Pure substances, mixtures and formulations: testing purity and purifying by filtration and crystallisation
By the end of this lesson you will be able to
- Define a pure substance in the chemical sense and distinguish it from the everyday meaning of “pure”, with particle diagrams.
- Use melting and boiling points, and the shape of a heating curve, to determine whether a sample is pure, and explain the effect of an impurity at the particle level.
- Describe a formulation and calculate the masses of its components from a recipe.
- Select and justify a separation method from the property that differs between the components, and outline filtration, evaporation and crystallisation.
- Calculate percentage purity, percentage recovery and the mass of crystals formed on cooling from solubility data.
You met this before
- An element contains only one kind of atom (copper, Cu; oxygen, O₂); a compound contains two or more elements chemically bonded in a fixed ratio (water, H₂O; sodium chloride, NaCl); a mixture contains substances that are not chemically bonded and can be in any proportion (air, sea water, brass).
- In a solid the particles vibrate about fixed positions in a regular arrangement; melting means they gain enough energy to break out of that arrangement and slide past each other.
- A solution is a solute dissolved in a solvent; it is saturated when no more solute will dissolve at that temperature; solubility is usually given in g of solute per 100 g of water and, for most solids, rises with temperature.
- Physical changes (dissolving, melting, evaporating) can be reversed by physical means and make no new substance — which is why mixtures can be separated without a chemical reaction.
- Percentage of a part: part ÷ whole × 100; 1 dm³ = 1 000 cm³; 1 kg = 1 000 g.
Sort into element, compound or mixture: air; distilled water; brass; sodium chloride; sea water; oxygen gas; carbon dioxide; tap water.
Reveal
Elements: oxygen gas. Compounds: distilled water, sodium chloride, carbon dioxide.
Mixtures: air (N₂, O₂, Ar, CO₂…), brass (copper and zinc in varying proportions), sea water (water and dissolved salts), tap water (water with dissolved minerals and a trace of chlorine). If you put tap water with the compounds, this lesson is for you.
Revise it in full: Year 1 · Everyday materials · Lesson 2 — Elements, compounds and mixtures, and what “pure” really means · Year 1 · Everyday materials · Lesson 4 — Separating mixtures: from the salt pan to the laboratory · Year 3 · Solutions, acids and bases · Lesson 1 — What a solution is
The puzzle
A regulator buys two boxes of paracetamol tablets from two market stalls. The tablets look identical, taste identical and carry the same printed name. Crushed, extracted and dried, both give a fine white powder.
In a melting-point tube, powder A melts sharply between 169 and 170 °C. Powder B starts to melt at 156 °C and is not fully liquid until 164 °C. The World Health Organization has estimated that about one medicine in ten sold in low- and middle-income countries is substandard or falsified. Which powder would you trust — and how can a thermometer tell you?
Predict which powder is pure paracetamol and write one sentence, about particles, explaining why the other one melts lower and over a range of temperatures.
Compare your answer
Powder A: pure paracetamol melts sharply at 169–170 °C. Powder B contains something else.
Foreign particles sit among the paracetamol molecules and spoil the regular arrangement of the solid, so less energy is needed to break it down; different parts of the solid hold different amounts of impurity, so they melt at different temperatures — a lower, wider melting range. By the end of this lesson you will be able to say that in examiner’s language and put numbers on it.
The science — structured notes
1.1 What “pure” means to a chemist
In everyday language “pure” means “nothing added”: pure orange juice, pure mountain water. In chemistry a pure substance is a single element or a single compound, containing only one kind of particle. Everything else is a mixture.
By the chemist’s definition “pure orange juice” is a mixture of water, sugars, citric acid, vitamins and pigments. “Pure mountain water” is a mixture of water with dissolved calcium, magnesium and hydrogencarbonate ions and dissolved air.
Particle diagrams make the test visual: one kind of particle (all atoms alike, or all molecules alike) means pure; two or more kinds means a mixture. Notice that a compound is pure even though it contains two elements — its particles are all identical.
“Pure gold jewellery marked 24 carat and 18 carat are both pure substances, because both are sold as gold.”
