HKDSE Chemistry Bridging Programe 1C

© 2009 Aristo Educational Press Ltd.14/F Lok's Industrial Building,204 Tsat Tsz Mui Road,North Point,Hong Kong.Tel.: 2811 2908Fax: 2565 6626Website: http://www.aristo.com.hk

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, ortransmitted in any form or by any means,electronic, mechanical, photo-copying, recording or otherwise, without the prior permission of Aristo Educational Press Ltd.

First published July, 2009

Chapter 11 Reactivity of metals 13

11.1 Different reactivities of metals 13

11.2 Comparing reactivity of common metals 13

11.3 The metal reactivity series 20

11.4 Chemical equations 21

11.5 Metal reactivity series and the tendency of metals to form positive ions 26

11.6 Displacement reactions of metals in aqueous solution 27

11.7 Ionic equations 28

11.8 Extraction of metals from their ores 31

Key terms 34Summary 35

Part III Metals

Chapter 10 Occurrence and extraction of metals 1

10.1 Uses of metals in our daily lives 1

10.2 Uses related to properties of metals 1

10.3 Occurrence of metals in nature 4

10.4 Extraction of metals from their ores 4

10.5 Discovery of metals 8

10.6 Conserving metals 10

Key terms 11Summary 12

Chapter 13 Corrosion of metals and their protection 61

13.1 Corrosion of metals 61

13.2 Rusting 61

13.3 Factors that speed up rusting 63

13.4 To observe rusting using rust indicator 64

13.5 Protecting iron from rusting 66

13.6 Socio-economic implications of rusting 70

13.7 Corrosion resistance of aluminium 70

Key terms 71

Summary 72

Chapter 12 Reacting masses 37

12.1 The mole concept in general 37

12.2 Percentage by mass of an element in a compound 42

12.3 Chemical formulae of compounds 44

12.4 Determination of empirical formulae 47

12.5 Determination of molecular formulae 49

12.6 Calculations based on equations 53

Key terms 58

Summary 59

1

Chapter 10 Occurrence and extraction of metals

Metals are very useful to us.

Metals have characteristic physical properties which make

them different from other materials such as wood, rock, glass

and plastics.(

)

10.1 Uses of metals in our daily lives 10.1

Table 10.1 lists the uses of some common metals.10.1

10.2 Uses related to properties of metals 10.2

electrical wires

Metal Reasons for use

Iron construction, transport

hard, strong, malleable and ductile;cheap

magnet magnetic

Copper excellent conductor of electricity,very ductile, corrosion resistant

water pipes non-poisonous, strong, malleableand ductile, corrosion resistant

cooking utensils excellent conductor of heat, non-poisonous, strong, malleable,corrosion resistant

overhead power cables

Aluminium very good conductor of electricity,low density, ductile; cheaper thancopper

N1

Uses

Note 1The world uses 9times more iron thanall the other metalsput together.

N2 &N3

Note 2Aluminium weighs only 34% as the same volume of iron.

(b) Aluminium cables are much lighter thancopper cables, thus the pylons supportingthem can be less sturdily built, with asaving in cost.

(c) Aluminium is notused in plumbingbecause it is difficultto weld aluminiumpipes by ordinarywelding.

Note 3(a) Aluminium is more than twice as conductive of electricity as

the same mass of copper.

light but strong, corrosion resistantaircraft body

jewellery, coinsSilver attractive silvery colour, corrosionresistant, malleable and ductile,moderately soft

2

Part III Metals

Aluminium saucepans, kitchen foil

very good conductor of heat, non-poisonous, corrosion resistant, verymalleable

soft drink cans non-poisonous, low density (henceconvenient to carry), corrosionresistant, very malleable

( )

window frames corrosion resistant, strong

N4

N4

supersonic aircraft

Titanium light but very strong, very corrosionresistant, malleable

jewellery, coins

Gold attractive golden yellow colour,extremely corrosion resistant (soalways shiny), most malleable andductile, quite soft when pure

( )

shield for heat, sunlight and radiation

excellent reflective quality

thermometersMercury liquid at room temperature, expandson heating, does not wet walls oftubes

N5

N6

Metal Uses Reasons for use

Note 4(a) The protective oxide

layer can be thickenedby anodizing.

