Answers to Cambridge IGCSE ® Chemistry Laboratory Practical Book Experimental skills and abilities Skills for scientific enquiry 1 (a) measuring cylinder (e) stopwatch (b) digital balance (f) digital timer (c) metre ruler (g) ruler (d) liquid in glass thermometer 2 (a) Time at least 10 oscillations with a stopwatch and divide the result by 10 to obtain the average time for one oscillation. (b) Weigh at least 20 pins on a balance and divide the result by 20 to obtain the average mass of one pin. 3 Device Accuracy metre ruler 1 mm vernier scale 0.1 mm stopwatch 0.5 s digital timer 1 ms digital balance typically 1 g or 0.1 g liquid in glass thermometer 1.0 °C 100 cm 3 measuring cylinder 1 cm 3 4 (a) 9.75 (b) 9.8 (c) 1 × 10 1 5 (a) (i) Effect of temperature on rate (ii) Light intensity, concentrations of solutions (iii) Manipulated variable Fixed variable Responding variable temperature light, concentration rate of reaction (b) 25 20 15 10 Temperature/°C 5 0 0 2 4 6 8 10 Rate of reaction (c) Increase in temperature will result in increase in rate of reaction. 1 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers to
Cambridge IGCSE ® Chemistry
Laboratory Practical Book
Experimental skills and abilities
Skills for scientific enquiry
1 (a) measuring cylinder (e) stopwatch
(b) digital balance (f) digital timer
(c) metre ruler (g) ruler
(d) liquid in glass thermometer
2 (a) Time at least 10 oscillations with a stopwatch and divide the result by 10 to obtain the average time for one oscillation.
(b) Weigh at least 20 pins on a balance and divide the result by 20 to obtain the average mass of one pin.
3
Device Accuracy
metre ruler 1 mm
vernier scale 0.1 mm
stopwatch 0.5 s
digital timer 1 ms
digital balance typically 1 g or 0.1 g
liquid in glass thermometer 1.0 °C
100 cm 3 measuring cylinder 1 cm 3
4 (a) 9.75 (b) 9.8 (c) 1 × 10 1
5 (a) (i) Effect of temperature on rate
(ii) Light intensity, concentrations of solutions
(iii)
Manipulated variable Fixed variable Responding variable
temperature light, concentration rate of reaction
(b) 25
20
15
10
Temperature/°C 5
0
0 2 4 6 8 10
Rate of reaction
(c) Increase in temperature will result in increase in rate of reaction.
1
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

1 The particulate nature of matter Answers 1.1 Rate of diffusion of ammonia and hydrogen chloride Method 1 To ensure that it is a fair test. 2 It would be too difficult to deal with the gases. The gases would begin to diffuse before the experiment started. 3 Ammonia is toxic. Concentrated hydrochloric acid is corrosive. Results and calculations Students should be able to show that the ammonia particles (molecules) have moved along the tube at a faster rate than the hydrogen chloride particles (molecules). Conclusions 1 The longer the tube, the longer the time taken for the white cloud of ammonium chloride to form. This is because the rate of diffusion will be slower for both gases. 2 The greater the diameter of the tube, the longer the time taken for the white cloud to form. 3 Temperature of the laboratory – the higher the temperature the faster the time; concentration of the gases used – the higher their concentration the faster the time; pressure – the higher the room pressure the slower the time. 4 Hydrogen chloride end. 5 The hydrogen chloride molecules are heavier and therefore have the slower rate of diffusion, so do not travel as far along the tube as the ammonia particles. Evaluation The experiment could be repeated to gain an average times and rates. Extension The rate of diffusion is not the same as the speed of the particles. The speed of the particles will be much greater than the rate of diffusion. The reason is that the rate of diffusion is lower due to the random collisions between the particles which prevent the particles from travelling in straight lines. 1.2 Reaction of potassium iodide with lead(II) nitrate Method 1 Because the particles in both solids are vibrating about a fixed point. For the reaction to occur their particles need to be forced to collide with one another. Shaking causes this to happen. 2 In case, when you are shaking the tube, you hit a bench or other solid object. 3 So that there was no contamination of each chemical. Results A yellow tinge should be visible on the solid particles, which takes a long time to form. When the two colourless solutions are mixed, a yellow precipitate (solid) forms. If left, the yellow precipitate sinks to the bottom of the tube and a colourless solution is visible above it. 2 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

1 The particulate nature of matter
Answers
1.1 Rate of diffusion of ammonia and hydrogen chloride
Method
1 To ensure that it is a fair test.
2 It would be too difficult to deal with the gases. The gases would begin to diffuse before
the experiment started.
3 Ammonia is toxic. Concentrated hydrochloric acid is corrosive.
Results and calculations
Students should be able to show that the ammonia particles (molecules) have moved along the tube at a faster rate
than the hydrogen chloride particles (molecules).
Conclusions
1 The longer the tube, the longer the time taken for the white cloud of ammonium chloride to form. This is because
the rate of diffusion will be slower for both gases.
2 The greater the diameter of the tube, the longer the time taken for the white cloud to form.
3 Temperature of the laboratory – the higher the temperature the faster the time; concentration of the gases used –
the higher their concentration the faster the time; pressure – the higher the room pressure the slower the time.
4 Hydrogen chloride end.
5 The hydrogen chloride molecules are heavier and therefore have the slower rate of diffusion, so do not travel as far
along the tube as the ammonia particles.
Evaluation
The experiment could be repeated to gain an average times and rates.
Extension
The rate of diffusion is not the same as the speed of the particles. The speed of the particles will be much greater than
the rate of diffusion. The reason is that the rate of diffusion is lower due to the random collisions between the particles
which prevent the particles from travelling in straight lines.
1.2 Reaction of potassium iodide with lead(II) nitrate
Method
1 Because the particles in both solids are vibrating about a fixed point. For the reaction to occur their particles
need to be forced to collide with one another. Shaking causes this to happen.
2 In case, when you are shaking the tube, you hit a bench or other solid object.
3 So that there was no contamination of each chemical.
Results
A yellow tinge should be visible on the solid particles, which takes a long time to form.
When the two colourless solutions are mixed, a yellow precipitate (solid) forms. If left, the yellow precipitate sinks to the
bottom of the tube and a colourless solution is visible above it.
2
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers Conclusions 1 A yellow tinge on the solid particles. 2 Two colourless solutions reacted to give another colourless solution and a yellow precipitate. 3 The reaction between the solutions. 4 The reaction in Procedure 2 occurred faster because, in solutions, the particles are able to move relative to one another. This allows them to collide more frequently with one another, causing the reaction to occur more rapidly. 5 Colourless. Evaluation In Procedure 1, the same mass, or, better, the same number of moles of each were used. In Procedure 2, the same volume of each solution, of the same concentration, could have been used. Extension Any one of: 1 The shaking could have been more vigorous, resulting in more collisions, and more successful collisions. 2 The solids could have been more finely crushed. With a larger surface area, more collisions and more successful collisions would occur. 3 The experiment could have been carried out at a higher temperature, so particles have more kinetic energy. More collisions and more successful collisions would occur. 1.3 Sublimation of iodine Method 1 Iodine is harmful. Its fumes are dangerous for the environment. 2 To prevent the majority of the iodine vapour from escaping. The iodine vapour returns to a solid on the large, cool surface area given by the glass wool. 3 No. The pressure build up in the tube when the iodine solid changed to iodine vapour would have caused it to be blown off. Results Much of the iodine will have moved from the bottom of the tube to the top. The black/violet solid iodine sublimed on heating to give violet iodine vapour. The iodine vapour returned to its solid form when it reached the cooler parts of the tube. Conclusions 1 Black/violet. 2 Violet. 3 At the top of the tube. The top of the tube is cooler than the bottom and so the iodine vapour sublimes to iodine solid. Evaluation A longer tube could have been used. Extension Any two from: arsenic, carbon and phosphorus (red). 3 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Conclusions Answers
1 A yellow tinge on the solid particles.
2 Two colourless solutions reacted to give another colourless solution and a yellow precipitate.
3 The reaction between the solutions.
4 The reaction in Procedure 2 occurred faster because, in solutions, the particles are able to move relative to
one another. This allows them to collide more frequently with one another, causing the reaction to occur
more rapidly.
5 Colourless.
Evaluation
In Procedure 1, the same mass, or, better, the same number of moles of each were used. In Procedure 2, the same
volume of each solution, of the same concentration, could have been used.
Extension
Any one of:
1 The shaking could have been more vigorous, resulting in more collisions, and more successful collisions.
2 The solids could have been more finely crushed. With a larger surface area, more collisions and more successful
collisions would occur.
3 The experiment could have been carried out at a higher temperature, so particles have more kinetic energy.
More collisions and more successful collisions would occur.
1.3 Sublimation of iodine
Method
1 Iodine is harmful. Its fumes are dangerous for the environment.
2 To prevent the majority of the iodine vapour from escaping. The iodine vapour returns to a solid on the large, cool
surface area given by the glass wool.
3 No. The pressure build up in the tube when the iodine solid changed to iodine vapour would have caused it to be
blown off.
Results
Much of the iodine will have moved from the bottom of the tube to the top.
The black/violet solid iodine sublimed on heating to give violet iodine vapour.
The iodine vapour returned to its solid form when it reached the cooler parts of the tube.
Conclusions
1 Black/violet.
2 Violet.
3 At the top of the tube. The top of the tube is cooler than the bottom and so the iodine vapour sublimes to
iodine solid.
Evaluation
A longer tube could have been used.
Extension
Any two from: arsenic, carbon and phosphorus (red).
3
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

2 Elements, compounds and Answers experimental techniques 2.1 Rock salt: an important raw material Method 1 To dissolve the salt. 2 To allow the sample to cool. To allow the earthy material to settle. 3 To allow the salt to crystallise slowly. If the solution was evaporated to dryness ‘spitting’ would occur and good shaped crystals would not be produced. Results and calculations During the experiment the solution filtered steadily. White cubic crystals of sodium chloride were produced on slow crystallisation. Conclusions Slow crystallisation will produce good crystals due to the particles in solution slowly coming together. Evaluation Given the time scale of one practical session, the results can be improved by taking more than one session. The earthy material residue can have more water added and more salt extracted. Carry out the process at least another two times. This will increase the amount of salt in solution to eventually crystallise. Extension 1 The evaporation process should be done very slowly. This is because sugar can easily char as it solidifies around the sides of the evaporating basin during the evaporating process. Also the crystallisation will require a lot longer for crystals to form from the concentrated solution and may need to be seeded with some pure sugar. 2 Because of the simplicity of the process it can be scaled up quite readily. Very large vats would be needed so that kg or tonne quantities can be produced. The vats can be supported on swivel supports. The vats would need electric stirrers in them. The heating of the solution would be by gas or electric heaters from below. Large quantities from the vats can be filtered through very large sieves. Slow evaporation of the filtrate from the collection vats would again be by either gas or electric heating from below. 2.2 Ascending chromatography Method 1 The inks present in the ball-point pen will separate also. 2 If they are not dry then the separation will not take place efficiently. 3 Improves separation of the mixture. (Some of the substances in the black felt-tip mixture are more soluble in ethanol than water.) 4 To maintain a constant atmosphere around the chromatogram. 5 So no smudging of spots takes place. Results and calculations The spots slowly appear from the mixture as the separation takes place. Over time the spots continue to separate (Figure 1) until the time you have allowed is up. A sample chromatogram is shown in Figure 1. 4 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

2 Elements, compounds and
Answers experimental techniques
2.1 Rock salt: an important raw material
Method
1 To dissolve the salt.
2 To allow the sample to cool. To allow the earthy material to settle.
3 To allow the salt to crystallise slowly. If the solution was evaporated to dryness ‘spitting’ would occur and good
shaped crystals would not be produced.
Results and calculations
During the experiment the solution filtered steadily. White cubic crystals of sodium chloride were produced on slow
crystallisation.
Conclusions
Slow crystallisation will produce good crystals due to the particles in solution slowly coming together.
Evaluation
Given the time scale of one practical session, the results can be improved by taking more than one session. The earthy
material residue can have more water added and more salt extracted. Carry out the process at least another two times.
This will increase the amount of salt in solution to eventually crystallise.
Extension
1 The evaporation process should be done very slowly. This is because sugar can easily char as it solidifies around the
sides of the evaporating basin during the evaporating process. Also the crystallisation will require a lot longer for
crystals to form from the concentrated solution and may need to be seeded with some pure sugar.
2 Because of the simplicity of the process it can be scaled up quite readily. Very large vats would be needed so that
kg or tonne quantities can be produced. The vats can be supported on swivel supports. The vats would need electric
stirrers in them. The heating of the solution would be by gas or electric heaters from below. Large quantities from
the vats can be filtered through very large sieves. Slow evaporation of the filtrate from the collection vats would
again be by either gas or electric heating from below.
2.2 Ascending chromatography
Method
1 The inks present in the ball-point pen will separate also.
2 If they are not dry then the separation will not take place efficiently.
3 Improves separation of the mixture. (Some of the substances in the black felt-tip mixture are more soluble in ethanol
than water.)
4 To maintain a constant atmosphere around the chromatogram.
5 So no smudging of spots takes place.
Results and calculations
The spots slowly appear from the mixture as the separation takes place. Over time the spots continue to separate
(Figure 1) until the time you have allowed is up. A sample chromatogram is shown in Figure 1.
4
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers chromatography dye R paper beaker dye Q solvent soaks up the paper dye P black ink solvent spot before during after Figure 1 Note: The R values will vary depending on the felt-tip pen used and solvent mixture. f Conclusions 1 Some dyes have a greater solubility than others in the different solvents. The dyes which have moved the greatest distance have the greatest solubility in the solvents. Their R values are also greater. f 2 The dye which was least soluble is the one which does not travel as far up the chromatogram – lowest R value. The f dye which was most soluble is the one which travels the furthest up the chromatogram – highest R value. f Evaluation Ensure the plastic bag is a good seal when over the apparatus. Repeat the chromatograms several times to ensure you have an overview of the results and average R values. f Extension 1 Different results will be obtained using different solvents such as propanone/water. Depending on the felt-tip pen used, the separation may be greater or less depending on how the dyes dissolve in the new solvent. 2 The substances to be separated do not have to be coloured. For example, amino acids obtained by hydrolysis of proteins are colourless. Colourless substances can be made visible by spraying the chromatogram with a locating agent such as ninhydrin. The locating agent will react with the colourless substances to form a coloured product. In other situations the position of the substances on the chromatogram may be located using ultraviolet light. 2.3 Elements, mixtures and compounds Method 1 Yes. 2 No. 3 It is an exothermic reaction. A new substance is being formed in the reaction. Results and calculations Compound Sulfur Iron Mixture (iron sulfide) appearance yellow grey dirty yellow very dark grey, almost black effect of magnet no effect attracted iron is separated from the no effect sulfur effect of water no effect no effect no effect no effect effect of dilute no effect bubbles of hydrogen gas bubbles of hydrogen gas bubbles of hydrogen sulfide hydrochloric acid are seen (effervescence); are seen (effervescence); gas are seen; smells of gives a ‘pop’ when tested gives a ‘pop’ when tested bad eggs and turns lead with a lighted spill with a lighted spill ethanoate paper black 5 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers
chromatography
paper dye R
beaker
dye Q
solvent soaks
up the paper dye P
black ink solvent
spot
before during after
Figure 1
Note: The R values will vary depending on the felt-tip pen used and solvent mixture.
f
Conclusions
1 Some dyes have a greater solubility than others in the different solvents. The dyes which have moved the greatest
distance have the greatest solubility in the solvents. Their R values are also greater.
f
2 The dye which was least soluble is the one which does not travel as far up the chromatogram – lowest R value. The
f
dye which was most soluble is the one which travels the furthest up the chromatogram – highest R value.
f
Evaluation
Ensure the plastic bag is a good seal when over the apparatus. Repeat the chromatograms several times to ensure you
have an overview of the results and average R values.
f
Extension
1 Different results will be obtained using different solvents such as propanone/water. Depending on the felt-tip pen
used, the separation may be greater or less depending on how the dyes dissolve in the new solvent.
2 The substances to be separated do not have to be coloured. For example, amino acids obtained by hydrolysis of
proteins are colourless. Colourless substances can be made visible by spraying the chromatogram with a locating
agent such as ninhydrin. The locating agent will react with the colourless substances to form a coloured product. In
other situations the position of the substances on the chromatogram may be located using ultraviolet light.
2.3 Elements, mixtures and compounds
Method
1 Yes.
2 No.
3 It is an exothermic reaction. A new substance is being formed in the reaction.
Results and calculations
Compound
Sulfur Iron Mixture (iron sulfide)
appearance yellow grey dirty yellow very dark grey, almost black
effect of magnet no effect attracted iron is separated from the no effect
sulfur
effect of water no effect no effect no effect no effect
effect of dilute no effect bubbles of hydrogen gas bubbles of hydrogen gas bubbles of hydrogen sulfide
hydrochloric acid are seen (effervescence); are seen (effervescence); gas are seen; smells of
gives a ‘pop’ when tested gives a ‘pop’ when tested bad eggs and turns lead
with a lighted spill with a lighted spill ethanoate paper black
5
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Conclusions Mixtures behave as the individual elements they are made up from. Compounds differ from the elements or mixtures in Answers that they are new substances and so behave differently . Evaluation Experiments should be reproducible. So the experiments should be repeated to show consistency of results. Also, other compounds their elements and mixtures should be tested. Extension 1 (a) Calcium and oxygen. (b) calcium + oxygen → calcium oxide 2Ca(s) + O (g) → 2CaO(s) 2 2 An element – a substance which cannot be further divided into simpler substances by chemical methods. A mixture – a system of two or more substances that can be separated by physical means. A compound – a substance formed by the combination of two or more elements in fixed proportions. 3 Atomic structure and bonding 3.1 Structure of substances Method 1 To ensure the eyes are shielded, in particular, from bromine and soot. 2 Bromine is toxic, as well as corrosive, and should only be used in a fume cupboard. Results and calculations Substance Appearance Inference Water colourless liquid small molecule polyethene translucent solid macromolecule pencil lead black solid macromolecule Bromine red liquid small molecule buckminster fullerene (soot) black powder macromolecule Nylon white solid macromolecule oxygen colourless gas small molecule sand brown solid macromolecule Conclusion The substances water, bromine and oxygen contain small molecules while the substances polyethene, pencil lead, buckminster fullerene, nylon and sand are macromolecular. Evaluation Other substances which are small molecules or macromolecules could be included in the table of substances. Extension 1 Simple molecular – a substance that possesses between one and a few hundred atoms per molecule. Giant molecular – a substance containing thousands of atoms per molecule. 6 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Conclusions
Mixtures behave as the individual elements they are made up from. Compounds differ from the elements or mixtures in
Answers
that they are new substances and so behave differently .
Evaluation
Experiments should be reproducible. So the experiments should be repeated to show consistency of results. Also, other
compounds their elements and mixtures should be tested.
Extension
1 (a) Calcium and oxygen.
(b) calcium + oxygen → calcium oxide
2Ca(s) + O (g) → 2CaO(s)
2
2 An element – a substance which cannot be further divided into simpler substances by chemical methods.
A mixture – a system of two or more substances that can be separated by physical means.
A compound – a substance formed by the combination of two or more elements in fixed proportions.
3 Atomic structure and bonding
3.1 Structure of substances
Method
1 To ensure the eyes are shielded, in particular, from bromine and soot.
2 Bromine is toxic, as well as corrosive, and should only be used in a fume cupboard.
Results and calculations
Substance Appearance Inference
Water colourless liquid small molecule
polyethene translucent solid macromolecule
pencil lead black solid macromolecule
Bromine red liquid small molecule
buckminster fullerene (soot) black powder macromolecule
Nylon white solid macromolecule
oxygen colourless gas small molecule
sand brown solid macromolecule
Conclusion
The substances water, bromine and oxygen contain small molecules while the substances polyethene, pencil lead,
buckminster fullerene, nylon and sand are macromolecular.
Evaluation
Other substances which are small molecules or macromolecules could be included in the table of substances.
Extension
1 Simple molecular – a substance that possesses between one and a few hundred atoms per molecule.
Giant molecular – a substance containing thousands of atoms per molecule.
6
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Answers Intramolecular bonds – forces which act within a molecule, for example covalent bonds. Intermolecular bonds – attractive forces which act between molecules, for example van der Waals’ forces. 2 Two examples: water and oxygen. O O O H H water molecule (H O) oxygen molecule (O ) 2 2 3 Two examples: polyethene and silicon(IV) oxide. There is an opportunity to add further CH groups at either end. 2 Si H H O C C Si O Si O Si O H H Si polyethene silicon(IV) oxide 3.2 Properties of ionic and covalent substances Method Because some of the substances may be harmful. Results and calculations Melting Solubility in Substance Appearance point/°C Boiling point/°C water/g/dm 3 Inference A (oxygen) colourless gas −218.80 −83.00 0.008 covalent B (sugar) white crystalline solid 186.00 decomposes 2115.0 covalent (or possibly ionic) C (sodium chloride) white crystalline solid 801.00 1413.00 350.0 ionic D (hexanol) colourless liquid −45.00 156.00 5.90 covalent E (potassium chloride) white crystalline solid 771.00 1500.00 342.00 ionic F (lead bromide) white powder 370.00 914.00 9.73 ionic G (hexane) colourless liquid −95.30 68.70 0.0095 covalent H (lithium chloride) White crystalline solid 614.00 1382.00 830.50 ionic Conclusion Substances C, E, F and H are ionic while substances A, B, D and G are covalent. Evaluation Extend the number of ionic and covalent substances to identify. Extension The ionic ones will conduct electricity and electrolysis will take place. 7 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers
Intramolecular bonds – forces which act within a molecule, for example covalent bonds.
Intermolecular bonds – attractive forces which act between molecules, for example van der Waals’ forces.
2 Two examples: water and oxygen.
O
O O
H H
water molecule (H O) oxygen molecule (O )
2 2
3 Two examples: polyethene and silicon(IV) oxide. There is an opportunity to add further CH groups at either end.
2
H H Si
O
C C
Si O Si O Si
H H O
Si
polyethene
silicon(IV) oxide
3.2 Properties of ionic and covalent substances
Method
Because some of the substances may be harmful.
Results and calculations
Melting Solubility in
Substance Appearance point/°C Boiling point/°C water/g/dm 3 Inference
A (oxygen) colourless gas −218.80 −83.00 0.008 covalent
B (sugar) white crystalline solid 186.00 decomposes 2115.0 covalent (or
possibly ionic)
C (sodium chloride) white crystalline solid 801.00 1413.00 350.0 ionic
D (hexanol) colourless liquid −45.00 156.00 5.90 covalent
E (potassium chloride) white crystalline solid 771.00 1500.00 342.00 ionic
F (lead bromide) white powder 370.00 914.00 9.73 ionic
G (hexane) colourless liquid −95.30 68.70 0.0095 covalent
H (lithium chloride) White crystalline solid 614.00 1382.00 830.50 ionic
Conclusion
Substances C, E, F and H are ionic while substances A, B, D and G are covalent.
Evaluation
Extend the number of ionic and covalent substances to identify.
Extension
The ionic ones will conduct electricity and electrolysis will take place.
7
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

