Chemistry Year 11 · Module 1 ⏱ ~30 min

Filtration and Crystallisation

In the 9th century, Arab chemist Jabir ibn Hayyan devised systematic methods for purifying substances — including filtration and crystallisation. His techniques, refined over 1200 years, are still used in every chemistry laboratory in the world today.

⚗️

📝 Choose how you work: type answers below, or work in your book.

📚 Know

  • The principle behind filtration and crystallisation
  • The key equipment used in each technique
  • Which mixture types each technique is suited for

🔗 Understand

  • Why particle size determines whether filtration works
  • Why solubility changes with temperature drive crystallisation
  • How to choose between techniques based on mixture properties

✅ Can Do

  • Describe filtration and crystallisation step-by-step
  • Compare the two techniques in a structured table
  • Apply the correct technique to a novel mixture scenario
📋

Key Definitions

filtrationSeparation of an insoluble solid from a liquid using a porous barrier (filter paper); separates by particle size.
crystallisationSeparation of a dissolved solid from solution by evaporating solvent until the solute precipitates as pure crystals.
residueThe solid material retained on the filter paper during filtration.
filtrateThe liquid that passes through the filter paper during filtration.
solubility curveA graph showing how the maximum solubility of a solute changes with temperature. Most solids become more soluble as temperature increases.
saturated solutionA solution containing the maximum possible amount of dissolved solute at a given temperature.

Core Content

🔬

Filtration

Principle

Filtration separates an insoluble solid from a liquid using a porous barrier (typically filter paper in a funnel). The liquid passes through the tiny pores; the solid particles are too large to pass through and are retained on the filter paper.

Separation basis: Particle size. The solid must be insoluble — if it dissolves in the liquid, it will pass through with the filtrate and cannot be collected this way.

Equipment and Procedure

Fold filter paper into cone shape → Place in glass funnel, moisten with solvent → Position funnel over conical flask (to collect filtrate) → Pour mixture gently down a glass rod into funnel → Allow liquid to drain through → Solid (residue) remains on filter paper → Liquid (filtrate) collects in flask below

What is separated

What you collectNameLocation
Insoluble solidResidueOn the filter paper
Liquid + dissolved substancesFiltrateIn the flask below

Limitations

🖼️
Diagram: Filtration Setup

Insert labelled diagram: filter paper folded in funnel, glass rod, conical flask below. Label residue, filtrate, funnel, filter paper, glass rod, conical flask. Show direction of liquid flow with arrows.

💎

Crystallisation

Principle

Crystallisation separates a dissolved solid (solute) from a solution by removing solvent until the solution becomes saturated, then cooling or continuing evaporation so that the excess solute comes out of solution as pure crystals. It exploits the fact that most solids become less soluble as temperature decreases.

Separation basis: Difference in solubility with temperature. As a hot saturated solution cools, it can no longer hold as much solute in solution — the excess crystallises out.

Procedure

Dissolve solid in minimum volume of hot solvent → Filter if any undissolved particles are present → Heat solution in evaporating basin until concentrated → Allow to cool slowly (slow cooling → larger, purer crystals) → Filter to collect crystals → Dry crystals (air-dry or low-temperature oven)

Why Slow Cooling Gives Better Crystals

Slow cooling allows solute particles to arrange themselves into an ordered lattice — producing larger, purer crystals. Rapid cooling traps impurities and produces small, impure crystals. This is why the technique is used for purification, not just separation.

Purity advantage: Crystallisation is a purification technique as well as a separation technique. Each crystallisation cycle removes impurities because the pure substance crystallises preferentially from the solution. Repeated crystallisation (recrystallisation) can achieve very high purity.
🖼️
Diagram: Crystallisation Setup

Insert labelled diagram: evaporating basin on tripod and gauze over Bunsen burner; separate diagram of crystals forming on cooling. Label: evaporating basin, solution, heat source, crystal formation on cooling.

