition. A compound has a definite composition by mass; a mixture has a variable one.
Homogeneous and heterogeneous mixtures
A mixture is homogeneous if it has a uniform composition and properties throughout and consists of a single phase; any sample taken from it has the same ratio of components. Solutions (salt in water), mixtures of miscible liquids (ethanol and water), air and alloys such as brass are all homogeneous. A mixture is heterogeneous if its composition is not uniform and two or more phases can be distinguished: sand in water, oil and water, or granite. The word describes the mixture, not whether its components are visible — a solution is homogeneous even though it contains two substances.
Whether two substances form a homogeneous mixture depends on the forces between their particles, which is the subject of Structure 2.2: ethanol and water mix in all proportions because both can form hydrogen bonds; hexane and water do not, and form two layers.
Ask two questions in order. Is there a fixed ratio held by bonds? If yes, it is a compound and only a reaction will separate it. If no, it is a mixture: is it one phase or several? The answer to the second question usually tells you the separation method. Two phases (a solid in a liquid) — filtration. One phase — a method based on a difference in boiling point, solubility or affinity for a stationary phase.
Exam focus · what the published papers show
A multiple-choice item in one session paired the type of mixture with the best method of separation, so that both had to be right to score. The published report says candidates knew the mixture should be separated by distillation, but some were unable to identify that the mixture is homogeneous. The chemistry they found harder was the classification, not the technique. Two miscible liquids form one phase, so the mixture is homogeneous — and it is precisely because they mix that filtration cannot separate them and distillation can.
Question. Which of the following are homogeneous mixtures? I an aqueous solution of sodium chloride; II a mixture of pentane and hexane; III a mixture of ethanol and water.
Work through them one at a time. I is a solution: the salt is dissolved, so there is a single phase. II is two alkanes of similar polarity, which are miscible — one phase. III is ethanol and water, miscible in all proportions — one phase.
Answer: all three. The item tests whether you know that “mixture” and “one phase” are compatible. None of the three is a compound: in none is there a fixed ratio or a bond between the components.
One published item states, as a correct option, that alloys are homogeneous mixtures of metals with other metals or non-metals. An alloy is a mixture, not a compound — its composition can vary — and it is homogeneous because it presents one phase. The same classification is assumed later when metallic bonding and alloys are examined in Structure 2.3 and 2.4.
2Separating mixtures: the prescribed techniques 1.1.1 SL + HL Strong
Six techniques are named in the syllabus: solvation, filtration, recrystallization, evaporation, distillation and paper chromatography. Each works because the components of the mixture differ in one physical property, and each is chosen by identifying that property. Learn the property, not just the apparatus: a question will describe an unfamiliar mixture and expect you to choose.
| Technique | Property exploited | Typical use |
|---|---|---|
| Solvation (dissolving) | Solubility in a chosen solvent | Dissolving salt away from sand before filtering |
| Filtration | Particle size: an undissolved solid is retained by the filter; the liquid (filtrate) passes through | Removing sand from salt solution; collecting crystals |
| Evaporation | Volatility of the solvent compared with the dissolved solid | Recovering salt from its solution |
| Recrystallization | Solubility that changes greatly with temperature, and differs between product and impurities | Purifying an impure solid product |
| Distillation (simple and fractional) | Boiling point | Water from sea water; ethanol from water; crude oil fractions |
| Paper chromatography | Relative affinity for a stationary phase and a mobile phase | Separating and identifying the dyes in an ink |
Solvation, filtration and evaporation
A mixture of salt and sand is separated in three steps, each exploiting one difference. Adding water dissolves the salt: its ions are surrounded by water molecules (solvation), while sand is insoluble. Filtration then retains the sand as the residue and lets the salt solution through as the filtrate. Evaporation of the filtrate removes the volatile water and leaves the non-volatile salt. Heating the solution only until crystals begin to form, and then leaving it to cool, gives better-formed crystals than evaporating to dryness — this is crystallization.
Recrystallization
Recrystallization purifies a solid product that is contaminated with small amounts of other substances. It depends on one property of the solvent: the product must be much more soluble in the hot solvent than in the cold. The impure solid is dissolved in the minimum volume of hot solvent; the hot solution is filtered to remove any insoluble impurity; the filtrate is allowed to cool slowly, so that the product crystallizes while soluble impurities, present in small amounts, stay in solution; the crystals are filtered off, rinsed with a little cold solvent and dried.
The two filtrations do opposite jobs. The hot filtration removes what did not dissolve — an impurity that is less soluble than the product. The cold filtration collects the crystals and leaves behind in the filtrate an impurity that is more soluble. The choice of solvent volume is a compromise: too much and some product remains dissolved even when cold, lowering the yield; too little and impurities also crystallize, lowering the purity.
Distillation
In simple distillation a solution is boiled; the vapour of the more volatile component passes into a condenser, is cooled and collected as the distillate. It separates a liquid from dissolved solids (pure water from sea water) or two liquids whose boiling points differ widely. When the boiling points are closer, fractional distillation is used: a fractionating column above the flask provides a large surface on which vapour repeatedly condenses and re-evaporates, so that the vapour becomes progressively richer in the more volatile component as it rises. The column has a temperature gradient — hottest at the bottom, coolest at the top — and fractions are collected over different temperature ranges. The industrial separation of crude oil is the large-scale example.
Paper chromatography
A spot of the mixture is placed on a base line drawn in pencil (ink would itself separate) near the bottom of a strip of chromatography paper. The paper is stood in a small depth of solvent, below the base line so that the spot does not simply dissolve into the solvent reservoir. The solvent — the mobile phase — rises through the paper by capillary action. The stationary phase is the water held in the cellulose fibres of the paper. Each component is continually distributed between the two phases: when in the mobile phase it moves, when held by the stationary phase it does not. A component with greater affinity for the mobile phase spends more time moving and travels further. When the solvent has nearly reached the top, its position — the solvent front — is marked.