3.3 Electrolysis of solutions Method Answers 1 An electric current is flowing and the substance is ionic. 2 So that there is no cross-contamination of substances being tested. Results and calculations Observation at cathode Observation at Solution Does the bulb light? (−ve) anode (+ve) Inference dilute sulfuric acid yes bubbles of a gas seen bubbles of a gas ionic (hydrogen) seen (oxygen) magnesium sulfate yes bubbles of a gas seen bubbles of a gas ionic (hydrogen) seen (oxygen) copper(II) sulfate yes becomes coated by a pink/ bubbles of a gas ionic brown solid (copper); seen (oxygen) bubbles of a gas seen (hydrogen) sugar solution no nothing nothing covalent Conclusion When an electric current is passed through a solution and the bulb lights then the dissolved substance is ionic . If the bulb does not light then it is covalent . The substances are broken down by the passage of an electric current – this is electrolysis . Evaluation Collect and test gases produced. Extension 1 Check melting points or boiling points or solubility, or all three. 2 Anode – the positive electrode. Cathode – the negative electrode. Electrode – a point where the electric current enters and leaves the electrolytic cell. Electrolysis – a process in which a chemical reaction is caused by the passage of an electric current. 4 Stoichiometry – chemical calculations 4.1 Determination of the formula of magnesium oxide Method 1 So that it sits on the bottom of the crucible where it will be heated most strongly. 2 If a coloured flame is used carbon (soot) will collect on the crucible and increase its mass. 3 The pipe-clay triangle allows the crucible to be heated more strongly and directly than using a gauze. 4 To allow oxygen gas, from the air, into the crucible so that the reaction can take place. 8 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

3.3 Electrolysis of solutions
Method
Answers
1 An electric current is flowing and the substance is ionic.
2 So that there is no cross-contamination of substances being tested.
Results and calculations
Observation at cathode Observation at
Solution Does the bulb light? (−ve) anode (+ve) Inference
dilute sulfuric acid yes bubbles of a gas seen bubbles of a gas ionic
(hydrogen) seen (oxygen)
magnesium sulfate yes bubbles of a gas seen bubbles of a gas ionic
(hydrogen) seen (oxygen)
copper(II) sulfate yes becomes coated by a pink/ bubbles of a gas ionic
brown solid (copper); seen (oxygen)
bubbles of a gas seen
(hydrogen)
sugar solution no nothing nothing covalent
Conclusion
When an electric current is passed through a solution and the bulb lights then the dissolved substance is ionic . If the
bulb does not light then it is covalent . The substances are broken down by the passage of an electric current – this is
electrolysis .
Evaluation
Collect and test gases produced.
Extension
1 Check melting points or boiling points or solubility, or all three.
2 Anode – the positive electrode.
Cathode – the negative electrode.
Electrode – a point where the electric current enters and leaves the electrolytic cell.
Electrolysis – a process in which a chemical reaction is caused by the passage of an electric current.
4 Stoichiometry – chemical
calculations
4.1 Determination of the formula of magnesium oxide
Method
1 So that it sits on the bottom of the crucible where it will be heated most strongly.
2 If a coloured flame is used carbon (soot) will collect on the crucible and increase its mass.
3 The pipe-clay triangle allows the crucible to be heated more strongly and directly than using a gauze.
4 To allow oxygen gas, from the air, into the crucible so that the reaction can take place.
8
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers 5 Magnesium oxide is formed as a white powder which easily escapes from the crucible. Allowing the lid to be removed for only a few seconds prevents too much being lost. 6 To ensure that all the magnesium has reacted with oxygen and so causing the maximum amount of magnesium oxide to be formed. Results and calculations 1 Mass of magnesium used = mass of crucible, lid and Mg − mass of crucible and lid. 2 Mass of oxygen gas reacting = final mass of crucible, MgO and lid − mass of crucible, lid and Mg. 3 Moles of Mg = mass/24. Moles of O = mass/16. Simplest ratio = moles of Mg: moles of O. Converted to a 1: x . Conclusions MgO. Evaluation 1 Nitrogen gas could react to form magnesium nitride, Mg N ; carbon dioxide could react to form magnesium oxide 3 2 and carbon. React with only pure oxygen gas, e.g. in a gas jar. 2 It could be carried out using a larger initial mass of magnesium ribbon. It could be repeated more than once. Extension Nitrogen gas and carbon dioxide. 3Mg + N → Mg N 2 3 2 2Mg + CO → 2MgO + C 2 4.2 Determination of the volume occupied by one mole of a gas Method 1 To ensure that all the magnesium oxide coating on the metal is removed. This allows the reaction with acid to happen straight away. 2 To ensure that all of the magnesium metal has reacted. Results and calculations 1 Moles of magnesium used = mass/ 24 = answer. 2 Moles of hydrogen gas which could be produced = answer to part 1 due to the 1:1 mole ratio in the balanced equation. 3 Volume produced if one mole of gas had been collected = volume collected (cm 3 )/moles of H = answer. 2 4 Volume = volume (cm 3 )/1000 = volume (dm 3 ) Conclusions 1 Should be close to 24 dm 3 . 2 Answers could include references to the different temperatures and pressures at which the experiment was carried out compared to the standard conditions of 25 o C (298 K) and 1 atmosphere of pressure. Evaluation 1 Could be repeated to get a number of results. Use a large initial mass of magnesium metal. 2 The gas could be collected by the downward displacement of water in a measuring cylinder or up-turned burette. 9 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers
5 Magnesium oxide is formed as a white powder which easily escapes from the crucible. Allowing the lid to be
removed for only a few seconds prevents too much being lost.
6 To ensure that all the magnesium has reacted with oxygen and so causing the maximum amount of magnesium
oxide to be formed.
Results and calculations
1 Mass of magnesium used = mass of crucible, lid and Mg − mass of crucible and lid.
2 Mass of oxygen gas reacting = final mass of crucible, MgO and lid − mass of crucible, lid and Mg.
3 Moles of Mg = mass/24.
Moles of O = mass/16.
Simplest ratio = moles of Mg: moles of O.
Converted to a 1: x .
Conclusions
MgO.
Evaluation
1 Nitrogen gas could react to form magnesium nitride, Mg N ; carbon dioxide could react to form magnesium oxide
3 2
and carbon. React with only pure oxygen gas, e.g. in a gas jar.
2 It could be carried out using a larger initial mass of magnesium ribbon. It could be repeated more than once.
Extension
Nitrogen gas and carbon dioxide.
3Mg + N → Mg N
2 3 2
2Mg + CO → 2MgO + C
2
4.2 Determination of the volume occupied by one mole
of a gas
Method
1 To ensure that all the magnesium oxide coating on the metal is removed. This allows the reaction with acid to
happen straight away.
2 To ensure that all of the magnesium metal has reacted.
Results and calculations
1 Moles of magnesium used = mass/ 24 = answer.
2 Moles of hydrogen gas which could be produced = answer to part 1 due to the 1:1 mole ratio in the balanced
equation.
3 Volume produced if one mole of gas had been collected = volume collected (cm 3 )/moles of H = answer.
2
4 Volume = volume (cm 3 )/1000 = volume (dm 3 )
Conclusions
1 Should be close to 24 dm 3 .
2 Answers could include references to the different temperatures and pressures at which the experiment was carried
out compared to the standard conditions of 25 o C (298 K) and 1 atmosphere of pressure.
Evaluation
1 Could be repeated to get a number of results. Use a large initial mass of magnesium metal.
2 The gas could be collected by the downward displacement of water in a measuring cylinder or up-turned burette. 9
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Extension Charles’s Law and Boyle’s Law. Answers 4.3 Determination of the percentage yield of a chemical reaction Method 1 To ensure that all the sodium carbonate solution had reacted. 2 A pipette is a more accurate way of measuring the volume of sodium carbonate added. A measuring cylinder is sufficient for the calcium nitrate as it is in excess and its volume does not need to be measured as accurately. 3 So that the mass is not higher due to the presence of water. 4 It would be difficult to make sure that all the calcium carbonate had been removed; some would be trapped in the holes of the filter paper. Results and calculations 1 Actual yield = mass of calcium carbonate obtained. 2 Moles of sodium carbonate used = 25 × 1/1000 = 0.025 moles 3 Moles of calcium carbonate produced will be the same, 0.025 moles, due to the 1:1 mole ratio shown by the equation. 4 Mass of calcium carbonate that should have been obtained = 0.025 × 100 = 2.5 g. 5 % yield = mass of calcium carbonate made × 100/2.5 = answer Conclusions 1 Comment on the percentage yield of the experiment. 2 Suitable comment and explanation. Evaluation Could have been repeated. Could have used more of the two chemicals to produce larger masses of calcium carbonate. Extension Ca 2+ (aq) + CO 2– (aq) → CaCO (s) 3 3 [1] for equation, [1] for correct state symbols. 5 Electricity and chemistry 5.1 Electrolysis of molten lead(II) bromide Method 1 Bromine and lead are toxic elements. 2 So that the ions are free to move. They can then be discharged at the different electrodes. Results and calculations 1 No. Ions are not free to move. 10 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Extension
Charles’s Law and Boyle’s Law.
Answers
4.3 Determination of the percentage yield of a chemical
reaction
Method
1 To ensure that all the sodium carbonate solution had reacted.
2 A pipette is a more accurate way of measuring the volume of sodium carbonate added. A measuring cylinder is
sufficient for the calcium nitrate as it is in excess and its volume does not need to be measured as accurately.
3 So that the mass is not higher due to the presence of water.
4 It would be difficult to make sure that all the calcium carbonate had been removed; some would be trapped in the
holes of the filter paper.
Results and calculations
1 Actual yield = mass of calcium carbonate obtained.
2 Moles of sodium carbonate used = 25 × 1/1000 = 0.025 moles
3 Moles of calcium carbonate produced will be the same, 0.025 moles, due to the 1:1 mole ratio shown by the
equation.
4 Mass of calcium carbonate that should have been obtained = 0.025 × 100 = 2.5 g.
5 % yield = mass of calcium carbonate made × 100/2.5 = answer
Conclusions
1 Comment on the percentage yield of the experiment.
2 Suitable comment and explanation.
Evaluation
Could have been repeated. Could have used more of the two chemicals to produce larger masses of calcium
carbonate.
Extension
Ca 2+ (aq) + CO 2– (aq) → CaCO (s)
3 3
[1] for equation, [1] for correct state symbols.
5 Electricity and chemistry
5.1 Electrolysis of molten lead(II) bromide
Method
1 Bromine and lead are toxic elements.
2 So that the ions are free to move. They can then be discharged at the different electrodes.
Results and calculations
1 No.
Ions are not free to move.
10
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers 2 Yes. The ions in the molten material are free to move to the electrodes. 3 Brown vapour. Bromide ions (−ve) are discharged at the anode (the +ve electrode). 4 Shiny solid. Lead ions (+ve) are discharged at the cathode (the −ve electrode). Conclusion Solid lead bromide(II) does not conduct electricity. However, when it is molten it does conduct and electrolysis takes place. Lead is produced at the cathode and bromine is produced at the anode. Evaluation Carry out chemical tests on the products to show that they contain bromine and lead. Extension 1 Electrolysis – a process in which a chemical reaction is caused by the passage of an electric current. Electrolyte – a substance which will carry electric current only when it is molten or dissolved. Electrodes – a point where the electric current enters and leaves the electrolytic cell. Inert – these are electrodes that do not react with the products of electrolysis, e.g. carbon, platinum. Cathode – the negative electrode. It is negatively charged because an excess of electrons move towards it. Cations – positive ions; these are attracted to the cathode. Anode – the positive electrode. It is positively charged because electrons are drawn away from it. Anions – negative ions; these are attracted to the anode. 2 At anode: bromide ion → bromine atom + electron Br − (l) → Br + e − At cathode: lead ion + electrons → lead atom Pb 2+ (l) + 2e − → Pb(l) 3 (a) Metal (b) A (c) Copper(II) oxide (d) Sodium chloride (or other metal chloride). 5.2 Electrolysis of water Method 1 All ions that would allow the water to conduct have been removed. 2 This is so the solution is not electrolysing during filling of the apparatus, which would change the overall volumes of gas produced. Results and calculations Approximate ratio of gases produced: cathode: anode 2:1 Effect of glowing splint with anode tube gas: relights it . The gas is oxygen . Effect of lighted splint with cathode tube gas: burns with a ‘pop’ . The gas is hydrogen . Do the results fit in with the known formula for water? Yes . Explain your answer. Formula is H O, which should give volume ratio of 2:1 on electrolysis . 11 2 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers
2 Yes.
The ions in the molten material are free to move to the electrodes.
3 Brown vapour.
Bromide ions (−ve) are discharged at the anode (the +ve electrode).
4 Shiny solid.
Lead ions (+ve) are discharged at the cathode (the −ve electrode).
Conclusion
Solid lead bromide(II) does not conduct electricity. However, when it is molten it does conduct and electrolysis takes
place. Lead is produced at the cathode and bromine is produced at the anode.
Evaluation
Carry out chemical tests on the products to show that they contain bromine and lead.
Extension
1 Electrolysis – a process in which a chemical reaction is caused by the passage of an electric current.
Electrolyte – a substance which will carry electric current only when it is molten or dissolved.
Electrodes – a point where the electric current enters and leaves the electrolytic cell.
Inert – these are electrodes that do not react with the products of electrolysis, e.g. carbon, platinum.
Cathode – the negative electrode. It is negatively charged because an excess of electrons move towards it.
Cations – positive ions; these are attracted to the cathode.
Anode – the positive electrode. It is positively charged because electrons are drawn away from it.
Anions – negative ions; these are attracted to the anode.
2 At anode:
bromide ion → bromine atom + electron
Br − (l) → Br + e −
At cathode:
lead ion + electrons → lead atom
Pb 2+ (l) + 2e − → Pb(l)
3 (a) Metal
(b) A
(c) Copper(II) oxide
(d) Sodium chloride (or other metal chloride).
5.2 Electrolysis of water
Method
1 All ions that would allow the water to conduct have been removed.
2 This is so the solution is not electrolysing during filling of the apparatus, which would change the overall
volumes of gas produced.
Results and calculations
Approximate ratio of gases produced: cathode: anode 2:1
Effect of glowing splint with anode tube gas: relights it .
The gas is oxygen .
Effect of lighted splint with cathode tube gas: burns with a ‘pop’ .
The gas is hydrogen .
Do the results fit in with the known formula for water? Yes .
Explain your answer. Formula is H O, which should give volume ratio of 2:1 on electrolysis . 11
2
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Conclusion Pure water is a very bad conductor of electricity. This is because it has very few ions present in the liquid. If dilute Answers sulfuric acid is added to water, ions are added and this makes the solution a conducting one. An electric current passes through this solution and electrolysis takes place. Hydrogen gas is produced at the cathode and oxygen gas is produced at the anode . The ratio of the volumes of the gases produced is 2:1 hydrogen to oxygen. This fits in with the formula for water being H O . 2 Evaluation The apparatus should be cleaned out with deionised water beforehand. This will clean out any ions left behind from previous experiments. Extension 1 At anode: hydroxide ions → water molecules + oxygen molecules + electrons 4OH − (aq) → 2H O(l) + O (g) + 4e − 2 2 At cathode: hydroxide ions + electrons → hydrogen molecules 4H + (aq) + 4e − → 2H (g) 2 2 The chlor-alkali industry The electrolysis of saturated sodium chloride solution (brine) is the basis of the chlor-akali industry. Three very important substances are produced in this electrolysis process – chlorine, sodium hydroxide and hydrogen. The electrolytic process is a very expensive one, requiring vast amounts of electricity. The process is economical only because all three products have a large number of uses. The major cell used is the membrane cell. This method is now preferred to the other two because it produces a purer product, it causes less pollution and it is cheaper to run. The brine is first purified to remove calcium, strontium and magnesium compounds by a process of ion exchange. The membrane cell is used continuously, with fresh brine flowing into the cell as the process breaks up the brine. The cell has been designed to ensure that the products do not mix. The ions in this concentrated sodium chloride solution are: from the water H + (aq), OH − (aq); from the sodium chloride Na + (aq), Cl − (aq). When the current flows, the chloride ions, Cl − (aq), are attracted to the anode. Chlorine gas is produced. This leaves a high concentration of sodium ions, Na + (aq), around the anode. The hydrogen ions, H + (aq), are attracted to the cathode and hydrogen gas is produced. This leaves a high concentration of hydroxide ions, OH − (aq), around the cathode. The sodium ions, Na + (aq), are drawn through the membrane, where they combine with the OH − (aq) to form sodium hydroxide, NaOH, solution. Electroplating cathode ( ) Electroplating is the process involving electrolysis to plate, or coat, one anode ( ) metal with another or a plastic with a metal. Often the purpose of object to be plated, e.g. spoon electroplating is to give a protective coating to the metal beneath. For example, bath taps are chromium plated to prevent corrosion, and at the same time are given a shiny, more attractive finish. The electroplating process is carried out in a cell such as the one shown here. This is often known as the ‘plating bath’ and it contains a suitable electrolyte, usually a solution of a metal salt. For silver plating, the electrolyte is a solution of a silver salt. The article to be plated is made anode made electrolyte the cathode in the cell so that the metal ions move to it when the current from the metal containing the being used metal being used is switched on. Plastics have been developed that are able to conduct for plating, for plating, electricity. For example, the plastic poly(pyrrole) can be electroplated in e.g. silver e.g. silver the same way as the metals. Figure 1 electroplating bath 12 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Conclusion
Pure water is a very bad conductor of electricity. This is because it has very few ions present in the liquid. If dilute
Answers
sulfuric acid is added to water, ions are added and this makes the solution a conducting one. An electric current passes
through this solution and electrolysis takes place. Hydrogen gas is produced at the cathode and oxygen gas is produced
at the anode . The ratio of the volumes of the gases produced is 2:1 hydrogen to oxygen. This fits in with the formula for
water being H O .
2
Evaluation
The apparatus should be cleaned out with deionised water beforehand. This will clean out any ions left behind from
previous experiments.
Extension
1 At anode:
hydroxide ions → water molecules + oxygen molecules + electrons
4OH − (aq) → 2H O(l) + O (g) + 4e −
2 2
At cathode:
hydroxide ions + electrons → hydrogen molecules
4H + (aq) + 4e − → 2H (g)
2
2 The chlor-alkali industry
The electrolysis of saturated sodium chloride solution (brine) is the basis of the chlor-akali industry. Three very
important substances are produced in this electrolysis process – chlorine, sodium hydroxide and hydrogen. The
electrolytic process is a very expensive one, requiring vast amounts of electricity. The process is economical only
because all three products have a large number of uses. The major cell used is the membrane cell. This method
is now preferred to the other two because it produces a purer product, it causes less pollution and it is cheaper
to run. The brine is first purified to remove calcium, strontium and magnesium compounds by a process of ion
exchange.
The membrane cell is used continuously, with fresh brine flowing into the cell as the process breaks up the brine.
The cell has been designed to ensure that the products do not mix. The ions in this concentrated sodium chloride
solution are: from the water H + (aq), OH − (aq); from the sodium chloride Na + (aq), Cl − (aq). When the current flows,
the chloride ions, Cl − (aq), are attracted to the anode. Chlorine gas is produced. This leaves a high concentration of
sodium ions, Na + (aq), around the anode.
The hydrogen ions, H + (aq), are attracted to the cathode and hydrogen gas is produced. This leaves a high
concentration of hydroxide ions, OH − (aq), around the cathode. The sodium ions, Na + (aq), are drawn through the
membrane, where they combine with the OH − (aq) to form sodium hydroxide, NaOH, solution.
Electroplating
cathode (  )
Electroplating is the process involving electrolysis to plate, or coat, one
anode (  )
metal with another or a plastic with a metal. Often the purpose of object to be plated,
e.g. spoon
electroplating is to give a protective coating to the metal beneath. For
example, bath taps are chromium plated to prevent corrosion, and at the
same time are given a shiny, more attractive finish. The electroplating
process is carried out in a cell such as the one shown here.
This is often known as the ‘plating bath’ and it contains a suitable
electrolyte, usually a solution of a metal salt. For silver plating, the
electrolyte is a solution of a silver salt. The article to be plated is made
anode made electrolyte
the cathode in the cell so that the metal ions move to it when the current from the metal containing the
being used metal being used
is switched on. Plastics have been developed that are able to conduct
for plating, for plating,
electricity. For example, the plastic poly(pyrrole) can be electroplated in e.g. silver e.g. silver
the same way as the metals. Figure 1 electroplating bath
12
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers 5.3 Electrolysis of brine Method 1 So that the chlorine gas produced can be identified by the colour changes that take place. 2 Chlorine gas is produced, which is a toxic substance. 3 This is so the solution is not electrolysing during filling of the apparatus, which would add the hazard of chlorine being produced during the set-up of the apparatus. Results and calculations Colour changes of the solution are as follows: The anode tube: initially: turns pink ; during the experiment: pink colour is bleached out (a blue colour is seen in the vicinity of the anode tube also) . The cathode tube: initially: original colour of universal indicator; during the experiment: original colour of universal indicator . Effect of lighted splint with cathode tube: gas burns with a ‘pop’ . The gas is hydrogen . Do the results fit in with the expectation of the electrolysis? Yes . Explain your answer . Three substances are produced: chlorine – bleaching agent, hydrogen gas, sodium hydroxide – an alkali . Conclusion Brine is a very good conductor of electricity. This is because it has a lot of ions present in the liquid. This makes the solution an electrolytic one. An electric current passes through this solution and electrolysis takes place. Hydrogen gas is produced at the cathode and chlorine gas is produced at the anode . Sodium hydroxide is left in solution. Evaluation To ensure the apparatus is free from other ions it should be washed out with the solution of brine to be used – several times. Extension 1 At anode: chloride ions → chlorine molecules + electrons 2Cl − (aq) → Cl (g) + 2e − 2 At cathode: hydrogen ions + electrons → hydrogen molecules 4H + (aq) + 4e – → 2H (g) 2 2 Hydrogen – making ammonia, rocket fuel, making hydrochloric acid Chlorine – water purification, making bleach, making plastics Sodium hydroxide – paper manufacture, soaps, aluminium extraction 13 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