(b) The surface oxide layercan be dyed to givevar ious attractivecolours.

Note 5(a) Titanium weighs only

60% as the same volumeof steel. Both are equallystrong.

(b) At high temperatures,titanium remains strongand its shape is notdistor ted. (Whenspeeding through air, asupersonic aircraftbecomes heated up byfriction with air.)

Note 6Today, tiny drops or thin sheets of gold can befound in nearly every telephone, computer,late-model car, automatic teller machine,video camera and liquid-crystal display.

electrical and electronic components best conductor of electricity,malleable and ductile, corrosionresistant

Table 10.1 Uses of some metals and their properties related to the uses. N7

Note 7Refer to ‘Supplementary information: Alloys’ in the Teacher’s Guide.

3


Class practice 10.1 10.1

The following table gives information on some metals (represented

by letters A to E).( A E )

Refer to the above information to answer the questions below:

(a) Which is the most widely used metal? Why?

(b) Which two metals are widely used in jewellery? Why?

(c) Which metal is the best conductor of heat? Give a reason why

we do not use this metal to make cooking utensils.

(d) Which metal is used to make aircraft bodies? Give a reason for

the choice.

(e) Which two metals are the best conductors of electricity? Give a

reason why one of these is used much more than the other.

(a)

(b)

(c)

(d)

(e)

MetalRelative

priceDensity

(g cm–3)

Meltingpoint

(°C)

Rank order inthermal conductivity

(1 = best)

(1 )

Rank order inelectrical conductivity

(1 = best)

(1 )

Other characteristics

Amost

expensive 19.3 1063 3 3

attractive golden yellow colour,quite soft, chemically inert

Bcheapest

7.9 1540 8 8 hard and strong

Cvery

expensive 10.5 961 1 1

attractive silvery colour, quitesoft, corrosion resistant

Dvery

expensive 4.5 1933 11 11strong, very corrosion resistant

Emedium

9.2 420 6 6poisonous, quite corrosionresistant

A10.1(a) B. It is hard and strong, yet very cheap.(b) A and C. They have a beautiful shiny appearance. A is chemically inert, and C is also corrosion resistant. They are both very expensive metals.(c) C. It is very expensive.(d) D. It is light (with a low density) but strong and corrosion resistant.(e) A and C. C is more widely used because it is cheaper.

4

Part III Metals

10.3 Occurrence of metals in nature 10.3

Most metals combine with other elements as compounds,

called ores.

Four important ores found in nature.

(a) Bauxite — the main ore of aluminium. It is mostly

aluminium oxide Al2O3.

(b) Copper pyrite — the main ore of copper. It is mostly

CuFeS2.

(c) Haematite — the main ore of iron. It is mostly iron(III)

oxide Fe2O3.

(d) Galena — the main ore of lead. It is mostly lead(II)

sulphide PbS.

A few metals (e.g. gold and platinum) are so unreactive

that they occur in the Earth in free state, as pure metal.

(a)

Al2O3

(b)

CuFeS2

(c)

(III) Fe2O3

(d)

(II) PbS

( )

10.4 Extraction of metals from their ores 10.4

Extraction methods

The process of separating a metal from its ore is called

extraction. There are three basic methods to extract a metal

from its ore:

1. heating the metal ore alone

2. heating the metal ore with carbon

3. electrolysis

Heating the metal ore alone

Some unreactive metals, mercury and silver, for example, can

be extracted from their ores by this method.

Mercury(II) oxide and silver oxide decompose on heating

to produce the metal and oxygen. We can use a glowing splint

to test for any oxygen evolved.

mercury(II) oxide mercury + oxygenred silvery

silver oxide silver + oxygenbrownish black silvery

1.

2.

3.

(II)

(II) +

+

N8

Note 8(a) Only those minerals which are worth mining to extract metals are called metal

ores. Thus a mineral may not necessarily be an ore.

(b) Most ores are mixtures of substances. For example, bauxite is about 75%Al2O3, 25% Fe2O3 (hence the brown colour); haematite is 85% or more Fe2O3;galena is only about 15% PbS.

