MODULE A2

Mixtures & Separations

Pure substances, solutions, colloids, solubility curves, and all 7 separation techniques โ€” sorted! ๐Ÿงช

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1. Pure Substances vs Mixtures

A pure substance contains only one type of particle (element or compound). It has sharp, fixed melting and boiling points. A mixture contains two or more substances not chemically combined.

PropertyPure SubstanceMixture
CompositionFixed & constantVariable
Melting/Boiling PointSharp & fixedBroad range
SeparationCannot be separated physicallyCan be separated by physical means
ExamplesWater (Hโ‚‚O), NaCl, Oโ‚‚Air, sea water, ink

Elements vs Compounds

Element: A pure substance made of only one type of atom. Cannot be broken down by chemical means. E.g. Fe, Oโ‚‚, Clโ‚‚.

Compound: A pure substance formed from two or more elements chemically bonded in fixed proportions. Has different properties from its elements. E.g. Hโ‚‚O, NaCl, COโ‚‚.

๐Ÿ’ก Memory trick: Elements are like individual LEGO colours. Compounds are LEGO bricks permanently glued together. Mixtures are bricks just sitting in the same bucket โ€” not glued!

2. Types of Mixtures

A. Homogeneous Mixtures (Solutions)

Uniform composition throughout โ€” you cannot see the individual components. Also called solutions.

B. Heterogeneous Mixtures

Non-uniform โ€” you can tell the components apart.

PropertySolutionColloidSuspension
Particle Size<1 nm1โ€“1000 nm>1000 nm
Visible to Eye?NoNoYes
Settles?NoNoYes
LightPasses throughScattered (Tyndall)Blocked
ExamplesSalt water, vinegarMilk, fog, jellyMuddy water, chalk in water
๐Ÿ”ฆ Tyndall Effect: Shine a beam of light through a colloid โ€” you can see the beam! This is because colloidal particles scatter the light. Try it with milk in water vs pure water.

3. Solubility

Solubility = the maximum mass (in grams) of solute that will dissolve in 100 g of solvent at a specific temperature. Units: g per 100 g water.

A saturated solution contains the maximum dissolved solute at that temperature. Any extra solute added remains undissolved.

Effect of Temperature

Solubility Calculation Formula

mass of solute = (solubility ร— mass of solvent) / 100
Example: Solubility of KNOโ‚ƒ at 60ยฐC = 110 g per 100 g water.
Mass dissolved in 250 g water = (110 ร— 250) / 100 = 275 g

Crystallisation from Cooling

When a saturated solution is cooled, the solubility decreases and excess solute crystallises out.

โš ๏ธ Worked Example: A saturated solution of KNOโ‚ƒ at 60ยฐC (solubility = 110 g/100g) is cooled to 20ยฐC (solubility = 31 g/100g). If it contains 100 g of water:
Mass at 60ยฐC = 110 g dissolved | Mass at 20ยฐC = 31 g dissolved
Mass of crystals formed = 110 โˆ’ 31 = 79 g

4. Separation Techniques

A. Filtration

Separates: Insoluble solid from liquid | Based on: Particle size

B. Evaporation

Separates: Dissolved solid from solution (solvent is lost) | Based on: Volatility

C. Crystallisation

Separates: Pure dissolved solid from solution | Based on: Differential solubility with temperature

D. Simple Distillation

Separates: Solvent from a solution | Based on: Different boiling points

E. Fractional Distillation

Separates: Miscible liquids with similar boiling points | Based on: Different boiling points

F. Separating Funnel

Separates: Immiscible liquids | Based on: Different densities

G. Paper Chromatography

Separates: Dissolved substances, especially dyes | Based on: Solubility in solvent AND attraction to paper

๐Ÿ“ Rf (retention factor) value:
Rf = distance moved by substance / distance moved by solvent front
Rf is always between 0 and 1. Pure compounds always give the same Rf in the same solvent.

5. Quick Reference: Choosing the Right Technique

TechniqueWhat it SeparatesProperty UsedWhat You Get
FiltrationInsoluble solid from liquidParticle sizeResidue + Filtrate
EvaporationDissolved solid from solutionVolatilitySolid only
CrystallisationPure dissolved solidSolubility vs tempPure crystals + solution
Simple DistillationSolvent from solutionBoiling pointPure liquid + solid
Fractional DistillationMiscible liquids (similar b.p.)Boiling pointSeparate liquids
Separating FunnelImmiscible liquidsDensitySeparate liquids
ChromatographyDissolved substances/dyesSolubility & attractionChromatogram + Rf values

โšก Paper Chromatography Simulation

Watch dyes separate as the solvent front travels up the chromatography paper. Each dye has a different Rf value based on its solubility and attraction to the paper.

Rf formula: Rf = distance moved by dye รท distance moved by solvent front
Each dye's Rf is constant for a given solvent โ€” like a fingerprint for the substance! ๐Ÿ”

๐Ÿ“ˆ Solubility Curves

Animated solubility curves for three common salts. Observe how solubility changes with temperature and use the tool below to calculate crystallisation.

๐Ÿงฎ Crystallisation Calculator

How many grams of crystals form when you cool a saturated solution?

๐Ÿƒ Flashcards โ€” Mixtures & Separations

Click the card to flip it! Use the arrows to navigate.