5.3 Electrolysis of brine Answers
Method
1 So that the chlorine gas produced can be identified by the colour changes that take place.
2 Chlorine gas is produced, which is a toxic substance.
3 This is so the solution is not electrolysing during filling of the apparatus, which would add the hazard of chlorine
being produced during the set-up of the apparatus.
Results and calculations
Colour changes of the solution are as follows:
The anode tube: initially: turns pink ; during the experiment: pink colour is bleached out (a blue colour is seen in the
vicinity of the anode tube also) .
The cathode tube: initially: original colour of universal indicator; during the experiment: original colour of universal
indicator .
Effect of lighted splint with cathode tube: gas burns with a ‘pop’ .
The gas is hydrogen .
Do the results fit in with the expectation of the electrolysis? Yes .
Explain your answer . Three substances are produced: chlorine – bleaching agent, hydrogen gas,
sodium hydroxide – an alkali .
Conclusion
Brine is a very good conductor of electricity. This is because it has a lot of ions present in the liquid. This makes the
solution an electrolytic one. An electric current passes through this solution and electrolysis takes place. Hydrogen gas is
produced at the cathode and chlorine gas is produced at the anode . Sodium hydroxide is left in solution.
Evaluation
To ensure the apparatus is free from other ions it should be washed out with the solution of brine to
be used – several times.
Extension
1 At anode:
chloride ions → chlorine molecules + electrons
2Cl − (aq) → Cl (g) + 2e −
2
At cathode:
hydrogen ions + electrons → hydrogen molecules
4H + (aq) + 4e – → 2H (g)
2
2 Hydrogen – making ammonia, rocket fuel, making hydrochloric acid
Chlorine – water purification, making bleach, making plastics
Sodium hydroxide – paper manufacture, soaps, aluminium extraction
13
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6 Chemical energetics Answers 6.1 Electrochemical cells: chemical energy to electrical energy Method 1 Copper. 2 The concentration of all the solutions is 1 mol/dm 3 . 3 To ensure that there are no impurities on their surfaces. Results Voltages should be near those shown in the table: [1] for close answer, [1] for using resolution shown on voltmeter throughout. Positive terminal Negative terminal Voltage/V copper lead 0.47 copper magnesium 2.71 copper iron 0.78 copper zinc 1.10 copper tin 0.48 Conclusions 1 Copper and magnesium. 2 Copper and lead. 3 0.00 V. 4 Magnesium is the most reactive, lead is the least reactive. 5 Lead, tin, iron, zinc, magnesium – 4 in correct order, all correct. 6 Mg → Mg 2+ + 2e − Evaluation The voltage could have been recorded a number of times to obtain average results. The solution of copper(II) sulfate could have been changed after each cell was finished with. Extension 1 2.24 V. 6.2 Calculating the energy of combustion of methanol and ethanol Method 1 Copper is a better conductor of heat than glass so more of the heat is transferred to the water. 2 To prevent the fuel in the burner from evaporating as it is carried around the laboratory to and from the balance. This would lead to incorrect mass readings for the mass of fuel burned. 3 Difficult to prevent incomplete combustion from occurring. Heat is lost to the surroundings. 4 The bottom of the calorimeter will be black, caused by incomplete combustion of the fuel producing carbon (soot). 14 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

6 Chemical energetics
Answers
6.1 Electrochemical cells: chemical energy to
electrical energy
Method
1 Copper.
2 The concentration of all the solutions is 1 mol/dm 3 .
3 To ensure that there are no impurities on their surfaces.
Results
Voltages should be near those shown in the table: [1] for close answer, [1] for using resolution shown on voltmeter
throughout.
Positive terminal Negative terminal Voltage/V
copper lead 0.47
copper magnesium 2.71
copper iron 0.78
copper zinc 1.10
copper tin 0.48
Conclusions
1 Copper and magnesium.
2 Copper and lead.
3 0.00 V.
4 Magnesium is the most reactive, lead is the least reactive.
5 Lead, tin, iron, zinc, magnesium – 4 in correct order, all correct.
6 Mg → Mg 2+ + 2e −
Evaluation
The voltage could have been recorded a number of times to obtain average results. The solution of copper(II)
sulfate could have been changed after each cell was finished with.
Extension
1 2.24 V.
6.2 Calculating the energy of combustion of methanol
and ethanol
Method
1 Copper is a better conductor of heat than glass so more of the heat is transferred to the water.
2 To prevent the fuel in the burner from evaporating as it is carried around the laboratory to and from the balance.
This would lead to incorrect mass readings for the mass of fuel burned.
3 Difficult to prevent incomplete combustion from occurring. Heat is lost to the surroundings.
4 The bottom of the calorimeter will be black, caused by incomplete combustion of the fuel producing
carbon (soot).
14
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers Results and calculations 1 Temperature rise in each experiment = maximum temperature of water – initial temperature of water. 2 Energy transferred = mass of water used × 4.2 × temperature rise in each experiment. 3 Mass of fuel burned in each experiment = final mass of burner, lid and fuel − initial mass of burner, lid and fuel. 4 Moles of methanol burned = mass of methanol burned/32. 5 Moles of ethanol burned = mass of ethanol burned/46. 6 Energy transferred by each fuel = answer to 2/answer to 4. 7 Energy of combustion = answers to 5/1000. Conclusions 1 It is an exothermic reaction, and heat is given out to the surroundings from the chemical reaction. 2 Non-standard conditions being used. The temperature and pressure are different from those used to produce data-book values. A large amount of the heat energy produced from the burner is lost to the environment, not all is transferred to the water. Evaluation Some sort of draught excluder could be placed around the apparatus. The flame height could be minimised, by shortening the wick length. The height of the calorimeter above the flame should be adjusted to give the best chance of heating the calorimeter most efficiently. Extension Information about the use of a bomb calorimeter should be found. 6.3 Determination of the energy change of a displacement reaction Method 1 It is a better heat insulator so less heat will escape. 2 Using a pipette is a more accurate way of measuring a volume of solution, the copper(II) sulfate will be totally used up in the reaction, it is not in excess, so an accurate volume is required for the calculations. 3 To ensure that all the copper(II) sulfate has reacted. Results and calculations Accurate recording of temperatures [1] to 0.5 °C. 1 Lines A, B, C, D drawn correctly on their graph. 2 Δ T recorded correctly from the graph, accurately. max 3 Energy produced (J) = 25 × 4.2 × Δ T = answer. max 4 Energy produced from one mole = answer to 3/0.025 = answer. Conclusions 1 Comment made about their result compared to data-book value. Sensible reason for the difference, e.g. heat loss to surroundings. 2 Zinc is a more reactive metal than iron and therefore a more vigorous reaction occurs, giving out more heat energy. 3 Copper(II) sulfate and tin would give a less exothermic reaction with a smaller energy change because tin is less reactive than both zinc and iron. 15 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations Answers
1 Temperature rise in each experiment = maximum temperature of water – initial temperature of water.
2 Energy transferred = mass of water used × 4.2 × temperature rise in each experiment.
3 Mass of fuel burned in each experiment = final mass of burner, lid and fuel − initial mass of burner, lid and fuel.
4 Moles of methanol burned = mass of methanol burned/32.
5 Moles of ethanol burned = mass of ethanol burned/46.
6 Energy transferred by each fuel = answer to 2/answer to 4.
7 Energy of combustion = answers to 5/1000.
Conclusions
1 It is an exothermic reaction, and heat is given out to the surroundings from the chemical reaction.
2 Non-standard conditions being used. The temperature and pressure are different from those used to produce
data-book values. A large amount of the heat energy produced from the burner is lost to the environment, not
all is transferred to the water.
Evaluation
Some sort of draught excluder could be placed around the apparatus.
The flame height could be minimised, by shortening the wick length.
The height of the calorimeter above the flame should be adjusted to give the best chance of heating the calorimeter
most efficiently.
Extension
Information about the use of a bomb calorimeter should be found.
6.3 Determination of the energy change of a
displacement reaction
Method
1 It is a better heat insulator so less heat will escape.
2 Using a pipette is a more accurate way of measuring a volume of solution, the copper(II) sulfate will be totally used
up in the reaction, it is not in excess, so an accurate volume is required for the calculations.
3 To ensure that all the copper(II) sulfate has reacted.
Results and calculations
Accurate recording of temperatures [1] to 0.5 °C.
1 Lines A, B, C, D drawn correctly on their graph.
2 Δ T recorded correctly from the graph, accurately.
max
3 Energy produced (J) = 25 × 4.2 × Δ T = answer.
max
4 Energy produced from one mole = answer to 3/0.025 = answer.
Conclusions
1 Comment made about their result compared to data-book value. Sensible reason for the difference, e.g. heat loss
to surroundings.
2 Zinc is a more reactive metal than iron and therefore a more vigorous reaction occurs, giving out more heat
energy.
3 Copper(II) sulfate and tin would give a less exothermic reaction with a smaller energy change because tin is less
reactive than both zinc and iron.
15
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Evaluation A lid could have been put on the polystyrene cup which would prevent heat from escaping to the surroundings. Answers Larger amounts of both chemicals could have been used which would give a larger temperature rise with a smaller error. Extension Extrapolation means extending the line on a graph to values beyond the experimental data collected to estimate the value of a variable. 7 Chemical reactions 7.1 How does changing surface area affect the rate of a reaction? Method To prevent acid spray from escaping from the flask which would make the loss in mass greater. Results and calculations 1 Table completed using the resolution shown by the balance. 2 Loss in mass worked out for each recorded result to the required accuracy. 3 Points plotted correctly, best fit curve/line through the point, correct labels on the axes, correct units shown on the axes. Conclusions 1 The experiment with the smaller marble chips. 2 Steeper line at the beginning of the experiment. 3 10 g of the smaller marble chips has a larger surface area than 10 g of the large chips. This allows a larger surface area for collisions to occur with acid particles. More collisions result from this giving more successful collisions and a faster rate of reaction. 4 Any three from: same temperature, same volume of acid used, same concentration of acid used, same mass of marble chips Evaluation A data logger could have been used to produce a continuous reading of the mass and to plot the graph after each reading. Extension Nanotubes have a very large surface area on which reactions can occur, leading to a different reaction pathway with a lower activation energy. 7.2 What is the effect of changing the temperature on the rate of reaction? Method 1 It would take much longer for it to be obscured. 2 So that the same amount of sulfur, each time, is needed to obscure it from sight. This allows the times to be 16 compared successfully. Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Evaluation
A lid could have been put on the polystyrene cup which would prevent heat from escaping to the surroundings.
Answers
Larger amounts of both chemicals could have been used which would give a larger temperature rise with a
smaller error.
Extension
Extrapolation means extending the line on a graph to values beyond the experimental data collected to estimate the
value of a variable.
7 Chemical reactions
7.1 How does changing surface area affect the rate of a
reaction?
Method
To prevent acid spray from escaping from the flask which would make the loss in mass greater.
Results and calculations
1 Table completed using the resolution shown by the balance.
2 Loss in mass worked out for each recorded result to the required accuracy.
3 Points plotted correctly, best fit curve/line through the point, correct labels on the axes, correct units shown on
the axes.
Conclusions
1 The experiment with the smaller marble chips.
2 Steeper line at the beginning of the experiment.
3 10 g of the smaller marble chips has a larger surface area than 10 g of the large chips. This allows a larger
surface area for collisions to occur with acid particles. More collisions result from this giving more successful
collisions and a faster rate of reaction.
4 Any three from: same temperature, same volume of acid used, same concentration of acid used, same mass of
marble chips
Evaluation
A data logger could have been used to produce a continuous reading of the mass and to plot the graph after each
reading.
Extension
Nanotubes have a very large surface area on which reactions can occur, leading to a different reaction pathway with a
lower activation energy.
7.2 What is the effect of changing the temperature on
the rate of reaction?
Method
1 It would take much longer for it to be obscured.
2 So that the same amount of sulfur, each time, is needed to obscure it from sight. This allows the times to be
16 compared successfully.
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers 3 It will increase the rate of reaction. 4 So that you are looking through the same depth of solution each time, ensuring a fair test. Results and calculations 1 Records the temperature and times accurately. 2 Rates worked out correctly from each time. 3 Points plotted correctly, best fit curve/line through the point, correct labels on the axes, correct units shown on the axes. Conclusions 1 A smaller time gives a faster rate. They are inversely proportional to one another. 2 As temperature increases the rate increases. 3 Two from: same volumes of acid and sodium thiosulfate, same concentrations of acid and sodium thiosulfate, same equipment/same size of flask used each time, same amount of swirling each time. 4 As the temperature increases the particles gain more kinetic energy, they move faster and collide more frequently with one another, which results in more successful collisions and a faster rate of reaction. Evaluation Use some type of data logger to accurately measure the formation of the same amount of sulfur being formed by looking through the reaction solution at a light. Repeat the experiment more times for each temperature in order to find an average time. Extension Research into Maxwell-Boltzmann distribution curves. 7.3 What is the effect of changing the concentration on the rate of a reaction? Method 1 It would take much longer for it to be obscured. 2 So that the same amount of sulfur, each time, is needed to obscure it from sight. This allows the times to be compared successfully. 3 It will increase the rate of reaction. 4 So that you are looking through the same depth of solution each time, ensuring a fair test. Results and calculations 1 Records the times accurately. 2 Rates worked out correctly from each time. 3 Points plotted correctly, best fit curve/line through the point, correct labels on the axes, correct units shown on the axes. Conclusions 1 The concentration is decreasing because water is being added. 2 Increasing the concentration increases the rate of the reaction. 3 Any two of: same total volume, same volume of acid, same concentration of acid, flask swirled only once each time, same temperature. 4 As concentration increases there are more particles in a given volume. This results in more collisions between the particles. Some of these extra collisions are successful collisions and so the rate increases. 17 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers
3 It will increase the rate of reaction.
4 So that you are looking through the same depth of solution each time, ensuring a fair test.
Results and calculations
1 Records the temperature and times accurately.
2 Rates worked out correctly from each time.
3 Points plotted correctly, best fit curve/line through the point, correct labels on the axes, correct units
shown on the axes.
Conclusions
1 A smaller time gives a faster rate. They are inversely proportional to one another.
2 As temperature increases the rate increases.
3 Two from: same volumes of acid and sodium thiosulfate, same concentrations of acid and sodium thiosulfate, same
equipment/same size of flask used each time, same amount of swirling each time.
4 As the temperature increases the particles gain more kinetic energy, they move faster and collide more frequently
with one another, which results in more successful collisions and a faster rate of reaction.
Evaluation
Use some type of data logger to accurately measure the formation of the same amount of sulfur being formed by
looking through the reaction solution at a light.
Repeat the experiment more times for each temperature in order to find an average time.
Extension
Research into Maxwell-Boltzmann distribution curves.
7.3 What is the effect of changing the concentration on
the rate of a reaction?
Method
1 It would take much longer for it to be obscured.
2 So that the same amount of sulfur, each time, is needed to obscure it from sight. This allows the times to be
compared successfully.
3 It will increase the rate of reaction.
4 So that you are looking through the same depth of solution each time, ensuring a fair test.
Results and calculations
1 Records the times accurately.
2 Rates worked out correctly from each time.
3 Points plotted correctly, best fit curve/line through the point, correct labels on the axes, correct units
shown on the axes.
Conclusions
1 The concentration is decreasing because water is being added.
2 Increasing the concentration increases the rate of the reaction.
3 Any two of: same total volume, same volume of acid, same concentration of acid, flask swirled only once each
time, same temperature.
4 As concentration increases there are more particles in a given volume. This results in more collisions between
the particles. Some of these extra collisions are successful collisions and so the rate increases.
17
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Evaluation One of: use some type of data logger to accurately measure the formation of the same amount of sulfur by looking Answers through the reaction solution at a light. Repeat the experiment more times for each concentration in order to find an average time. Extension First order simply means that there is a direct relationship between the concentration of a reactant and the rate. If the concentration of a reactant doubles, the rate will also double. 8 Acids, bases and salts 8.1 Hydrated salts: how much water do they contain? Method 1 So the frame can be directed at the bottom of the crucible and the sample will be heated most strongly. 2 If a coloured flame is used carbon (soot) will collect on the crucible and increase its mass. 3 The steam (water) needs to be allowed to evaporate away. 4 To ensure that all the water had been driven off and that only anhydrous magnesium sulfate remained. Results and calculations 1 All masses recorded to the accuracy of the balance used. 2 Mass of hydrated magnesium sulfate = mass of crucible and hydrated MgSO − mass of crucible correctly 4 calculated. 3 Mass of water removed = mass of crucible and hydrated MgSO − mass after final heating. 4 4 Mass of anhydrous magnesium sulfate = mass after final heating − mass of crucible = answer. 5 Moles of anhydrous magnesium sulfate = answer to 4/84 = answer. 6 Number of moles of water removed = answer to 3/18 = answer. 7 x = moles of water/moles of anhydrous MgSO = answer. 4 8 Formula of hydrated magnesium sulfate = MgSO . x H O. 4 2 Conclusions 1 Comment about how their result compares to the actual formula of MgSO .7H O. 4 2 2 Magnesium sulfate could have been lost from the crucible as a powder whilst being heated. Evaluation It could be repeated more than once using the same starting mass of hydrated magnesium sulfate to get an average. A larger mass of hydrated magnesium sulfate could have been used to have larger mass readings and, therefore, a lower percentage error in the readings. Extension Anhydrous magnesium sulfate is used as a drying agent to remove traces of water from substances. 18 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Evaluation
One of: use some type of data logger to accurately measure the formation of the same amount of sulfur by looking
Answers
through the reaction solution at a light.
Repeat the experiment more times for each concentration in order to find an average time.
Extension
First order simply means that there is a direct relationship between the concentration of a reactant and the rate.
If the concentration of a reactant doubles, the rate will also double.
8 Acids, bases and salts
8.1 Hydrated salts: how much water do they contain?
Method
1 So the frame can be directed at the bottom of the crucible and the sample will be heated most strongly.
2 If a coloured flame is used carbon (soot) will collect on the crucible and increase its mass.
3 The steam (water) needs to be allowed to evaporate away.
4 To ensure that all the water had been driven off and that only anhydrous magnesium sulfate remained.
Results and calculations
1 All masses recorded to the accuracy of the balance used.
2 Mass of hydrated magnesium sulfate = mass of crucible and hydrated MgSO − mass of crucible correctly
4
calculated.
3 Mass of water removed = mass of crucible and hydrated MgSO − mass after final heating.
4
4 Mass of anhydrous magnesium sulfate = mass after final heating − mass of crucible = answer.
5 Moles of anhydrous magnesium sulfate = answer to 4/84 = answer.
6 Number of moles of water removed = answer to 3/18 = answer.
7 x = moles of water/moles of anhydrous MgSO = answer.
4
8 Formula of hydrated magnesium sulfate = MgSO . x H O.
4 2
Conclusions
1 Comment about how their result compares to the actual formula of MgSO .7H O.
4 2
2 Magnesium sulfate could have been lost from the crucible as a powder whilst being heated.
Evaluation
It could be repeated more than once using the same starting mass of hydrated magnesium sulfate to get an average.
A larger mass of hydrated magnesium sulfate could have been used to have larger mass readings and, therefore, a lower
percentage error in the readings.
Extension
Anhydrous magnesium sulfate is used as a drying agent to remove traces of water from substances.
18
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers 8.2 Determination of the concentration of a solution of hydrochloric acid Method 1 So that the correct volume can be read from the scale. 2 To see the colour change more clearly. 3 Phenolphthalein is a weak acid so putting different numbers of drops into each experiment would give different results. 4 So that the end-point/colour change is not missed giving an inaccurate volume needed for the point of neutralisation. Results and calculations 1 Results recorded. All burette readings to two decimal places. 2 Average volume of acid used calculated correctly using the three results within 0.10 cm 3 of one another to two decimal places. 3 Moles of sodium hydroxide used = 25 × 1/1000 = 0.025 moles. 4 Moles of HCl which would react = 0.025 moles due to the 1:1 mole relationship in the balanced chemical equation. 5 Concentration of acid = 0.025 × 1000/average volume = answer. Conclusions Comparison with range of acid combinations given in the apparatus and chemicals list. Evaluation Use of a pH probe to find the end-point at exactly pH 7. Extension They can be used to find the acidity of soil samples, the pH of acids used in the flavouring of foods, e.g. citric acid, any suitable example. 8.3 Preparation of hydrated magnesium sulfate Method 1 To mix the reactants together. It increases the rate of reaction. 2 The production of carbon dioxide gas causes the effervescence. 3 No more carbon dioxide was being produced because the reaction had stopped. 4 It means that the magnesium carbonate was in excess which ensures that all of the sulfuric acid had reacted. 5 To leave some water which will be taken into the crystals as they are formed as water of crystallisation. 6 Prevents impurities landing in the magnesium sulfate as the crystals form. Results Colourless crystals. Conclusions 1 Soluble. 2 Anhydrous magnesium sulfate. A white powder. 3 Water is part of the crystals and gets locked inside the crystals between the ions for the crystals to form their correct shapes. 19 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

