Separation and Purity
Select a separation method from physical properties and use measured data to assess purity.
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The core idea
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Learning objectives
- describe methods of separation and purification for the components of mixtures, to include: — use of a suitable solvent, filtration and crystallisation or evaporation
- describe methods of separation and purification for the components of mixtures, to include: — distillation and fractional distillation (see also 8.1(b))
- describe methods of separation and purification for the components of mixtures, to include: — paper chromatography
- suggest suitable separation and purification methods, given information about the substances involved in the following types of mixtures: — solid-solid
- suggest suitable separation and purification methods, given information about the substances involved in the following types of mixtures: — solid-liquid
- suggest suitable separation and purification methods, given information about the substances involved in the following types of mixtures: — liquid-liquid (miscible)
- interpret paper chromatograms including comparison with ‘known’ samples (the use of Rf values is not required)
- deduce from given melting point and boiling point data the identities of substances and their purity.
Choose a separation method from a difference in physical properties. State what passes through, what remains, and how the required substance is recovered.
1. Definition
Separation divides a mixture without forming new substances. Purification removes unwanted substances. A pure substance contains one substance and has characteristic physical properties.
2. Key Ideas
- Filtration separates an insoluble solid from a liquid.
- Crystallisation obtains a dissolved solid without heating the solution to dryness.
- Simple distillation obtains a solvent from a solution.
- Fractional distillation separates miscible liquids with different boiling points.
- Paper chromatography separates soluble substances because they travel differently with a solvent.
- A pure substance melts or boils sharply at its characteristic temperature.
3. Detailed Explanations
Match method to evidence
Use filtration when particle size and insolubility differ. Use crystallisation when a solid is soluble and may decompose on strong heating. Use distillation when volatility or boiling point differs.
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For a solid–solid mixture, choose a solvent that dissolves only one solid. Add the solvent, stir, filter, and wash and dry the insoluble residue. Recover the dissolved solid from the filtrate by crystallisation or evaporation. For a solid–liquid mixture, filter if the solid is insoluble; if it is dissolved, use crystallisation, evaporation or distillation according to the product required. For miscible liquids, use fractional distillation when their boiling points differ.
Choose evaporation when the solvent is not required and the dissolved solid is stable on heating. Choose crystallisation when crystals are wanted or heating to dryness could decompose the solid. Choose simple distillation when the solvent itself must be collected.
Carry out simple and fractional distillation
For simple distillation, heat the solution so that the more volatile liquid boils. Its vapour enters the condenser, cools and changes back to liquid. Collect this liquid as the distillate; the less volatile dissolved substance remains in the flask. Keep the thermometer bulb level with the side arm so it measures the vapour entering the condenser. Cooling water enters the lower condenser port and leaves from the upper port, keeping the water jacket full.
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For fractional distillation, place a fractionating column between the heated flask and condenser. Repeated condensation and vaporisation in the column allow vapour richer in the lower-boiling liquid to reach the condenser first. Use this method for miscible liquids, especially when their boiling points are close enough that simple distillation would separate them poorly.
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Do not heat a closed apparatus. The receiver remains open to the air so pressure cannot build up.
Set up and read paper chromatography
In chromatography, place the sample on a pencil baseline above the solvent. The solvent moves through the paper and carries components different distances. Here, interpret the number of spots and their positions compared with known samples. Where your syllabus also requires calculated R_f values, the methods of separation and purification lesson teaches that calculation.
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Judge purity
An impurity usually lowers and broadens a solid’s melting range. A pure liquid boils at a sharp characteristic temperature at a stated pressure. One chromatogram spot supports purity only for the solvent and locating conditions used.
4. Common Mistakes
- Filtration cannot remove a dissolved solute.
- Evaporation to dryness is not the safest way to obtain every soluble solid.
- The chromatography baseline must be pencil, not ink.
- The sample spot must begin above the solvent level.
- “High melting point” does not prove purity; use a sharp characteristic value.
5. Exam Tips
Exam question 1: Separate with the required product in mindCore
A mixture contains sand, salt and water. Describe how to obtain dry sand and pure water.
Show Answer
Filter the mixture. Wash and dry the residue to obtain sand. Distil the filtrate and collect the condensed solvent as pure water.
Use the pattern property difference → method → named apparatus action → product obtained.
Exam question 2: Use melting data to identify purityCore
Pure solid A melts at 80 °C and pure solid B melts at 122 °C. A sample melts sharply at 121–122 °C. Another sample melts over 116–121 °C. Identify the first sample and comment on the second.
Show Answer
The first sample is consistent with pure B because it melts sharply at B’s characteristic temperature. The second sample is impure B: its melting range is lower and broader.
