Mixtures, Solutions & Separation · Grade 6 Science
Chapter: Matter, Mixtures & Water PropertiesMixtures, Solutions & Separation
Learning goals
- I can explain why the components of a mixture can be separated but those of a compound cannot.
- I can use solubility figures to predict how much of a solid will dissolve.
- I can choose a separation method by naming the property difference it exploits.
Readings for this unit
Before this unit
Read the chapter opening and the named observation or diagram. Clarify one word or unit that blocks understanding. Recall the relevant earlier measurement, matter, or energy idea.
Study the unit across three 35–45-minute class sessions. Use the worked examples after an attempt and the supplied records when actual equipment is not available.
The air in this room is a mixture: about 78 per cent nitrogen, 21 per cent oxygen, almost 1 per cent argon and about 0.04 per cent carbon dioxide, with water vapour varying from almost nothing to a few per cent. Nothing holds those gases together and nothing fixes the proportions, which is why the carbon dioxide figure has risen from about 0.028 per cent in 1800 to about 0.042 per cent now.
Sea water is a mixture too, carrying roughly 35 g of dissolved salts in every litre. So is the sand on the beach, so is milk, and so is almost everything you meet outside a laboratory. A pure substance with nothing else in it is the rare case rather than the normal one.
Because the components keep their own properties, every mixture can in principle be taken apart, and the method you choose depends on which property the components differ in. That single sentence organises the whole of this lesson.
Build your understanding
Where this fits
The previous lesson distinguished changes that make a new substance from changes that do not. This one deals with substances that are mixed together without reacting at all, which is the state most materials are found in. The techniques here, especially filtering and evaporating, are used again in chapter 7 when water quality is sampled and tested.
Use three flexible 40-45-minute sessions: what makes something a mixture; dissolving and how much will dissolve; matching separation methods to property differences, then the practice set, revision and the exit check. Workshop, review, quiz and exams have separate time.
Part 1 · Parts that stay themselves
Stir iron filings into sand and you have a mixture. The iron is still iron, still grey, still magnetic and still of density 7.87 g/cm3, and the sand is still sand. You can mix them in any proportion you like, 1 g of iron to 99 g of sand or the other way round, because nothing fixes the ratio. That is the signature of a mixture: the components keep every property they had, and the proportions are free. It is also the reason a magnet pulls the iron straight back out, since the property that made the iron magnetic was never surrendered when you stirred it in.
Compare that with iron sulfide, which is what you get if you heat iron and sulfur together until they react. The product is a dull black solid that a magnet ignores, and no amount of stirring, filtering or magnet-waving will get the iron back; separating it needs further chemical reactions. The proportions are fixed too, because the iron and sulfur combine in a set ratio by mass rather than in whatever amounts you happened to put in. A mixture can be unmixed by physical means and a compound cannot, and that is the practical difference between the two. Air makes the same point on a larger scale. Because it is a mixture rather than a compound, its components can be pulled apart physically: industrial plants chill air until it liquefies below minus 190 degrees Celsius and then warm it slowly, so nitrogen boils off first at minus 196 degrees Celsius and oxygen follows at minus 183. That is distillation, and it works only because nothing in air is chemically joined to anything else.
Pause and explain
Why can a magnet separate iron from a sand mixture but not from iron sulfide?
Check your reasoning after answering
Because mixing leaves properties intact while reacting does not. In the mixture the iron is still iron and still magnetic, so a magnet acts on it directly. In iron sulfide the iron has combined with sulfur to form a new substance that is not magnetic, so there is no longer any iron present for the magnet to attract.
Part 2 · How much will actually dissolve
When a solid dissolves, the solid is the solute, the liquid is the solvent and the result is a solution. The solute has not disappeared and has not reacted: its particles have separated and spread evenly among the solvent particles, which is why a solution looks uniform and why the mass is conserved exactly. Add 10 g of salt to 100 g of water and you have 110 g of salt solution, and boiling the water away returns close to 10 g of salt. There is a limit, though. At 20 degrees Celsius, 100 mL of water will dissolve about 36 g of salt and no more; beyond that the extra simply sits at the bottom and the solution is said to be saturated.
Different solids have wildly different limits, and the limits respond differently to temperature. Sugar is far more soluble than salt: about 204 g will dissolve in 100 mL of water at 20 degrees Celsius, rising to something like 487 g at 100 degrees Celsius. Salt barely shifts at all, going from about 36 g to only about 39 g over the same range. That contrast has a practical consequence you can predict without being told: cooling a hot saturated sugar solution forces a great deal of sugar back out as crystals, which is how rock candy is made, while cooling a salt solution produces almost nothing.