8.2 Determination of the concentration of a solution of Answers
hydrochloric acid
Method
1 So that the correct volume can be read from the scale.
2 To see the colour change more clearly.
3 Phenolphthalein is a weak acid so putting different numbers of drops into each experiment would give different
results.
4 So that the end-point/colour change is not missed giving an inaccurate volume needed for the point of neutralisation.
Results and calculations
1 Results recorded. All burette readings to two decimal places.
2 Average volume of acid used calculated correctly using the three results within 0.10 cm 3 of one another to two
decimal places.
3 Moles of sodium hydroxide used = 25 × 1/1000 = 0.025 moles.
4 Moles of HCl which would react = 0.025 moles due to the 1:1 mole relationship in the balanced chemical equation.
5 Concentration of acid = 0.025 × 1000/average volume = answer.
Conclusions
Comparison with range of acid combinations given in the apparatus and chemicals list.
Evaluation
Use of a pH probe to find the end-point at exactly pH 7.
Extension
They can be used to find the acidity of soil samples, the pH of acids used in the flavouring of foods, e.g. citric acid, any
suitable example.
8.3 Preparation of hydrated magnesium sulfate
Method
1 To mix the reactants together. It increases the rate of reaction.
2 The production of carbon dioxide gas causes the effervescence.
3 No more carbon dioxide was being produced because the reaction had stopped.
4 It means that the magnesium carbonate was in excess which ensures that all of the sulfuric acid had reacted.
5 To leave some water which will be taken into the crystals as they are formed as water of crystallisation.
6 Prevents impurities landing in the magnesium sulfate as the crystals form.
Results
Colourless crystals.
Conclusions
1 Soluble.
2 Anhydrous magnesium sulfate. A white powder.
3 Water is part of the crystals and gets locked inside the crystals between the ions for the crystals to form their
correct shapes.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Evaluation You could leave the solution obtained after the filtration to evaporate slowly to form crystals. This would ensure Answers sufficient water would be present to form perfect crystals. Extension 1 Reacting dilute sulfuric acid with: magnesium metal, magnesium oxide, magnesium hydroxide. 2 H SO + Mg → MgSO + H O 2 4 4 2 H SO + MgO → MgSO + H O 2 4 4 2 H SO + Mg(OH) → MgSO + 2H O 2 4 2 4 2 9 The Periodic Table 9.1 Reactions of the Group 1 metals Method 1 To prevent the possibility of the hot metals from spitting out of the water towards the students. 2 So that it does not start to react with the oxygen or water in the air. 3 The shiny surface rapidly tarnishes. This is because the metal reacts rapidly with water and oxygen in the environment. 4 So that the reaction can be controlled easily/so that the reaction will not be too vigorous. 5 Removing the oil exposes the metal’s surface so that the reaction starts more quickly when it is put into the water. 6 Because the metals float and move around the surface of the water – there is less chance, if placed in the centre of the trough, that they will reach the side. If they do reach the side, the exothermic nature of the reaction can cause the glass trough to crack. Results Each row completed correctly. Colour of solution pH of the Colour of Observations when added to formed with universal solution Metal metal water indicator solution formed lithium silvery lithium floats on the surface and blue 10–12 moves around as a gas is produced potassium silvery potassium floats on the surface violet 12–14 and moves around rapidly, orange flames are observed; the metal melts sodium silvery sodium floats around on the blue/violet 12 surface and moves around rapidly; the metal melts Conclusions 1 Because they float on water they have a density lower than that of water. 2 Potassium. 3 Lithium. 4 Lithium, sodium, potassium. 5 When metals react they lose electrons. As you go down Group 1, the metal atoms get bigger and they lose electrons more easily due to the lessening of the attraction between the positive nucleus and the outer shell electron. 20 6 2K + 2H O → 2KOH + H . 2 2 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Evaluation
You could leave the solution obtained after the filtration to evaporate slowly to form crystals. This would ensure
Answers
sufficient water would be present to form perfect crystals.
Extension
1 Reacting dilute sulfuric acid with: magnesium metal, magnesium oxide, magnesium hydroxide.
2 H SO + Mg → MgSO + H O
2 4 4 2
H SO + MgO → MgSO + H O
2 4 4 2
H SO + Mg(OH) → MgSO + 2H O
2 4 2 4 2
9 The Periodic Table
9.1 Reactions of the Group 1 metals
Method
1 To prevent the possibility of the hot metals from spitting out of the water towards the students.
2 So that it does not start to react with the oxygen or water in the air.
3 The shiny surface rapidly tarnishes. This is because the metal reacts rapidly with water and oxygen in the
environment.
4 So that the reaction can be controlled easily/so that the reaction will not be too vigorous.
5 Removing the oil exposes the metal’s surface so that the reaction starts more quickly when it is put into
the water.
6 Because the metals float and move around the surface of the water – there is less chance, if placed in the centre of
the trough, that they will reach the side. If they do reach the side, the exothermic nature of the reaction can cause
the glass trough to crack.
Results
Each row completed correctly.
Colour of solution pH of the
Colour of Observations when added to formed with universal solution
Metal metal water indicator solution formed
lithium silvery lithium floats on the surface and blue 10–12
moves around as a gas is produced
potassium silvery potassium floats on the surface violet 12–14
and moves around rapidly, orange
flames are observed; the metal
melts
sodium silvery sodium floats around on the blue/violet 12
surface and moves around rapidly;
the metal melts
Conclusions
1 Because they float on water they have a density lower than that of water.
2 Potassium.
3 Lithium.
4 Lithium, sodium, potassium.
5 When metals react they lose electrons. As you go down Group 1, the metal atoms get bigger and they lose electrons
more easily due to the lessening of the attraction between the positive nucleus and the outer shell electron.
20 6 2K + 2H O → 2KOH + H .
2 2
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers 7 All three metals react to give the relevant metal hydroxide. These are soluble in water forming aqueous hydroxide ions. 8 They are far too reactive. 9 Francium. The trend is that, as you go down the group, the metals get more reactive and francium is at the bottom of Group 1. Evaluation 1 Comment using something from the results. Extension 1 The videos are available on www.youtube.com. 2 Three suitable uses. 9.2 Halogen displacement reactions Method 1 Bromine is corrosive and iodine will stain your fingers. 2 The halogens themselves are too dangerous to use. Chlorine is a toxic gas and bromine is a very toxic and corrosive liquid. 3 Too reactive. 4 It is very rare and is radioactive. 5 Immiscible. Results For completing each row, one mark. Halogen Colour in water Colour in cyclohexane chlorine pale green/colourless pale green/colourless bromine orange/red more intense orange/red iodine brown/yellow violet Chlorine water Bromine water Iodine water colour after shaking with sodium iodide solution brown brown colour of each layer with cyclohexane upper violet violet lower brown/yellow brown/yellow colour after shaking with sodium bromide solution orange/red brown colour of each layer with cyclohexane upper intense orange/red brown lower orange/red brown/yellow colour after shaking with sodium chloride solution orange/red brown/yellow colour of each layer with upper intense orange/red violet cyclohexane lower orange/red brown/yellow Conclusions 1 Three. 2 Cl + 2NaI → 2NaCl + I 2 2 Cl + 2NaBr → 2NaCl + Br 2 2 Br + 2NaI → 2NaBr + I 2 2 3 They have been reduced; they have gained electrons. 21 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers
7 All three metals react to give the relevant metal hydroxide. These are soluble in water forming aqueous
hydroxide ions.
8 They are far too reactive.
9 Francium. The trend is that, as you go down the group, the metals get more reactive and francium is at the
bottom of Group 1.
Evaluation
1 Comment using something from the results.
Extension
1 The videos are available on www.youtube.com.
2 Three suitable uses.
9.2 Halogen displacement reactions
Method
1 Bromine is corrosive and iodine will stain your fingers.
2 The halogens themselves are too dangerous to use. Chlorine is a toxic gas and bromine is a very toxic and
corrosive liquid.
3 Too reactive.
4 It is very rare and is radioactive.
5 Immiscible.
Results
For completing each row, one mark.
Halogen Colour in water Colour in cyclohexane
chlorine pale green/colourless pale green/colourless
bromine orange/red more intense orange/red
iodine brown/yellow violet
Chlorine water Bromine water Iodine water
colour after shaking with sodium iodide solution brown brown
colour of each layer with cyclohexane upper violet violet
lower brown/yellow brown/yellow
colour after shaking with sodium bromide solution orange/red brown
colour of each layer with cyclohexane upper intense orange/red brown
lower orange/red brown/yellow
colour after shaking with sodium chloride solution orange/red brown/yellow
colour of each layer with upper intense orange/red violet
cyclohexane lower orange/red brown/yellow
Conclusions
1 Three.
2 Cl + 2NaI → 2NaCl + I
2 2
Cl + 2NaBr → 2NaCl + Br
2 2
Br + 2NaI → 2NaBr + I
2 2
3 They have been reduced; they have gained electrons.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

4 Least reactive: iodine, bromine, chlorine: most reactive. 5 Fluorine – colourless, astatine – brown/black. 6 Fluorine. It has the smallest atoms and is able to attract the incoming electron most strongly. Answers Evaluation 1 Yes, chlorine showed two displacement reactions, bromine one and iodine none. 2 The pure halogens could have been used. Extension 1 Any suitable use for each of the five halogens or their compounds. 2 Starch. Starch turns blue/black in the presence of iodine. 9.3 Using transition metal ions as catalysts Method 1 To avoid cross-contamination of the solutions. The reaction may start sooner than expected. 2 In one mole of iron(III) sulfate, Fe (SO ) , there are 2 moles of Fe 3+ (aq) ions. In one mole of the other catalysts 2 4 3 being used there is only one mole of the transition metal ion. 3 Potassium peroxodisulfate is one of the reactants so the reaction will not start until it is added. 4 To ensure a fair test. Swirling will help to mix the chemicals, increasing the rate of reaction. Results and calculations 1 Recording results to the resolution allowed by the stopwatch used. 2 Calculating rates correctly for the six experiments. 3 Suitable scale for both axes. Labelling of axes. Correct units used on axes. Graph drawn correctly. Conclusions 1 States the compounds which increased the rate, from their results. 2 States the compound which had the greatest effect on increasing the rate, from their results. 3 States any of the compounds which decreased the rate, from their results – inhibitors. 4 A catalyst must not be changed chemically at the end of the reaction. 5 A catalyst increases the rate of reaction by providing an alternative route from reactants to products with a lower activation energy. Because of the lower activation energy, more of the collisions become successful collisions, increasing the rate of the reaction. Evaluation Repeat for each catalyst. This would allow an average time to be calculated and would allow anomalous results to be identified. Some sort of white background, e.g. paper or white tile, could be used behind the boiling tube in which the reaction is occurring to make it easier to spot the blue/black colour of the indicator. Extension Ammonia gas – finely divided iron. Sulfuric acid – vanadium(V) oxide. Margarine – nickel. Nitric acid – platinum. 22 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