N9Note 9Pure aluminium oxide is white. The bauxite ore shown is brownbecause it contains appreciable amounts of iron(III) oxide.

N10

Note 10(a) If a gas relights a glowing splint, we usually

conclude that the gas is oxygen. However,another possibility would be dinitrogen oxide N2O.

(b) In comparison, a burning splint should be used totest for hydrogen in the ‘pop’ sound test.

5


Heating the metal ore with carbon (carbon reduction)

Extraction of iron from its ore — haematite (iron(III)oxide)

Iron can be extracted from its ore haematite. At high

temperatures, carbon can react with iron(III) oxide in

haematite. In the process, carbon removes the oxygen from

iron(III) oxide to form iron.

iron(III) oxide + carbon iron + carbon dioxide

Extraction of lead from its ore — galena (lead(II)sulphide)

Extraction can be divided into two stages:

The first stage is to heat the ore galena (PbS) in air to form

lead(II) oxide:

lead(II) sulphide + oxygen lead(II) oxide + sulphur dioxide

The lead(II) oxide formed in the reaction is then heated

with carbon and is changed to lead.

lead(II) oxide + carbon lead + carbon dioxide

Extraction of metals by electrolysis

Extraction of metals by electrolysis is the most expensive

method but it is the only effective method for some reactive

metals.

During electrolysis, purified ore is melted; then electricity is

passed through the ore to separate the metal from it.

Aluminium is a reactive metal, and can be separated from its

ore (bauxite) only by electrolysis (Figure 10.1). The process is as

follows:

electricity

aluminium oxide aluminium + oxygen(from bauxite)

N11

N12

Note 11There is a competition for oxygen. At high temperatures, carbon holds on to oxygen more tightly than leaddoes, hence carbon can take oxygen away from lead(II) oxide. Besides carbon, carbon monoxide andhydrogen can also reduce oxides of metals low in the reactivity series. For example,

CuO(s) + CO(g) Cu(s) + CO2(g);PbO(s) + H2(g) Pb(s) + H2O(�)

()

( ( I I I ))

(III)

(III)

(III) + +

( (II))

(PbS)

(II)

(II) + (II) +

(II)

(II) + +

( ) ( 10.1)

+ ( )

Note 12Bunsen flame is not hot enough to give an obvious result in the extraction reaction. However, trace of ironcan be detected when a magnet is placed near the resultant reaction mixture.

6

Part III Metals

Figure 10.1 Electrolysis of aluminium oxide.

positive (+) electrodes (carbon)(+) ( )

negative (–) electrode(–)

molten aluminium

molten aluminium oxide and cryolite

tapping hole

Aluminium oxide has a very high melting point (2047°C). Cryolite(Sodium aluminium fluoride, Na3AlF6) is added to lower the meltingpoint to about 950°C.

Learning tip

(2047°C)( Na3AlF6)

950°C

Common ores of some metals and their methods of

extraction are listed in Table 10.2.10 .2N13

Note 13Common names of ores need not be memorized.

7


silver glance (or as freeelement)

()

copper pyrite

MetalYear of

discovery

Main ore/sourcefrom which metal

is obtained

/

Major metalliccompound

in the ore/source

/

Usual method of extraction

Potassium1807 A.D.

in deposits and seawater

electrolysis of molten oreSodium 1807 A.D.

common salt in seawater; rock salt

Calcium 1808 A.D. limestone

Magnesium 1808 A.D. magnesite

Aluminium1827 A.D. bauxite

Zinc

1000 A.D.

zinc blende Step 1:heat in air

sulphide oxide

Step 2:heat with carbon

oxide metal

Iron

3000 B. C.

3000

ancient(B. C.)

( )

5000 B. C.

5000

haematite

Lead galena similar to that of Zn

Copper

Mercury cinnabar

Silver displacement from solution

( )

mechanical separation(to get rid of impurities)

( )Gold

(as free element) (

)

Ag2S

CuFeS2

KCl

NaCl

CaCO3

MgCO3

Al2O3. 2H2O

ZnS

Fe2O3

PbS

HgS

Au

Mostdifficult

Eas

e of

ext

ract

ion

Easiest

heat withoxide metal

carbon

heat insulphide metal

air

Table 10.2 Methods of extraction of some metals from common ores.

