Topic: Separating techniques

1:03 understand how the results of experiments involving the dilution of coloured solutions and diffusion of gases can be explained

Diffusion is the spreading out of particles in a gas or liquid. There is a net movement of particles from areas of high concentration to areas of low concentration until a uniform concentration is achieved.

 

i) dilution of coloured solutions

Dissolving potassium manganate(VII) in water demonstrates that the diffusion in liquids is very slow because there are only small gaps between the liquid particles into which other particles diffuse.

The random motion of particles cause the purple colour to eventually be evenly spread out throughout the water.

Adding more water to the solution causes the potassium manganate(VII) particles to spread out further apart therefore the solutions becomes less purple. This is called dilution.

 

ii) diffusion experiments

When ammonia gas and hydrogen chloride gas mix, they react together to form a white solid called ammonium chloride.

ammonia                  +              hydrogen chloride                 –>            ammonium chloride

NH3(g)                     +              HCl(g)                                     –>            NH4Cl(s)

A cotton wool pad was soaked in ammonia solution and another was soaked in hydrogen chloride solution. The two pads were then put into opposite ends of a dry glass tube at the same time.

The white ring of ammonium chloride forms closer to the hydrochloric acid end because ammonia particles are lighter than hydrogen chloride particles and therefore travel faster.

Even though these particles travel at several hundred metres per second, it takes about 5 min for the ring to form. This is because the particles move in random directions and will collide with air particles in the tube.

1:04 know what is meant by the terms: solvent, solute, solution, saturated solution

When a solid dissolves in a liquid:

  • the substance that dissolves is called the solute
  • the liquid in which it dissolves is called the solvent
  • the liquid formed is a solution
  • a saturated solution is a solution into which no more solute can be dissolved

 

1:05 (Triple only) know what is meant by the term solubility in the units g per 100g of solvent

Solubility is defined in terms of the maximum mass of a solute that dissolves in 100g of solvent. The mass depends on the temperature.

For example, the solubility of sodium chloride (NaCl) in water at 25⁰C is about 36g per 100g of water.

1:06 (Triple only) understand how to plot and interpret solubility curves

The solubility of solids changes as temperature changes. This can be plotted on a solubility curve.

Image result for solubility curve

The salts shown on this graph are typical: the solubility increases as temperature increases.

For example, the graph above shows that in 100g of water at 50⁰C the maximum mass of potassium nitrate (KNO₃) which will dissolve is 80g.

However, if the temperature were 80⁰C a mass of 160g of potassium nitrate (KNO₃) would dissolve in 100g of water.

1:07 (Triple only) practical: investigate the solubility of a solid in water at a specific temperature

At a chosen temperature (e.g. 40⁰C) a saturated solution is created of potassium nitrate (KNO₃) for example.

Some of this solution (not any residual solid) is poured off and weighed. The water is then evaporated from this solution to leave a residue of potassium nitrate which is then weighed.

The difference between the two measured masses is the mass of evaporated water.

The solubility, in grams per 100g of water, is equal to 100 times the mass of potassium nitrate residue divided by the mass of evaporated water.

 solubility (g/100g) = \frac{mass Of Solute}{mass Of Solvent} \times 100

1:09 understand that a pure substance has a fixed melting and boiling point, but that a mixture may melt or boil over a range of temperatures

Pure substances, such as an element or a compound, melt and boil at fixed temperatures.

However, mixtures melt and boil over a range of temperatures.

Example: although pure water boils at 100⁰C, the addition of 10g of sodium chloride (NaCl) to 1000cm³ of water will raise the boiling point to 100.2⁰C.

Example: although pure water melts at 0⁰C, the addition of 10g of sodium chloride (NaCl) to 1000cm³ of water will lower the melting point to -0.6⁰C.

1:10 describe these experimental techniques for the separation of mixtures: simple distillation, fractional distillation, filtration, crystallisation, paper chromatography

Simple distillation

This method is used to separate a liquid from a solution. For example: separating water from salt water.

The salt water is boiled. The water vapour condenses back into a liquid when passed through the condenser. The salt is left behind in the flask.

Note: cold water is passed into the bottom of the condenser and out through the top so that the condenser completely fills up with water.

 

Fractional distillation

This method is used to separate a mixture of different liquids that have different boiling points. For example, separating alcohol from a mixture of alcohol and water.

Water boils at 100oC and alcohol boils at 78oC. By using the thermometer to carefully control of temperature of the column, keeping it at 78oC, only the alcohol remains as vapour all the way up to the top of the column and passes into the condenser.

The alcohol vapours then condense back into a liquid.

 

 

Filtration

This method is used to separate an insoluble solid from a liquid. For example: separating sand from a mixture of sand and water.