Reveal answer
False. 24 carat gold is (very nearly) pure gold, 99.9 % Au. 18 carat means 18 parts of gold in 24 — 75 % gold by mass, alloyed with silver and copper to make it harder.
An alloy is a mixture of metals, so 18 carat gold is a mixture; it melts over a range of temperatures instead of at gold’s 1 064 °C.
1.2 Melting and boiling points: the thermometer as a purity test
A pure substance melts and boils at fixed, sharp temperatures (at a stated pressure). Pure water freezes at 0 °C and boils at 100 °C at 101.3 kPa; pure paracetamol melts at 169–170 °C. These values are properties of the substance and can be looked up to identify it.
An impurity changes both. It lowers the melting point and widens it into a range: impurity particles spoil the regular arrangement of the solid, so less energy is needed to break it down, and regions with more impurity melt first.
A dissolved solid raises the boiling point: the solute particles take up part of the liquid surface and hold on to water molecules, so the liquid must be hotter before enough of them escape. Sea water, with about 35 g of dissolved salts in every kilogram, freezes at about −1.9 °C and boils at about 100.6 °C.
The flat part of the pure curve is where the heat goes into separating particles, not into making them move faster — so the temperature stops rising until all the solid has melted. The impure sample has no flat part: as it melts, the liquid gets richer in impurity and the melting temperature keeps sliding.
Pure paracetamol melts at 169–170 °C. A quality-control laboratory records: batch 1, 169–170 °C; batch 2, 162–167 °C; batch 3, 150–163 °C. Rank the batches by purity and justify.
Compare with the data-book value. Batch 1 matches it and melts over only 1 °C.
Look at the start and the width. Batch 2 starts 7 °C low and melts over 5 °C; batch 3 starts 19 °C low and melts over 13 °C.
Answer. Purest: batch 1, then batch 2, then batch 3 — the lower the start and the wider the range, the more impurity.
Limit. A melting point shows that an impurity is there, not what it is or exactly how much; chromatography (Lesson 3) identifies it.
A trader sells a coin as “pure silver”. A jeweller melts a filing from its edge: it begins to melt at 802 °C and is fully liquid at 893 °C. Pure silver melts at 962 °C. What can you conclude?
Start of melting. 802 °C is 160 °C below the melting point of silver, so the coin is not pure silver.
Range. It melts over 91 °C, the signature of a mixture.
Answer. The coin is an alloy — these are the melting temperatures of sterling silver, 92.5 % silver and 7.5 % copper, a sensible coin metal because it is harder than pure silver. “Pure” on the label is the everyday meaning.
A student distils a liquid from a river sample. The distillate boils at exactly 100.0 °C at 101.3 kPa and leaves no residue when a drop evaporates on a clean watch glass. A second sample, taken straight from the river, boils at 100.2 °C and leaves a white ring. Explain what each result shows.
Reveal answer
The distillate is pure water: it boils sharply at water’s boiling point at that pressure, and no dissolved solid is left when it evaporates.
The river water is a mixture: dissolved solids raise its boiling point (by 0.2 °C here) and remain as a white residue — mostly calcium and magnesium salts — when the water evaporates.
1.3 Formulations: mixtures designed on purpose
Not every mixture is an impure version of something. A formulation is a mixture designed as a useful product, made by mixing its components in carefully measured quantities, each chosen for a purpose. Medicines, paints, fertilisers, alloys, cosmetics, cleaning products and fuels are all formulations.
A paint, for example, has a pigment for colour, a binder that forms the film, a solvent that keeps it liquid and additives that stop it settling or growing mould.
| WHO oral rehydration salts (per litre of solution) | Mass / g | Purpose |
|---|---|---|
| Sodium chloride, NaCl | 2.6 | replaces sodium lost in diarrhoea |
| Glucose, anhydrous | 13.5 | carries sodium (and so water) across the gut wall |
| Potassium chloride, KCl | 1.5 | replaces potassium |
| Trisodium citrate dihydrate | 2.9 | corrects the acidity of the blood |
| Total | 20.5 | dissolved in clean water to make 1.00 dm³ |
This formulation has saved millions of children’s lives since the 1970s, and the exact proportions matter: too much salt makes the solution draw water out of the body, too little glucose and the sodium is not absorbed.