4 Least reactive: iodine, bromine, chlorine: most reactive.
5 Fluorine – colourless, astatine – brown/black.
Answers 6 Fluorine. It has the smallest atoms and is able to attract the incoming electron most strongly.
Evaluation
1 Yes, chlorine showed two displacement reactions, bromine one and iodine none.
2 The pure halogens could have been used.
Extension
1 Any suitable use for each of the five halogens or their compounds.
2 Starch. Starch turns blue/black in the presence of iodine.
9.3 Using transition metal ions as catalysts
Method
1 To avoid cross-contamination of the solutions. The reaction may start sooner than expected.
2 In one mole of iron(III) sulfate, Fe (SO ) , there are 2 moles of Fe 3+ (aq) ions. In one mole of the other catalysts
2 4 3
being used there is only one mole of the transition metal ion.
3 Potassium peroxodisulfate is one of the reactants so the reaction will not start until it is added.
4 To ensure a fair test. Swirling will help to mix the chemicals, increasing the rate of reaction.
Results and calculations
1 Recording results to the resolution allowed by the stopwatch used.
2 Calculating rates correctly for the six experiments.
3 Suitable scale for both axes. Labelling of axes. Correct units used on axes. Graph drawn correctly.
Conclusions
1 States the compounds which increased the rate, from their results.
2 States the compound which had the greatest effect on increasing the rate, from their results.
3 States any of the compounds which decreased the rate, from their results – inhibitors.
4 A catalyst must not be changed chemically at the end of the reaction.
5 A catalyst increases the rate of reaction by providing an alternative route from reactants to products with a lower
activation energy. Because of the lower activation energy, more of the collisions become successful collisions,
increasing the rate of the reaction.
Evaluation
Repeat for each catalyst. This would allow an average time to be calculated and would allow anomalous results to be
identified. Some sort of white background, e.g. paper or white tile, could be used behind the boiling tube in which the
reaction is occurring to make it easier to spot the blue/black colour of the indicator.
Extension
Ammonia gas – finely divided iron.
Sulfuric acid – vanadium(V) oxide.
Margarine – nickel.
Nitric acid – platinum.
22
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

10 Metals Answers 10.1 Metal displacement reactions Method 1 To ensure that there are no impurities, e.g. oxide coatings, on the surface of the metal which might hinder its reaction. 2 They are too reactive. 3 So that all sides of the metal are able to react with the solution. It may increase the rate by providing a larger surface area. 4 Some of the reactions may be slow and may need time to show any sign of reaction. Results Correct completion of each row, one mark. Metal nitrate solutions Metals Magnesium Copper Iron Tin Lead Zinc Magnesium ✓ ✓ ✓ ✓ ✓ Copper ✗ ✗ ✗ ✗ ✗ Iron ✗ ✓ ✓ ✓ ✗ Tin ✗ ✓ ✗ ✓ ✗ Lead ✗ ✓ ✗ ✗ ✗ Zinc ✗ ✓ ✓ ✓ ✓ Conclusions 1 Magnesium. 2 Copper. 3 Magnesium, zinc, iron, tin, lead, copper. 4 Mg + CuSO → MgSO + Cu 4 4 5 The more reactive metal is being oxidised because it is losing electrons. 6 Cu → Cu 2+ + 2e − 7 They are being reduced because the ions are gaining electrons. 8 Ag + + e − → Ag 9 The more reactive metal is the one which is able to lose its electrons most easily. Evaluation 1 Comment. 2 The reaction needed a longer time to occur OR there was no obvious colour change but the reaction did occur. Any suitable answer. 3 Any two suitable answers, e.g. give the reaction longer than 5 minutes before recording the results. Repeat the reaction more than once. Extension No. In halogen displacement reactions the more reactive halogen gains electrons and is reduced, whereas in metal displacement reactions the more reactive metal is oxidised and loses electrons. 23 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

10 Metals Answers
10.1 Metal displacement reactions
Method
1 To ensure that there are no impurities, e.g. oxide coatings, on the surface of the metal which might hinder its
reaction.
2 They are too reactive.
3 So that all sides of the metal are able to react with the solution. It may increase the rate by providing a larger surface
area.
4 Some of the reactions may be slow and may need time to show any sign of reaction.
Results
Correct completion of each row, one mark.
Metal nitrate solutions
Metals Magnesium Copper Iron Tin Lead Zinc
Magnesium ✓ ✓ ✓ ✓ ✓
Copper ✗ ✗ ✗ ✗ ✗
Iron ✗ ✓ ✓ ✓ ✗
Tin ✗ ✓ ✗ ✓ ✗
Lead ✗ ✓ ✗ ✗ ✗
Zinc ✗ ✓ ✓ ✓ ✓
Conclusions
1 Magnesium.
2 Copper.
3 Magnesium, zinc, iron, tin, lead, copper.
4 Mg + CuSO → MgSO + Cu
4 4
5 The more reactive metal is being oxidised because it is losing electrons.
6 Cu → Cu 2+ + 2e −
7 They are being reduced because the ions are gaining electrons.
8 Ag + + e − → Ag
9 The more reactive metal is the one which is able to lose its electrons most easily.
Evaluation
1 Comment.
2 The reaction needed a longer time to occur OR there was no obvious colour change but the reaction did occur.
Any suitable answer.
3 Any two suitable answers, e.g. give the reaction longer than 5 minutes before recording the results. Repeat the
reaction more than once.
Extension
No. In halogen displacement reactions the more reactive halogen gains electrons and is reduced, whereas in metal
displacement reactions the more reactive metal is oxidised and loses electrons.
23
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

10.2 Rusting of iron After setting up the experiment, it is best to leave the tubes for a week before results are recorded. Answers Method 1 Boiling water removes dissolved air, and oxygen, from the water. Covering the water with a layer of oil prevents air from re-dissolving into the water from the atmosphere. 2 By placing a drying agent, anhydrous calcium sulfate, in the tube which will absorb any water which is present in the tube. Results Observations recorded in the table. Tube Observation 1 water and oxygen brown/orange rust can be observed on the iron nail 2 without oxygen no rust observed 3 without water no rust observed 4 with salt solution a lot of brown/orange rust observed; more than in the tube with water and oxygen 5 with zinc no rust observed Conclusions 1 In tubes 1 and 4. 2 Because both water and oxygen/air were present. 3 Tubes 2, 3 and 5. In Tube 2 there was no oxygen/air. In Tube 3 there was no water, both need to be present for rusting to occur. In Tube 5 the iron nail was in contact with a more reactive metal which reacted with the oxygen/ air and water in preference to the iron. 4 In Tube 4. 5 Oxidation, loss of electrons. Fe → Fe 3+ + e − 6 Reduction, gain of electrons. O + 4e − → 2O 2− 2 Evaluation Suitable answer. Extension 1 Corrosion. 2 Any two more reactive metals, e.g. magnesium, calcium. 10.3 Metal reactivity Method Any three of: same mass of metal, same concentration of acid, same volume of acid, swirled each tube once, same temperature, all the metals were powders. Results Observations and times recorded correctly for each metal. 24 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