The mixture is poured into the filter paper. The sand does not pass through and is left behind (residue) but the water passes through the filter paper and is collected in the conical flask (filtrate).

 

 

Crystallisation

This method is used to obtain a salt which contains water of crystallisation from a salt solution. For example: hydrated copper sulfate crystals (CuSO4.5H2O(s)) from copper sulfate solution (CuSO4(aq)).

Gently heat the copper sulfate solution in an evaporating basin until a hot saturated solution forms. Leave in a warm place to allow the hydrated copper sulfate crystals to form. Remove the crystals by filtration and wash with distilled water. Dry by leaving in a warm place.

If instead the solution is heated until all the water evaporates, you would produce a powder of anhydrous copper sulfate (CuSO4(s)).

 

Paper chromatography

This method can be used to separate the parts of a mixture into their components. For example, the different dyes in ink can all be separated and identified.

The coloured mixture to be separated (e.g. a food dye) is dissolved in a solvent like water or ethanol and carefully spotted onto the chromatography paper on the baseline, which is drawn in pencil so it doesn’t ‘run or smudge’.

The paper is carefully dipped into the solvent and suspended so the baseline is above the liquid solvent, otherwise all the spots would dissolve in the solvent. The solvent is absorbed into the paper and rises up it as it soaks into the paper. The choice of solvent depends on the solubility of the dye. If the dye does not dissolve in water then normally an organic solvent (e.g. ethanol) is used.

As the solvent rises up the paper it will carry the dyes with it. Each different dye will move up the paper at different rates depending on how strongly they stick to the paper and how soluble they are in the solvent.

1:11 understand how a chromatogram provides information about the composition of a mixture

Paper chromatography can be used to investigate the composition of a mixture.

A baseline is drawn on the paper. The mixture is spotted onto the baseline alongside known or standard reference materials. The end of the paper is then put into a solvent which runs up the paper and through the spots, taking some or all of the dyes with it.

Different dyes will travel different heights up the paper.

The resulting pattern of dyes is called a chromatogram.

In the example shown, the mixture is shown to contain the red, blue and yellow dyes. This can be seen because these dots which resulted from the mixture have travelled the same distance up the paper as have the red, blue and yellow standard reference materials.

1:12 understand how to use the calculation of Rf values to identify the components of a mixture

When analysing a chromatogram, the mixture being analysed is compared to standard reference materials by measuring how far the various dyes have travelled up the paper from the baseline where they started.

For each dye, the Rf value is calculated. To do this, 2 distances are measured:

  • The distance between the baseline and the dye
  • The distance between the baseline and the solvent front, which is how far the solvent has travelled from the baseline

The Rf value is calculated as follows:

 R_f=\frac{distance\:of\:dye\:from\:baseline }{distance\:of\:solvent\:front\:from\:baseline}

If the Rf value of one of the components of the mixture equals the Rf value of one of the standard reference materials then that component is know to be that reference material. 

Note that because the solvent always travels at least as far as the highest dye, the Rf value is always between 0 and 1.

Dyes which are more soluble will have higher Rf values than less soluble dyes. In other words, more soluble dyes move further up the paper. The extreme case of this is for insoluble dyes which don’t move at all (Rf value = 0). The other aspect affecting how far a dye travels is the affinity that dye has for the paper (how well it ‘sticks’ to the paper).

1:13 practical: investigate paper chromatography using inks/food colourings

  1. A pencil line (baseline) is drawn 1cm from the bottom of the paper. Pencil will not dissolve in the solvent, but if ink were used instead it might dissolve and interfere with the results of the chromatography.
  2. A spot of each sample of dye is dropped at different points along the baseline.
  3. The paper is suspended in a beaker which contains a small amount of solvent. The bottom of the paper should be touching the solvent, but the baseline with the dyes should be above the level of the solvent. This is important so the dyes don’t simply dissolve into the solvent in the beaker.
  4. A lid should cover the beaker so the atomosphere becomes saturated with the solvent. This is so the solvent does not evaporate from the surface of the paper.
  5. When the solvent has travelled to near the top of the paper, the paper is removed from the solvent and a pencil line drawn (and labelled) to show the level the solvent reached up the paper. This is called the solvent front.
  6. The chromatogram is then left to dry so that all the solvent evaporates.

Common solvents are water or ethanol. The choice of solvent depends on whether most of the dyes are soluble in that solvent.

1:25 write word equations and balanced chemical equations (including state symbols): for reactions studied in this specification and for unfamiliar reactions where suitable information is provided

Example:

Sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).

Word equation:

     sodium + water –> sodium hydroxide + hydrogen

Writing the chemical equation

A chemical equation represents what happens in terms of atoms in a chemical reaction.

