HomeLearning HubIB DP ChemistryS1.1 Particulate nature of matter
S1.1

Introduction to the particulate nature of matter

Structure 1 · Models of the particulate nature of matter · SL and HL

🎯What you need to be able to do

  • Distinguish between the properties of elements, compounds and mixtures.
  • Explain how solvation, filtration, recrystallization, evaporation, distillation and paper chromatography separate mixtures, and choose the right one for a given mixture.
  • Understand the difference between homogeneous and heterogeneous mixtures.
  • Distinguish the states of matter using the kinetic molecular theory, and use the state symbols (s), (l), (g) and (aq) in equations.
  • Name all six changes of state and say which are endothermic.
  • Interpret changes in physical properties and temperature during a change of state.
  • Convert between the Celsius and Kelvin scales, and explain what temperature measures.

📚The chemistry

Elements, compounds and mixtures

An element is a substance that cannot be chemically broken down into anything simpler; there are about 118 of them and they are what the periodic table lists. A compound contains atoms of different elements chemically bonded in a fixed ratio. A mixture contains more than one element or compound in no fixed ratio, not chemically bonded, and therefore separable by physical means.

Three consequences follow, and questions test all three:

  • A compound’s properties bear no necessary relation to those of its elements. Sodium is a metal that explodes in water; chlorine is a poisonous green gas; sodium chloride is table salt. A mixture, by contrast, keeps the properties of its components — iron filings mixed with sulfur are still magnetic.
  • A compound has a fixed composition by mass. Water is 11.2% hydrogen whether it came from a tap or a comet. A mixture can be any proportion you like.
  • Separating a compound requires a chemical change (electrolysis, thermal decomposition); separating a mixture requires only a physical one.

Mixtures are homogeneous if the composition is uniform throughout and there is a single phase — salt water, air, an alloy, a solution of ethanol in water. They are heterogeneous if the composition varies and separate phases are visible — sand in water, oil and water, granite.

“Air is a compound.” Air is a homogeneous mixture: its composition is not fixed (it varies with humidity, altitude and pollution), its components are not chemically bonded to one another, and it can be separated by fractional distillation, which is a physical process. Alloys are the same argument in a different setting — see S2.4.

Choosing a separation technique

Every separation technique exploits a difference in a physical property. Naming the property is what earns the mark; naming the technique alone rarely does.

  • Solvation (dissolving) — difference in solubility in a chosen solvent. Add water to a salt/sand mixture and only the salt dissolves.
  • Filtration — difference in particle size, separating an insoluble solid from a liquid. The solid caught in the paper is the residue; the liquid that passes is the filtrate.
  • Evaporation — recovers a dissolved solid by removing the solvent. Only works if you want the solid and the solid is thermally stable.
  • Distillation — difference in boiling point. Simple distillation recovers a solvent from a solution; fractional distillation separates two liquids whose boiling points differ, using a fractionating column so that the mixture repeatedly condenses and re-evaporates up its length.
  • Recrystallization — difference in how solubility varies with temperature. Dissolve the impure solid in the minimum volume of hot solvent, then cool: the product crystallises out while the impurities, present in much smaller amounts, stay in solution. This is the standard purification step in any organic preparation, and the purity of what you get is judged by a sharp melting point.
  • Paper chromatography — difference in relative attraction to a mobile and a stationary phase. Treated fully in S2.2, because the attractions involved are intermolecular forces.

The kinetic molecular theory

The kinetic molecular theory is a model: matter consists of particles in constant motion, and the state of a substance is decided by the balance between the kinetic energy of those particles and the forces of attraction between them.

  • Solid (s) — particles closely packed in a fixed arrangement, vibrating about fixed positions. Fixed shape, fixed volume, essentially incompressible.
  • Liquid (l) — particles still close together but free to move past one another. Fixed volume, takes the shape of the container, very slightly compressible.
  • Gas (g) — particles far apart, moving rapidly and randomly, with negligible attraction between them. No fixed shape or volume, highly compressible.

The fourth state symbol, (aq), is not a state of matter at all: it means dissolved in water. Getting state symbols right matters more than it looks — whether a species is (aq) or (s) decides whether it appears in an equilibrium expression (R2.3) and whether it can conduct electricity (S2.1).

The six changes of state

Melting solid → liquid
endothermic
Freezing liquid → solid
exothermic
Vaporization liquid → gas
(evaporation or boiling) · endothermic
Condensation gas → liquid
exothermic
Sublimation solid → gas
endothermic
Deposition gas → solid
exothermic

Evaporation and boiling are not the same thing. Evaporation happens at any temperature, only at the surface, and only for those particles in the high-energy tail of the distribution that have enough energy to escape the attractions holding them in the liquid — which is why evaporation cools what is left behind. Boiling happens at one specific temperature, throughout the bulk of the liquid, when its vapour pressure equals the external pressure.