A clinic needs 250 cm³ of oral rehydration solution. Calculate the mass of sodium chloride needed and the percentage by mass of glucose in the dry powder.
Scale the recipe. 250 cm³ is 0.250 dm³.
mass of NaCl = 2.6 g dm⁻³ × 0.250 dm³
mass of NaCl = 0.65 g
Percentage of glucose. % glucose = 13.5 g ÷ 20.5 g × 100
% glucose = 65.9 % of the powder by mass
A student says a formulation is “just an impure substance”. Explain why a chemist would disagree, using a fertiliser labelled NPK 10 : 20 : 10 as your example.
Reveal answer
An impure substance is a pure substance contaminated with something unwanted. A formulation is a mixture whose every component is wanted, and present in a measured, stated proportion because each has a job.
NPK 10 : 20 : 10 is formulated to supply nitrogen (for leaf growth), phosphorus (roots and flowers) and potassium in fixed proportions — the numbers are percentages by mass of N, P₂O₅ and K₂O. Change the proportions and the product does a different job.
1.4 Choosing a separation method: find the property that differs
Every separation exploits a property that is different for the components. Name that property and the method follows; examiners reward the property as the justification.
| Mixture | Property that differs | Method |
|---|---|---|
| Insoluble solid in a liquid (sand in water) | particle size; solubility | filtration (or decanting if the solid settles) |
| Soluble solid in a solvent, solid wanted (salt in water) | volatility: the solvent evaporates, the solute does not | evaporation, or crystallisation for purer crystals |
| Solvent wanted from a solution (water from sea water) | boiling point | simple distillation (Lesson 2) |
| Miscible liquids (ethanol and water; crude oil) | boiling point, when they are close | fractional distillation (Lesson 2) |
| Dissolved coloured or colourless substances (dyes, amino acids) | attraction to a stationary phase vs a mobile phase | chromatography (Lesson 3) |
| Immiscible liquids (oil and water) | density; they form layers | separating funnel |
| Iron filings in sulfur | magnetism | magnet |
“To get pure salt from rock salt (salt mixed with sand and clay), I will filter the rock salt to remove the sand, then distil the filtrate to collect the salt.”
Reveal answer
Two errors. Rock salt must first be dissolved in warm water — dry rock salt cannot be filtered. And distillation collects the solvent (water) as the distillate; the salt stays in the flask.
The right sequence: grind, dissolve in warm water and stir, filter (sand and clay stay as the residue), then evaporate or crystallise the filtrate to obtain the salt.
1.5 Filtration, evaporation and crystallisation
Filtration passes a mixture through filter paper whose pores are smaller than the solid particles but larger than dissolved particles and solvent molecules. The trapped solid is the residue; the liquid that passes through is the filtrate. The pore size is what decides what is trapped — the point of the May 2021 Chemistry question on well-water filters.
Evaporation to dryness drives off all the solvent and leaves every dissolved solid — wanted and unwanted — as a crust. Crystallisation is gentler and gives a purer product: heat the solution only until it is saturated (a drop on a cold glass rod forms crystals), then leave it to cool.
As it cools, the solubility falls, the excess solute comes out as crystals, and soluble impurities, present in small amounts, stay dissolved. Filter off the crystals, rinse with a little cold water and dry them between filter papers or in a warm oven.
The solubility of potassium nitrate is 110 g per 100 g of water at 60 °C and 32 g per 100 g of water at 20 °C. A saturated solution in 50.0 g of water at 60 °C is cooled to 20 °C. Calculate the mass of crystals formed.
Dissolved at 60 °C. 110 g × 50.0 ÷ 100 = 55.0 g
Can stay dissolved at 20 °C. 32 g × 50.0 ÷ 100 = 16.0 g
Crystals. 55.0 g − 16.0 g = 39.0 g of potassium nitrate crystals
Sense check. 71 % of the solid comes out: a steep solubility curve makes cooling an efficient purification. Sodium chloride’s curve is almost flat, so salt is purified by evaporating most of the water instead.