Rf = distance moved by the centre of the spot ÷ distance moved by the solvent front
Both distances are measured from the base line. Rf has no units and can never exceed 1, because a spot cannot travel further than the solvent carrying it. An Rf value identifies a substance only for a stated solvent and paper: the same substance gives different values in different solvent systems.
Exam focus · what the published papers show
Of everything in Structure 1.1, recrystallization draws the most pointed examiner comment. On a structured question about it, one report records a mean of 0.52 marks out of 3 for the opening part, and observes that students seemed to be totally unfamiliar with the technique, with many blanks, or random guesses, on the purification of solids through recrystallization. Another report is blunter: it is also critical that students have hands-on practical experience of all the experimental techniques, such as recrystallization.
The same report identifies what separated those who scored: students who had studied recrystallization answered this well, recognizing that the solubility of the solute needed to vary greatly with temperature. That single idea is the whole method.
Any three of: dissolve in minimum volume of hot solvent; hot solution filtered; solution allowed to cool slowly; crystals filtered off; crystals rinsed with a small quantity of cold solvent.
The published note: Apply ECF if both M1 & M2 not awarded because "hot" omitted. One adjective governs two of the three marks. A second scheme adds that steps must relate to the process of recrystallization and that examiners may not award filter or centrifuge twice unless separate functions specified — writing “filter” twice without saying what each filtration removes earns one mark, not two.
Question. Outline how impurities with differing solubilities are separated from the desired product. [2] Deduce how using too much, and too little, solvent would affect the yield and purity. [2]
The published answers: Less soluble impurity removed: filtering hot solution; More soluble impurity removed: removing/filtering crystals from cold solution.
Solvent volume. Too much solvent and some product stays dissolved even when cold: low/decreased yield. Too little and the impurities also saturate and crystallize with the product: low/decreased purity. The scheme also accepts the converse for each — increased purity for the first, decreased yield for the second — because the trade-off runs both ways.
The published scheme wants two distinct ideas. M1, the mechanism: continuous cycle of adsorption and desorption/dissolution, or solute moves when in solvent AND does not move when on paper; the note also allows answers in terms of an equilibrium/partitioning. M2, the discrimination: the components have different attraction to mobile phase AND stationary phase, or are separated based on solubilities in/affinities to the two phases.
A report on this item says most candidates scored M1 but could not explain the second half. Naming both phases is what turns one mark into two.
One report notes that weaker students also included the length below the baseline in the measurement, and another that candidates lost marks by overlooking the solvent front. Measure both distances from the base line, and to the centre of the spot.
Question. Black ink was run in two solvent systems of propan-2-ol and water: system 1 was 20 % propan-2-ol, system 2 was 80 %. The red dye gave Rf = 0.566 in system 1 and 0.451 in system 2. Deduce which dye is more polar, red or blue.
Fix the solvent first. More water means a more polar mobile phase. A component that travels further in the more aqueous system is better dissolved by a polar solvent — it is the more polar component.
Then apply it. The published answer identifies blue as more polar, because blue travels further with higher concentration water, with the reason higher concentration water mixture more polar. The scheme wants a named substance and the property of the solvent that justifies it.
3States of matter and the kinetic molecular theory 1.1.2 SL + HL Strong
The kinetic molecular theory is a model: matter consists of particles (atoms, molecules or ions) in constant random motion, and the differences between solids, liquids and gases are differences in how strongly the particles attract one another compared with how much kinetic energy they have. Every observable property of a state is explained by that picture, and examination questions ask for exactly that explanation — the observable property linked to its particle-level cause.
| Solid | Liquid | Gas | |
|---|---|---|---|
| Shape | Fixed | Takes the shape of the container | Fills the container |
| Volume | Fixed; incompressible | Fixed; almost incompressible | Variable; easily compressed |
| Arrangement | Particles close together in a regular arrangement | Particles close together, irregular arrangement | Particles far apart, random arrangement |
| Movement | Vibrate about fixed positions | Move past one another | Move rapidly in straight lines between collisions |
| Attractions | Strong | Weaker than in the solid, but significant | Negligible |
| Kinetic energy | Lowest | Intermediate | Highest |
A liquid flows because its particles, though touching, are free to move past one another; it cannot be compressed because the particles are already in contact and there is no empty space to close up. A gas is easily compressed because most of its volume is empty space. Each property has its reason, and the reason is always about arrangement, movement or attraction.
Changes of state
Six changes of state are named in the syllabus. Melting (s → l), vaporization (l → g) and sublimation (s → g) require energy to overcome attractions between particles: they are endothermic. Freezing (l → s), condensation (g → l) and deposition (g → s) release energy as attractions form: they are exothermic. Vaporization occurs in two ways. Evaporation takes place only at the surface and at any temperature, because some surface particles always have enough energy to escape; boiling takes place throughout the liquid, at the temperature at which the vapour pressure of the liquid equals the external pressure.
State symbols
Chemical equations carry state symbols: (s) solid, (l) liquid, (g) gas and (aq) aqueous — dissolved in water. The fourth is not a state of matter in its own right, which is why “aqueous” and “liquid” are not interchangeable: NaCl(aq) is a solution of sodium chloride in water; NaCl(l) is molten sodium chloride above 801 °C. Written equations for changes of state consist of nothing but the formula and two state symbols, for example H2O(l) → H2O(g) for vaporization and I2(s) → I2(g) for sublimation.
Exam focus · what the published papers show
M1 liquids flow / shape not fixed AND molecules/particles free to move. M2 non-compressible / fixed volume AND molecules/particles are close/touching/strongly attracted to each other.
And the note that decides the grade: Award [1] for two correct macroscopic properties. Two properties with no explanation earn one mark out of two. The word AND in each line is doing the work — every property must be joined to its particle reason.