Pause and explain
Sugar's solubility rises from about 204 g to 487 g per 100 mL between 20 and 100 degrees Celsius, while salt's rises only from 36 g to 39 g. What does this predict about cooling each solution?
Check your reasoning after answering
That cooling a hot saturated sugar solution will deposit a large mass of crystals, roughly 283 g per 100 mL, because the liquid can no longer hold what it held when hot. A saturated salt solution cooled the same way can only shed about 3 g per 100 mL, so almost nothing appears. Crystallisation by cooling works well for sugar and poorly for salt.
Part 3 · Every method uses a difference
Filtration separates by particle size: sand grains are far too large to pass through the pores of filter paper while dissolved particles pass straight through, so filtering a sand and salt-water mixture leaves sand in the paper and clear salt water in the beaker. Evaporation separates by boiling point, in the crude sense that the solvent leaves as a vapour and the solute does not; boiling off the water leaves the salt behind as crystals. A magnet separates by magnetism, and decanting simply separates by density, pouring a lighter liquid off a settled solid.
Distillation is the one worth understanding properly, because it separates two liquids that are thoroughly mixed. Ethanol boils at 78 degrees Celsius and water at 100, so heating a mixture carefully to just above 78 drives off mostly ethanol vapour, which is then cooled in a condenser and collected as liquid. Chromatography uses a different property again: how strongly each component sticks to the paper compared with how readily it travels in the solvent, which is why a single black ink spot separates into several coloured bands as the solvent rises up the paper. It is worth noticing that these methods are not laboratory curiosities. Filtration and settling are the first two stages of almost every municipal water treatment works, distillation separates crude oil into petrol, diesel and the rest, and chromatography is what a laboratory uses to find out which substances are present in a blood or water sample. The same handful of property differences does nearly all the work.
Pause and explain
Filtration will not separate salt from water. Which property difference does it rely on, and why does salt water defeat it?
Check your reasoning after answering
Filtration relies on a difference in particle size, trapping anything too large to pass through the pores of the paper. Dissolved salt has separated into individual particles spread among the water, and these are far smaller than the pores, so the salt travels through with the water. Separating them needs a method based on boiling point instead.
Part 4 · Choosing in the right order
Take a mixture of iron filings, sand and salt, and the method follows from asking which component differs from the others in a way you can act on. Iron is the only magnetic component, so a magnet removes it first and leaves the other two untouched. Then add water and stir: salt dissolves and sand does not, so that step has turned one hard problem into two easy ones. Filter the result and the sand stays in the paper while the salt water passes through. Finally evaporate the salt water and the salt is left as crystals.
Order matters more than students expect. If you add the water first, the iron filings get wet and begin to rust, and you have introduced a chemical change into what was meant to be a physical separation. If you evaporate before filtering, the salt crystallises among the sand and you are back where you started. A good plan names the property used at each stage, which is also the quickest way to check that your plan can work at all: if you cannot say what difference a step exploits, the step will not separate anything. That test also tells you when a mixture cannot be separated with the equipment you have. Two liquids whose boiling points differ by only a degree or two will not separate cleanly by simple distillation, and two powders of the same size and the same solubility will not separate by filtering or dissolving at all. Recognising that early is more useful than attempting a method that was never going to work.
Pause and explain
Why should the magnet be used before water is added to an iron, sand and salt mixture?
Check your reasoning after answering
Because water starts a chemical change the separation does not want. Wet iron filings begin to rust, turning iron into iron oxide, which is no longer the substance you set out to recover and is not attracted to a magnet in the same way. Removing the iron while everything is dry keeps the whole process physical and reversible.
Apply the whole unit · Plan the separation
A beaker holds 200 mL of water in which 20 g of salt is dissolved, together with 15 g of sand and some small pieces of cork floating on top. Cork has a density of about 0.24 g/cm3.
Your task
Write a separation plan, naming the property difference used at each stage.
What your response should show
- Orders the steps so that no step undoes an earlier one
- Names a specific property difference for each of the three stages
- Predicts the recovered masses from the quantities given
- Accounts for a realistic shortfall rather than claiming perfect recovery
Response guidance — after your attempt
Decide the order of steps and say what leaves the mixture at each one.
Name the property difference each step depends on.
State the mass you expect to recover for the salt and explain any shortfall.