N14

N15

Note 15Much silver and somegold are also recoveredfrom by-products in theextraction of other metals.

Note 14Copper and mercury are quiteunreactive. Therefore in thecontrolled heating of their sulphidesin air, the sulphur par t wil l beoxidized to sulphur dioxide, leavingthe metals behind, e.g.

HgS + O2 Hg + SO2

8

Part III Metals

( )

(a) (b) (c)

(d) (e) (f)

(g)

10.2

Suggest an extraction method to obtain each of the following

metals from their ores. Write an appropriate word equation if

applicable.

(a) aluminium (b) sodium (c) iron (d) silver

(e) platinum (f) magnesium (g) lead

Class practice 10.2

A10.2(a) Electrolysis.

electrolysisaluminium oxide aluminium + oxygen

(b) Electrolysis.electrolysis

sodium chloride sodium + chlorine

(c) Heating the metal ore with carbon.heat with carbon

haematite iron + carbon dioxide

(d) Displacement from solution containing silver.

(e) Physical separation.

(f) Electrolysis.electrolysis

magnesium chloride magnesium + chlorine

(g) Heating the metal ore in air and then with carbonheat in air

lead(II) sulphide Lead(II) oxide + sulphurdioxide

heat with carbonLead(II) oxide lead + carbon dioxide

10.5 Discovery of metals 10.5

Factors affecting the discovery and use of metals

Abundance of metals in the Earth’s crust

The percentage by mass of elements in the Earth’s crust is

shown in Figure 10.2.

Although gold and silver are the earliest metals discovered

by humans, they are not massively used. This is because of the

low abundance of gold and silver in the Earth’s crust.

Ease of extraction of metals

The most abundant metal in the Earth’s crust is aluminium.

Although iron is less abundant than aluminium in the Earth’s

crust, it is easier to be extracted from its ore. That’s why it was

more widely used than aluminium.

10.2

Figure 10.2 Percentage abundance bymass of elements in the Earth’s crust.

oxygen 46.6%

silicon 27.7%

all other elements

1.5% magnesium

2.1%potassium

2.6%

calcium

3.6%

sodium

2.8%

iron

5.0%

aluminium 8.1%

N16

Note 16If all the gold available in the world today were melted to make a cube, the cube would be 15 –18 m on eachside. It has been estimated that the total gold resource not yet extracted is only 50% of the existing amount.

N17

Note 17A few metals are obtained from the sea or made by scientists.

9


Factors affecting the availability of metals

In fact, the availability (and hence price) of a metal depends

mainly on:

1. the abundance of the metal in the Earth’s crust

2. the ease of mining its ore and the cost

3. the ease of extracting the metal from its ore and the cost of

extraction

( )

1.

2.

3. N18

Note 18Another factor affecting the price of a metal is its supply and demand.

Example 10.1Discovery of metals

(a) Some information about the extraction methods of fourdifferent metals

A, B, C and D are listed as follows:

(a) A B C D

(i) Arrange the metals in the order of ease ofextraction from their ores, the easiest first.

(ii) Arrange the reactivity of the metals in ascendingorder.

(iii) Deduce the order of discovery of the metals, theearliest first.

(b) Aluminium is the most abundant metal in the Earth’scrust, yet it is more expensive than iron. Explain.

Solution

(a) (i) A, D, B, C

(ii) The extraction methods are indication of thereactivity of metals. Thus, the ascending order ofreactivity of the four metals is the same as theorder of ease of extraction of these metals. That is,A, D, B, C.

(iii) The order of discovery of metals relates directly tothe reactivity and hence the ease of extraction ofthe metals. Therefore, the correct order is A, D, B,C.

(i)( )

(ii)

(iii) ()

(b)

(a) (i) A D B C

(ii)

( )A D B C

(iii)

A D BC

10.1

Metal

A

B

C

D

Extraction method

Mechanical separation

Carbon reduction

Electrolysis

Heating the metal ore in air

cont'd

10

Part III Metals

(b) Though aluminium is the most abundant metal, muchof it exists in clay (mainly silicates). It is not economicalto extract aluminium from clay. In fact, we extractaluminium from bauxite by electrolysis. Muchelectricity is needed and electricity is expensive.