This is the most repeated examiner complaint attached to Structure 1.1. On one question, a report records a mean of 0.54 out of 2 and explains: Most gained no marks for this question because they did not realise that there was a change of state. The report on the other version of the paper is more specific — candidates missed the change of state that occurs when the liquid propane is heated from 200 K to 250 K and they simply referred to expansion and the gas laws.
The habit that prevents it: before answering any question about heating or cooling, compare the state symbols (or the temperatures with the melting and boiling points). If the state changes, the change of state is the answer.
Most published schemes carry the general instruction Ignore missing or incorrect state symbols in an equation unless directed otherwise in the "Notes" column — but individual notes override it routinely, and reports return to the omission repeatedly, advising teachers to encourage students to report correctly balanced chemical equations with corresponding state symbols. In one thermochemistry question the source of error was to overlook the state symbol and use the value for liquid water instead of gaseous. Write state symbols every time.
Question. Which are correct? I molecules gain kinetic energy and temperature increases; II added energy overcomes hydrogen bonds between molecules; III molecules gain sufficient energy to move from fixed positions.
During a change of state the temperature does not change — the energy supplied separates the particles rather than speeding them up — so I is false. II is what melting is for water: energy breaks hydrogen bonds holding the lattice. III describes the outcome: particles fixed in position become free to move.
Answer: II and III only. Statement I is the trap because it is half true: the molecules do gain energy, but as potential energy of separation, not kinetic energy, so the thermometer reading does not change.
4Temperature, kinetic energy and heating curves 1.1.3 SL + HL Repeated
The temperature, T, in kelvin (K) is a measure of the average kinetic energy, Ek, of the particles. The kelvin is the SI unit of temperature and has the same incremental value as the Celsius degree: a rise of 1 °C is a rise of 1 K.
T (K) = θ (°C) + 273 the conversion is an addition, never a multiplication
The two scales differ in where they start. The Celsius zero is the freezing point of water, a convenient reference with no physical significance for particles. The kelvin zero, absolute zero (0 K = −273 °C), is the temperature at which the particles would have their minimum possible kinetic energy. Because average kinetic energy is directly proportional to the absolute temperature, doubling the kelvin temperature doubles the average kinetic energy — but doubling a Celsius temperature does not.
Not every particle in a sample moves at the same speed. At any temperature there is a spread of kinetic energies, and temperature measures their average. At the same temperature, particles of different substances have the same average kinetic energy; heavier particles therefore move more slowly on average (Ek = ½mv2). The shape of this distribution is developed in Reactivity 2.2.
Introduction to the particulate nature of matter
Heat the sample and watch where the energy goes. Inside a state it raises kinetic energy and the temperature climbs; at a change of state it goes into potential energy and the temperature stops.
The plateaus are the whole point. During melting and boiling the temperature does not rise even though energy is still going in, because that energy is separating particles rather than speeding them up.
Particles are drawn with speeds sampled from the kinetic model at the temperature you set: mean speed rises with √T. Arrangement and spacing change at each transition.
Heating curve. The marker shows where your temperature sits on the path from solid to gas.
Interpreting a heating curve
When a pure solid is heated at a steady rate, its temperature rises until it reaches the melting point, then stays constant while it melts, rises again through the liquid range, stays constant again while it boils, and finally rises as the gas is heated. On the sloping sections the energy supplied increases the average kinetic energy of the particles, so the temperature rises. On the flat sections the energy is used to overcome the attractions between particles — it increases their potential energy — and the average kinetic energy, and therefore the temperature, stays the same. The boiling plateau is longer than the melting plateau because far more energy is needed to separate particles completely than to loosen them into a liquid. A cooling curve is the mirror image: the plateaus appear at the same temperatures, and energy is released during freezing and condensation.
A sharp melting point is therefore evidence of purity: an impure solid melts over a range of temperatures, usually below the melting point of the pure substance, which is why melting point determination is used to check the success of a recrystallization.
Exam focus · what the published papers show
Question. What happens to the average kinetic energy of the particles in a gas when the absolute temperature is doubled?
Because Ek is proportional to T in kelvin, it doubles. The word doing the work is absolute. Heating a gas from 30 °C to 60 °C is not a doubling: in kelvin it is 303 K to 333 K, a factor of 1.10, so the average kinetic energy rises by about a tenth.
Any question that doubles, halves or takes a ratio of temperatures is a kelvin question. Convert first, then take the ratio. The error is invisible in your working and fatal to your answer, and it recurs in gas-law questions in Structure 1.5, where the same conversion decides the whole calculation.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.
Which is a homogeneous mixture?
Which statements about mixtures are correct?
I. The components may be elements or compounds.
II. All components must be in the same phase.
III. The components retain their individual properties.
Which statement describes all homogeneous mixtures?
Which is correct?
Which equation represents sublimation?
Which equation represents the deposition of iodine?
Which best describes the particles in a gas when the temperature rises from 23 °C to 46 °C?
Sodium thiosulfate solution reacts with dilute hydrochloric acid to form a precipitate of sulfur at room temperature.
Na2S2O3(aq) + 2HCl(aq) → S(s) + SO2(g) + 2NaCl(aq) + X
(a) Identify the formula and state symbol of X. [1]
Solutions and mark-scheme guidance · Set 1A
1A.1 C
Ethanol and water are miscible in all proportions and form a single phase. Oil and water form two layers; sand and chalk are insoluble solids, so both B and D are heterogeneous.
1A.2 B
A mixture may contain elements (air contains N2 and O2) or compounds, and each component keeps its own properties. II is false: heterogeneous mixtures (sand in water) contain more than one phase.
1A.3 A
Homogeneous means uniform composition: any sample drawn from it has the same ratio. B and D describe a compound, not a mixture; C is false because mixtures can be separated by physical methods.
1A.4 C
Both kinds of mixture exist, and because the components are not bonded to one another each keeps its own chemical properties — they are not averaged.
