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S4 Chemistry | Separation and Purity Lesson-Prep Guide (for tutors) | Chromatography, distillation, recrystallisation
Students can recite distillation, filtration and chromatography but freeze when asked why one is chosen; Rf values and yields then get mixed up as decimals and percentages. This lesson turns the choice into four criteria and three calculations.
Objectives — Students can: (1) choose a separation method from solubility, boiling point, particle size or adsorption; (2) calculate Rf from a chromatogram and identify a substance by running a standard alongside; (3) judge purity from yield, titration assay and melting point range.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Dissolving and filtration | How salt is recovered from brine | Evaporate or distil |
| Percentages | What is 18.5 over 25.0 as a percentage | 74.0% |
| Titration mole ratio | Mole ratio of sodium carbonate to HCl | 1 to 2 |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Four criteria | Solubility, boiling point, particle size, adsorption | Name the criterion first |
| 8-20 min | Chromatography | Solute 4.2 cm, solvent front 6.0 cm | Rf 0.70 |
| 20-32 min | Distillation and funnels | Ethanol 78.4 C, water 100.0 C | 21.6 C apart |
| 32-44 min | Recrystallisation | 25.0 g crude, 18.5 g recovered | 74.0% |
| 44-54 min | Assay and melting point | 2.50 g sample, 24.0 cm3 of 1.00 M HCl | 50.9% pure |
| 54-60 min | Exit check | Three questions collected | Independent |
03 Teaching script (read it out)
Scene 1: Ask what differs before asking which method
Separation methods are not a list to memorise; they map four differences onto four techniques. Different solubility means crystallisation or extraction; different boiling point means distillation; different particle size means filtration; different adsorption between phases means chromatography.
Sand and salt: sand is insoluble, so dissolve, filter the sand, then evaporate the filtrate. Recovering 4.6 g of 5.0 g sand gives 92%, and 1.7 g of 2.0 g salt gives 85%, for 90% overall.
Name where the loss goes: sticking to filter paper and beaker walls, spitting during evaporation, and solute left in the mother liquor. Answer that, not vague error.
Sand and salt: sand is insoluble, so dissolve, filter the sand, then evaporate the filtrate. Recovering 4.6 g of 5.0 g sand gives 92%, and 1.7 g of 2.0 g salt gives 85%, for 90% overall.
Name where the loss goes: sticking to filter paper and beaker walls, spitting during evaporation, and solute left in the mother liquor. Answer that, not vague error.
Have students run oil and water through the four criteria and justify a funnel over distillation.
Scene 2: Rf is a ratio, not a distance
Rf is the solute distance over the solvent front: 4.2 over 6.0 is 0.70, and the second spot at 2.4 cm gives 0.40, a gap of 0.30, which is why they separate.
Rf always lies between 0 and 1; a value above 1 means the two distances were swapped. Rf shifts with solvent, temperature and paper, so identification means running a standard on the same sheet, not quoting a table.
In a separating funnel, layering follows density: ether at 0.713 g/cm3 floats on water at 0.998, so the organic layer is on top. Answer with density, not colour.
Rf always lies between 0 and 1; a value above 1 means the two distances were swapped. Rf shifts with solvent, temperature and paper, so identification means running a standard on the same sheet, not quoting a table.
In a separating funnel, layering follows density: ether at 0.713 g/cm3 floats on water at 0.998, so the organic layer is on top. Answer with density, not colour.
Have students explain why Rf only means something on the same sheet.
Scene 3: Purity needs three pieces of evidence
First, yield: 18.5 g from 25.0 g is 74.0%; below seventy percent, check dissolution loss against incomplete crystallisation.
Second, assay by titration: 2.50 g of impure sodium carbonate needs 24.0 cm3 of 1.00 M HCl, so 0.024 mol of acid means 0.012 mol of carbonate, which is 1.27 g and 50.9% purity.