Answer
๐Ÿ‘† Click card to flip

โ“ Quiz โ€” Mixtures & Separations

๐Ÿ”ข Worked Example โ€” Solubility & Crystallisation

Problem: The solubility of potassium nitrate (KNOโ‚ƒ) is 110 g per 100 g water at 60ยฐC and 31 g per 100 g water at 20ยฐC. A saturated KNOโ‚ƒ solution contains 250 g of water at 60ยฐC. It is then cooled to 20ยฐC.

Type the answer (number only, or type show to reveal):

Step 1: Mass dissolved at 60ยฐC

Using the solubility formula: mass = (solubility ร— mass of water) / 100
At 60ยฐC, solubility = 110 g/100 g water. Mass of water = 250 g. What mass is dissolved?

Step 2: Mass that can remain dissolved at 20ยฐC

At 20ยฐC, solubility = 31 g/100 g water. Same 250 g of water. What mass stays in solution at 20ยฐC?

Step 3: Mass of crystals formed

The crystals that form = mass dissolved at 60ยฐC minus the mass that remains at 20ยฐC. Calculate this.

Step 4: Explain what happens to the remaining solution

After cooling, the solution at 20ยฐC is still _______. Type the word that describes the solution.

๐Ÿ”— Matching โ€” Techniques & Descriptions

Click a term on the left, then click its matching description on the right. Green = correct, Red = try again!

Term / Concept

Description / Definition

๐Ÿ“ CSEC-Style Questions

Past-paper style questions with full mark schemes. Click to reveal!

Q1. A student is given a mixture of sand, salt, and water. Describe fully how the student can obtain (i) dry sand, and (ii) pure dry salt. [6 marks] +
Mark Scheme

1 Filter the mixture through filter paper in a funnel.

2 Sand is collected as the residue on the filter paper; rinse with distilled water and dry in an oven. (Dry sand โœ“)

3 The filtrate (salt water) is collected in a beaker.

4 Heat the filtrate gently in an evaporating dish until saturated (small crystals appear at edges), then remove from heat.

5 Allow to cool slowly โ€” salt crystals form.

6 Filter off the crystals, pat dry with filter paper and leave to dry. (Pure salt โœ“)

Award: 1 mark per correct step (max 6). Accept evaporation to dryness for step 4โ€“6 but note crystals may decompose.

Q2. The solubility of copper sulfate at 80ยฐC is 55 g per 100 g water and at 20ยฐC is 20 g per 100 g water. A saturated solution at 80ยฐC contains 200 g of water. Calculate the mass of crystals that form when cooled to 20ยฐC. [4 marks] +
Mark Scheme

1 At 80ยฐC: mass dissolved = (55 ร— 200) รท 100 = 110 g โœ“

2 At 20ยฐC: mass that remains = (20 ร— 200) รท 100 = 40 g โœ“

3 Mass of crystals = 110 โˆ’ 40 = 70 g โœ“

4 Units: grams (g) โœ“ [accept g of CuSOโ‚„]

Q3. Distinguish between a solution, a colloid and a suspension. Give ONE example of each. [6 marks] +
Mark Scheme

1 Solution: Homogeneous mixture; particle size <1 nm; particles do not settle; transparent. Example: salt water / vinegar. โœ“โœ“

2 Colloid: Particle size 1โ€“1000 nm; particles do not settle; scatters light (Tyndall effect). Example: milk / fog. โœ“โœ“

3 Suspension: Particle size >1000 nm; particles settle on standing; opaque. Example: muddy water / chalk in water. โœ“โœ“

Award 2 marks each: 1 for characteristic, 1 for example.

Q4. In a paper chromatography experiment, a dye moves 7.5 cm and the solvent front moves 12.5 cm. (a) Calculate the Rf value. (b) A second experiment using the same solvent gives Rf = 0.60 for a known pure dye. What can you conclude? [4 marks] +
Mark Scheme

1 Rf = distance moved by substance รท distance moved by solvent front โœ“

2 Rf = 7.5 รท 12.5 = 0.60 โœ“

3 The Rf of the dye (0.60) matches the known pure dye (0.60) โœ“

4 Conclusion: The unknown dye is (likely) the same substance as the known pure dye. โœ“

Note: Same Rf in same solvent = likely same substance, but not 100% conclusive โ€” use multiple solvents for certainty.

Q5. State the separation technique you would use to obtain (i) pure ethanol from an ethanol-water mixture, (ii) iron filings from sand, (iii) kerosene from water. Explain your choice in each case. [6 marks] +
Mark Scheme

1 (i) Fractional distillation โ€” because ethanol and water are miscible liquids with close but different boiling points (78ยฐC and 100ยฐC). โœ“โœ“

2 (ii) Using a magnet (magnetic attraction) โ€” iron is magnetic; sand is not. โœ“โœ“

3 (iii) Separating funnel โ€” because kerosene and water are immiscible liquids with different densities; they form two separate layers. โœ“โœ“

โญ Key Formulas & Concepts

Solubility Formula

mass = (solubility ร— mass solvent) / 100

Solubility in g per 100 g water

Rf Value

Rf = dist (substance) / dist (solvent front)

Always between 0 and 1

Crystallisation

crystals = dissolved(Tโ‚) โˆ’ dissolved(Tโ‚‚)

Tโ‚ > Tโ‚‚ (cooling down)

Tyndall Effect

Colloids scatter light โ†’ beam visible

Solutions are transparent

7 Techniques

Filter ยท Evap ยท Crystal ยท S.Dist ยท F.Dist ยท Sep.Funnel ยท Chrom

Know what each separates!

Mixture Types

Solution < 1nm ยท Colloid 1-1000nm ยท Suspension >1000nm

Particle size determines type

๐Ÿ“š Resources