1A.5 B
Sublimation is solid → gas for one substance, with no chemical change. C is the reverse process, deposition; A and D are chemical reactions.
1A.6 B
Deposition is gas → solid. A is condensation, C is vaporization and D is sublimation.
1A.7 B
In kelvin the change is 296 K → 319 K, a factor of only 1.08, so the average kinetic energy increases but does not double (A). C would follow only from doubling the kelvin temperature. Doubling the Celsius reading is the trap the item is built on.
1A.8 [1]
H2O(l) ✔ — the scheme requires both the formula and the state symbol, and states Do not accept H2O (aq). Water is the solvent, not something dissolved in water, so it cannot be “aqueous”.
Two teaching hours at both levels, outcomes 1.2.1 and 1.2.2, and one additional higher level hour, outcome 1.2.3.
Guiding question: How do the nuclei of atoms differ?
Structure 1.2 · The nuclear atom
5The nuclear atom and its symbol 1.2.1 SL + HL Strong
At the start of the twentieth century the atom was pictured as a sphere of positive charge with electrons embedded in it. In 1909 Geiger and Marsden, working with Rutherford, fired alpha particles — helium nuclei, He2+ — at a very thin gold foil. Almost all passed straight through; a few were deflected through large angles, and about one in several thousand bounced back. Rutherford's interpretation is the nuclear model: nearly all the mass and all the positive charge of an atom are concentrated in a tiny, dense nucleus, and the rest of the atom is mostly empty space occupied by the electrons.
Atoms contain a positively charged, dense nucleus composed of protons and neutrons (together, nucleons). Negatively charged electrons occupy the space outside the nucleus. The nucleus is about 10−15 m across; the atom about 10−10 m.
| Particle | Location | Relative mass | Relative charge |
|---|---|---|---|
| Proton | Nucleus | 1 | +1 |
| Neutron | Nucleus | 1 | 0 |
| Electron | Outside the nucleus | negligible (about 1/1840) | −1 |
Two numbers identify an atom. The atomic number, Z, is the number of protons; it defines the element — every atom with 29 protons is copper. The mass number, A, is the number of protons plus neutrons. Because the mass of the electron is negligible, the mass number counts nucleons only. In a neutral atom the number of electrons equals Z; an ion differs from its atom only in the number of electrons — never in protons or neutrons. A positive ion has lost electrons; a negative ion has gained them.
number of neutrons = A − Z · number of electrons = Z − charge
The behaviour of the three particles in an electric field follows from their charge and mass. A beam of neutrons passes undeflected. Protons are deflected towards the negative plate, and electrons towards the positive plate — and far more strongly, because for the same charge the electron has a much smaller mass.
Exam focus · what the published papers show
A published report on a one-mark symbol question records a surprising number of incorrect answers (50%) due to inverted A and Z — half of all the wrong answers were the right numbers in the wrong positions. The report suggests a cause: this may be explained in the way atomic numbers and relative atomic masses are shown in the periodic table of the data booklet. The two-second check: A is always the larger number, because A = Z + N and no atom has a negative number of neutrons.
Question. State the number of each type of subatomic particle in the potassium ion, 4119K+. [1]
| protons | 19 — the atomic number, which is what makes it potassium |
| electrons | 18 — 19 protons less the single positive charge |
| neutrons | 22 — A − Z = 41 − 19 |
This is examined constantly and answered well: one report calls it probably the best answered question on the paper. The only failure is subtracting the charge from the wrong quantity.
A. Deduce the nuclear symbol for an ion of nickel-58 with 26 electrons. [1] “Nickel” fixes Z = 28 — the question does not give it; you find it in the periodic table. “-58” is A. The charge is protons minus electrons, 28 − 26 = +2. Published answer: 5828Ni2+. The charge is the part candidates forget, because the question never uses the word.
B. Deduce the nuclear symbol of the isotope of thallium containing 122 neutrons. [1] Thallium fixes Z = 81, so A = 81 + 122 = 203: 20381Tl. This is the question on which half the errors were inversions — 81203Tl would claim an atom with 203 protons.
One item shows a beam of particles passing between charged plates and asks which path belongs to which particle. A published report records that some teachers thought the question required knowledge of the mass spectrometer; the examiners replied that it did not require this knowledge to work out the answer. Charge and mass are enough.
6Isotopes and relative atomic mass 1.2.2 SL + HL Strong
Isotopes are atoms of the same element with different numbers of neutrons: the same atomic number, different mass numbers. Chlorine-35 (17 p, 18 n) and chlorine-37 (17 p, 20 n) are isotopes of chlorine.
Chemical behaviour is decided by electrons, and isotopes of an element have the same number and arrangement of electrons. Isotopes are therefore chemically identical: they undergo the same reactions and form the same bonds, molecular shapes and dipole moments. They differ only in mass, and therefore only in physical properties that depend on mass: density, melting and boiling point (slightly), rate of diffusion and the relative molecular mass of their compounds. Heavy water, D2O, boils at 101.4 °C and is about 11 % denser than H2O, yet reacts in the same way. Some isotopes are radioactive; their unstable nuclei decay, a nuclear property that does not change their chemistry, which is why isotopic tracers can follow atoms through a reaction mechanism.
Ask of every property: does it depend on mass? Density, rate of diffusion, boiling point and relative molecular mass do, so they differ between isotopes. Reactions, bonding, charge, electron configuration and molecular shape do not. A published report advises teachers to emphasise the difference between physical and chemical properties, and how this relates to the properties of different isotopes of an element.
Relative atomic mass from isotopic abundance
Most elements occur naturally as a mixture of isotopes. The relative atomic mass, Ar, of an element is the weighted mean mass of its atoms relative to one-twelfth of the mass of an atom of carbon-12. It is weighted because each isotope contributes in proportion to its abundance, which is why Ar values are rarely whole numbers. In calculations the mass number is used for the mass of each isotope.