Step 1: To write a chemical equation we need to know the chemical formulae of the substances.

     Na + H2O –> NaOH + H2

Step 2: The next step is to balance the equation: write a large number before each compound so the number of atoms of each element on the left hand side (reactants) matches the number on the right (products). This large number is the amount of each compound or element.

During this balancing stage the actual formulas for each compound must not be changed. Only the number of each compound changes.

     2Na + 2H2O –> 2NaOH + H2

If asked for an equation, the chemical equation must be given.

 

State symbols are used to show what physical state the reactants and products are in.

State symbolsPhysical state
(s)Solid
(l)Liquid
(g)Gas
(aq)Aqueous solution (dissolved in water)

Example:

A solid piece of sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).

     2Na(s) + 2H2O(l) –> 2NaOH(aq) + H2(g)

2:10 understand how to determine the percentage by volume of oxygen in air using experiments involving the reactions of metals (e.g. iron) and non-metals (e.g. phosphorus) with air

The following 3 experiments can be used to determine that oxygen (O2) makes up approximately 20% by volume of air.

Copper

The copper is in excess and uses up the oxygen to form copper oxide (CuO).

All the oxygen in the air is therefore used up, and so the volume of the air decreases by about 20% (the percentage of oxygen in air).

 

Iron

The iron reacts with the oxygen in the air (rusting).

As long as the iron and water are in excess, the total volume of air enclosed by the apparatus decreases by about a fifth (20%) over several days.

 

Phosphorus

The phosphorus is lit with a hot wire.

It reacts with the oxygen in the air and causes the water level in the bell jar to rise by about 20%.

 

2:14 Practical: determine the approximate percentage by volume of oxygen in air using a metal or a non-metal

The following 3 experiments can be used to determine that oxygen (O2) makes up approximately 20% by volume of air.

Copper

The copper is in excess and uses up the oxygen to form copper oxide (CuO).

All the oxygen in the air is therefore used up, and so the volume of the air decreases by about 20% (the percentage of oxygen in air).

 

Iron

The iron reacts with the oxygen in the air (rusting).

As long as the iron, oxygen and water are all in excess, the total volume of air enclosed by the apparatus decreases by about a fifth (20%) over several days.

 

Phosphorus

The phosphorus is lit with a hot wire.

It reacts with the oxygen in the air and causes the water level in the bell jar to rise by about 20%.

 

2:32 know that bases can neutralise acids

Metal oxides, metal hydroxides and ammonia (NH₃) are called bases.

Bases neutralise acids by combining with the hydrogen ions in them.

The key reaction is:

   acid             +             base             →            salt             +             water

An example of this is:

   sulfuric acid   +   copper oxide   →   copper sulfate   +   water

   H₂SO₄          +          CuO          →          CuSO₄          +          H₂O

2:42 practical: prepare a sample of pure, dry hydrated copper(II) sulfate crystals starting from copper(II) oxide

Excess Solid Method:

Preparing pure dry crystals of copper sulfate (CuSO4) from copper oxide (CuO) and sulfuric acid (H2SO4)

StepExplanation
Heat acid (H2SO4) in a beakerSpeeds up the rate of reaction
Add base (CuO) until in excess (no more copper oxide dissolves) and stir with glass rodNeutralises all the acid
Filter the mixture using filter paper and funnelRemoves any excess copper oxide
Heat the filtered solution (CuSO4)Hot saturated solution forms
Allow the solution to cool so that hydrated crystals formCopper(II) sulfate is less soluble in cold water
Remove the crystals by filtration and wash with distilled waterRemoves any impurities
Dry by leaving in a warm placeEvaporates the water

 

Select a set of flashcards to study:

     Terminology

     Skills and equipment

     Remove Flashcards

Section 1: Principles of chemistry

      a) States of matter

      b) Atoms

      c) Atomic structure

     d) Relative formula masses and molar volumes of gases

     e) Chemical formulae and chemical equations

     f) Ionic compounds

     g) Covalent substances

     h) Metallic crystals

     i) Electrolysis

 Section 2: Chemistry of the elements

     a) The Periodic Table

     b) Group 1 elements: lithium, sodium and potassium

     c) Group 7 elements: chlorine, bromine and iodine

     d) Oxygen and oxides

     e) Hydrogen and water

     f) Reactivity series

     g) Tests for ions and gases

Section 3: Organic chemistry

     a) Introduction

     b) Alkanes

     c) Alkenes

     d) Ethanol

Section 4: Physical chemistry

     a) Acids, alkalis and salts

     b) Energetics

     c) Rates of reaction

     d) Equilibria

Section 5: Chemistry in industry

     a) Extraction and uses of metals

     b) Crude oil

     c) Synthetic polymers

     d) The industrial manufacture of chemicals