Use the curves above. A solution of 30 g of sodium chloride in 100 g of water at 60 °C is cooled to 20 °C. How much salt crystallises? Explain why salt producers at the Sua Pan in Botswana use the Sun rather than cooling to make salt from brine.
Reveal answer
None: the solubility of sodium chloride at 20 °C is about 36 g per 100 g of water, more than the 30 g present, so the solution is still unsaturated.
Because salt’s solubility hardly changes with temperature, cooling brine gives almost no crystals. Removing the solvent does: solar evaporation in shallow ponds concentrates the brine until it is saturated, and every further gram of water that evaporates leaves about 0.36 g of salt behind as crystals.
1.6 Percentage purity and percentage recovery
Two numbers describe how good a purification was. Percentage purity says how much of a sample is the wanted substance. Percentage recovery says how much of the wanted substance you managed to collect from what was there at the start.
percentage purity = mass of pure substance in the sampletotal mass of the sample × 100
percentage recovery = mass of pure substance collectedmass of that substance at the start × 100
25.0 g of rock salt is dissolved in warm water, filtered, and the filtrate is evaporated. The dry, clean salt weighs 21.3 g. Calculate the percentage purity of the rock salt, and state one reason the true value may be different.
% purity = 21.3 g ÷ 25.0 g × 100
% purity = 85.2 %
Limit. Some salt stays on the filter paper and the glassware, which makes the answer too low; soluble impurities (such as calcium sulfate) are left in the product when the water evaporates, which makes it too high. Rinsing the residue and crystallising instead of evaporating reduces both errors.
A brine contains 150 g dm⁻³ of dissolved solids, of which 88.0 % by mass is sodium chloride. 2.00 m³ of brine is evaporated in a solar pond. The salt harvested weighs 275 kg, and analysis shows it contains 252 kg of sodium chloride. Calculate the percentage purity of the harvested salt and the percentage recovery of the sodium chloride.
Volume. 2.00 m³ is 2 000 dm³.
dissolved solids = 150 g dm⁻³ × 2 000 dm³
dissolved solids = 300 000 g
dissolved solids = 300 kg
NaCl at the start = 0.880 × 300 kg
NaCl at the start = 264 kg
% purity = 252 kg ÷ 275 kg × 100
% purity = 91.6 %
% recovery = 252 kg ÷ 264 kg × 100
% recovery = 95.5 %
Meaning. Almost all the salt was collected, but 23 kg of the product is something else — the magnesium and calcium salts that crystallise with it, which a producer removes by stopping the evaporation before they come out.
Compound Z dissolves 8.0 g per 100 g of water at 90 °C and 0.50 g per 100 g at 20 °C; its impurity, Q, is very soluble at both temperatures.
A 5.00 g crude sample, 92.0 % Z, is recrystallised from the minimum volume of water at 90 °C and cooled to 20 °C. Calculate the maximum mass of pure Z that can be collected and the percentage recovery, and explain why the impurity does not crystallise.
Reveal answer
Mass of Z = 0.920 × 5.00 g
Mass of Z = 4.60 g
Minimum water at 90 °C = 4.60 g ÷ 8.0 g × 100 g
Minimum water at 90 °C = 57.5 g
Z still dissolved at 20 °C = 0.50 g × 57.5 ÷ 100
Z still dissolved at 20 °C = 0.29 g
Z collected = 4.60 g − 0.29 g
Z collected = 4.31 g
% recovery = 4.31 ÷ 4.60 × 100
% recovery = 93.7 %
Q (0.40 g) is very soluble, so 57.5 g of cold water is far from saturated with it; Q stays in the filtrate. The price of purity is the 0.29 g of Z that also stays in solution — using more water than the minimum would lose more.
Try it — how much does an impurity lower a melting point?
The lab heats a sample in a beaker over a burner with a thermometer in it. You choose the pure substance (water ice with salt, stearic acid with palmitic acid, or naphthalene with camphor), the percentage by mass of impurity mixed in (0 to 20 %) and the heating rate.