Ar = Σ (isotope mass × abundance) ÷ Σ (abundance) with percentage abundances, divide the total by 100
Question. A sample of argon is 0.34 % 36Ar, 0.06 % 38Ar and 99.6 % 40Ar. Calculate its relative atomic mass to two decimal places. [2]
Ar = (0.0034 × 36) + (0.0006 × 38) + (0.996 × 40) = 0.1224 + 0.0228 + 39.84 = 39.99
Reasonableness check. Over 99 % of the sample has mass 40, so the answer must lie just below 40. The tabulated value (39.95) is lower because natural argon has a slightly different composition from this sample.
Question. Gallium (Ar = 69.72) consists of two stable isotopes, Ga-69 and Ga-71. What is the relative abundance of Ga-71?
69(1 − x) + 71x = 69.72 → 69 + 2x = 69.72 → x = 0.36
Answer: 36 %. The options offer both 36 % and 64 %, so the item really tests which isotope you solved for. A shortcut: the fraction of the heavier isotope is (69.72 − 69) ÷ (71 − 69) — its distance from the lighter isotope divided by the gap. Reports record 66 % and 54 % correct in the two versions of the paper.
Question. In naturally occurring sulfur the relative abundance of 36S is 0.0100 %. Calculate the number of atoms of this isotope in 1.00 g of natural sulfur. [2]
n(36S) = (0.0100/100) × (1.00/32.07) = 3.12 × 10−6 mol; N = 3.12 × 10−6 × 6.02 × 1023 = 1.88 × 1018 atoms
Here the data-booklet value is the right one: the question is about natural sulfur, so 32.07 is correct. The rule is not “never use the table”; it is “use the table for the natural element, and the given abundances for a stated sample”. Only about a quarter of candidates completed this part.
The nuclear atom — isotopes and relative atomic mass
Relative atomic mass is a weighted mean, not an average of the isotope masses. Change the abundances and watch Ar move towards whichever isotope you made more common.
Natural abundances are the accepted values for terrestrial samples. Ar is computed as Σ(abundance × isotope mass) ÷ 100 every time you move a slider.
Mass spectrum of the singly charged ions. Bar height is relative abundance; position is m/z.
Exam focus · what the published papers show
In every recent session a scheme has asked for Ar from an isotopic composition, and the notes column names the periodic-table value and refuses it: argon 39.99 (Do not accept the data booklet value), sulfur 32.29 (Do not accept 32.07 which is the data booklet value), nickel 58.77, thallium 204.40 and bromine 79.99, each with the tabulated value refused.
A report on the last confirms the behaviour it targets: the most common mistake was quoting the relative atomic mass found in the data booklet (79.90). A question that gives you abundances describes a particular sample; the answer must come from those numbers.
Schemes carry the note M2 can only be awarded for answer with two decimal places, and the stems ask for two decimal places. Reports show what this costs: some lost the second mark for not giving the final answer to two decimal places, and weaker candidates sometimes rounded the answer to remove the decimal places. Half the marks are lost after the chemistry is finished.
When a paper asked candidates to contrast the atomic structures of two bromine isotopes, a report observed that many gave a definition of isotopes, without comparing them. A contrast names what differs and what does not: both have 35 protons and 35 electrons; 79Br has 44 neutrons and 81Br has 46.
Question. Which property is the same for water containing tritium (3H) and water containing hydrogen (1H)? Density; dipole moment of a molecule; boiling point; relative molecular mass.
Eliminate on mass: relative molecular mass, density and boiling point all depend on it. The dipole moment depends on the electron distribution across the O–H bonds, which is identical. Answer: dipole moment. A report diagnoses the wrong answers: they may result from associating dipole moment with size.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.
Which shows the number of subatomic particles in 31P3−?
| Protons | Neutrons | Electrons | |
|---|---|---|---|
| A. | 15 | 16 | 18 |
| B. | 15 | 16 | 12 |
| C. | 16 | 31 | 15 |
| D. | 31 | 31 | 15 |
In which set do all the species contain more electrons than neutrons?
Which quantities are different between two species represented by the notation 12852Te and 12853I−?
Which species has two more neutrons than electrons?
Bromine consists of two stable isotopes that exist in approximately a 1 : 1 ratio. The relative atomic mass, Ar, of bromine is 79.90. Which are the stable isotopes of bromine?
What is the relative atomic mass of a sample of chlorine containing 70 % of the 35Cl isotope and 30 % of the 37Cl isotope?
Fast moving helium nuclei (4He2+) were fired at a thin piece of gold foil with most passing undeflected but a few deviating largely from their path.
(a) Suggest what can be concluded about the gold atom from this experiment. [2]
Most 4He2+ passing straight through: …
Very few 4He2+ deviating largely from their path: …
Magnesium is a group 2 metal which exists as a number of isotopes and forms many compounds.
(a) State the nuclear symbol notation, AZX, for magnesium-26. [1]
(b) Mass spectroscopic analysis of a sample of magnesium gave the following results: Mg-24 78.60 %; Mg-25 10.11 %; Mg-26 11.29 %. Calculate the relative atomic mass, Ar, of this sample of magnesium to two decimal places. [2]
Naturally occurring silver is composed of two stable isotopes, 107Ag and 109Ag. The relative atomic mass of silver is 107.87. Show that isotope 107Ag is more abundant. [1]
Dinitrogen monoxide, N2O, causes depletion of ozone in the stratosphere. Different sources of N2O have different ratios of 14N : 15N.