On the other hand, we extract iron from haematite byheating haematite with carbon in a blast furnace. Thecost is relatively low.

➲ Try Chapter Exercise Q21

(b)(

)

➲ 21

10.6 Conserving metals 10.6

Need for conserving metals

Ores (the main sources of metals) in the Earth’s crust are limited

in amount and non-renewable.

It is clear that conserving metals is important.

Ways of conserving metals

Four possible ways to conserve metals are (The 4 ‘R’s of

conservation):

1. Reusing metal articles

2. Replacing metals by other materials

3. Reducing the use of metal articles

4. Recycling used metals

Reusing metal articles

However, this method of conserving metals is still not very

popular at present.

Replacing metals by other materials

New materials have already been used to replace some metals.

For example, PVC (a plastic) can replace iron or copper in

making water pipes; optical fibres replace copper telephone

cables. However, we should remember that plastics are made

from petroleum and the supply of petroleum is also limited.

( )

(4

R )

1. (reuse)

2. (replace)

3. (reduce)

4. (recycle)

PVC ( )

N19

Note 19Up till now, there is no economically feasible, large-scale method for extracting aluminium from clay.

N20Note 20People are accustomed to the throw-away style of using things. Theyhave yet to learn that world resources are limited and there is an urgentneed to conserve materials.

11


Reducing the use of metal articles

To conserve metals and protect our environment, the most

effective way is to avoid producing metal wastes. If this is

unavoidable, the amount of metal waste should be kept to a

minimum.

Recycling used metals

Recycling metals means melting down used metals and using

them again.

Importance of recycling

1. Metal resources are limited. However, demand for metals

keeps on increasing. Recycling is an effective way of

conserving metals.

2. Recycling saves energy and other resources (e.g.

electricity, water, fuels). Thus, recycling also reduces the

cost of metals.

3. Recycling helps to reduce pollution. Recycling of metals

can reduce pollution due to metal waste. Also, recycling

would cause less metal ores to be mined. This would

minimize the environmental problems arising from mining

and extraction from ores.

4. Recycling promotes public awareness of resource scarcity.

1.

2. (

)

3.

4.

N21

Note 21The demand for metals is ever increasing due to increasing worldpopulation and rising living standards. For example, in 1900, about15 000 tonnes of aluminium were produced; today, the amount isabout 30 000 000 tonnes a year, an increase of 2000 times.

Note 22Mining would produce a lot of harmful chemicals, rocks, mud and dust, and noise at the same time.The extraction of metals also produces pollutants. For example, extracting copper from copper pyrites(mainly CuFeS2) releases sulphur dioxide. Besides, mining leaves very large holes on the ground,which are unsightly and wasteful of land.

N22

KK ee yy tt ee rr mm ss

1. bauxite 4

Page

2. copper pyrite 4

3. extraction 4

6. recycling 10

7. replacing 10

8. reusing 10

4. galena 4

5. haematite 4

12

Part III Metals

SS uu mm mm aa rr yy10.1 Uses of metals in our daily lives

1. Metals are very useful to us and they are used in many different ways in our daily lives.

10.2 Uses related to properties of metals

2. Uses of metals are closely related to their availability, and .Properties usually taken into consideration include: appearance, strength, hardness, density,malleability, ductility and corrosion resistance. (Refer to Table 10.1 on p.1.)

10.3 Occurrence of metals in nature

3. Only a few metals (e.g. silver, and ) occur in elemental formin nature. Most metals exist as in ores.

10.4 Extraction of metals from their ores

4. Metals can be obtained from by a process called .

5. Metals can be extracted from their ores by(a) heating the metal ore (b) heating the metal ore with (c)

10.5 Discovery of metals

6. Different metals were discovered at different time in human history. Factors affecting thediscovery and uses of a metal include (a) of the metal in the Earth’s crust(b) of extraction of the metal.