The reading is the temperature at which melting starts. The numbers are a simplified model — a straight-line fall per per cent of impurity — which is close to real behaviour for small amounts of impurity. Use stearic acid, whose real melting point is 69 °C.
Suppose your means for stearic acid at 2 °C/min were: 0 % → 69.0; 4 % → 65.4; 8 % → 61.8; 12 % → 58.2; 16 % → 54.6; 20 % → 51.0 °C.
(a) Calculate the gradient of the line and state its unit.
(b) Determine whether the melting temperature is proportional to the percentage of impurity, and whether the fall in melting temperature is.
(c) A second student, heating at 5 °C per minute, got 69.9 °C at 0 % and 51.9 °C at 20 %. Name the type of error and explain its cause.
Reveal answer
(a) Gradient = (51.0 − 69.0) °C ÷ (20 − 0) %
Gradient = −18.0 ÷ 20
Gradient = −0.90 °C per %
(b) The temperature is not proportional to the percentage: the line does not pass through the origin (69.0 °C at 0 %), and doubling the impurity from 4 to 8 % does not double the temperature. It is linear.
The fall below 69.0 °C is proportional: 3.6 °C at 4 %, 7.2 °C at 8 %, 14.4 °C at 16 % — doubling the impurity doubles the fall.
(c) Systematic error: both readings are 0.9 °C too high. At a fast heating rate the heat cannot spread into the sample quickly enough, so the solid lags behind the thermometer: by the time the first liquid is seen, the thermometer already reads higher than the temperature at which melting began. It shifts every point by a similar amount, so the gradient is almost unchanged but the intercept is wrong.
Common mistakes to avoid
- Using the everyday meaning of “pure” — in a chemistry answer, pure means one substance only; “natural” or “nothing added” is not the test.
- Saying an impurity raises the melting point — it lowers it and widens it into a range; a dissolved solid raises the boiling point.
- Calling a compound a mixture — water contains two elements but is a pure compound; its molecules are identical.
- Filtering something that is dissolved — dissolved particles pass through the filter paper; filtration only removes insoluble solids.
- Writing “distil to collect the salt” — distillation collects the solvent; the solute stays in the flask.
- Evaporating a hydrated salt to dryness — it destroys the crystals’ water of crystallisation; heat to saturation and let it crystallise.
- Confusing purity and recovery — purity is about what is in your product; recovery is about how much of what was there you collected.
- “Amount of salt” — say the mass of salt (in g), or the volume of solution (in cm³).
Study smart: how this is examined
Study routine. Today: redraw the four particle boxes and the two heating curves from memory, then do the rock-salt and potassium-nitrate EXAMPLES without looking. In two days: write the separation table’s “property” column from memory and do the Stretch check.
In a week: run the purity lab as a full investigation, then the Think task and the quiz.
How we found out — the chemist who named the poisons in the food
In 1820 Friedrich Accum, a German chemist working in London, published A Treatise on Adulterations of Food, and Culinary Poisons. Using the tests of his day he showed that bread was whitened with alum, pickles were greened with copper salts, sweets were coloured with lead and mercury compounds, and tea was re-dried from used leaves.
The book sold out within a month, and its title page carried a line from the Bible: “There is death in the pot.” Accum named the traders who had been convicted — which made him powerful enemies — and he left England the next year after a scandal.
The evidence that finally changed the law came from a microscope. Between 1851 and 1854 the physician Arthur Hill Hassall examined thousands of samples bought in ordinary London shops and reported in The Lancet that most coffee contained chicory, much of the “mustard” was flour and turmeric, and many coloured sweets contained lead chromate and other poisonous pigments.
Because he named the shops and printed the results, Parliament held an inquiry, and the Adulteration of Food and Drink Act followed in 1860; later Acts in the 1870s made public analysts a regular part of local government.
The idea that a substance can be identified and its purity checked by a measurable property — a melting point, a crystal shape, later a chromatogram or a spectrum — is the foundation of every modern food and medicine regulator. The regulator in this lesson’s puzzle, with a melting-point tube, is doing what Accum and Hassall did.
