Separation and Purification Worked ExamplesIGCSE Chemistry 0620

(a) X is the solid that stays in the filter paper: the residue. Y is the liquid that passes through: the filtrate.

(b) Charcoal is insoluble in water, so X is charcoal. Sodium chloride dissolves, so Y is aqueous sodium chloride (sodium chloride solution).

(c) Stirring and warming make the sodium chloride dissolve faster and completely, so none is left mixed with the charcoal.

(d) Heat Y in an evaporating basin to evaporate the water. The sodium chloride is left as a solid (it can be crystallised by evaporating some of the water and leaving it to cool, then filtering and drying the crystals).

The salt is dissolved. To get crystals, some of the water is evaporated until the solution is saturated, and then it is left to cool so crystals form. This is crystallisation. The answer is B.

A is wrong: filtration separates an insoluble solid from a liquid; a dissolved salt passes straight through the filter paper. C collects the water (the solvent), not crystals of the salt. D separates liquids with different boiling points.

(a) P is a condenser; Q is a conical flask (the receiver).

(b) Connection 2, at the lower end. Water entering at the bottom fills the condenser jacket completely, so the whole tube is cooled and all the steam condenses (water entering at the top would drain out and leave the jacket partly empty).

(c) $100$ °C — the boiling point of water. The bulb is level with the side arm so it measures the temperature of the vapour that is entering the condenser, which tells you what is distilling.

(d) The distillate in Q is colourless (pure water). The flask contains a more concentrated, darker blue solution; if heated for long enough, solid copper(II) sulfate is left.

A dissolved impurity lowers the melting point (and makes it melt over a range) and raises the boiling point.

Lower than $-98$ °C means more negative: $-101$ °C is lower than $-98$ °C, but $-96$ °C is higher. Higher than $65$ °C: $67$ °C.

The answer is D. B comes from treating $-96$ as lower than $-98$ (it is 2 degrees warmer). A and C have a lower boiling point, which a dissolved solid does not cause.

(a) Sample A is pure: (1) it changes state at a single, constant temperature (flat section), and (2) that temperature is $69$ °C, matching the data book melting point.

(b) Sample B has no flat section: it solidifies gradually over a range of temperatures (about $66$ °C down to $61$ °C), all below $69$ °C. An impurity lowers the melting point and makes the substance melt over a range.

(c) Between $3$ and $8$ minutes liquid J is freezing (turning into a solid), so liquid and solid are both present.

As the particles form the solid, energy is released. This replaces the energy being lost to the surroundings, so the temperature stays constant until all the liquid has solidified.

Pure water boils at exactly $100$ °C (at normal pressure) — a dissolved impurity would raise the boiling point. The likely impurity here contains sulfate ions, and the sulfate test (acidify, add aqueous barium nitrate) gives no change if none are present. The answer is A.

B and C: anhydrous copper(II) sulfate turning blue only shows that water is present — it would turn blue with impure water too, so it says nothing about purity.

D: a white precipitate means sulfate ions are present, so the water is not pure.

(a) A is a fractionating column.

(b) (1) The alcohols are flammable, so a naked flame must be avoided. (2) A water bath gives gentle, even heating that is easy to control, so the fractions separate well.

(c) Methanol distils first because it has the lowest boiling point. The thermometer reads about $65$ °C while it is being collected.

(d) A water bath cannot heat the flask above $100$ °C (the boiling point of water). Butan-1-ol boils at $118$ °C, above $100$ °C, so it never boils and remains in the flask.

Add water to the mixture and stir: only R dissolves. Filter. The filtrate is a solution of R; the residue is S + T.

Obtain R by crystallisation: heat the filtrate to evaporate some of the water, leave to cool so crystals form, filter them off and dry them.

Dry the residue, then stir it with propanone (the suitable solvent for S): S dissolves, T does not. Filter again.

The residue is T: wash it with a little propanone and leave to dry.

Obtain S by evaporating the propanone from the filtrate. Because propanone is flammable, use a warm water bath, not a naked flame, in a fume cupboard or well-ventilated room.

(a) The first flat section is at $65$ °C, the boiling point of methanol. Methanol has the lower boiling point, so it reaches the top of the column first.

(b) The temperature is constant at $65$ °C from $0$ to $17$ cm³, so this distillate is pure methanol.

(c) The methanol is running out, so the vapour reaching the thermometer is a mixture of methanol and water vapour. As less methanol is left, the temperature rises until only water is distilling, at $100$ °C.

(d) The change-over is in the middle of the rise, at about $20$ cm³, so the mixture contained about $20$ cm³ of methanol.

(a) The zinc and carbonate ions form the insoluble salt; sodium sulfate stays in solution.

(b) Dissolve each solid in distilled water in separate beakers, stirring until dissolved.

Mix the two solutions in a beaker and stir: a white precipitate of zinc carbonate forms.

Filter the mixture using a filter funnel and filter paper. The zinc carbonate is the residue.

Wash the residue in the filter paper with distilled water to remove the soluble sodium sulfate.

Dry the residue by leaving it in a warm place (or an oven at low temperature) or pressing between filter papers. Do not heat strongly — zinc carbonate decomposes on strong heating.

(c) The mass would be too high: sodium sulfate solution left on the residue leaves solid sodium sulfate behind when it dries, so the product is impure.

(a) Undissolved solid left over shows the solution is saturated — it contains the maximum mass of solute that can dissolve at $40$ °C.

(b) If the solution cooled, some potassium chloride would crystallise out, so the mass dissolved would be too low.

(c) Find the mass of dissolved solid and the mass of water separately.

The water is the solution minus the dissolved solid.

Scale to $100$ g of water.

(d) Heating to constant mass makes sure all the water has been removed; otherwise the mass of solid would be too high.

(a) Amount of magnesium sulfate, converting cm³ to dm³:

One mole of crystals forms from each mole of $\text{MgSO}_4$. $M_r$ of $\text{MgSO}_4\cdot 7\text{H}_2\text{O} = 24 + 32 + 64 + 7(18) = 246$.

(b) $M_r$ of $\text{MgSO}_4 = 24 + 32 + 64 = 120$.

The powder ($2.41$ g) matches anhydrous magnesium sulfate, not the hydrated crystals ($4.92$ g). Heating to dryness drove off the water of crystallisation.

(c) Heat the solution to evaporate some of the water until it reaches the crystallisation point (a drop on a cold glass rod forms crystals). Leave it to cool so crystals form, filter them off, wash with a little cold distilled water and dry between filter papers.

(d) Percentage yield:

Some magnesium sulfate stays dissolved in the solution left after the crystals are filtered off (the solution is still saturated when cold), and a little is lost when washing.