10.2 Rusting of iron
Answers After setting up the experiment, it is best to leave the tubes for a week before results are recorded.
Method
1 Boiling water removes dissolved air, and oxygen, from the water. Covering the water with a layer of oil prevents air
from re-dissolving into the water from the atmosphere.
2 By placing a drying agent, anhydrous calcium sulfate, in the tube which will absorb any water which is present in
the tube.
Results
Observations recorded in the table.
Tube Observation
1 water and oxygen brown/orange rust can be observed on the iron nail
2 without oxygen no rust observed
3 without water no rust observed
4 with salt solution a lot of brown/orange rust observed; more than in
the tube with water and oxygen
5 with zinc no rust observed
Conclusions
1 In tubes 1 and 4.
2 Because both water and oxygen/air were present.
3 Tubes 2, 3 and 5. In Tube 2 there was no oxygen/air. In Tube 3 there was no water, both need to be present for
rusting to occur. In Tube 5 the iron nail was in contact with a more reactive metal which reacted with the oxygen/
air and water in preference to the iron.
4 In Tube 4.
5 Oxidation, loss of electrons. Fe → Fe 3+ + e −
6 Reduction, gain of electrons. O + 4e − → 2O 2−
2
Evaluation
Suitable answer.
Extension
1 Corrosion.
2 Any two more reactive metals, e.g. magnesium, calcium.
10.3 Metal reactivity
Method
Any three of: same mass of metal, same concentration of acid, same volume of acid, swirled each tube once, same
temperature, all the metals were powders.
Results
Observations and times recorded correctly for each metal.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers Conclusions 1 Copper. 2 Magnesium. 3 Copper, lead, iron, zinc, magnesium. 4 Zn + 2HCl → ZnCl + H 2 2 [1] for formulae of reactants, [1] for formulae of products, [1] for correctly balanced. Evaluation It could be repeated to gain an average time for each metal. The powders used could be newly obtained/fresh, some may have had an oxide coating which slowed down their reaction. Any suitable answers. Extension 1 Any three more reactive metals, e.g. potassium, sodium, lithium, calcium. 2 Any suitable, e.g. platinum, gold, silver. 11 Air and water 11.1 The active part of the air Method 1 Oxygen is used up. 2 Copper is oxidised to copper(II) oxide. 3 The mass of copper is added to the mass of oxygen in copper(II) oxide. 4 Copper + oxygen → copper(II) oxide 2Cu(s) + O (g) → 2CuO(s) 2 Results and calculations Initial colour of the copper is pink/brown . Final colour of the copper after heating in air is black . Mass of glass tube + copper at start = 50.00 g Mass of glass tube + copper oxide after heating = 50.15 g Difference in mass = 0.15 g Initial volume of air in the syringe = 100 cm 3 Final volume of gas left in the syringe after the apparatus has cooled = 79 cm 3 Volume change = 21 cm 3 21 × 100 The percentage, %, of oxygen gas in the air = = 21% 100 What happened to the candle in the beaker when the remaining gas from the cooled syringe was squeezed gently onto it? It goes out . Conclusion 21%. Evaluation Use a more reactive metal (e.g. calcium) to react with the oxygen in the air, but note that it will increase the hazards. 25 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Conclusions Answers
1 Copper.
2 Magnesium.
3 Copper, lead, iron, zinc, magnesium.
4 Zn + 2HCl → ZnCl + H
2 2
[1] for formulae of reactants, [1] for formulae of products, [1] for correctly balanced.
Evaluation
It could be repeated to gain an average time for each metal. The powders used could be newly obtained/fresh,
some may have had an oxide coating which slowed down their reaction. Any suitable answers.
Extension
1 Any three more reactive metals, e.g. potassium, sodium, lithium, calcium.
2 Any suitable, e.g. platinum, gold, silver.
11 Air and water
11.1 The active part of the air
Method
1 Oxygen is used up.
2 Copper is oxidised to copper(II) oxide.
3 The mass of copper is added to the mass of oxygen in copper(II) oxide.
4 Copper + oxygen → copper(II) oxide
2Cu(s) + O (g) → 2CuO(s)
2
Results and calculations
Initial colour of the copper is pink/brown .
Final colour of the copper after heating in air is black .
Mass of glass tube + copper at start = 50.00 g
Mass of glass tube + copper oxide after heating = 50.15 g
Difference in mass = 0.15 g
Initial volume of air in the syringe = 100 cm 3
Final volume of gas left in the syringe after the apparatus has cooled = 79 cm 3
Volume change = 21 cm 3
21 × 100
The percentage, %, of oxygen gas in the air = = 21%
100
What happened to the candle in the beaker when the remaining gas from the cooled syringe was squeezed gently
onto it? It goes out .
Conclusion
21%.
Evaluation
Use a more reactive metal (e.g. calcium) to react with the oxygen in the air, but note that it will increase the hazards.
25
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Extension 1 Component % Answers Nitrogen 78.08 Neon 0.002 Oxygen 20.95 Helium 0.000 5 Argon 0.93 Krypton 0.000 1 Carbon dioxide 0.04 Xenon plus minute amounts of other gases 0.000 01 2 glucose + oxygen → carbon dioxide + water + energy C H O (aq) + 6O (g) → 6CO (g) + 6H O(l) + energy 6 12 6 2 2 2 3 By the fractional distillation of dry and carbon-dioxide free liquid air. 11.2 Making a fertiliser Method 1 Overflow can take place and also adds errors to the titration figure. 2 So you can see the end-point of the titration better. 3 To make sure the crystals are their actual colour. 4 It was put aside to cool. Then slow crystallisation can take place. If you evaporate to dryness, spitting occurs and the crystals will not form properly. Results and calculations This will vary. Conclusion Crystals of the fertiliser ammonium sulfate can be made by the titration of aqueous ammonia with dilute sulfuric acid. Evaluation Repeat the titration several times and take the average of the results to ensure that the titre is as accurate as possible. Extension 1 The first titration would be the rough one. The titrations would be repeated and the average of the three values within 0.1 cm 3 would be taken. The concentration of the ammonia solution can then be found using: M V M V 1 1 2 2 = M M acid alkali where: M = concentration of the acid used 1 V = average volume of acid used (cm 3 ) 1 M = number of moles of acid shown in the chemical equation acid M = concentration of the alkali used 2 V = volume of the alkali used (cm 3 ) 2 M = number of moles of alkali shown in the chemical equation. alkali 2 A ‘weak alkali’, such as ammonia, produces fewer hydroxide ions when it dissolves in water than a strong soluble base of the same concentration. It is only partially ionised. It has a lower pH than a strong base, but still above 7. ammonia + water L ammonium ions + hydroxide ions NH (g) + H O(l) L NH + (aq) + OH − (aq) 3 2 4 The ammonia molecules react with the water molecules to form ammonium ions and hydroxide ions. However, fewer ammonia molecules do this, so only a low concentration of hydroxide ions is produced. 26 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Extension
1 Component %
Answers
Nitrogen 78.08 Neon 0.002
Oxygen 20.95 Helium 0.000 5
Argon 0.93 Krypton 0.000 1
Carbon dioxide 0.04 Xenon plus minute amounts of other gases 0.000 01
2 glucose + oxygen → carbon dioxide + water + energy
C H O (aq) + 6O (g) → 6CO (g) + 6H O(l) + energy
6 12 6 2 2 2
3 By the fractional distillation of dry and carbon-dioxide free liquid air.
11.2 Making a fertiliser
Method
1 Overflow can take place and also adds errors to the titration figure.
2 So you can see the end-point of the titration better.
3 To make sure the crystals are their actual colour.
4 It was put aside to cool. Then slow crystallisation can take place. If you evaporate to dryness, spitting occurs and
the crystals will not form properly.
Results and calculations
This will vary.
Conclusion
Crystals of the fertiliser ammonium sulfate can be made by the titration of aqueous ammonia with dilute
sulfuric acid.
Evaluation
Repeat the titration several times and take the average of the results to ensure that the titre is as accurate as possible.
Extension
1 The first titration would be the rough one. The titrations would be repeated and the average of the three values
within 0.1 cm 3 would be taken. The concentration of the ammonia solution can then be found using:
M V M V
1 1 = 2 2
M M
acid alkali
where:
M = concentration of the acid used
1
V = average volume of acid used (cm 3 )
1
M = number of moles of acid shown in the chemical equation
acid
M = concentration of the alkali used
2
V = volume of the alkali used (cm 3 )
2
M = number of moles of alkali shown in the chemical equation.
alkali
2 A ‘weak alkali’, such as ammonia, produces fewer hydroxide ions when it dissolves in water than a strong soluble
base of the same concentration. It is only partially ionised. It has a lower pH than a strong base, but still above 7.
ammonia + water L ammonium ions + hydroxide ions
NH (g) + H O(l) L NH + (aq) + OH − (aq)
3 2 4
The ammonia molecules react with the water molecules to form ammonium ions and hydroxide ions. However,
fewer ammonia molecules do this, so only a low concentration of hydroxide ions is produced.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers 3 Formula is (NH ) SO . RFM = 116. 4 2 4 2 × 14 % N = × 100 = 21.21% 116 11.3 The effects of acid rain Method 1 Because it is a dilute solution of acid and so more of it is required to show the effect of time for reaction. 2 Some of the reactions will be slow. This means observations are necessary over a period of time. Results and calculations Material Observation after 2 minutes Suitability of the material for construction copper no reaction suitable limestone bubbles of a gas produced unsuitable* brick some bricks produce a few bubbles of gas. suitable* lead no reaction suitable steel small amount of bubbles of gas produced (only a few) suitable zinc bubbles of gas produced and zinc dissolves unsuitable aluminium no reaction suitable concrete a few bubbles of gas produced suitable* * Limestone and concrete in very large lumps show a very slow reaction. Conclusion 1 Copper, lead, steel, aluminium. 2 Limestone, concrete, brick, zinc. Evaluation Add some other materials, such as cast iron and plated metals. Use nitric acid as well as sulfuric acid in the acid rain. Extension 1 This is due to the fact that the materials are used in large lumps, e.g. limestone, brick, concrete, which slows the reaction down considerably. Also metals like zinc are used because they are relatively cheap when used in galvanising. 2 Reduce the amount of fossil fuels burned and turn to renewables like solar, wind and tidal power. Also reduce the amount of rotting vegetation by cutting down on food waste. 12 Sulfur 12.1 Sulfuric acid: a useful quantitative analytical chemical Method 1 Overflow can take place and also adds errors to the titration figure. 2 So you can see the end-point of the titration better. 3 To ensure a good mixing of the reactants. 4 Because the first value is carried out not knowing where the end-point will be. 27 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers
3 Formula is (NH ) SO . RFM = 116.
4 2 4
2 × 14
% N = × 100 = 21.21%
116
11.3 The effects of acid rain
Method
1 Because it is a dilute solution of acid and so more of it is required to show the effect of time for reaction.
2 Some of the reactions will be slow. This means observations are necessary over a period of time.
Results and calculations
Material Observation after 2 minutes Suitability of the material for construction
copper no reaction suitable
limestone bubbles of a gas produced unsuitable*
brick some bricks produce a few bubbles of gas. suitable*
lead no reaction suitable
steel small amount of bubbles of gas produced (only a few) suitable
zinc bubbles of gas produced and zinc dissolves unsuitable
aluminium no reaction suitable
concrete a few bubbles of gas produced suitable*
* Limestone and concrete in very large lumps show a very slow reaction.
Conclusion
1 Copper, lead, steel, aluminium.
2 Limestone, concrete, brick, zinc.
Evaluation
Add some other materials, such as cast iron and plated metals. Use nitric acid as well as sulfuric acid in the acid rain.
Extension
1 This is due to the fact that the materials are used in large lumps, e.g. limestone, brick, concrete, which slows
the reaction down considerably. Also metals like zinc are used because they are relatively cheap when used in
galvanising.
2 Reduce the amount of fossil fuels burned and turn to renewables like solar, wind and tidal power. Also reduce
the amount of rotting vegetation by cutting down on food waste.
12 Sulfur
12.1 Sulfuric acid: a useful quantitative analytical chemical
Method
1 Overflow can take place and also adds errors to the titration figure.
2 So you can see the end-point of the titration better.
3 To ensure a good mixing of the reactants.
4 Because the first value is carried out not knowing where the end-point will be.
27
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations Answers Rough 1 2 4 final burette 25.3 25.5 25.3 25.4 reading/cm 3 initial burette 0.0 0.0 0.0 0.0 reading/cm 3 volume of acid 25.3 25.5 25.3 25.4 used/cm 3 1 From this balanced equation it can be seen that: 1 mole reacts with 2 moles. average titration × molarity (25.5 + 25.3 + 25.4/3) × 0.05 = = 0.0013 moles 1000 1000 Since 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide, the number of moles of sodium hydroxide which will react with 2 × 0.0013 moles of sulfuric acid = 0.0026 moles which is found in 25 cm 3 of solution. The concentration of the alkali is: 1000 moles of alkali in 25 cm 3 × = 0.10 mol/dm 3 25 0.05 × 25.4 M × 25.0 2 = 2 1 2 M = 0.10 mol/dm 3 2 Conclusion 0.10 M. Evaluation Carry out more titrations to ensure you have an accurate average for the volume of sulfuric acid added. Extension 1 Having obtained an accurate value for the volume of sulfuric acid required to neutralise the 25 cm 3 of sodium hydroxide solution, now mix these values of volumes together. This is carried out without the phenolphthalein indicator. Then take a sample of this mixture, now sodium sulfate solution, in an evaporating basin and slowly evaporate on a water bath. Once the solution is quite concentrated (small volume), place to one side to allow the crystals to form. 2 Equation: sulfuric acid + sodium hydroxide → sodium sulfate + water H SO (aq) + 2NaOH(aq) → Na SO (aq) + H O(l) 2 4 2 4 2 Moles of sulfuric acid used in the titration: 0.0013 moles = moles of sodium sulfate produced Mass of sodium sulfate produced = 0.0013 × RFM sodium sulfate = 0.0013 × 142 = 0.1846 g 12.2 Concentrated sulfuric acid Method 1 Concentrated sulfuric acid is a very dangerous acid. Noxious fumes may be produced. 2 If the acid gets onto skin, it will badly damage the skin within seconds. 28 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations
Answers Rough 1 2 4
final burette 25.3 25.5 25.3 25.4
reading/cm 3
initial burette 0.0 0.0 0.0 0.0
reading/cm 3
volume of acid 25.3 25.5 25.3 25.4
used/cm 3
1 From this balanced equation it can be seen that: 1 mole reacts with 2 moles.
average titration × molarity (25.5 + 25.3 + 25.4/3) × 0.05
= = 0.0013 moles
1000 1000
Since 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide, the number of moles of sodium hydroxide
which will react with 2 × 0.0013 moles of sulfuric acid = 0.0026 moles which is found in 25 cm 3 of solution.
The concentration of the alkali is:
1000
moles of alkali in 25 cm 3 × = 0.10 mol/dm 3
25
0.05 × 25.4 M × 25.0
2 = 2
1 2
M = 0.10 mol/dm 3
2
Conclusion
0.10 M.
Evaluation
Carry out more titrations to ensure you have an accurate average for the volume of sulfuric acid added.
Extension
1 Having obtained an accurate value for the volume of sulfuric acid required to neutralise the 25 cm 3 of sodium
hydroxide solution, now mix these values of volumes together. This is carried out without the phenolphthalein
indicator. Then take a sample of this mixture, now sodium sulfate solution, in an evaporating basin and slowly
evaporate on a water bath. Once the solution is quite concentrated (small volume), place to one side to allow the
crystals to form.
2 Equation:
sulfuric acid + sodium hydroxide → sodium sulfate + water
H SO (aq) + 2NaOH(aq) → Na SO (aq) + H O(l)
2 4 2 4 2
Moles of sulfuric acid used in the titration: 0.0013 moles = moles of sodium sulfate produced
Mass of sodium sulfate produced = 0.0013 × RFM sodium sulfate
= 0.0013 × 142
= 0.1846 g
12.2 Concentrated sulfuric acid
Method
1 Concentrated sulfuric acid is a very dangerous acid. Noxious fumes may be produced.
2 If the acid gets onto skin, it will badly damage the skin within seconds.
28
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers Results and calculations Observation with Substance Formula concentrated sulfuric acid Inference Equation for reaction taking place sugar C H O turns brown and then black carbon left sucrose (sugar) → carbon + water 6 12 6 lots of steam and fumes elements of C H O (s) → 12C(s) +11H O(l) 12 22 11 2 produced water, H and O, removed by acid volume of substance tested quadruples – at least! gets very hot copper(II) CuSO .5H O turns from blue to white lots water of hydrated → anhydrous + water 4 2 sulfate of steam produced crystallisation is copper(II) copper(II) hydrate removed volume of substance tested sulfate sulfate shrinks a little substance is now CuSO .5H O(s) → CuSO (s) +5H O(l) anhydrous 4 2 4 2 gets very hot Conclusion Concentrated sulfuric acid is an excellent dehydrating agent. It will remove water from salt hydrates such as hydrated copper(II) sulfate or the elements of water from substances such as sugar. When this happens the reactions are very exothermic . Evaluation Test other salts such as iron(II) sulfate or other sugars and starch. Extension 1 hydrated cobalt(II) chloride → anhydrous cobalt(II) chloride + water CoCl .6H O(s) → CoCl (s) + 6H O(l) 2 2 2 2 2 ethene (C H ) 2 4 12.3 Properties of dilute sulfuric acid Dilute sulfuric acid is a strong acid. It, therefore, completely dissociates in aqueous solution. water sulfuric acid hydrogen ions + sulfate ions water H SO (aq) 2H + (aq) + SO 2– (aq) 2 4 4 The normal salts produced in the experiments in this practical are sulfates. Method 1 To trap the gas inside so it can be tested. 2 If the temperature rises, this suggests that a reaction has taken place. 3 To pass the gas through the limewater. 4 To speed up the reaction. 29 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations Answers
Observation with
Substance Formula concentrated sulfuric acid Inference Equation for reaction taking place
sugar C H O turns brown and then black carbon left sucrose (sugar) → carbon + water
6 12 6
lots of steam and fumes elements of C H O (s) → 12C(s) +11H O(l)
12 22 11 2
produced water, H and O,
removed by acid
volume of substance tested
quadruples – at least!
gets very hot
copper(II) CuSO .5H O turns from blue to white lots water of hydrated → anhydrous + water
4 2
sulfate of steam produced crystallisation is
copper(II) copper(II)
hydrate removed
volume of substance tested sulfate sulfate
shrinks a little substance is now
CuSO .5H O(s) → CuSO (s) +5H O(l)
anhydrous 4 2 4 2
gets very hot
Conclusion
Concentrated sulfuric acid is an excellent dehydrating agent. It will remove water from salt hydrates such as hydrated
copper(II) sulfate or the elements of water from substances such as sugar. When this happens the reactions are very
exothermic .
Evaluation
Test other salts such as iron(II) sulfate or other sugars and starch.
Extension
1 hydrated cobalt(II) chloride → anhydrous cobalt(II) chloride + water
CoCl .6H O(s) → CoCl (s) + 6H O(l)
2 2 2 2
2 ethene (C H )
2 4
12.3 Properties of dilute sulfuric acid
Dilute sulfuric acid is a strong acid. It, therefore, completely dissociates in aqueous solution.
sulfuric acid water hydrogen ions + sulfate ions
H SO (aq) water 2H + (aq) + SO 2– (aq)
2 4 4
The normal salts produced in the experiments in this practical are sulfates.
Method
1 To trap the gas inside so it can be tested.
2 If the temperature rises, this suggests that a reaction has taken place.
3 To pass the gas through the limewater.
4 To speed up the reaction.
29
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations Reaction with dilute Name of salt Answers Substance sulfuric acid Inference produced magnesium effervescence; ‘pops’ with reaction occurs; a salt and magnesium sulfate lighted splint hydrogen produced sodium hydroxide no observable reaction but reaction has taken place to sodium sulfate temperature rises produce a salt and water sodium carbonate effervescence; turns reaction occurs; a salt, sodium sulfate limewater cloudy white carbon dioxide and water produced copper(II) oxide no reaction unless warmed; copper(II) sulfate produced copper(II) sulfate solution goes blue along with water Conclusion Dilute sulfuric acid is a strong acid. It reacts with MAZIT metals such as magnesium to produce a salt and release hydrogen gas. When it reacts with sodium hydroxide the neutralisation reaction forms the sodium sulfate and water only. In the reaction with sodium carbonate it forms sodium sulfate and releases the gas carbon dioxide. In the final experiment with copper(II) oxide, on warming, copper(II) sulfate a blue hydrate salt is produced. These are general reactions of strong acids. Another example of a strong acid would be hydrochloric acid. Evaluation Other substances can be tested, e.g. iron, zinc, as per the reaction with magnesium. Compare with reactions of a weak acid, such as ethanoic acid. Extension 1 Normal salt – a substance formed when all the replaceable hydrogen of an acid is completely replaced by metal ions or the ammonium ion. Acid salt – a substance formed when only some of the replaceable hydrogen of an acid is replaced by metal ions or the ammonium ion. Neutralisation reaction – the process by which the acidity or alkalinity of a substance is cancelled out. MAZIT metals – the metals magnesium, aluminium zinc, iron, tin. 2 (a) with magnesium: magnesium + sulfuric acid → magnesium sulfate + hydrogen Mg(s) + H SO (aq) → MgSO (aq) + H (g) 2 4 4 2 (b) with sodium hydroxide sodium hydroxide + sulfuric acid → sodium sulfate + water 2NaOH(aq) + H SO (aq) → Na SO (aq) + 2H O(g) 2 4 2 4 2 (c) with sodium carbonate sodium carbonate + sulfuric acid → sodium sulfate + carbon dioxide + water Na CO (s) + H SO (aq) → Na SO (aq) + CO (g) + H O(g) 2 3 2 4 2 4 2 2 (d) with copper(II) oxide copper oxide + sulfuric acid → copper(II) sulfate + water CuO(s) + H SO (aq) → CuSO (aq) + H O(g) 2 4 4 2 3 (a) with magnesium Mg(s) + 2H + (aq) → Mg(s) + H (g) 2 (b) with sodium hydroxide H + (aq) + OH − (aq) → H O(l) 2 30 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations
Reaction with dilute Name of salt
Answers
Substance sulfuric acid Inference produced
magnesium effervescence; ‘pops’ with reaction occurs; a salt and magnesium sulfate
lighted splint hydrogen produced
sodium hydroxide no observable reaction but reaction has taken place to sodium sulfate
temperature rises produce a salt and water
sodium carbonate effervescence; turns reaction occurs; a salt, sodium sulfate
limewater cloudy white carbon dioxide and water
produced
copper(II) oxide no reaction unless warmed; copper(II) sulfate produced copper(II) sulfate
solution goes blue along with water
Conclusion
Dilute sulfuric acid is a strong acid. It reacts with MAZIT metals such as magnesium to produce a salt and release
hydrogen gas. When it reacts with sodium hydroxide the neutralisation reaction forms the sodium sulfate and water
only. In the reaction with sodium carbonate it forms sodium sulfate and releases the gas carbon dioxide. In the final
experiment with copper(II) oxide, on warming, copper(II) sulfate a blue hydrate salt is produced.
These are general reactions of strong acids. Another example of a strong acid would be hydrochloric acid.
Evaluation
Other substances can be tested, e.g. iron, zinc, as per the reaction with magnesium. Compare with reactions of a weak
acid, such as ethanoic acid.
Extension
1 Normal salt – a substance formed when all the replaceable hydrogen of an acid is completely replaced by metal
ions or the ammonium ion.
Acid salt – a substance formed when only some of the replaceable hydrogen of an acid is replaced by metal ions or
the ammonium ion.
Neutralisation reaction – the process by which the acidity or alkalinity of a substance is cancelled out.
MAZIT metals – the metals magnesium, aluminium zinc, iron, tin.
2 (a) with magnesium:
magnesium + sulfuric acid → magnesium sulfate + hydrogen
Mg(s) + H SO (aq) → MgSO (aq) + H (g)
2 4 4 2
(b) with sodium hydroxide
sodium hydroxide + sulfuric acid → sodium sulfate + water
2NaOH(aq) + H SO (aq) → Na SO (aq) + 2H O(g)
2 4 2 4 2
(c) with sodium carbonate
sodium carbonate + sulfuric acid → sodium sulfate + carbon dioxide + water
Na CO (s) + H SO (aq) → Na SO (aq) + CO (g) + H O(g)
2 3 2 4 2 4 2 2
(d) with copper(II) oxide
copper oxide + sulfuric acid → copper(II) sulfate + water
CuO(s) + H SO (aq) → CuSO (aq) + H O(g)
2 4 4 2
3 (a) with magnesium
Mg(s) + 2H + (aq) → Mg(s) + H (g)
2
(b) with sodium hydroxide
H + (aq) + OH − (aq) → H O(l)
2
30
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers (c) with sodium carbonate CO 2− (aq) + 2H + (aq) → CO (g) + H O(l) 3 2 2 (d) with copper(II) oxide O 2− (s) + 2H + (aq) → H O(l) 2 13 Inorganic carbon chemistry 13.1 Limestone: a useful resource Method 1 Calcium oxide is corrosive and calcium hydroxide is an irritant. 2 Alkali produced – calcium hydroxide. 3 Calcium carbonate was produced when carbon dioxide was blown through the calcium hydroxide solution. Results and calculations Procedure step Observation inference 3 white-powder coating appears on the new substance produced upon heating – calcium surface of the limestone oxide 10 indicator solution goes blue alkali produced 11 goes cloudy white carbon dioxide from breath is interacting with calcium hydroxide solution Conclusion When limestone is heated strongly it is converted to calcium oxide (quicklime). When water is added to the calcium oxide , a new substance is produced called calcium hydroxide (slaked lime) which is an alkali . When carbon dioxide is blown through the solution it forms calcium carbonate. Evaluation Repeat the experiments a number of times to show the results are reproducible. Also, compare the results with actual substances and how they behave. Extension 1 (a) calcium carbonate L calcium oxide + carbon dioxide CaCO (s) L CaO(s) + CO (g) 3 2 (b) calcium oxide + water → calcium hydroxide CaO(s) + H O(l) → Ca(OH) (s) 2 2 (c) calcium hydroxide + carbon dioxide → calcium carbonate + water Ca(OH) (aq) + CO (g) → CaCO (s) + H O(l) 2 2 3 2 2 carbonate ion (from CaCO ) + hydrogen ion (from acid soils) → carbon dioxide + water 3 CO 2− (s) + 2H + (aq) → CO (g) + H O(l) 3 2 2 3 Thermal decomposition – the breakdown of a substance under the influence of heat. Neutralisation reaction – the process in which the acidity or alkalinity of a substance is cancelled out. Flue gas desulfurisation – the process by which sulfur dioxide gas is removed from the waste gases of power stations by passing them through calcium hydroxide slurry. Alkali – a soluble base which produces OH − (aq) ions in water. 31 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers
(c) with sodium carbonate
CO 2− (aq) + 2H + (aq) → CO (g) + H O(l)
3 2 2
(d) with copper(II) oxide
O 2− (s) + 2H + (aq) → H O(l)
2
13 Inorganic carbon chemistry
13.1 Limestone: a useful resource
Method
1 Calcium oxide is corrosive and calcium hydroxide is an irritant.
2 Alkali produced – calcium hydroxide.
3 Calcium carbonate was produced when carbon dioxide was blown through the calcium hydroxide solution.
Results and calculations
Procedure step Observation inference
3 white-powder coating appears on the new substance produced upon heating – calcium
surface of the limestone oxide
10 indicator solution goes blue alkali produced
11 goes cloudy white carbon dioxide from breath is interacting with
calcium hydroxide solution
Conclusion
When limestone is heated strongly it is converted to calcium oxide (quicklime). When water is added to the calcium
oxide , a new substance is produced called calcium hydroxide (slaked lime) which is an alkali . When carbon dioxide is
blown through the solution it forms calcium carbonate.
Evaluation
Repeat the experiments a number of times to show the results are reproducible. Also, compare the results with actual
substances and how they behave.
Extension
1 (a) calcium carbonate L calcium oxide + carbon dioxide
CaCO (s) L CaO(s) + CO (g)
3 2
(b) calcium oxide + water → calcium hydroxide
CaO(s) + H O(l) → Ca(OH) (s)
2 2
(c) calcium hydroxide + carbon dioxide → calcium carbonate + water
Ca(OH) (aq) + CO (g) → CaCO (s) + H O(l)
2 2 3 2
2 carbonate ion (from CaCO ) + hydrogen ion (from acid soils) → carbon dioxide + water
3
CO 2− (s) + 2H + (aq) → CO (g) + H O(l)
3 2 2
3 Thermal decomposition – the breakdown of a substance under the influence of heat.
Neutralisation reaction – the process in which the acidity or alkalinity of a substance is cancelled out.
Flue gas desulfurisation – the process by which sulfur dioxide gas is removed from the waste gases of power
stations by passing them through calcium hydroxide slurry.
Alkali – a soluble base which produces OH − (aq) ions in water.
31
Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

13.2 Does the food we eat contain carbon? Method Answers 1 To ensure the foodstuffs decompose to carbon. 2 Carbon dioxide, CO . 2 3 The carbon in the foodstuffs reacts with oxygen in air. Results and calculations Food Effect of heat Effect of extracted gas on limewater Does the food contain carbon? bread eventually turns black turns limewater slightly cloudy white yes as the food chars rice eventually turns black turns limewater slightly cloudy white yes as the food chars cornflakes eventually turns black turns limewater slightly cloudy white yes as the food chars macaroni eventually turns black turns limewater slightly cloudy white yes as the food chars spaghetti eventually turns black turns limewater slightly cloudy white yes as the food chars Conclusion From the results of my experiments I can say that the foods I tested contain carbon . Evaluation Include other foodstuffs. At times during the experiments allow more air into the test. Extension 1 carbon + oxygen → carbon dioxide C(s) + O (g) → CO (g) 2 2 2 Aerobic respiration – this is respiration that takes place using the oxygen from the air. During the process, carbon dioxide and energy are released. 3 Hydrocarbons – these are substances that contain carbon and hydrogen only. Carbon cycle – the continuous process by which carbon is exchanged between organisms and the environment. It involves photosynthesis, respiration and decay of animals, plants and other organisms, as well as the burning of fossil fuels. Photosynthesis – the chemical process by which green plants synthesise their carbon compounds from atmospheric carbon dioxide, using light as the energy source and chlorophyll as the catalyst. 13.3 Carbon dioxide Method 1 Because one of the products is insoluble calcium sulfate. This would prevent further reaction of the acid with the limestone. 2 Carbon dioxide is not very soluble in water, otherwise it could not be collected in this way. 3 To ensure the gas does not escape from the test tube. 32 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