Where it is used now
- Medicines quality control. Every batch of a medicine is checked against a pharmacopoeia standard: identity, purity (often by melting point and chromatography) and the mass of active ingredient per tablet. National medicines regulators test samples taken from pharmacies and markets.
- Salt and soda ash at Sua Pan, Botswana. Brine pumped from beneath the pan is concentrated by the Sun in large evaporation ponds so that salt crystallises out, and soda ash (sodium carbonate) is also produced from the brine — crystallisation on an industrial scale.
- Sugar refineries in Eswatini, Mauritius, Brazil and India crystallise sucrose from concentrated cane juice, spin off the syrup that holds the impurities, and recrystallise for white sugar.
- Formulated products everywhere. Oral rehydration salts in every clinic; fertilisers blended to a stated N : P : K ratio; paints; the petrol and diesel that are blended to a specification at every refinery.
- Semiconductor silicon must be 99.9999999 % pure (nine nines) for computer chips — a single impurity atom per billion silicon atoms matters.
Science now — and the scientist you need to be
What is happening in 2026
The WHO has run a global surveillance and monitoring system for substandard and falsified medical products since 2013, and national regulators report suspect products to it. Increasingly, inspectors screen medicines in the field with handheld spectrometers that compare a tablet’s infrared or Raman “fingerprint” with a reference library in seconds, sending only suspicious samples to a laboratory.
In factories, purity is watched while the product is being made rather than afterwards: probes inside crystallisers measure crystal size and concentration continuously, and software adjusts the cooling rate — a field called process analytical technology.
The next decade will bring more of this and more automation. But every automatic reading still rests on the ideas of this lesson: one substance has one set of properties, and an impurity changes them in a measurable way.
The mindshift
What you need to be able to do
Classify substances with particle diagrams; use a melting range to judge purity and explain it with particles; calculate a formulation and a crystallisation yield; choose a separation and justify it with a property; report percentage purity and recovery with the reasons they might be wrong.
See an example
The soil stays in the filter as the residue; the filtrate is cloudy at first if fine clay passes through (pores too large). After evaporation, white cubic crystals of salt form in the saucer, with a brownish rim where the dissolved soil substances collect — evidence that evaporation to dryness traps soluble impurities.
A good assistant answer: most table salt is a formulation — sodium chloride with a small amount of potassium iodide or iodate (for health) and an anti-caking agent — so it is not chemically pure, even though it is more than 97 % NaCl.
Key terms
| Pure substance | A single element or a single compound, containing only one kind of particle; it has a sharp, fixed melting point and boiling point. |
| Mixture | Two or more substances not chemically combined, in any proportions; separable by physical methods. |
| Formulation | A mixture designed as a useful product, with each component present in a measured quantity for a purpose. |
| Melting range | The temperatures between the first liquid appearing and the last solid disappearing; wide and low for an impure sample. |
| Filtration | Separating an insoluble solid from a liquid by passing the mixture through filter paper. |
| Residue | The solid left in the filter paper. |
| Filtrate | The liquid (or solution) that passes through the filter paper. |
| Crystallisation | Obtaining crystals of a solute by making its solution saturated and letting it cool or evaporate slowly. |
| Water of crystallisation | Water that is part of a crystal’s structure, as in CuSO₄·5H₂O. |
| Percentage purity | Mass of pure substance ÷ total mass of the sample × 100. |
| Percentage recovery | Mass of substance collected ÷ mass of that substance at the start × 100. |
Think like a scientist
Criterion C A student purified crude aspirin (pure aspirin melts at 135–136 °C) by recrystallisation from hot ethanol and water. After each recrystallisation she dried the crystals, weighed them and measured the melting range three times. Her results:
| Number of recrystallisations | Mass of crystals / g | Melting starts, trials 1, 2, 3 / °C | Melting ends (mean) / °C |
|---|---|---|---|
| 0 (crude) | 10.00 | 121, 123, 122 | 131 |
| 1 | 8.40 | 131, 130, 131 | 135 |
| 2 | 7.60 | 134, 134, 128 | 136 |
| 3 | 7.05 | 135, 134, 135 | 136 |
(a) Identify the anomalous reading and suggest a specific cause. [2]
(b) Calculate the mean temperature at which melting starts for each sample, excluding the anomaly, and the melting range. [3]
(c) Calculate the percentage of the crude mass recovered after three recrystallisations, and the percentage lost in the first. [2]
(d) Describe the trend in purity and in mass, and explain the trade-off using the idea of solubility. [3]
(e) The student concludes: “The crude aspirin was 70.5 % pure.” Evaluate this conclusion. [2]
(f) Suggest one improvement to the method and one extension. [2]
Model answer
(a) 128 °C, trial 3 after two recrystallisations — 6 °C below the other two trials (134 °C) [1].