7. Factors affecting the of a metal include(a) abundance of the metal in the Earth’s crust(b) ease of mining its ore and the cost(c) ease of extracting the metal from its ore and the cost

10.6 Conserving metals

8. Metal resources are limited and there is a need for .

9. Four methods are used to conserve metals:(a) metal articles(b) metals by other materials(c) the use of metal articles(d) used metals

10. metals means melting down used metals and using them again.Recycling

conservation

availability

ease

abundance

electrolysis

carbon

alone

ores extraction

compounds

gold platinum

properties prices

Reusing

Replacing

Reducing

Recycling

13

Chapter 11 Reactivity of metals

11.1 Different reactivities of metals 11.1

Reactivity is the readiness to react.

To compare reactivity of metals, we usually base on three

factors:

1. The lowest temperature at which the reaction starts

2. The rate (speed) of reaction

3. The amount of heat energy given out during reaction

N1Note 1Ask students why it is unfair and misleading to compare the reactivity of magnesium and ironin the following way:

Add a magnesium ribbon to water at 25°C; add ironpowder to dilute hydrochloric acid at 80°C; observewhich metal gives out gas bubbles faster.

1.

2.

3.

11.2 Comparing reactivity of common metals 11.2

Reactions of metals with air

Most metals combine with oxygen to form oxides when they are

heated in air. The reactions of some common metals with air are

summarized in Table 11.1.11.1

Potassium gentle heating It burns vigorously with alilac (pale purple) flame toproduce an orange powder.

potassium + oxygenpotassium superoxide (KO2)

(KO2)

Calcium strong heating It burns quite vigorouslywith a brick-red flame toproduce a white powder.

calcium + oxygencalcium oxide (CaO)

(CaO)

Sodium gentle heating It burns vigorously with agolden yellow flame toproduce a white powder.

sodium + oxygensodium oxide (Na2O)

(Na2O)

Magnesium strong heating It burns with a very brightwhite flame to produce awhite powder.

magnesium + oxygenmagnesium oxide (MgO)

(MgO)

B

U

R

N

Metal Conditions for reaction Observation Word equation

N2

K(s) + O2(g) KO2(s)

4Na(s) + O2(g) 2Na2O(s)

2Ca(s) + O2(g) 2CaO(s)

2Mg(s) + O2(g) 2MgO(s)

N3

Note 2(a) A little calcium nitride Ca3N2 (pale yellow) is also

formed. This point need not be mentioned tostudents.

(b) The equations provided in Section 11.2 only appear in the Teacher’sEdition. After learning equations, students should, in Activity, turn allthe word equations in Section 11.2 into chemical equations.

Note 3A little magnesium nitride Mg3N2

(pale yellow) is also formed. This

point need not be mentioned tostudents.

14

Part III Metals

Aluminium strong heating Aluminium powder burns togive out much heat; a whitepowder forms.

aluminium + oxygenaluminium oxide (Al2O3)

(Al2O3)

Iron strong heating Iron powder burns withyellow showery sparks toproduce a black solid.

iron + oxygen iron(II)iron(III) oxide (Fe3O4)

(II III)(Fe3O4)

Zinc strong heating Zinc powder burns to giveout some heat; a powder(yellow when hot, whitewhen cold) forms.

()

zinc + oxygen zinc oxide(ZnO)

(ZnO)

B

U

R

N

Lead strong heating It melts to silvery balls; apowder (orange when hot,yellow when cold) is seen onthe surface.

()

lead + oxygen lead(II)oxide (PbO)

(II) (PbO)

Mercury very strong heating A red powder forms on thesurface.

mercury + oxygenmercury(II) oxide (HgO)

(II) (HgO)

Copper very strong heating Its surface turns black. copper + oxygencopper(II) oxide (CuO)

(II) (CuO)

DO

NOT

BURN

Silver—

No apparent change even onvery strong heating. —

Gold —No apparent change even onvery strong heating. —

Platinum —No apparent change even onvery strong heating. —

NO

REACTION


Table 11.1 Reactions of some common metals with air.

4Al(s) + 3O2(g) 2Al2O3(s)

2Zn(s) + O2(g) 2ZnO(s)

3Fe(s) + 2O2(g) Fe3O4(s)

2Pb(s) + O2(g) 2PbO(s)

2Cu(s) + O2(g)2CuO(s)

2Hg(�) + O2(g)2HgO(s)

N4

N4


Note 4Zinc oxide has a certain structure (which is white) at room temperature; ithas a different structure (which is yellow) at high temperatures. Thedifferent colours of lead(II) oxide at different temperatures can also beexplained by difference in structure.

