Exploring Mixtures and their Separation · Lesson 13 of 13
Chapter Summary and Practice
“Every mixture has a separation strategy—except lemonade after someone drinks it.”
• Connect mixture type with observable properties. • Recall all percentage concentration formulas. • Choose a separation method from a distinguishing property. • Solve concentration and solubility problems step by step. • Plan multi-stage separation sequences.
Mixtures surround us in food, air, water, medicines, materials and living systems. Their components are not joined in one universal way, so no single separation method works for every mixture. The successful method is chosen by asking which useful property differs: particle size, density, solubility, boiling point, movement through paper, magnetism or ability to sublime.
Detailed Chapter Summary
A homogeneous mixture has uniform composition, while a heterogeneous mixture is non-uniform at some scale. A solution is homogeneous and contains particles smaller than 1 nm. A suspension contains particles larger than 1000 nm that may be visible, settle and be removed by filtration. A colloid contains particles about 1–1000 nm; it appears uniform, does not settle ordinarily and scatters light.
Concentration describes how much solute is present in a stated amount of solvent or solution. Mass by mass percentage compares solute mass with total solution mass. Mass by volume percentage compares solute mass with final solution volume. Volume by volume percentage compares solute volume with final solution volume.
Solubility is the maximum amount that dissolves in a fixed solvent quantity at a stated temperature. A saturated solution has reached this maximum. Crystallization uses a solubility change to obtain a pure solid. Slow cooling generally gives particles more time to form larger, better-shaped crystals.
Distillation recovers a liquid by vaporisation and condensation. Simple distillation can separate a liquid from a dissolved solid or miscible liquids with a boiling-point difference of at least about 25 °C. Fractional distillation is used when boiling points are closer. Paper chromatography separates soluble components that move at different rates through paper.
A separating funnel uses immiscibility and density to separate liquid layers. Sublimation separates a sublimable solid from a non-subliming solid. Centrifugation uses rapid spinning and density differences. Coagulation converts fine suspended particles into larger clumps that settle and can be filtered.
The Tyndall effect makes a beam visible when colloidal or suspension particles scatter light. In a colloid, the dispersed phase is distributed through a dispersion medium. An emulsion is a liquid-in-liquid colloid and may be oil-in-water or water-in-oil.
Important Concepts and Formulas
| Question to Ask | Useful Method |
|---|---|
| Can a pure solid form because solubility changes with temperature? | Crystallization |
| Can a valuable liquid be vaporised and condensed? | Distillation |
| Do dissolved components move differently through paper? | Paper chromatography |
| Are two liquid layers immiscible and different in density? | Separating funnel |
| Does one solid sublime while the other does not? | Sublimation |
| Can denser particles be moved outward by rapid spinning? | Centrifugation |
| Can fine suspended particles be made to clump? | Coagulation followed by settling and filtration |
At a Glance
• Solutions are homogeneous; suspensions and colloids are heterogeneous. • Particle size explains visibility, settling, filtration and light scattering. • Concentration percentages must use the correct total in the denominator. • Crystallization, distillation and chromatography separate homogeneous mixtures using different properties. • Separating funnels, sublimation, centrifugation and coagulation suit particular heterogeneous mixtures. • The Tyndall effect distinguishes true solutions from particle-containing mixtures.
Revise, Reflect, Refine
Problem
A dry mixture contains 75 g sugar, 420 g flour and 5 g sodium hydrogencarbonate. Express each component by mass percentage.
- 1.Total mass = 75 + 420 + 5 = 500 g.
- 2.Sugar = (75 ÷ 500) × 100 = 15% m/m.
- 3.Flour = (420 ÷ 500) × 100 = 84% m/m.
- 4.Sodium hydrogencarbonate = (5 ÷ 500) × 100 = 1% m/m.
- 5.The values sum to 100%, confirming the calculation.
Problem
Brass contains 70% copper by mass. Find copper and zinc in 120 g brass.
- 1.Copper mass = (70 ÷ 100) × 120 g = 84 g.
- 2.The remaining percentage is zinc: 100% − 70% = 30%.
- 3.Zinc mass = (30 ÷ 100) × 120 g = 36 g.
- 4.Check: 84 g + 36 g = 120 g.
Problem
A uses 20 g sugar with 80 g water; B uses 20 g sugar with 100 g water; C uses 30 g sugar with 80 g water. Find each mass percentage and identify the most concentrated.