Third, melting point: 122 to 124 C against a literature value of 122 C; a narrow range near the literature value means pure, while a wide or depressed range means impurity.
Distillation rests on numbers too: ethanol at 78.4 C and water at 100.0 C differ by 21.6 C, which is wide enough to collect a narrow fraction.
Second, assay by titration: 2.50 g of impure sodium carbonate needs 24.0 cm3 of 1.00 M HCl, so 0.024 mol of acid means 0.012 mol of carbonate, which is 1.27 g and 50.9% purity.
Third, melting point: 122 to 124 C against a literature value of 122 C; a narrow range near the literature value means pure, while a wide or depressed range means impurity.
Distillation rests on numbers too: ethanol at 78.4 C and water at 100.0 C differ by 21.6 C, which is wide enough to collect a narrow fraction.
Warn that yield and purity answer different questions and must be reported separately.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | Rf and separation | Solute distance over front distance | 0.70 and 0.40 |
| Calculation | Recrystallisation yield | Actual over theoretical times 100 | 74.0% |
| Calculation | Assay by titration | Moles of acid, then carbonate, then mass, then percent | 50.9% |
| Explanation | Choose a method | State the difference, then the technique | Any of four criteria |
Sanity check (teach this): Rf must lie between 0 and 1: 0.70 is valid, while 1.43 means the distances are inverted.
A yield cannot exceed 100%: 74.0% is plausible, and more means wet product or a tared container error.
Purity scales with titre volume, so a larger volume must give a larger percentage.
Using a 1 to 1 ratio halves the mass to 0.64 g and the purity to 25.4%, which is obviously off.
A melting range of 122 to 124 C sits on the literature value and is narrow, so the solid is pure.
A yield cannot exceed 100%: 74.0% is plausible, and more means wet product or a tared container error.
Purity scales with titre volume, so a larger volume must give a larger percentage.
Using a 1 to 1 ratio halves the mass to 0.64 g and the purity to 25.4%, which is obviously off.
A melting range of 122 to 124 C sits on the literature value and is narrow, so the solid is pure.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Rf for a solute at 4.2 cm with a front at 6.0 cm | 0.70 | Divide distances, keep units |
| Yield from 25.0 g crude to 18.5 g recovered | 74.0% | Actual over theoretical |
| Mass of sodium carbonate in a 2.50 g sample | 1.27 g | The 1 to 2 ratio is essential |
| Overall recovery of a sand and salt mixture | 90% | Add both recoveries over total mass |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Inverting Rf | Answers 1.43 | Rf is below 1; above 1 means it is upside down. |
| Reporting yield as purity | Says the purity is 74.0% | Yield is how much you got; purity is how much of it is the target. |
| Ignoring the 1 to 2 ratio | Gets 25.4% purity | Carbonate takes two protons, so halve the moles. |
| Calling a wide range pure | Says 122 to 132 C is pure | A wide range means impurity, which lowers and broadens the melting point. |
07 Homework (paper, section and question numbers)
- Secondary Chemistry chapter index — Separation and purity unit, all sections
- Titration and pH chapter — The titration basis for assay
- Lesson-prep hub — Compare other chapters
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Definition and two values of Rf | 0.70 and 0.40 | Wrong means redo the division |
| Apparatus and reason for oil and water | Separating funnel, different densities | Wrong means return to act one |
| Mole ratio for the assay | 1 to 2 | Two wrong means repeat act three |
09 Tutor FAQ
Why run a standard rather than look up Rf?
Rf varies with paper, solvent ratio, temperature and saturation, so a side-by-side standard is the only fair comparison.
Why cool before filtering in recrystallisation?
Cooling lowers solubility so the product crystallises while impurities stay in the mother liquor; poor cooling costs yield.
Can simple distillation separate liquids 5 C apart?
Barely. It needs repeated vapour-liquid equilibria in a fractionating column; simple distillation usually wants 20 C or more.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