Likely cause: the crystals for that trial were not fully dry, so traces of solvent acted as an impurity; or the tube was heated too quickly and the start was misjudged — a specific cause, not “a mistake” [1].
(b) Crude: (121 + 123 + 122) ÷ 3 = 122 °C; range 122–131 °C (9 °C) [1].
One: (131 + 130 + 131) ÷ 3 = 130.7 °C, which is 131 °C to the thermometer’s precision; range 131–135 °C (4 °C).
Two: (134 + 134) ÷ 2 = 134 °C; range 134–136 °C (2 °C).
Three: (135 + 134 + 135) ÷ 3 = 134.7 °C, which is 135 °C; range 135–136 °C (1 °C) [1 for both, 1 for rounding to whole degrees].
(c) Recovery = 7.05 g ÷ 10.00 g × 100
Recovery = 70.5 % [1].
Lost in the first = (10.00 − 8.40) ÷ 10.00 × 100
Lost in the first = 16.0 % [1].
(d) Each recrystallisation raised the start of melting (122 → 131 → 134 → 135 °C) and narrowed the range (9 → 4 → 2 → 1 °C), approaching the pure value of 135–136 °C: purity rose [1].
But the mass fell each time (10.00 → 8.40 → 7.60 → 7.05 g), with smaller gains in purity [1]. Some aspirin always stays dissolved in the cold solvent and is lost in the filtrate, along with the impurities; each extra purification buys a little purity for a certain loss of product [1].
(e) Not valid: 70.5 % is the recovery, not the purity. Aspirin was lost in the filtrate at every stage, so the crude sample contained more than 7.05 g of aspirin and was more than 70.5 % pure [1].
The melting data show the crude sample was clearly impure, but a melting range cannot give an exact percentage; a titration or chromatography with a calibration would be needed [1].
(f) Improvement: dry the crystals to constant mass in a desiccator before weighing and before the melting-point test, so solvent does not lower the melting point or add mass [1].
Extension: investigate how the volume of solvent used (the IV) affects the percentage recovery, keeping the crude mass fixed [1].
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
Select the pure substances from this list and give a reason: steel; distilled water; air; copper; brine; carbon dioxide.
Reveal answer
Distilled water, copper and carbon dioxide [1]; each contains only one kind of particle (one element or one compound), whereas steel, air and brine are mixtures of two or more substances [1].
A sample of naphthalene melts between 74 and 78 °C. Pure naphthalene melts at 80 °C. Explain what the data show about the sample.
Reveal answer
The sample is impure: it melts below 80 °C and over a range of 4 °C rather than sharply [1].
The impurity particles disrupt the regular arrangement of naphthalene molecules in the solid, so less energy is needed to separate them and different regions melt at different temperatures [1].
Outline how to obtain dry, pure copper(II) sulfate crystals from a mixture of copper(II) sulfate powder and insoluble copper(II) oxide.
Reveal answer
Add warm water and stir so that the copper(II) sulfate dissolves; filter — the copper(II) oxide stays as the residue [1].
Heat the blue filtrate gently until it is saturated (crystals form on a cold glass rod dipped in it), then leave it to cool so crystals form [1].
Filter off the crystals, rinse with a little cold water and dry them between filter papers — do not evaporate to dryness, which would remove the water of crystallisation [1].