- 1.A: total mass = 20 + 80 = 100 g; percentage = 20%.
- 2.B: total mass = 20 + 100 = 120 g; percentage = (20 ÷ 120) × 100 ≈ 16.67%.
- 3.C: total mass = 30 + 80 = 110 g; percentage = (30 ÷ 110) × 100 ≈ 27.27%.
- 4.C has the greatest solute mass per 100 g solution and is therefore the most concentrated.
Problem
Separate sand, common salt and naphthalene.
- 1.Use sublimation first: heat gently so naphthalene sublimes and deposits on a cool surface.
- 2.Add water to the remaining sand and salt; salt dissolves while sand does not.
- 3.Filter the mixture; sand remains as residue and salt solution passes as filtrate.
- 4.Use evaporation or crystallization on the filtrate to recover common salt.
- 5.The order matters because each step isolates one component using a different property.
Problem
Choose among simple or fractional distillation for water–acetone, acetone–alcohol, alcohol–chloroform and alcohol–benzene.
- 1.Water and acetone differ by 44 °C, so simple distillation is suitable.
- 2.Acetone and alcohol differ by 22 °C, so fractional distillation is more suitable.
- 3.Alcohol and chloroform differ by 17 °C, so fractional distillation is more suitable.
- 4.Alcohol and benzene differ by only 2 °C, so fractional distillation is required for useful separation.
- 5.Water and salt are handled by simple distillation when water must be recovered.
Problem
Potassium nitrate solubility is 62 g per 100 g water at 40 °C. Find the amount for a saturated solution in 50 g water.
- 1.Write the ratio: 62 g solute for 100 g water.
- 2.Scale the solvent from 100 g to 50 g by multiplying by 50 ÷ 100.
- 3.Required solute = 62 × 50 ÷ 100 = 31 g.
- 4.Therefore, 31 g potassium nitrate is required.
| Mixture | Method | Reason |
|---|---|---|
| Mud from muddy water | Coagulation, settling and filtration | Fine particles form larger removable clumps |
| Plasma from blood components | Centrifugation | Components differ in density |
| Naphthalene and sand | Sublimation | Only naphthalene sublimes |
| Chalk powder and common salt | Dissolution followed by filtration and crystallization | Salt dissolves in water; chalk does not |
| Common salt and water | Evaporation, crystallization or distillation | Choice depends on which component must be recovered |
| Oil and water | Separating funnel | Liquids are immiscible and differ in density |
| Flower pigments | Paper chromatography | Pigments move at different rates |
Quiz
Which expression represents Mass by mass percentage?
Which expression represents Mass by volume percentage?
Which statement correctly applies to the lesson “Chapter Summary and Practice”?
Which additional statement also correctly applies to the lesson “Chapter Summary and Practice”?
Which further statement also correctly applies to the lesson “Chapter Summary and Practice”?
Practice Problems
- Explain why milk is heterogeneous even though it appears uniform.
- Compare evaporation, crystallization and distillation by what is recovered.
- A hot saturated solution holds 95 g solute and retains 68 g after cooling. Find the mass deposited.
- Explain why solutions do not show the Tyndall effect.
- Identify the dispersed phase and dispersion medium in blood in a general particle-and-liquid description.
- Design a labelled separating-funnel procedure for cooking oil and water.
- Explain why sewage treatment may use sedimentation, coagulation and filtration in sequence.
The Journey Beyond
Investigate light scattering using safe colloids and suspensions under adult supervision. Grow crystals of common salt, sugar or another approved substance and compare their shapes with a magnifying glass. Use paper chromatography to investigate pigments in green and red leaves or components in food colours.
Design a separation game in which each challenge gives a mixture and a set of observable properties. Another project could model a simple outdoor distillation arrangement for obtaining cleaner water, with careful attention to condensation and safe heating. Waste sorting, water treatment and recovery of valuable materials from discarded batteries show how separation science supports a cleaner and more sustainable world.
The Quest Continues …
Blood performs many coordinated functions and no simple mixture can reproduce all of them. The continuing question is whether materials can be designed that safely carry out some essential functions of blood for many different patients.
A separation method is not selected because a mixture “looks dirty” or because a method is familiar. It is selected because the components differ in a specific, useful physical property.
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Tyndall Effect
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