6. Worked Examples
Modelled example 1
Recover a Soluble Solid
Problem
Study the worked solution
Concentrate the solution
Method
Heat gently to evaporate some water until nearly saturated.Reason
Removing some solvent prepares the solution to crystallise on cooling without heating the salt to dryness.Working
Gentle evaporation → nearly saturated solution.Form and collect crystals
Method
Cool, then filter the crystals.Reason
Solubility falls on cooling and the crystals become a solid residue.Working
Cool → crystallise → filter.Purify and dry
Method
Wash with a little cold distilled water and dry between filter papers.Reason
Cold water removes adhering solution while limiting crystal loss.Working
Washed, dry copper(II) sulfate crystals.
Guided practice 2
Choose Between Simple and Fractional Distillation
Problem
Choose and justify
Hints
Hint 1: mixture type
Hint 2: boiling evidence
View solution step by step
Choose the method
Method
Use fractional distillation.Reason
P and Q are miscible liquids with different, fairly close boiling points.Working
Mixture type + boiling-point evidence → fractional distillation.Explain the column
Method
Use a fractionating column before the condenser.Reason
Repeated condensation and vaporisation make the vapour reaching the condenser richer in the lower-boiling liquid P.Working
P distils first near 78 °C.
Guided practice 3
Interpret Chromatogram Spots
Problem
Separate both conclusions
Hints
Hint 1: count
Hint 2: compare
View solution step by step
Use spot count
Method
Infer that the unknown is a mixture.Reason
It separates into two visible components under these conditions.Working
Two spots → mixture.Use alignment
Method
State that its components are consistent with A and B.Reason
Each unknown spot aligns with one reference in the same chromatogram.Working
Contains components consistent with A and B.
Common misconception 4
Do Not Filter a Dissolved Solute
Learner plan
Track the dissolved salt
View solution step by step
Reject filtration
Method
State that dissolved salt passes through the filter.Reason
Its particles are not an insoluble solid trapped by filter paper.Working
Residue: none of the dissolved salt; filtrate: salt solution.Choose crystallisation
Method
Concentrate and cool the solution, then collect and dry crystals.Reason
The method uses the solute’s changing solubility as solvent is removed and temperature falls.Working
Crystallisation recovers the salt.
Examiner practice 5
Recover Sand and Pure Water
Examination question
Write the method in order
View solution step by step
Separate sand
1 markMethod
Filter the mixture.Reason
Sand is insoluble and remains as residue.Working
Sand residue; salt solution filtrate.Finish the sand
1 markMethod
Wash the residue and dry it.Reason
Washing removes adhering salt solution.Working
Dry sand obtained.Recover water
2 marksMethod
Distil the filtrate and collect the condensed liquid.Reason
Water vaporises and condenses while dissolved salt remains in the flask.Working
Distillate: pure water.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Award method points only in a workable order.
Challenge 6
Choose Methods for an Ink Mixture
Two-goal transfer
Match method to goal
Hints
Hint 1: water
Hint 2: dyes
View solution step by step
Recover the solvent
Method
Use simple distillation and collect the condensed water.Reason
Water is volatile while the dyes remain in the flask.Working
Pure-water product: distillate.Analyse the dyes
Method
Use paper chromatography and count separated visible spots.Reason
Dyes move different distances with the solvent under the chosen conditions.Working
Number of spots → number of visible separated components under those conditions.
7. Mind Stretchers
Mind stretcher 1: Question one-spot evidenceExtension
Why does one spot not prove that a sample is pure in every possible test?
Show Answer
Two components can travel together in one solvent or may not be visible with one locating method. A different solvent or locating condition may separate them.
Mind stretcher 2: Explain repeated separationExtension
Why does a fractionating column improve the separation of miscible liquids with fairly close boiling points?
Show Answer
Repeated condensation and vaporisation enrich the rising vapour in the more volatile liquid before it reaches the condenser.
Mind stretcher 3: Recover both solidsExtension
A mixture contains soluble salt and insoluble sand. Describe how to obtain both substances dry, starting with a suitable solvent.
Show Answer
Add water and stir so the salt dissolves. Filter the mixture, then wash and dry the sand residue. Concentrate the salt solution, cool it to crystallise the salt, filter the crystals, wash them with a little cold water and dry them.
8. Quiz
Before attempting the assessment, check that you can:
- select a method from solubility, boiling point or particle-size evidence;
- use a suitable solvent to separate two solids and recover both products;
- distinguish residue, filtrate, distillate and crystals;
- explain a paper chromatography setup and chromatogram;
- use melting and boiling behaviour as purity evidence;
- state the limits of one purity test.
Practise this: for each new mixture, write four parts before naming a method: the mixture type, the physical-property difference, the required product and the apparatus action.