13.2 Does the food we eat contain carbon?
Method
Answers
1 To ensure the foodstuffs decompose to carbon.
2 Carbon dioxide, CO .
2
3 The carbon in the foodstuffs reacts with oxygen in air.
Results and calculations
Food Effect of heat Effect of extracted gas on limewater Does the food contain carbon?
bread eventually turns black turns limewater slightly cloudy white yes
as the food chars
rice eventually turns black turns limewater slightly cloudy white yes
as the food chars
cornflakes eventually turns black turns limewater slightly cloudy white yes
as the food chars
macaroni eventually turns black turns limewater slightly cloudy white yes
as the food chars
spaghetti eventually turns black turns limewater slightly cloudy white yes
as the food chars
Conclusion
From the results of my experiments I can say that the foods I tested contain carbon .
Evaluation
Include other foodstuffs. At times during the experiments allow more air into the test.
Extension
1 carbon + oxygen → carbon dioxide
C(s) + O (g) → CO (g)
2 2
2 Aerobic respiration – this is respiration that takes place using the oxygen from the air. During the process, carbon
dioxide and energy are released.
3 Hydrocarbons – these are substances that contain carbon and hydrogen only.
Carbon cycle – the continuous process by which carbon is exchanged between organisms and the environment.
It involves photosynthesis, respiration and decay of animals, plants and other organisms, as well as the burning of
fossil fuels.
Photosynthesis – the chemical process by which green plants synthesise their carbon compounds from
atmospheric carbon dioxide, using light as the energy source and chlorophyll as the catalyst.
13.3 Carbon dioxide
Method
1 Because one of the products is insoluble calcium sulfate. This would prevent further reaction of the acid with the
limestone.
2 Carbon dioxide is not very soluble in water, otherwise it could not be collected in this way.
3 To ensure the gas does not escape from the test tube.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers Results and calculations Test on the carbon dioxide Observation Inference colour colourless smell no smell what happens to the burning goes out does not support combustion splint pH of the solution should be around 5 acidic gas effect of limewater turns it cloudy white useful test for carbon dioxide burning magnesium white powder and black specks supports combustion of strongly produced burning substances burning candle flame goes out denser than air and does not support combustion density experiments gas can be poured denser than air Conclusion It has no colour. It has no smell. It will not support combustion . Its solution has a pH of 5 which makes it an acidic gas. It turns limewater cloudy white . This is used as a chemical test for carbon dioxide. It is denser than air. Evaluation Tests with the gas should be repeated to show they are reproducible. Demonstrate that sulfuric acid is not the acid to use to produce the gas efficiently. Extension 1 (a) A reaction which involves the two processes of reduction and oxidation. (b) Oxidised – magnesium, reduced – carbon dioxide. 2 An acid such as carbonic acid which produces a low concentration of H + (aq) in water solution. It is only partially ionised. 3 Fire extinguishers, carbonated drinks, refrigerants, special smoke effects (less so as this can be dangerous), heat transfer agent. 14 Organic chemistry 1 14.1 Is methane a hydrocarbon? Method 1 Limewater contains water. The cobalt chloride paper is a test for the presence of water. 2 Plastic one would melt. 3 So the glass is not cracked. 33 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations Answers
Test on the carbon dioxide Observation Inference
colour colourless
smell no smell
what happens to the burning goes out does not support combustion
splint
pH of the solution should be around 5 acidic gas
effect of limewater turns it cloudy white useful test for carbon dioxide
burning magnesium white powder and black specks supports combustion of strongly
produced burning substances
burning candle flame goes out denser than air and does not
support combustion
density experiments gas can be poured denser than air
Conclusion
It has no colour.
It has no smell.
It will not support combustion .
Its solution has a pH of 5 which makes it an acidic gas.
It turns limewater cloudy white . This is used as a chemical test for carbon dioxide.
It is denser than air.
Evaluation
Tests with the gas should be repeated to show they are reproducible. Demonstrate that sulfuric acid is not the acid to
use to produce the gas efficiently.
Extension
1 (a) A reaction which involves the two processes of reduction and oxidation.
(b) Oxidised – magnesium, reduced – carbon dioxide.
2 An acid such as carbonic acid which produces a low concentration of H + (aq) in water solution. It is only partially
ionised.
3 Fire extinguishers, carbonated drinks, refrigerants, special smoke effects (less so as this can be dangerous), heat
transfer agent.
14 Organic chemistry 1
14.1 Is methane a hydrocarbon?
Method
1 Limewater contains water. The cobalt chloride paper is a test for the presence of water.
2 Plastic one would melt.
3 So the glass is not cracked.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations 1 Turns from blue to pink. This is caused by the effect of the water produced by burning a substance that contains Answers hydrogen. 2 Turns cloudy white. This is caused by the effect of the carbon dioxide produced by burning a substance that contains carbon burning. 3 Elements present are C and H. Conclusion When the gases produced by burning methane in air are passed though the apparatus, various observations can be made. The cobalt chloride paper tests for water and turns from a blue colour to a pink colour. This shows the presence of the element hydrogen in methane. The limewater tests for carbon dioxide and goes white . This shows the presence of the element carbon in methane. Overall this shows that methane and, hence, other alkanes contain the elements carbon and hydrogen only. They are, therefore, hydrocarbons . Evaluation Repeat the experiment with other hydrocarbons such as hexane, looking for the same results. Extension 1 Complete combustion – a chemical reaction in which a substance reacts rapidly with a plentiful supply of oxygen with the production of heat and light. Hydrocarbon – a substance that contains carbon and hydrogen only. 2 (a) In complete combustion of methane, one of the products is carbon dioxide, whilst in the case of incomplete combustion carbon monoxide is produced. (b) methane + oxygen → carbon monoxide + water CH (g) + O (g) → CO(g) + 2H O(g) 4 2 2 3 Sulfur dioxide will turn acidified potassium manganate(VII) solution from purple to colourless. 14.2 Difference between alkanes and alkenes Method 1 The hydrocarbons may affect your skin. 2 Vapours are harmful. Results and calculations Shake with bromine in an Effect of a Substance Appearance Smell organic solvent lighted splint hexane colourless liquid petroleum smell no change burns cleanly hexene colourless liquid sweet petroleum smell decolourises the bromine in burns with more an organic solvent of a smoky flame Conclusion It is possible to identify an alkene using the fact that it will decolourise bromine dissolved in an organic solvent. In this reaction the bromine adds across the double bond in the alkene. The reaction is an example of an addition reaction. 34 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations
1 Turns from blue to pink. This is caused by the effect of the water produced by burning a substance that contains
Answers
hydrogen.
2 Turns cloudy white. This is caused by the effect of the carbon dioxide produced by burning a substance that
contains carbon burning.
3 Elements present are C and H.
Conclusion
When the gases produced by burning methane in air are passed though the apparatus, various observations can be made.
The cobalt chloride paper tests for water and turns from a blue colour to a pink colour. This shows the presence of the
element hydrogen in methane.
The limewater tests for carbon dioxide and goes white . This shows the presence of the element carbon in methane.
Overall this shows that methane and, hence, other alkanes contain the elements carbon and hydrogen only. They are,
therefore, hydrocarbons .
Evaluation
Repeat the experiment with other hydrocarbons such as hexane, looking for the same results.
Extension
1 Complete combustion – a chemical reaction in which a substance reacts rapidly with a plentiful supply of oxygen
with the production of heat and light.
Hydrocarbon – a substance that contains carbon and hydrogen only.
2 (a) In complete combustion of methane, one of the products is carbon dioxide, whilst in the case of incomplete
combustion carbon monoxide is produced.
(b) methane + oxygen → carbon monoxide + water
CH (g) + O (g) → CO(g) + 2H O(g)
4 2 2
3 Sulfur dioxide will turn acidified potassium manganate(VII) solution from purple to colourless.
14.2 Difference between alkanes and alkenes
Method
1 The hydrocarbons may affect your skin.
2 Vapours are harmful.
Results and calculations
Shake with bromine in an Effect of a
Substance Appearance Smell organic solvent lighted splint
hexane colourless liquid petroleum smell no change burns cleanly
hexene colourless liquid sweet petroleum smell decolourises the bromine in burns with more
an organic solvent of a smoky flame
Conclusion
It is possible to identify an alkene using the fact that it will decolourise bromine dissolved in an organic solvent. In this
reaction the bromine adds across the double bond in the alkene. The reaction is an example of an addition reaction.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

ANSWERS Evaluation Test other alkanes and alkenes. Compare, therefore, a greater selection. Extension 1 (a) Used in the manufacture of ethanol. It is formed when water (as steam) is added across the double bond in ethene. For this reaction to take place, the reactants have to be passed over a catalyst of phosphoric(V) acid (absorbed on silica pellets) at a temperature of 300 °C and pressure of 60 atmospheres. (b) Ethane is produced. This reaction is called hydrogenation. The conditions necessary for this reaction to take place are a temperature of 200 °C in the presence of a nickel or platinum catalyst. 14.3 Hydrocarbons can form isomers Method Hexane. The molecules can approach each other closely due to the ‘linear’ nature of the molecules. This will maximise the weak intermolecular forces. Results and calculations 1 Drawings (3D) and names of the five isomers of C H : 6 14 (a) Hexane H H H H H H H C C C C C C H H H H H H H H H H H H H H C C C C C C H H H H H H H (b) 2-Methylpentane H H H H H H C C C C C H H C H H H H H H (c) 3-Methylpentane H H H H H H C C C C C H H H C H H H H H (d) 2,2-Dimethylbutane H H H H C H H H C C C C H H C H H H H H 35 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Evaluation ANSWERS
Test other alkanes and alkenes. Compare, therefore, a greater selection.
Extension
1 (a) Used in the manufacture of ethanol. It is formed when water (as steam) is added across the double bond in
ethene. For this reaction to take place, the reactants have to be passed over a catalyst of phosphoric(V) acid
(absorbed on silica pellets) at a temperature of 300 °C and pressure of 60 atmospheres.
(b) Ethane is produced. This reaction is called hydrogenation. The conditions necessary for this reaction to take
place are a temperature of 200 °C in the presence of a nickel or platinum catalyst.
14.3 Hydrocarbons can form isomers
Method
Hexane. The molecules can approach each other closely due to the ‘linear’ nature of the molecules. This will maximise
the weak intermolecular forces.
Results and calculations
1 Drawings (3D) and names of the five isomers of C H :
6 14
(a) Hexane
H H H H H H
H C C C C C C H
H H H H H H
H H H H H H
H C C C
C C C H
H H H H H H
(b) 2-Methylpentane
H H H H H
H C C C C C H
H C H H H
H H H
(c) 3-Methylpentane
H H H H H
H C C C C C H
H H C H H
H H H
(d) 2,2-Dimethylbutane
H H H
H C H H
H C C C C H
H C H H
H H H
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(e) 2,3-Dimethylbutane H H H ANSWERS H H C H H C C C C H H C H H H H H Conclusion There are five isomers of hexane. There is one which has no branch chains while the other four have branched chains. The molecule with a straight chain has the highest melting and boiling points. This is the case because the forces of attraction between the molecules are at a maximum when there are no branched chains. Evaluation Try to draw the 3D structures of the molecules to see how they look in reality. Extension 1 Due to the presence of many isomers in a variety of organic molecules. 2 position of functional group Position of functional group 1 2 3 4 5 H C=CHCH CH CH 2 2 2 3 1-pentene 1 2 3 4 5 CH CH=CHCH CH 3 2 3 2-pentene 15 Organic chemistry 2 15.1 Organic structures and functional groups Method 1 (b). They have the same functional group but (b) is a bigger molecule and has a greater RMM. So the intermolecular forces will be larger. 2 Because they have the same functional group. Results and calculations 1 3D drawing: (a) Ethanoic acid (b) Propanoic acid 36 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

(e) 2,3-Dimethylbutane
H H H
ANSWERS
H H C H
H C C C C H
H C H H
H H H
Conclusion
There are five isomers of hexane. There is one which has no branch chains while the other four have branched chains.
The molecule with a straight chain has the highest melting and boiling points. This is the case because the forces of
attraction between the molecules are at a maximum when there are no branched chains.
Evaluation
Try to draw the 3D structures of the molecules to see how they look in reality.
Extension
1 Due to the presence of many isomers in a variety of organic molecules.
2 position of functional group Position of functional group
1 2 3 4 5
H 2 C=CHCH 2 CH 2 CH 3
1-pentene
1 2 3 4 5
CH 3 CH=CHCH 2 CH 3
2-pentene
15 Organic chemistry 2
15.1 Organic structures and functional groups
Method
1 (b). They have the same functional group but (b) is a bigger molecule and has a greater RMM. So the intermolecular
forces will be larger.
2 Because they have the same functional group.
Results and calculations
1 3D drawing:
(a) Ethanoic acid
(b) Propanoic acid
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ANSWERS (c) Ethanol (d) Ethyl ethanoate 2 (e) 120 o (f) 109°28’ 3 Number of Total number Number of Number elements *Relative of atoms single covalent of double Molecule present formula mass present bonds covalent bonds A ethanoic acid 3 60 8 6 1 B propanoic acid 3 74 11 9 1 C ethanol 3 46 9 8 0 D ethyl ethanoate 3 88 14 12 1 *To calculate the relative formula mass (RFM) you need to add together all the relative atomic masses (RAM) of the atoms present in a molecule: RAM: H = 1, C = 12, O = 16. Conclusion Organic molecules have a 3D structure. Their melting and boiling points depend on this 3D structure, RMM and functional groups present. The bond angles found in organic molecules are the same throughout organic chemistry. Evaluation Build a series of other molecules. Perhaps those that contain nitrogen, such as amino acids. Extension 1 (a), (b) – carboxylic acid (COOH), (c) – alcohol (OH), (d) – ester (–OR). 2 (a), (b) – carboxylic acid, (c) – alcohol, (d) – ester. 3 Ethyl propanoate. 15.2 Nylon rope trick Method 1 The chemicals involved in the experiment are harmful to skin. 2 In case of splashing. 3 To produce two separate layers. You do not want mixing. 4 To leave so that the solvents evaporate, giving nylon. 37 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

ANSWERS
(c) Ethanol
(d) Ethyl ethanoate
2 (e) 120 o
(f) 109°28’
3
Number of Total number Number of Number
elements *Relative of atoms single covalent of double
Molecule present formula mass present bonds covalent bonds
A ethanoic acid 3 60 8 6 1
B propanoic acid 3 74 11 9 1
C ethanol 3 46 9 8 0
D ethyl ethanoate 3 88 14 12 1
*To calculate the relative formula mass (RFM) you need to add together all the relative atomic masses (RAM) of the
atoms present in a molecule: RAM: H = 1, C = 12, O = 16.
Conclusion
Organic molecules have a 3D structure. Their melting and boiling points depend on this 3D structure, RMM
and functional groups present. The bond angles found in organic molecules are the same throughout organic
chemistry.
Evaluation
Build a series of other molecules. Perhaps those that contain nitrogen, such as amino acids.
Extension
1 (a), (b) – carboxylic acid (COOH), (c) – alcohol (OH), (d) – ester (–OR).
2 (a), (b) – carboxylic acid, (c) – alcohol, (d) – ester.
3 Ethyl propanoate.
15.2 Nylon rope trick
Method
1 The chemicals involved in the experiment are harmful to skin.
2 In case of splashing.
3 To produce two separate layers. You do not want mixing.
4 To leave so that the solvents evaporate, giving nylon.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations 1 Answers 40 mL 30 20 10 2 It is a white slimy looking solid. Looks quite soft. Conclusion The polymer nylon is made by the mixing of 1,6-diaminohexane and sebacoyl chloride. During this reaction the small molecule hydrogen chloride is produced (or condensed out). The name of this type of reaction is condensation polymerisation. Evaluation Increase the amount of chemicals to produce a larger sample. Use a better set of apparatus. [Photo credit © Charles D. Winters/Science Photo Library] Extension 1 Monomer – a simple molecule, such as ethene, which can be polymerised. Polymer – a substance possessing very large molecules consisting of repeated units or monomers. Polymers, therefore, have a very large relative molecular mass. H N(CH ) NHOC (CH ) COOH – this is the amide link 2 2 6 2 4 2 (a) Ethane-1,2-diol, benzene-1,4-dicarboxylic acid (b) Uses of nylon: making ropes, clothing Uses of terylene: making clothing, in mixed plastics, e.g. with PVC to form braided terylene-reinforced PVC for hosepipes. 38 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Results and calculations
1
Answers
40 mL
30
20
10
2 It is a white slimy looking solid. Looks quite soft.
Conclusion
The polymer nylon is made by the mixing of 1,6-diaminohexane and sebacoyl chloride. During this reaction the
small molecule hydrogen chloride is produced (or condensed out). The name of this type of reaction is condensation
polymerisation.
Evaluation
Increase the amount of chemicals to produce a larger sample. Use a better set of apparatus.
[Photo credit © Charles D. Winters/Science Photo Library]
Extension
1 Monomer – a simple molecule, such as ethene, which can be polymerised.
Polymer – a substance possessing very large molecules consisting of repeated units or monomers. Polymers,
therefore, have a very large relative molecular mass.
H N(CH ) NHOC (CH ) COOH – this is the amide link
2 2 6 2 4
2 (a) Ethane-1,2-diol, benzene-1,4-dicarboxylic acid
(b) Uses of nylon: making ropes, clothing
Uses of terylene: making clothing, in mixed plastics, e.g. with PVC to form braided terylene-reinforced PVC for
hosepipes.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers 15.3 Properties of dilute ethanoic acid Method 1 To ensure gas does not escape. 2 A temperature rise indicates that a chemical reaction has taken place. 3 To easily pass gas through limewater. 4 It acts as a catalyst for reaction. Results and calculations Substance Reaction with dilute ethanoic acid Inference magnesium effervescence, gas ‘pops’ with lighted splint hydrogen gas produced sodium hydroxide temperature rises chemical reaction takes place – neutralisation sodium carbonate effervescence, gas turns limewater cloudy white carbon dioxide gas produced ethanol sweet smelling substance produce an ester was formed Conclusion Dilute ethanoic acid is a weak acid. It reacts with MAZIT metals such as magnesium to produce a salt and releases hydrogen gas. When it reacts with sodium hydroxide, the neutralisation reaction forms sodium ethanoate and water only. In the reaction with sodium carbonate, it forms sodium ethanoate and releases the gas carbon dioxide . In the final experiment with ethanol, ethyl ethanoate, an ester, is produced. Evaluation Test the acid with other MAZIT metals. Compare reactions with those that take place with propanoic acid, another member of the carboxylic acids. Extension 1 (a) with magnesium ethanoic acid + magnesium → magnesium ethanoate + water 2CH COOH(aq) + Mg(s) → (CH COO) Mg(aq) + H (g) 3 3 2 2 (b) with sodium hydroxide ethanoic acid + sodium hydroxide → sodium ethanoate + water CH COOH(aq) + NaOH(aq) → CH COONa(aq) + H O(l) 3 3 2 (c) with sodium carbonate ethanoic acid + sodium carbonate → sodium ethanoate + carbon dioxide + water 2CH COOH(aq) + Na CO (aq) → 2CH COONa(aq) + CO (g) + H O(l) 3 2 3 3 2 2 (d) with ethanol ethanoic acid + ethanol ← ethyl ethanoate + water CH COOH(l) + C H OH(l) ← CH COOC H (aq) + H O(l) 3 2 5 3 2 5 2 2 (a) with magnesium Mg(s) + 2H + (aq) → Mg 2+ (aq) + H (g) 2 (b) with sodium hydroxide H + (aq) + OH − (aq) → H O(l) 2 (c) with sodium carbonate CO 2− (aq) + 2H + (aq) → CO (g) + H O(l) 3 2 2 3 Weak acid – an acid which produces a low concentration of H + (aq) in water solution, for example ethanoic acid. It is only partially ionised. Neutralisation reaction – the process in which the acidity or alkalinity of a substance is cancelled out. MAZIT metals – the metals magnesium, aluminium, zinc, iron, tin Homologous series – a series of compounds in which each member differs from the next by a specific number and kind of atom. These compounds have the same general formula and similar properties. 39 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