15


Figure 11.1 summarizes the reactivity of common metals

with air (oxygen).11 . 1

( )

Appearance of metals and storage methods

We can roughly predict the reactivity of an unfamiliar metal from

its appearance and the method that has been used to store it.

Reactions of metals with water

Action of potassium on water

If we add a small piece of potassium to water, it moves about

quickly on the water surface with a hissing sound, burning

with a lilac flame.

potassium + water potassium hydroxide solution + hydrogen

The resultant solution is alkaline because potassium

hydroxide is formed. It will turn red litmus blue.

+ +

Potassium , K

burn

Calcium , Ca

no reaction

Aluminium , Al

Iron , Fe

Sodium , Na

do not burn

Magnesium , Mg

Zinc , Zn

Lead , Pb

Mercury , Hg

Copper , Cu

Silver , Ag

Gold , Au

Platinum , Pt

Metal Reaction with air

Figure 11.1 The reactivity of commonmetals towards air (oxygen).

( )

Reactivity of metalswith air decreases

N5

Note 5An explosion will occur if a large piece of potassium (or sodium) is added to water.

2K(s) + 2H2O(�) 2KOH(aq) + H2(g)

16

Part III Metals

Action of sodium on water

Sodium reacts with water in a similar way as potassium, but

less vigorously.

When we add a small piece of sodium to water, it moves

about on the surface of water. It burns with a golden yellow flame

and becomes smaller and smaller.

The resultant solution is alkaline because sodium hydroxide is

formed:

sodium + water sodium hydroxide solution + hydrogen

Action of calcium on water

Unlike potassium and sodium, calcium is denser than water.

When we drop small pieces of calcium into water, they sink to

the bottom. Colourless gas bubbles are evolved at a moderate

rate. The gas can be collected as shown in Figure 11.2. When

the gas is tested with a burning splint, it burns with a ‘pop’

sound. This shows that the gas is hydrogen.

N6

Note 6If the sodium ball happens to stick to the wall of the containerwhile moving around, it will burn with a golden yellow flame.

2Na(s) + 2H2O(�) 2NaOH(aq) + H2(g)

+ +

11.2

( )

Figure 11.2 The reaction ofcalcium with cold water.

hydrogen gas

water invertedfunnel inverted funnel

calcium

hydrogen gas

calcium

water

A milky suspension is produced as the white calcium

hydroxide formed is only slightly soluble in water.

calcium + water calcium hydroxide + hydrogenCa(s) + 2H2O(�) Ca(OH)2(s) + H2(g)

+ +

17


Action of magnesium, aluminium, zinc and iron onsteam

Magnesium has almost no reaction with cold water. It reacts

slowly with hot water to give magnesium hydroxide (only

slightly soluble in water) and hydrogen.

magnesium + water magnesium hydroxide + hydrogen

With steam, however, the reaction is vigorous. We can

carry out the experiment using a set-up as shown in Figure

11.3.

Mg(s) + 2H2O(�) Mg(OH)2(s) + H2(g)

( )

+ +

11.3

Figure 11.3 The reaction of heatedmagnesium with steam.

wet sand magnesium ribbon

heat

deliverytube

hydrogen

water

trough

With strong heating, the water in the wet sand turns into

steam. The steam then reacts with magnesium to give an intense

white light. A white solid product, magnesium oxide, is formed.

magnesium + steam magnesium oxide + hydrogen

Zinc and iron do not react with cold or hot water. Heated

zinc or iron reacts with steam in a similar way as magnesium

does. However, the reaction is less vigorous for zinc, and even

less for iron.

zinc + steam zinc oxide + hydrogen

iron + steam iron(II) iron(III) oxide + hydrogen

Aluminium does not seem to react with steam. This is

because aluminium metal is usually covered with a very thin

layer of aluminium oxide. The oxide layer protects the metal

from reaction. If the protective oxide layer is removed by some

method, the aluminium obtained would be more reactive than

zinc, but less reactive than magnesium.

Lead, copper, mercury, silver and gold, even if heated

strongly, have no reaction with steam. Figure 11.4 summarizes

the reactivity of common metals towards water.