(i) State one analytical technique that could be used to determine the ratio of 14N : 15N. [1]
(ii) A sample of gas was enriched to contain 2 % by mass of 15N with the remainder being 14N. Calculate the relative molecular mass of the resulting N2O. [2]
(iii) Predict, giving two reasons, how the first ionization energy of 15N compares with that of 14N. [2]
Rhenium, Re, was the last element with a stable isotope to be isolated. The stable isotope of rhenium contains 110 neutrons. State the nuclear symbol notation AZX for this isotope. [1]
Solutions and mark-scheme guidance · Set 1B
1B.1 A
Phosphorus has Z = 15 (found from the periodic table). Neutrons = 31 − 15 = 16. A 3− charge means three electrons gained: 15 + 3 = 18. B subtracts the charge instead of adding it.
1B.2 C
14N3−: 10 electrons, 7 neutrons. 16O2−: 10 electrons, 8 neutrons. 11C: 6 electrons, 5 neutrons. All three have more electrons than neutrons. Neutral 14N and 16O have equal numbers, and 11C4+ has only 2 electrons.
1B.3 D
Te: 52 p, 76 n, 52 e. I−: 53 p, 75 n, 54 e. All three counts differ; only the mass number is shared. The item punishes anyone who assumes equal mass numbers mean equal neutrons.
1B.4 C
Na+: 12 neutrons, 10 electrons — a difference of two. Li+: 3 n, 2 e. Be2+: 5 n, 2 e. Ca2+: 22 n, 18 e.
1B.5 A
With a 1 : 1 ratio the weighted mean is the simple average of the two masses, which must be about 79.9: only 79 and 81 average to 80. The others average to 80.5, 79 and 79.5.
1B.6 C
(0.70 × 35) + (0.30 × 37) = 24.5 + 11.1 = 35.6. B is the familiar data-booklet value for natural chlorine, which has a different composition.
1B.7 [2]
Most pass straight through: most of the atom is empty space OR the nucleus is very small compared with the atom ✔. Very few deviate: the nucleus is positive «and repels the He2+» OR dense/heavy OR very small ✔. The scheme notes: Do not accept the same reason for both M1 and M2, and Do not accept only “nucleus repels 4He2+ particles” for M2 — say what property of the nucleus causes the repulsion.
1B.8 [3]
(a) 2612Mg ✔
(b) Ar = (24 × 78.60 + 25 × 10.11 + 26 × 11.29) ÷ 100 = 24.3269 = 24.33 ✔✔. Award [2] for correct final answer. Do not accept data booklet value (24.31).
1B.9 [1]
Ar is closer to 107 than to 109 / less than 108, the average of 107 and 109, «so more 107Ag» ✔. The scheme also accepts a calculation giving more than 50 % 107Ag: 107(1 − x) + 109x = 107.87 gives x = 0.435, so 56.5 % is 107Ag.
1B.10 [5]
(i) mass spectrometry/MS ✔
(ii) Ar(N) = (98 × 14 + 2 × 15) ÷ 100 = 14.02 ✔; Mr = (14.02 × 2) + 16.00 = 44.04 ✔
(iii) Any two: the same, because they have the same nuclear charge/number of protons ✔; neutrons have no charge and do not affect the attraction ✔; the same electron configuration/shielding ✔. «Same» need only be stated once. The chemistry of isotopes, applied to an ionization energy.
1B.11 [1]
18575Re ✔ — Z = 75 from the periodic table; A = 75 + 110 = 185.
Structure 1.2 · Additional higher level
7Mass spectra and relative atomic mass 1.2.3 HL only Strong
Relative atomic masses are measured, not calculated from theory, and the instrument that measures them is the mass spectrometer. A gaseous sample is ionized, the ions are accelerated, separated according to their mass-to-charge ratio, m/z, and detected. The output is a mass spectrum: a plot of relative abundance (or intensity) against m/z. The operational details of the instrument are not assessed; what you must be able to do is read the spectrum.
The horizontal axis is m/z; the vertical axis is relative abundance. For singly charged ions — the assumption in these questions — m/z is numerically the mass of the ion, so for an element each peak marks one isotope, and its height gives that isotope's relative abundance.
Reading a spectrum therefore gives both the identity of the isotopes (from the positions of the peaks) and their relative abundances (from the heights). The relative atomic mass follows from the same weighted mean as before: multiply each m/z by its abundance, add, and divide by the total abundance. When heights are given as relative intensities rather than percentages — the tallest peak set to 100 — the division must be by the sum of the heights, not by 100.
Question. A mass spectrum shows two peaks: m/z 10 at 19.9 % and m/z 11 at 80.1 %. What is the relative atomic mass? Options 10.0, 10.2, 10.5, 10.8.
Ar = (10 × 0.199) + (11 × 0.801) = 1.99 + 8.811 = 10.80
You can also reason without arithmetic: the taller peak is the heavier one, so the answer must be nearer 11 than 10. 10.5 is the answer for equal abundances and catches anyone who averages the masses instead of weighting them. Reports on these items record 90 % and 75 % correct.
Elements that form diatomic molecules
Chlorine and bromine exist as diatomic molecules, and their mass spectra show peaks for the atom ions (Cl+) and for the molecular ions (Cl2+). With two isotopes, a molecule can be made in three ways — light + light, light + heavy and heavy + heavy — so there are three molecular peaks. The middle combination can form in two ways (light-then-heavy and heavy-then-light), so its probability is doubled. For chlorine, with isotopes in approximately a 3 : 1 ratio, the molecular peaks at m/z 70, 72 and 74 are in the ratio 9 : 6 : 1; for bromine, with nearly equal abundances, the peaks at 158, 160 and 162 are close to 1 : 2 : 1.
Question. Chlorine is diatomic and contains 75 % 35Cl and 25 % 37Cl. Predict the full mass spectrum.
| 35 + 35 | m/z 70: 0.75 × 0.75 = 0.5625 → relative height 9 |
| 35 + 37 and 37 + 35 | m/z 72: 2 × 0.75 × 0.25 = 0.375 → 6 |
| 37 + 37 | m/z 74: 0.25 × 0.25 = 0.0625 → 1 |
The molecular peaks are 70 : 72 : 74 = 9 : 6 : 1, and the full spectrum also shows the atom-ion peaks at 35 and 37 in the ratio 3 : 1. A graph showing only one set, or showing 72 as the tallest, is wrong.