A 12.0 g sample of impure benzoic acid gives 10.2 g of pure benzoic acid. Calculate the percentage purity of the sample.
Reveal answer
% purity = 10.2 g ÷ 12.0 g × 100 [1]
% purity = 85.0 % [1]
Using the WHO oral rehydration recipe (2.6 g NaCl, 13.5 g glucose, 1.5 g KCl, 2.9 g trisodium citrate per dm³), calculate the total mass of powder in a sachet designed to make 500 cm³, and the mass of glucose in it.
Reveal answer
Total = 20.5 g dm⁻³ × 0.500 dm³
Total = 10.25 g, about 10.3 g [1]
Glucose = 13.5 g dm⁻³ × 0.500 dm³
Glucose = 6.75 g [1]
Sea water is filtered and the filtrate is tested. Explain why the filtrate still boils above 100 °C at 101.3 kPa, and name a method that would give water boiling at exactly 100 °C.
Reveal answer
Filtration removes only insoluble particles; the dissolved salt ions are much smaller than the pores and pass through into the filtrate [1].
The dissolved ions raise the boiling point, because the liquid must be hotter before water molecules escape fast enough to boil [1].
Simple distillation: only the water evaporates and condenses, so the distillate is pure water [1].
7. The solubility of potassium chloride is 34.0 g per 100 g of water at 20 °C and 51.3 g at 80 °C.
A student dissolves 40.0 g of crude KCl (containing 2.0 g of sand and 1.5 g of very soluble sodium nitrate) in the minimum mass of water at 80 °C, filters it hot, and cools the filtrate to 20 °C. Calculate the mass of water needed, the mass of KCl crystals formed, and the percentage recovery; state where each impurity ends up.
Reveal answer
Mass of KCl = 40.0 − 2.0 − 1.5
Mass of KCl = 36.5 g
Water needed at 80 °C = 36.5 ÷ 51.3 × 100
Water needed at 80 °C = 71.2 g [1]
KCl still dissolved at 20 °C = 34.0 × 71.2 ÷ 100
KCl still dissolved at 20 °C = 24.2 g
Crystals = 36.5 − 24.2
Crystals = 12.3 g [1]
% recovery = 12.3 ÷ 36.5 × 100
% recovery = 33.7 % [1]
Sand: removed as the residue in the hot filtration. Sodium nitrate: stays dissolved in the cold filtrate. The low recovery reflects KCl’s shallow solubility curve — cooling is a poor way to recover it [1].
8. Two white powders, P and Q, both melt at 122 °C. When they are ground together in equal masses, the mixture melts between 95 and 110 °C. What can you conclude about P and Q, and why is this “mixed melting point” test more convincing evidence than comparing the two melting points alone?
Reveal answer
P and Q are different substances [1]: if they were the same compound, mixing them would make no difference and the mixture would still melt sharply at 122 °C [1].
Instead each acts as an impurity in the other, lowering and widening the melting point [1].
Two different compounds can have the same melting point by coincidence (benzoic acid melts at 122 °C, and so do other substances), so equal melting points do not prove identity; a mixed melting point that stays sharp and unchanged does [1].
Summary
- A pure substance is one element or one compound — one kind of particle; everything else is a mixture. A formulation is a mixture designed on purpose, each component measured for a job.
- Pure substances melt and boil sharply at fixed temperatures; impurities lower and widen the melting point and dissolved solids raise the boiling point — a thermometer is a purity test.
- Choose a separation by the property that differs: particle size and solubility (filtration), volatility (evaporation, crystallisation), boiling point (distillation), attraction to a stationary phase (chromatography).
- Crystallisation gives purer crystals than evaporation to dryness and keeps water of crystallisation; the mass of crystals from cooling = solubility at the high temperature minus solubility at the low, scaled to the mass of water.
- Percentage purity = pure mass ÷ sample mass × 100; percentage recovery = mass collected ÷ mass present at the start × 100; purity and recovery trade against each other.
Next time → Distillation: simple and fractional distillation, the ethanol–water curve you will measure yourself, and how a refinery splits crude oil into the fractions that fuel the world.