⚖️

Comparing Filtration and Crystallisation

FeatureFiltrationCrystallisation
Separation basisParticle sizeSolubility change with temperature
What is separatedInsoluble solid from liquidDissolved solid from solution
Key equipmentFilter paper, funnel, conical flaskEvaporating basin, heat source, funnel (for final step)
ResultResidue + filtrateDry crystals + mother liquor
Purification?Limited — removes insoluble impurities onlyYes — each cycle improves purity
Cannot separateDissolved solutes from each otherInsoluble solids from liquids
Key decision rule: Is the solid dissolved or undissolved? If undissolved → filtration. If dissolved → crystallisation. This is the first question to ask when choosing a technique.

Worked Examples

1

Worked Example 1 — Stepwise: selecting a technique for a given mixture

A student has a beaker of muddy water (water + sand + dissolved salt). Describe how to obtain: (a) pure dry sand, and (b) pure dry salt. Justify your choice of technique in each case.
1
Identify what is dissolved vs undissolved
Sand is insoluble in water — it does not dissolve. Salt (NaCl) is soluble in water — it is fully dissolved. These two properties determine which technique applies to each substance.
2
Part (a) — Obtaining pure dry sand
Use filtration. Pour the muddy water through filter paper in a funnel. The sand (insoluble, large particles) is trapped as residue on the filter paper. The filtrate (water + dissolved salt) passes through. Rinse the residue with distilled water to remove any remaining salt, then dry.
3
Part (b) — Obtaining pure dry salt
First filter to remove the sand, collecting the filtrate (salt water). Then use crystallisation: heat the filtrate in an evaporating basin until saturated, cool slowly, filter off the crystals, and dry. The salt crystallises out of solution as temperature decreases.
4
Justify technique choices
Filtration is used for sand because it is insoluble (separation by particle size). Crystallisation is used for salt because it is dissolved (separation by solubility change with temperature). Neither technique alone can separate both components — the two techniques must be used in sequence.
Answer
(a) Filtration — sand is insoluble; retained as residue, rinsed and dried. (b) Crystallisation of the filtrate — salt is dissolved; crystallises on cooling. Techniques are applied sequentially.
2

Worked Example 2 — Stepwise: explaining why crystallisation purifies

A student dissolves impure copper sulfate (CuSO₄) in hot water, then allows the solution to cool slowly. They collect large blue crystals and filter off the remaining solution. Explain why the crystals are purer than the original sample.
1
What happens when a hot solution cools?
As temperature decreases, solubility decreases. The solution can no longer hold all the dissolved CuSO₄ — the excess comes out of solution as solid crystals.
2
Why does crystallisation select for the pure substance?
Crystal growth is a highly ordered process. CuSO₄ molecules (or ions) fit neatly into the crystal lattice; impurities have different shapes and sizes and are mostly excluded from the lattice. The crystal lattice preferentially incorporates pure CuSO₄.
3
Where do the impurities go?
Most impurities remain in the solution (the "mother liquor"). When the crystals are filtered off and washed with a small amount of cold solvent, the remaining impurities on the crystal surface are removed.
Answer
As the hot solution cools, CuSO₄ preferentially crystallises because its ions fit into the ordered crystal lattice, while impurities are excluded. The impurities remain in the mother liquor. Filtering and washing the crystals removes surface impurities, yielding a purer product than the original sample.
⚠️

Common Mistakes

Trying to filter a dissolved solid. If the solid is dissolved, it will pass straight through the filter paper with the liquid. Filtration only works on insoluble solids. If you want to recover a dissolved solid, you need crystallisation or evaporation.
Rapid cooling during crystallisation. Fast cooling produces small, impure crystals with impurities trapped inside. Slow cooling allows an ordered lattice to form, producing large, pure crystals. Exam questions often ask you to explain why slow cooling is preferred.
Confusing residue and filtrate. Residue = solid on the filter paper. Filtrate = liquid that drips through. These terms are frequently reversed in student responses. Remember: "resi-due" stays on top; "fil-trate" goes through.