Heating curves and the temperature plateaus

Heat a pure solid steadily and plot temperature against time. The graph rises, then holds flat at the melting point, rises again, holds flat at the boiling point, then rises. The flat sections are the part that gets examined:

During a change of state the energy supplied does not raise the temperature, because it is being used to overcome the forces of attraction between particles rather than to increase their average kinetic energy. Since temperature measures average kinetic energy, the temperature cannot change while the potential energy is changing. The plateau at boiling is longer than the plateau at melting, because separating particles completely takes far more energy than merely freeing them to move past each other.

“The temperature stays constant because the heat is being shared.” No. It stays constant because the energy goes into potential energy — breaking intermolecular attractions — and not into kinetic energy. The distinction between kinetic and potential energy is the whole content of that plateau, and answers that do not name it do not score.

Temperature and the Kelvin scale

Temperature, in kelvin, is a measure of the average kinetic energy of the particles in a sample. Not the total energy, and not the energy of any individual particle — at any instant the particles have a wide spread of speeds (see the Maxwell–Boltzmann distribution in R2.2).

The kelvin is the SI unit and has the same size increment as the degree Celsius, so the conversion is an offset with no scaling:

\[ T\,/\,\mathrm{K} \;=\; \theta\,/\,^{\circ}\mathrm{C} \;+\; 273.15 \]

Zero kelvin, absolute zero, is the temperature at which particles would have minimum kinetic energy. Because the Kelvin scale starts there, kelvin temperatures are proportional to average kinetic energy: doubling the kelvin temperature doubles the average kinetic energy, which is untrue of Celsius. That is exactly why every gas equation in S1.5 demands kelvin.

✏️Worked example

A student is given 8.0 g of a solid mixture of sodium chloride, sand and iodine, and is asked to recover all three components. Iodine sublimes at 114 °C; sand is insoluble in water; sodium chloride is soluble.
(a) Describe, in order, how the three components could be separated, naming the physical property exploited at each step.
(b) State the temperature at which iodine sublimes, in kelvin, to the appropriate precision.
(c) The recovered sodium chloride is heated until it melts at 801 °C. Explain why the temperature stays at 801 °C while the solid is melting, even though heating continues.

(a) Take the steps in an order that removes one component cleanly each time.

  • Step 1 — warm the mixture gently in a covered vessel with a cold surface above it. The iodine sublimes and then deposits as crystals on the cold surface. Property exploited: iodine turns directly to a gas well below the temperature at which either of the other two changes state.
  • Step 2 — add water and stir. The sodium chloride dissolves and the sand does not. Property exploited: difference in solubility in water.
  • Step 3 — filter. The sand is the residue retained by the paper; the salt solution is the filtrate. Property exploited: difference in particle size between the undissolved solid and the solution.
  • Step 4 — evaporate the filtrate (or distil it, if you also want the water back). The solvent leaves and solid sodium chloride remains. Property exploited: the very large difference in boiling point between water and sodium chloride.

(b) \( T = 114 + 273.15 = 387.15 \). The Celsius value is given to the nearest degree, so the answer cannot be more precise than that: 387 K. Note there is no degree symbol and no “degrees kelvin”.

(c) While the sodium chloride is melting, the energy supplied is used to overcome the electrostatic forces of attraction between the ions in the lattice — it increases the potential energy of the particles. Temperature measures the average kinetic energy of the particles, and the average kinetic energy is not changing, so the temperature does not rise. Only when the last of the solid has melted does further heating go into kinetic energy again, and the temperature resumes climbing.

Check it. Sanity-check the order in (a) by asking what would break if you swapped two steps. Adding water first would work, but you would then have to recover iodine from a wet residue by subliming it out of damp sand — messier, and iodine is slightly soluble, so you would lose some. Filtering before dissolving would achieve nothing at all, since everything is solid. For (b), check the direction: kelvin values are always larger than the Celsius value for ordinary temperatures, so 387 > 114 is right; an answer smaller than 114 means you subtracted.
Quoting 387.15 K, and writing “387 °K”. Two separate errors that both appear in the same line. Adding 273.15 to a value known only to the nearest degree does not make it known to two decimal places — carry the precision of the original measurement. And the kelvin takes no degree symbol: it is 387 K, never 387 °K. Examiners take these off routinely, and they cost nothing to avoid.

📝Practise

Work through these on paper, then reveal the answer. Each targets a different objective from the list above.