Exam focus · what the published papers show
A one-mark question asks which technique determines the relative proportions of the isotopes of an element. The scheme accepts mass spectrometry OR mass spectroscopy OR mass spectrum OR MS, and reports record 70 % and 80 % correct. Mass spectrometry answers “how heavy?”, never “what shape?” — it cannot distinguish enantiomers, and it is not infrared spectroscopy.
The scheme wants the two operations: multiply the relative intensity by the m/z value of each isotope ✔; add the products / find the weighted average ✔. Award [1 max] for stating “m/z values of isotopes AND relative abundance/intensity” but not stating these need to be multiplied. Naming the data without saying what you do with it earns half.
One report records that only 38% of the candidates attributed the peaks with m/z larger than the molar mass to heavier isotopes of the elements. A small peak one or two units above the molecular ion is almost always a heavier isotope — 13C in an organic compound, or 37Cl or 81Br. Nothing has gained mass; a minority of the molecules were always heavier. A related report notes that a significant number seem to believe the molecular ion is always the tallest peak: it is the peak of highest m/z in the main cluster, not necessarily the tallest.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.
Which technique is used to detect the isotopes of an element?
Calcium carbide, CaC2, is an ionic solid. (b) Describe how the relative atomic mass of a sample of calcium could be determined from its mass spectrum. [2]
Solutions and mark-scheme guidance · Set 1C
1C.1 A
Isotopes differ only in mass, so only a technique that separates by mass can detect them.
1C.2 B
The peaks are about 79 % at 24, 10 % at 25 and 11 % at 26: (24 × 0.79) + (25 × 0.10) + (26 × 0.11) = 24.3. The answer must be close to 24 because nearly four-fifths of the atoms have mass 24; 25.0 is the unweighted average.
1C.3 B
(23 × 0.80) + (28 × 0.20) = 18.4 + 5.6 = 24.0. Unusually for such a spectrum the isotopes are five mass units apart, so the weighting shifts the answer a whole unit from the major isotope.
1C.4 A
The relative molecular mass comes from the three molecular-ion peaks only, weighted by their intensities and divided by the sum of the intensities (52 + 100 + 48 = 200), not by 100 and not by the sum of the m/z values (B). A gives 31 992 ÷ 200 = 159.96. C and D wrongly include the atom-ion peaks, which belong to fragments, not to Br2.
1C.5 [1]
(63 × 69 + 65 × 31) ÷ 100 ✔ = 63.62 — reading the peak heights as 69 % and 31 %. The scheme also accepts the algebraic route: 65x + 63(1 − x) = 63.62, giving x = 0.31 (31 %) and 1 − x = 0.69 (69 %).
1C.6 [2]
Multiply the relative intensity by the m/z value of each isotope ✔; sum these products and divide by the total intensity / find the weighted average ✔. Award [1 max] for stating “m/z values of isotopes AND relative abundance/intensity” but not stating these need to be multiplied.
Review · Structure 1.1 and 1.2
8Misconceptions, the examiner’s view, and the question types
- “A solution is not a mixture because you cannot see the parts.” Why it is wrong: a mixture is defined by the absence of bonds and of a fixed ratio, not by visibility. Correct model: a solution is a homogeneous mixture. Consequence: the classification half of a two-part item is lost.
- “The temperature rises while a solid melts because energy is being supplied.” During a change of state the energy increases the potential energy of the particles; average kinetic energy, and so temperature, is constant.
- “Doubling the temperature in °C doubles the kinetic energy.” Kinetic energy is proportional to the kelvin temperature only.
- “(aq) and (l) mean the same thing.” (aq) means dissolved in water; water itself is (l). A scheme refuses H2O(aq).
- “Isotopes have different chemical properties because they have different masses.” Chemistry is decided by electrons; isotopes differ only in mass-dependent physical properties.
- “Ar is the average of the isotope masses.” It is a weighted mean; the unweighted average is a standard distractor.
- “The data-booklet Ar is always the answer.” For a stated sample, the answer comes from the given abundances; schemes refuse the tabulated value.
- HL “The molecular ion is always the tallest peak.” It is the peak of highest m/z in the main cluster.
Evidence base: the IB Diploma chemistry subject reports quoted in the exam-focus sections of this page.
Answered well: counting protons, neutrons and electrons in atoms and ions (probably the best answered question on the paper); reading a relative atomic mass from a simple two-isotope spectrum (90 % and 75 % correct); naming mass spectrometry as the technique for isotopic composition (70–80 % correct); and the first, mechanistic, mark for explaining chromatography.
Found difficult: (1) recrystallization, where many students appeared totally unfamiliar with the technique and a three-mark part averaged 0.52; (2) recognising a change of state in a heating question, where answers turned to expansion and gas laws instead; (3) classifying a mixture as homogeneous when choosing a separation method; (4) giving Ar to two decimal places and resisting the data-booklet value; (5) writing nuclear symbols with A and Z the right way round; (6) the second chromatography mark, which needs the difference in attraction to both phases; (7) at HL, attributing peaks above the molecular ion to heavier isotopes.
What successful answers did: joined each observable property to its particle-level cause in one sentence; checked the state symbols before choosing an explanation; wrote the weighted-mean expression explicitly and kept the precision the question asked for; and named what was being compared rather than giving a definition.