📓 Copy Into Your Books

📖 Key Facts

  • Filtration: separates insoluble solid from liquid (particle size)
  • Crystallisation: separates dissolved solid via solubility change
  • Residue = solid on filter paper; filtrate = liquid through
  • Slow cooling = larger, purer crystals

🔑 Technique Choice

  • Solid undissolved → filtration
  • Solid dissolved → crystallisation
  • Both present → filter first, then crystallise
  • Need high purity → recrystallise multiple times

🎯 Exam Approach

  • Step 1: Is the solid dissolved or not?
  • Step 2: Name the technique and its principle
  • Step 3: List equipment and steps in order
  • Step 4: Explain what you collect and where

⚠️ Exam Traps

  • Dissolved solid ≠ filterable
  • Residue stays on paper; filtrate goes through
  • Slow cooling → purer crystals (exam favourite)
  • Crystallisation purifies; filtration doesn't always

Activities

⚖️ Activity 1 — Compare

Side-by-Side Technique Comparison

For each scenario below, identify which technique applies and explain why. Then compare the two techniques directly in Part C.

Scenario A

A student has a mixture of chalk powder (CaCO₃, insoluble in water) in water. They want to collect the dry chalk. Which technique should they use? Describe the steps.

✏️ Answer in your book
Scenario B

A student has a solution of potassium nitrate (KNO₃) in water. They want to collect pure KNO₃ crystals. Which technique should they use? Describe the steps.

✏️ Answer in your book
Part C — Direct Comparison

In two to three sentences, explain the key difference between filtration and crystallisation in terms of (i) what property of the mixture each technique exploits, and (ii) what you collect at the end of each process.

✏️ Answer in your book
🌍 Activity 2 — Apply to Novel Context

Apply to an Unfamiliar Scenario

Use your understanding of both techniques to answer these novel contexts. These go beyond standard examples — think carefully about what is dissolved and what is not.

Novel Context 1

A mining company extracts gold ore from the ground. The ore is a mixture of gold particles (insoluble in water) and sodium chloride (dissolved in the water trapped in the ore). Outline a two-step separation procedure to obtain both pure gold particles and pure NaCl crystals from this ore sample.

✏️ Answer in your book
Novel Context 2

A student attempts to purify salt by filtering a salt solution through filter paper, then collecting the residue on the filter paper as their "pure salt". Explain what is wrong with this method and what they should do instead.

✏️ Answer in your book

Multiple Choice

Multiple Choice Questions

Click to check. One attempt only.

1. Which property of a mixture determines whether filtration can be used to separate it?

A
The colour of the solid component
B
Whether the solid is soluble or insoluble in the liquid
C
The boiling point of the liquid component
D
The electrical conductivity of the solid

2. During filtration of muddy water, what does the filtrate contain?

A
Only the mud (insoluble solids)
B
Only pure water with no dissolved substances
C
Water and any dissolved substances — but not the insoluble mud
D
The mud and the dissolved substances, but not the water

3. Why does slow cooling during crystallisation produce larger, purer crystals than rapid cooling?

A
Slow cooling allows an ordered crystal lattice to form, excluding impurities; rapid cooling traps impurities inside the crystal
B
Slow cooling produces more solvent evaporation, increasing the yield of crystals
C
Rapid cooling causes the solute to decompose before it can crystallise
D
Slow cooling keeps the temperature below the melting point of the crystal, preventing re-dissolution

4. A student wants to obtain pure copper sulfate (CuSO₄) crystals from a solution that also contains a small amount of insoluble sand. What is the correct sequence of steps?

A
Crystallise the solution first, then filter to remove sand from the crystals
B
Filter the solution to collect the copper sulfate as residue, then dry
C
Evaporate the solution completely to dryness, then scrape out the product
D
Filter first to remove the insoluble sand, then crystallise the filtrate to obtain CuSO₄ crystals

5. The following data shows the solubility of potassium nitrate (KNO₃) at different temperatures:

Temperature (°C)Solubility (g per 100 mL water)
2031
4064
60110
80169

A student prepares a saturated solution at 80°C containing 100 g of KNO₃ in 100 mL water, then cools it to 20°C. Approximately how many grams of KNO₃ crystallise out?