1. Iron filings and powdered sulfur are mixed in the ratio 7 : 4 by mass and then heated strongly to form iron(II) sulfide. State two ways in which the product differs from the mixture.
(i) In the mixture the components keep their own properties, so the iron can still be removed with a magnet and the sulfur can still be dissolved out with a suitable solvent; in the compound the iron and sulfur are chemically bonded, so neither works — iron(II) sulfide is not magnetic. (ii) The mixture can be made in any ratio and can be separated by physical means; the compound has a fixed ratio (1 : 1 by moles) and can only be broken down chemically. A third acceptable difference: the compound is a single substance with one sharp melting point, whereas the mixture melts over a range and each component melts separately.
2. Classify each as an element, a compound, a homogeneous mixture or a heterogeneous mixture: (a) brass, (b) carbon dioxide, (c) graphite, (d) muddy river water, (e) vinegar.
(a) Homogeneous mixture — brass is an alloy of copper and zinc in no fixed ratio, uniform throughout. (b) Compound — carbon and oxygen chemically bonded in a fixed 1 : 2 ratio. (c) Element — graphite is an allotrope of carbon, one element only. (d) Heterogeneous mixture — suspended solids give more than one visible phase and non-uniform composition. (e) Homogeneous mixture — a solution of ethanoic acid in water, single phase, composition not fixed.
3. Ethanol boils at 78 °C and water at 100 °C. Explain how a mixture of the two could be separated, and why simple distillation gives a poorer separation than fractional distillation.
Both are liquids that are completely miscible, so filtration is useless; the separation must exploit the difference in boiling point, so use distillation. Heat the mixture: the vapour is richer in the more volatile component, ethanol, which distils over first and is condensed and collected. The problem is that the vapour above a mixture is never pure — it always contains some water — so a single evaporation and condensation gives an impure distillate. A fractionating column provides a large surface area on which the rising vapour repeatedly condenses and re-evaporates. Each of those cycles is effectively another distillation, so the vapour reaching the top is far richer in ethanol than one cycle could make it.
4. A crude organic solid is purified by recrystallization. Explain why the minimum volume of hot solvent is used, and why the mixture is then cooled slowly.
The minimum volume of hot solvent is used so that the solution is saturated at the higher temperature. If excess solvent is used, then on cooling the solution may still be unsaturated and little or no product crystallises — the yield collapses. Slow cooling allows the crystal lattice to grow in an orderly way, so that only molecules of the product fit into the growing crystal and impurities are excluded, remaining in the mother liquor. Rapid cooling traps impurities inside the solid and gives small, less pure crystals. The purity of the recovered solid is then checked by taking a melting point: a pure substance melts sharply over a narrow range, an impure one melts lower and over a wider range.
5. Sketch and describe the heating curve obtained when a pure solid is heated at a constant rate from below its melting point to above its boiling point.
Temperature on the y-axis, time on the x-axis. The curve consists of five sections: a rise (solid warming), a horizontal plateau at the melting point, a rise (liquid warming), a longer horizontal plateau at the boiling point, and a final rise (gas warming). On the sloping sections, energy increases the average kinetic energy of the particles, so the temperature rises. On the plateaus, energy increases potential energy by overcoming forces of attraction between particles, so the temperature is constant even though heating continues. The boiling plateau is longer than the melting plateau because vaporization requires the attractions to be overcome completely, whereas melting only loosens the fixed arrangement.
6. A gas sample is at 27 °C. (a) Convert this to kelvin. (b) The sample is heated until the average kinetic energy of its particles has doubled. State the new temperature in °C. (c) Explain why the calculation in (b) cannot be done on the Celsius scale.
(a) \( T = 27 + 273.15 = 300\ \mathrm{K} \) (300.15 K, but the data justify 3 significant figures). (b) Average kinetic energy is proportional to the absolute temperature, so doubling it means doubling the kelvin temperature: \( 2 \times 300 = 600\ \mathrm{K} \), which is \( 600 - 273 = \mathbf{327\ ^{\circ}C} \). (c) The Celsius scale has an arbitrary zero — the freezing point of water — not the point of minimum kinetic energy. Ratios on it are therefore meaningless: doubling 27 °C to 54 °C corresponds to going from 300 K to 327 K, an increase of only 9% in kinetic energy, not 100%. Only a scale whose zero is absolute zero makes temperature proportional to average kinetic energy.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • PhET — States of Matter, which lets you heat and cool a box of particles and watch the arrangement and speed change; the best single way to make the heating-curve plateau obvious.
  • RSC Learn Chemistry — practical protocols for recrystallization, distillation and melting-point determination, written for school laboratories.
  • Your data booklet — not external, but open it now and find the periodic table and the constants page, so that you know where they are before you need them.