Six question types cover most of these two topics, and each has a fixed opening move.
| If the question asks… | …then |
|---|---|
| Classify this mixture / choose the method | Fixed ratio and bonds? Compound. Otherwise one phase or two? Then name the property the method exploits. |
| Explain a property of a state | State the property and the particle reason — arrangement, movement or attraction — in one sentence. |
| What happens on heating? | Check for a change of state first. Plateau → potential energy; slope → kinetic energy. |
| Protons, neutrons, electrons | Z from the periodic table; N = A − Z; electrons = Z − charge. A is the larger number. |
| Calculate Ar | Weighted mean from the given abundances; two decimal places if asked; never the table value. |
| Read a mass spectrum HL | Peak position = isotope mass; height = abundance; divide by the sum of the heights; diatomic → three molecular peaks. |
9Quick check
- A mixture of two miscible liquids with different boiling points is to be separated. State the type of mixture, the method, and the property the method exploits.
- Give the steps of a recrystallization that would earn full marks, and state which word two of those marks depend on.
- Explain, using the kinetic molecular theory, why a liquid cannot be compressed.
- Name the change of state shown by I2(s) → I2(g), and the reverse change.
- A gas is heated from 25 °C to 50 °C. By what factor does the average kinetic energy of its particles increase?
- State the number of protons, electrons and neutrons in 3717Cl−.
- An element is 60.0 % mass number 24 and 40.0 % mass number 26. Calculate its relative atomic mass to two decimal places.
- Copper has Ar = 63.55 and two isotopes, Cu-63 and Cu-65. Deduce the percentage of Cu-63.
- HL An element has peaks at m/z 62 (69.2 %) and 64 (30.8 %). Calculate its relative atomic mass to two decimal places.
- HL A diatomic element X2 has two isotopes in a 1 : 1 ratio. Predict the ratio of the three molecular peaks and explain the middle one.
| 1 | Homogeneous; fractional distillation, which exploits the difference in boiling point. Miscible liquids form one phase, which is why filtration cannot work. |
| 2 | Dissolve in the minimum volume of hot solvent; filter the hot solution; cool slowly; filter off the crystals; rinse with a little cold solvent. Two marks depend on hot. |
| 3 | A liquid has a fixed volume because its particles are already close together/touching and attract one another strongly, so there is no space to close up. The property alone is worth at most one mark. |
| 4 | Sublimation; the reverse, I2(g) → I2(s), is deposition. |
| 5 | 298 K → 323 K: a factor of 1.08, not 2. |
| 6 | 17 protons, 18 electrons, 20 neutrons. |
| 7 | (24 × 0.600) + (26 × 0.400) = 14.40 + 10.40 = 24.80 — “24.8” would lose the precision mark. |
| 8 | 63x + 65(1 − x) = 63.55 → x = 0.725: 72.5 % Cu-63. |
| 9 | (62 × 0.692) + (64 × 0.308) = 42.904 + 19.712 = 62.62. |
| 10 | 1 : 2 : 1. The middle mass arises two ways (light + heavy and heavy + light); each outer peak only one way. |
10Summary and knowledge organiser
Essential knowledge
- Elements cannot be broken down chemically; compounds contain elements bonded in a fixed ratio and have their own properties; mixtures have no fixed ratio, no bonds between components, and are separated physically.
- Homogeneous mixtures are uniform and single-phase (solutions, alloys, air); heterogeneous mixtures are not.
- Separation methods exploit one property each: solubility (solvation), particle size (filtration), volatility (evaporation), temperature-dependent solubility (recrystallization), boiling point (distillation), affinity for two phases (chromatography, Rf = distance of spot ÷ distance of solvent front).
- The kinetic molecular theory explains the properties of solids, liquids and gases from particle arrangement, movement and attraction; melting, vaporization and sublimation are endothermic; freezing, condensation and deposition exothermic.
- Temperature in kelvin is proportional to the average kinetic energy of the particles; T(K) = θ(°C) + 273. During a change of state temperature is constant.
- The atom has a small, dense, positive nucleus of protons and neutrons, surrounded by electrons of negligible mass; Z = protons, A = protons + neutrons.
- Isotopes have the same Z, different A; the same chemical properties, different mass-dependent physical properties. Ar is the abundance-weighted mean relative to 12C.
- HL A mass spectrum gives isotope masses (peak positions) and abundances (heights); diatomic elements show three molecular peaks.
Examination checklist
- Classify the mixture and choose the method; name the property exploited.
- Every explanation of a state: observable property and particle reason.
- Look for a change of state before explaining any heating result.
- Convert to kelvin before any ratio of temperatures.
- Nuclear symbols: A on top (larger), Z below, charge top right.
- Ar: weighted mean from the given data, to the precision asked, never the table value.
| Outcome | Key facts and relationships | Must-remember distinctions and common errors |
|---|---|---|
| Classifying matter 1.1.1 | Element · compound (fixed ratio, bonded) · mixture (no fixed ratio). Homogeneous = one phase. | Solutions and alloys are homogeneous mixtures, not compounds. |
| Separation 1.1.1 | Filtration, evaporation, recrystallization, distillation, chromatography; Rf = a ÷ b, from the base line. | Recrystallization: minimum hot solvent; two filtrations remove different impurities. |
| States 1.1.2 | Arrangement, movement, attraction; six changes of state; (s), (l), (g), (aq). | Property AND reason. (aq) ≠ (l). |
| Temperature 1.1.3 | T(K) = θ(°C) + 273; Ek ∝ T(K); plateaus on heating curves. | Plateau: potential energy rises, T constant. |
| Nuclear atom 1.2.1 | p: 1, +1 · n: 1, 0 · e: negligible, −1. N = A − Z; electrons = Z − charge. | A and Z inverted in half the wrong answers. |
| Isotopes, Ar 1.2.2 | Ar = Σ(mass × abundance) ÷ Σ(abundance). | Two decimal places; data-booklet value refused for a stated sample. |
| HL Mass spectra 1.2.3 | Position = m/z = isotope mass; height = abundance. Cl2 70 : 72 : 74 ≈ 9 : 6 : 1; Br2 ≈ 1 : 2 : 1. | Divide by the sum of heights. Peaks above M+ are heavier isotopes. |