A
31 g
B
64 g
C
69 g
D
100 g

Short Answer

📝

Short Answer Questions

6. Describe the principle of filtration and explain why it cannot be used to separate sodium chloride from a sodium chloride solution. 3 MARKS

✏️ Answer in your book

7. A student dissolves 80 g of potassium chloride in 100 mL of hot water, then allows the solution to cool to room temperature. They observe crystals forming. Explain why crystals form on cooling, referring to solubility and saturation. 3 MARKS

✏️ Answer in your book

8. A chemist has a sample of impure table salt (NaCl with some coloured pigment impurities that are also soluble in water). Evaluate whether a single crystallisation step will produce chemically pure NaCl, and suggest how the chemist could improve the purity of their final product. 4 MARKS

✏️ Answer in your book

✅ Comprehensive Answers

⚖️ Activity 1 — Compare

A: Filtration. Steps: fold filter paper, place in funnel over conical flask, pour chalk mixture down a glass rod into funnel, allow water (filtrate) to drain through, collect chalk (residue) on filter paper, rinse with distilled water, dry residue.

B: Crystallisation. Steps: heat KNO₃ solution in evaporating basin to concentrate it, cool slowly to allow crystals to form, filter off crystals, dry on filter paper.

C: Filtration exploits particle size — the insoluble solid is too large to pass through the filter paper — and you collect the solid as residue and liquid as filtrate. Crystallisation exploits the decrease in solubility with temperature — the dissolved solid comes out of solution as crystals when cooled — and you collect dry crystals as the product.

🌍 Activity 2 — Apply to Novel Context

Novel Context 1: Step 1 — Filtration: pour the ore/water mixture through filter paper to collect gold particles as residue (gold is insoluble). The filtrate contains the NaCl solution. Step 2 — Crystallisation: heat the filtrate in an evaporating basin to concentrate it, cool slowly, filter off NaCl crystals, dry them.

Novel Context 2: The method is wrong because salt (NaCl) is soluble in water — it dissolves and passes straight through the filter paper with the liquid. There would be no salt residue to collect. The student should instead use crystallisation: heat the salt solution in an evaporating basin to evaporate the water, allow to cool slowly so salt crystallises out, then filter and dry the crystals.

❓ Multiple Choice

1. B — Filtration separates insoluble solids from liquids; solubility is the key factor.

2. C — Filtrate = everything that passes through (water + dissolved substances). Insoluble mud is trapped as residue.

3. A — Slow cooling allows ordered lattice formation, excluding impurities. Rapid cooling traps them inside.

4. D — Filter first (remove sand), then crystallise the filtrate (obtain CuSO₄). Reversing the order would embed sand into crystals.

5. C — At 80°C, 100 g dissolves in 100 mL. At 20°C, only 31 g can remain dissolved. So 100 − 31 = 69 g crystallises out.

📝 Short Answer Model Answers

Q6 (3 marks): Filtration separates mixtures based on particle size — insoluble solid particles are too large to pass through filter paper, while the liquid and any dissolved substances pass through (1 mark). In a sodium chloride solution, the NaCl is fully dissolved — it exists as individual Na⁺ and Cl⁻ ions dispersed throughout the water (1 mark). These ions are far too small to be trapped by filter paper; they simply pass through with the water, so filtration cannot separate them from the solution (1 mark).

Q7 (3 marks): As the solution cools, the solubility of KCl decreases — less KCl can remain dissolved at lower temperatures (1 mark). The solution becomes saturated and then supersaturated as it cools — it contains more dissolved KCl than can be held in solution at that temperature (1 mark). The excess KCl can no longer remain dissolved and comes out of solution as solid crystals (precipitates) (1 mark).

Q8 (4 marks): A single crystallisation step will improve purity but will not produce chemically pure NaCl (1 mark). Because the impurities are also soluble in water, some will remain in solution during the first crystallisation and some may be incorporated into the crystal surface (1 mark). To improve purity: the chemist should perform recrystallisation — dissolve the crystals again in minimum hot water, then allow to cool and crystallise again (1 mark). Each recrystallisation cycle further reduces the impurity level because the NaCl crystallises preferentially; after 2–3 cycles, purity will be significantly higher (1 mark).

Mark lesson as complete

Tick when you've finished all activities and checked your answers.

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