First skim off the cork, which floats because its density of 0.24 g/cm3 is far below the roughly 1.0 g/cm3 of the salt water; the property difference is density. Second, filter the remaining mixture: the sand grains are too large to pass through the filter paper while the dissolved salt passes through with the water, so the difference exploited is particle size, and about 15 g of sand should be recovered from the paper once it is dried. Third, evaporate the filtrate by gentle heating; the water leaves as vapour and the salt, which does not vaporise at these temperatures, stays behind, so the difference is boiling point. I would expect close to 20 g of salt, but slightly less in practice, because a little salt solution soaks into the filter paper and some crystals are lost when scraping the dish. Reporting 18 or 19 g would be a normal result rather than a sign of error.
Scientific references for this unit
- Royal Society of Chemistry: separation techniques and solubility data
- National Oceanic and Atmospheric Administration: why the ocean is salty
- NOAA Global Monitoring Laboratory: the changing composition of the atmosphere
- National Institute of Standards and Technology: solubility and phase change data
Parent / teacher teaching notes
Begin with the literal observation, quantity, or diagram relationship. Ask the learner to explain why the evidence supports the claim, check a relevant unit or boundary, then discuss an alternative. The supplied numbers are illustrative, not real observations. A parent can read aloud and allow a spoken rehearsal before writing. Judge science and evidence rather than political, faith, or family agreement. Use the complete data option whenever an actual procedure is unavailable or unsuitable.
The key idea
In a mixture every component keeps its own properties, which is exactly why mixtures can be separated. Each separation method works by exploiting one property the components do not share, so choosing a method means first finding the difference you can use.
Worked examples
Example 1 · A two-stage mixture model
- State the filter’s modeled role.
- Add water and salt mass after filtration.
- Identify the residue and the water pathway.
Example 2 · Transfer to a different case
- Identify the distinction and conditions in this new case.
- Use the relevant mechanism or quantity; preserve units and the chosen boundary.
- Explain why the answer fits and what further evidence would be needed.
Example 3 · Identify solute, solvent, and concentration · a focused example
- State the literal condition or observation in the example.
- Apply the distinction explained in this teaching part.
- Check whether the claim stays within the given evidence.
Avoid these mistakes
- Misleading claim: A dissolved substance has ceased to exist.
- Misleading claim: Removing visible particles automatically produces safe pure water.
- Omitting units, measurement timing, or a model boundary makes the evidence hard to interpret.
- Treating illustrative classroom data as real field measurements or copying a model is not an independent explanation.
Check your understanding
Does dissolved salt disappear as matter?
Discuss the answer
No. Its particles remain in the solution, contributing mass and properties even though they are not individually visible.
Is the second solution necessarily more concentrated?
Discuss the answer
No. The solute and volume both double, leaving the same ratio under this definition.
Why is the filtered liquid still a mixture?
Discuss the answer
Dissolved salt passes through with the water. Removing visible sand does not establish pure water.
Does recovering less sand prove sand was destroyed?
Discuss the answer
No. Identify retained material, loss pathways, and measurement uncertainty before assigning a cause.
Sustained scientific explanation and revision
These activities are scheduled in the parent plan. Save a first response, a correction, and the reason for the change in the science notebook.
Activity 1 · Mixtures, Solutions & Separation · Model and explain
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Mixtures, Solutions & Separation, study Parts 3–4 and attempt the worked case: Track the substances and mass at each stage, with one limitation. Explain the steps before opening the model.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
After removing 3 g sand, 105 g of salt solution remains in the idealized model. After water enters the air, 5 g salt remains. The water is transferred, not destroyed. Real recovery may be incomplete; these assumptions do not show that the liquid was safe to drink.
Activity 2 · Mixtures, Solutions & Separation · Practice, revise and exit
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Mixtures, Solutions & Separation, complete practice questions 1–6 and both choice checks, then answer independently: A filter removes visible dirt. Does that prove it removes dissolved chemicals or microbes? Save a correction explaining one changed claim or method.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
No. Filter pore size and other treatment properties matter. Visible clarity alone cannot establish removal of dissolved substances or drinking safety.
Words to know
- solution
- A mixture with dissolved particles distributed through a solvent at the particle scale.
- solute
- The substance dissolved in a solution.
- evidence
- A relevant observation, measurement, or record used to evaluate a claim.
- system boundary
- The chosen limit specifying what is included and which exchanges cross it.
- uncertainty
- A stated limit in measurement or knowledge, distinct from a careless mistake.
In the real world
Cities make tap water safe by separating out impurities — screening and filtering to trap solids, then settling and treating the water — and dry coastal regions use evaporation to pull fresh water out of salty seawater.
Exit check
A filter removes visible dirt. Does that prove it removes dissolved chemicals or microbes?
Check your explanation after answering
No. Filter pore size and other treatment properties matter. Visible clarity alone cannot establish removal of dissolved substances or drinking safety.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.