15.3 Properties of dilute ethanoic acid Answers
Method
1 To ensure gas does not escape.
2 A temperature rise indicates that a chemical reaction has taken place.
3 To easily pass gas through limewater.
4 It acts as a catalyst for reaction.
Results and calculations
Substance Reaction with dilute ethanoic acid Inference
magnesium effervescence, gas ‘pops’ with lighted splint hydrogen gas produced
sodium hydroxide temperature rises chemical reaction takes place – neutralisation
sodium carbonate effervescence, gas turns limewater cloudy white carbon dioxide gas produced
ethanol sweet smelling substance produce an ester was formed
Conclusion
Dilute ethanoic acid is a weak acid. It reacts with MAZIT metals such as magnesium to produce a salt and releases
hydrogen gas. When it reacts with sodium hydroxide, the neutralisation reaction forms sodium ethanoate and water
only. In the reaction with sodium carbonate, it forms sodium ethanoate and releases the gas carbon dioxide . In the final
experiment with ethanol, ethyl ethanoate, an ester, is produced.
Evaluation
Test the acid with other MAZIT metals. Compare reactions with those that take place with propanoic acid, another
member of the carboxylic acids.
Extension
1 (a) with magnesium
ethanoic acid + magnesium → magnesium ethanoate + water
2CH COOH(aq) + Mg(s) → (CH COO) Mg(aq) + H (g)
3 3 2 2
(b) with sodium hydroxide
ethanoic acid + sodium hydroxide → sodium ethanoate + water
CH COOH(aq) + NaOH(aq) → CH COONa(aq) + H O(l)
3 3 2
(c) with sodium carbonate
ethanoic acid + sodium carbonate → sodium ethanoate + carbon dioxide + water
2CH COOH(aq) + Na CO (aq) → 2CH COONa(aq) + CO (g) + H O(l)
3 2 3 3 2 2
(d) with ethanol
ethanoic acid + ethanol ← ethyl ethanoate + water
CH COOH(l) + C H OH(l) ← CH COOC H (aq) + H O(l)
3 2 5 3 2 5 2
2 (a) with magnesium Mg(s) + 2H + (aq) → Mg 2+ (aq) + H (g)
2
(b) with sodium hydroxide H + (aq) + OH − (aq) → H O(l)
2
(c) with sodium carbonate CO 2− (aq) + 2H + (aq) → CO (g) + H O(l)
3 2 2
3 Weak acid – an acid which produces a low concentration of H + (aq) in water solution, for example ethanoic acid. It
is only partially ionised.
Neutralisation reaction – the process in which the acidity or alkalinity of a substance is cancelled out.
MAZIT metals – the metals magnesium, aluminium, zinc, iron, tin
Homologous series – a series of compounds in which each member differs from the next by a specific number
and kind of atom. These compounds have the same general formula and similar properties.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

16 Experimental chemistry Answers 16.1 Missing labels from reagent bottles: what a problem! Method a – test for chloride, b – test for carbonate, c – test for sulfate. Results and calculations Unknown solution Test Observation Inference A (sodium carbonate) a effervescence not a chloride but could be a carbonate colourless solution b effervescence could be a carbonate colourless solution c colourless solution it is not a sulfate B (sodium chloride) a white precipitate it is a chloride b colourless solution not a carbonate c colourless solution not a sulfate C (sodium sulfate) a colourless solution not a chloride b colourless solution not a carbonate c white precipitate it is a sulfate Conclusion Solution A is sodium carbonate . Solution B is sodium chloride . Solution C is sodium sulfate . Evaluation Leave the chloride precipitate in light – it should go grey. Add a little acid to the sulfate to check the precipitate does not dissolve in an acid. (If it had, this would have shown it to be a sulfite.) Extension 1 A – CO 2− (aq) + 2H + (aq) → CO (g) + H O(l) 3 2 2 B – Cl − (aq) + Ag + (aq) → AgCl(s) C – SO 2− (aq) + Ba 2+ (aq) → BaSO (s) 4 4 2 Carry out a series of flame tests to identify the metal ions present: Potassium – lilac Copper – green Calcium – brick red 3 Sulfite present. 40 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

16 Experimental chemistry
Answers
16.1 Missing labels from reagent bottles: what a
problem!
Method
a – test for chloride, b – test for carbonate, c – test for sulfate.
Results and calculations
Unknown solution Test Observation Inference
A (sodium carbonate) a effervescence not a chloride but
could be a carbonate
colourless solution
b effervescence could be a carbonate
colourless solution
c colourless solution it is not a sulfate
B (sodium chloride) a white precipitate it is a chloride
b colourless solution not a carbonate
c colourless solution not a sulfate
C (sodium sulfate) a colourless solution not a chloride
b colourless solution not a carbonate
c white precipitate it is a sulfate
Conclusion
Solution A is sodium carbonate .
Solution B is sodium chloride .
Solution C is sodium sulfate .
Evaluation
Leave the chloride precipitate in light – it should go grey.
Add a little acid to the sulfate to check the precipitate does not dissolve in an acid. (If it had, this would have shown
it to be a sulfite.)
Extension
1 A – CO 2− (aq) + 2H + (aq) → CO (g) + H O(l)
3 2 2
B – Cl − (aq) + Ag + (aq) → AgCl(s)
C – SO 2− (aq) + Ba 2+ (aq) → BaSO (s)
4 4
2 Carry out a series of flame tests to identify the metal ions present:
Potassium – lilac
Copper – green
Calcium – brick red
3 Sulfite present.
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers 16.2 Using flame colours to identify unknown metal ions Method 1 Hottest part of flame – cleans best. 2 So that we take even more care about its use. The acid is corrosive and should be handled very carefully. Eye protection on all the time! 3 They could spit and crackle, and be thrown in all directions. Results and calculations Solution Flame colour Metal ion present A brick red calcium B blue/white (but could just be pale blue) lead C green (turquoise) copper D lilac (or pale purple) potassium E yellow with little sparks iron(III) F pale green barium G intense yellow sodium Conclusion Metal ions can be identified by carrying out flame tests. These tests are straightforward in nature. Evaluation Take all the class experimental results and compare all the results. This will give a better overall set of results for the class. Extension Carry out experiments using aqueous ammonia and/or sodium hydroxide solutions to identify the metal ions. 16.3 How pure is your water supply? Method 1 Looking for slight changes after each drop. 2 To see whether any precipitates dissolve. Results and calculations Appearance of Effect of the addition of Effect of the addition of dilute sodium sample at start dilute hydrochloric acid hydroxide to excess A colourless solution white precipitate which dissolves in excess B no change to the pale blue pale blue precipitate which does not solution dissolve in excess C effervescence colourless solution D no change to the pale green pale green precipitate which does not solution dissolve in excess E colourless solution white precipitate which does dissolve in excess 41 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

16.2 Using flame colours to identify unknown metal ions Answers
Method
1 Hottest part of flame – cleans best.
2 So that we take even more care about its use. The acid is corrosive and should be handled very carefully. Eye
protection on all the time!
3 They could spit and crackle, and be thrown in all directions.
Results and calculations
Solution Flame colour Metal ion present
A brick red calcium
B blue/white (but could just be pale blue) lead
C green (turquoise) copper
D lilac (or pale purple) potassium
E yellow with little sparks iron(III)
F pale green barium
G intense yellow sodium
Conclusion
Metal ions can be identified by carrying out flame tests. These tests are straightforward in nature.
Evaluation
Take all the class experimental results and compare all the results. This will give a better overall set of results for
the class.
Extension
Carry out experiments using aqueous ammonia and/or sodium hydroxide solutions to identify the metal ions.
16.3 How pure is your water supply?
Method
1 Looking for slight changes after each drop.
2 To see whether any precipitates dissolve.
Results and calculations
Appearance of Effect of the addition of Effect of the addition of dilute sodium
sample at start dilute hydrochloric acid hydroxide to excess
A colourless solution white precipitate which dissolves in excess
B no change to the pale blue pale blue precipitate which does not
solution dissolve in excess
C effervescence colourless solution
D no change to the pale green pale green precipitate which does not
solution dissolve in excess
E colourless solution white precipitate which does dissolve in
excess
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Conclusion Water samples may be tested for the presence of metal ions by adding dilute hydrochloric acid. If a carbonate is present Answers then bubbles of the gas carbon dioxide will be seen. The samples may also be tested for the presence of metal ions which may be harmful using sodium hydroxide . White or a coloured precipitate may be produced, from which an indication of the metal ion present can be obtained. If excess of dilute sodium hydroxide is used then this will also help in the identification of specific metal ions. Evaluation Include test for carbon dioxide gas. Also use flame test to compare results. Extension 1 Slightly different results can be obtained from those tests with sodium hydroxide. For example, in the case of iron a dirty green precipitate is produced. 2 Bubble the gas through limewater. If it turns cloudy white it is carbon dioxide. Practical test past exam questions 1 (c) table of results initial readings completed correctly [1] final readings completed correctly [1] all readings to 1 decimal place [1] differences completed correctly [1] readings comparable to supervisor’s [2] [6] (d) pink [1] to colourless [1] (not clear) [2] (e) neutralisation/exothermic [1] (f) (i) C/3 smallest B/2 largest [1] one correct = 1 [1] (ii) order is C/3 A/1 B/2 [2] one correct = 1 [2] (g) experiment 2 is twice the volume of experiment 1 or converse [1] (h) twice value from table result for experiment 3 [1] cm 3 [1] [2] (i) use a pipette/burette [1] (j) effect: none [1] reason: no change in concentration/temperature has no effect on quantities or moles/only affects speed [1] [2] (k) any correct method that would work – precise details not needed reagents [1] method [1] result [1] [3] using same method with different acids = 0 e.g. to sodium hydroxide add named acid measure temperature change largest change = strongest/more concentrated solution e.g. to sodium hydroxide add named (excess) metal salt solution filter precipitate largest mass = strongest/more concentrated solution [Total: 21] 2 (a) green [1] (b) green [1] (precipitate) turns brown/rusty at top [1] [2] (c) effervescence/fizz/bubbles [1] glowing splint [1] glows brighter/relights/burns [1] [3] (d) green precipitate [1] [1] (e) no reaction/change [1] [1] (f) white [1] precipitate [1] [2] (g) blue [1] [1] 42 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Conclusion
Water samples may be tested for the presence of metal ions by adding dilute hydrochloric acid. If a carbonate is present
Answers
then bubbles of the gas carbon dioxide will be seen. The samples may also be tested for the presence of metal ions
which may be harmful using sodium hydroxide . White or a coloured precipitate may be produced, from which an
indication of the metal ion present can be obtained. If excess of dilute sodium hydroxide is used then this will also help
in the identification of specific metal ions.
Evaluation
Include test for carbon dioxide gas. Also use flame test to compare results.
Extension
1 Slightly different results can be obtained from those tests with sodium hydroxide. For example, in the case of iron a
dirty green precipitate is produced.
2 Bubble the gas through limewater. If it turns cloudy white it is carbon dioxide.
Practical test past exam questions
1 (c) table of results
initial readings completed correctly [1]
final readings completed correctly [1] all readings to 1 decimal place [1]
differences completed correctly [1]
readings comparable to supervisor’s [2] [6]
(d) pink [1] to colourless [1] (not clear) [2]
(e) neutralisation/exothermic [1]
(f) (i) C/3 smallest B/2 largest [1] one correct = 1 [1]
(ii) order is C/3 A/1 B/2 [2] one correct = 1 [2]
(g) experiment 2 is twice the volume of experiment 1 or converse [1]
(h) twice value from table result for experiment 3 [1] cm 3 [1] [2]
(i) use a pipette/burette [1]
(j) effect: none [1]
reason: no change in concentration/temperature has no effect on quantities or moles/only
affects speed [1] [2]
(k) any correct method that would work – precise details not needed
reagents [1] method [1] result [1] [3]
using same method with different acids = 0
e.g. to sodium hydroxide add named acid
measure temperature change
largest change = strongest/more concentrated solution
e.g. to sodium hydroxide add named (excess) metal salt solution
filter precipitate
largest mass = strongest/more concentrated solution
[Total: 21]
2 (a) green [1]
(b) green [1] (precipitate)
turns brown/rusty at top [1] [2]
(c) effervescence/fizz/bubbles [1]
glowing splint [1] glows brighter/relights/burns [1] [3]
(d) green precipitate [1] [1]
(e) no reaction/change [1] [1]
(f) white [1] precipitate [1] [2]
(g) blue [1] [1]
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers (h) green/white/blue [1] according to supervisor’s precipitate [1] turns green [1] [3] (i) fizz/bubbles/ammonia [1] litmus/indicator paper blue/purple [1] [2] (j) iron [1] (II) [1] sulfate [1] [3] (k) transition metal nitrate [2] 3 (c) Table of results for Experiments 1–4 mass of solids correctly recorded [1] initial and maximum temperature boxes correctly completed [1] temperature differences correct [1] temperature changes increasing [1] [4] (d) results for Experiment 5 initial and final temperatures completed [1] temperature change completed correctly and shown as negative [1] [2] (e) appropriate scale on y -axis which uses at least half of the grid [1] all points correctly plotted [2], −1 for any incorrect (ignore origin) best fit straight line graph drawn with a ruler [1] [4] (f) (i) value from graph [1] extrapolation shown clearly [1] [2] (ii) value from graph [1] shown clearly [1] [2] (g) endothermic [1] (h) lower temperature change [1] greater volume/more acid [1] [2] note: halved = 2 marks (i) room temperature/initial temperature from table [1] reaction: finished [1] [2] (j) advantage, e.g. comparability of results/fair test [1] (ignore reference to accuracy or reliability) disadvantage, e.g. reaction not finished/temperature still changing [1] [2] 4 (a) table of results for Experiment 1 initial final and difference volume boxes completed correctly [1] readings to 1 decimal place [1] comparable to supervisor’s [1] ±2 cm 3 [3] (b) table of results for Experiment 2 initial and final boxes completed correctly [1] difference box completed correctly [1] comparable to supervisor’s [1] ±5 cm 3 [3] (c) colourless [1] pink/magenta [1] [2] (d) neutralisation/acid–base [1] [1] (e) 2 × volume for Experiment 1 from table [1] cm 3 [1] [2] (allow 1 mark for double the volume) (f) (i) reacts with [1] neutralises the acid [1] less sodium hydroxide needed [1] [2] (ii) volume in (e) – volume added in Experiment 2, [1] e.g. 50–17 correct value, [2] e.g. 33 [2] (iii) estimate based on (ii) answer to (ii)/3 divided into 50 × 0.1, e.g. 0.5 (allow 1 mark for 0.45–0.6 g) [1] (g) no effect [1] reason – reaction not affected by temperature/volumes/concentrations the same [1] [2] 43 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Answers
(h) green/white/blue [1] according to supervisor’s precipitate [1]
turns green [1] [3]
(i) fizz/bubbles/ammonia [1]
litmus/indicator paper blue/purple [1] [2]
(j) iron [1] (II) [1]
sulfate [1] [3]
(k) transition metal
nitrate [2]
3 (c) Table of results for Experiments 1–4
mass of solids correctly recorded [1]
initial and maximum temperature boxes correctly completed [1]
temperature differences correct [1]
temperature changes increasing [1] [4]
(d) results for Experiment 5
initial and final temperatures completed [1]
temperature change completed correctly and shown as negative [1] [2]
(e) appropriate scale on y -axis which uses at least half of the grid [1]
all points correctly plotted [2], −1 for any incorrect (ignore origin)
best fit straight line graph drawn with a ruler [1] [4]
(f) (i) value from graph [1] extrapolation shown clearly [1] [2]
(ii) value from graph [1]
shown clearly [1] [2]
(g) endothermic [1]
(h) lower temperature change [1]
greater volume/more acid [1] [2]
note: halved = 2 marks
(i) room temperature/initial temperature from table [1]
reaction: finished [1] [2]
(j) advantage, e.g. comparability of results/fair test [1]
(ignore reference to accuracy or reliability)
disadvantage, e.g. reaction not finished/temperature still changing [1] [2]
4 (a) table of results for Experiment 1
initial final and difference volume boxes completed correctly [1]
readings to 1 decimal place [1]
comparable to supervisor’s [1] ±2 cm 3 [3]
(b) table of results for Experiment 2
initial and final boxes completed correctly [1]
difference box completed correctly [1]
comparable to supervisor’s [1] ±5 cm 3 [3]
(c) colourless [1] pink/magenta [1] [2]
(d) neutralisation/acid–base [1] [1]
(e) 2 × volume for Experiment 1 from table [1] cm 3 [1] [2]
(allow 1 mark for double the volume)
(f) (i) reacts with [1] neutralises the acid [1]
less sodium hydroxide needed [1] [2]
(ii) volume in (e) – volume added in Experiment 2, [1] e.g. 50–17
correct value, [2] e.g. 33 [2]
(iii) estimate based on (ii) answer to (ii)/3 divided into 50 × 0.1, e.g. 0.5
(allow 1 mark for 0.45–0.6 g) [1]
(g) no effect [1]
reason – reaction not affected by temperature/volumes/concentrations the same [1] [2]
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

(h) (i) more accurate [1] than a measuring cylinder [1] (not more accurate than a burette = 0) [2] (ii) no effect/advantage [1] Answers not measuring temperature changes/no temperature difference [1] [2] Alternative to Practical past exam questions 1 (a) (i) (gas) syringe [1] (ii) arrow shown under copper [1] (b) spatula [1] (c) black [1] (d) to return to room or initial temperature [1] correct volume of gas [1] [2] 2 (a) pestle [1], mortar [1] [2] (b) stir/mix/shake or heat/boil [1] (c) diagram showing funnel [1] indication of filter paper [1] (labels are not necessary) [2] (d) by heat (evaporation) [1] to crystallising point (or description of this process) [1] in fume cupboard [1] [2] (e) melting point or chromatography [1] 3 (a) (i) silver/grey metal [1] (ii) white [1] (b) oxygen [1] (c) to let air/oxygen enter or to make sure all magnesium reacted [1] (d) 2 from: error in weighing loss of magnesium oxide some magnesium unreacted [2] 4 (a) any of: trough, tub, beaker, bowl, water bath, basin [1] (b) arrow to space in boiling tube above liquid in water trough [1] (c) any of: to absorb, contain, store, trap, hold the alkane/liquid [1] (d) one of: gives large surface area, acts as catalyst, speeds up reaction [1] (e) test with bromine (water or in organic solvent) [1] Result: goes colourless [1] [2] 44 Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

(h) (i) more accurate [1] than a measuring cylinder [1]
(not more accurate than a burette = 0) [2]
Answers (ii) no effect/advantage [1]
not measuring temperature changes/no temperature difference [1] [2]
Alternative to Practical past exam questions
1 (a) (i) (gas) syringe [1]
(ii) arrow shown under copper [1]
(b) spatula [1]
(c) black [1]
(d) to return to room or initial temperature [1]
correct volume of gas [1] [2]
2 (a) pestle [1], mortar [1] [2]
(b) stir/mix/shake or heat/boil [1]
(c) diagram showing funnel [1]
indication of filter paper [1] (labels are not necessary) [2]
(d) by heat (evaporation) [1]
to crystallising point (or description of this process) [1]
in fume cupboard [1] [2]
(e) melting point or chromatography [1]
3 (a) (i) silver/grey metal [1]
(ii) white [1]
(b) oxygen [1]
(c) to let air/oxygen enter or to make sure all magnesium reacted [1]
(d) 2 from: error in weighing
loss of magnesium oxide
some magnesium unreacted [2]
4 (a) any of: trough, tub, beaker, bowl, water bath, basin [1]
(b) arrow to space in boiling tube above liquid in water trough [1]
(c) any of: to absorb, contain, store, trap, hold the alkane/liquid [1]
(d) one of: gives large surface area, acts as catalyst, speeds up reaction [1]
(e) test with bromine (water or in organic solvent) [1]
Result: goes colourless [1] [2]
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Cambridge IGCSE Chemistry Laboratory Practical Book © Doug Wilford & Bryan Earl 2015

Cambridge IGCSE Chemistry Laboratory Practical Book Answer