N7

Note 7The aluminium oxide layer is so thin that themetal still looks shiny and silvery white.

Note 8We can remove the protective oxide layer onaluminium by rubbing it with a solution ofmercury(II) chloride and hydrochloric acid.

N8

Mg(s) + H2O(g) MgO(s) + H2(g)

Zn(s) + H2O(g) ZnO(s) + H2(g)

3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)

+ +

+ +

+ (II III) +

11.4

18

Part III Metals

Potassium , Kmetals react with coldwater

heated metals reactwith steam

Calcium , Ca

Aluminium , Al

Iron , Fe

Sodium , Na

heated metals do notreact with water orsteam

Magnesium , Mg

Zinc , Zn

Lead , Pb

Mercury , Hg

Copper , Cu

Silver , Ag

Gold , Au

Platinum , Pt

Metal Reaction with water

metal + watermetal hydroxide +hydrogen

metal + steammetal oxide +hydrogen

—

Equation

Figure 11.4 The reactivity ofcommon metals towards liquidwater/steam.

Reactivityof metalstowardswaterdecreases

N9Note 9Point out to students thatheated magnesium, aluminium,zinc and iron react with steamto form metal oxides, nothydroxides.

(Rb)

11.1

Rubidium (Rb) reacts with water to form hydrogen and a

compound. Is this compound rubidium oxide or rubidium

hydroxide? Why?

Class practice 11.1

A11.1Rubidium hydroxide. As can be seen from Figure 11.4, very reactive metals (e.g. K, Na, Ca) react with water toform a metal hydroxide and hydrogen; fairly reactive metals form a metal oxide and hydrogen. Rubidium should bemore reactive than potassium, since it is lower than potassium in Group I. (The reactivity of Group I elementsincreases down the group.) Thus rubidium should form rubidium hydroxide, not rubidium oxide.

Reactions of metals with dilute hydrochloric acid anddilute sulphuric acid

Figure 11.5 shows the reactivity of common metals towards

dilute hydrochloric acid or dilute sulphuric acid.

11.5N10

Note 10Dilute sulphuric acid forms insoluble sulphates (such as CaSO4, PbSO4) which would stop the reaction aftera short while; dilute nitric acid is not a typical acid towards metals (as it has oxidizing properties).

Potassium or sodium reacts with dilute acids explosively, so NEVERattempt to perform this experiment in a school laboratory.

Learning tip

19


Potassium , K explosive reaction

reacts with acid, moreslowly down the series

Calcium , Ca

Aluminium , Al

Iron , Fe

Sodium , Na

very slow reaction

no reaction

Magnesium , Mg

Zinc , Zn

Lead , Pb

Mercury , Hg

Copper , Cu

Silver , Ag

Gold , Au

Platinum , Pt

Metal Reaction with dilute acid

metal + hydrochloricacid metal chloride +hydrogenormetal + sulphuricacidmetal sulphate +hydrogen

—

Equation

Figure 11.5 The reactivity ofcommon metals towards dilutehydrochlor ic acid/dilutesulphuric acid.

Reactivityof metals towards dilutehydrochloric/sulphuric aciddecreases

Reactive metals react with dilute hydrochloric acid and

sulphuric acid to give salts and hydrogen.

✘ Zinc reacts with both concentrated and dilute sulphuric acid togive out hydrogen gas.

✔ Zinc reacts with concentrated sulphuric acid to give outsulphur dioxide rather than hydrogen gas. (Details will bediscussed in Chapter 31.)

Check your concept

✘

✔(

)

20

Part III Metals

11.3 The metal reactivity series 11.3

By comparing their reactions with air, water and dilute

hydrochloric acid, we can arrange common metals in order of

reactivity. The list is called the metal reactivity series (Figure

11.6).(

11.6)

Potassium , K

Sodium , Na

Calcium , Ca

Magnesium , Mg

Aluminium , Al

Zinc , Zn

Iron , Fe

Lead , Pb

Copper , Cu

Mercury , Hg

Silver , Ag

Platinum , Pt

Gold , Au

most reactive

decreasing reactivity

least reactive

N11

Note 11There are over 80 metals. The reactivity series shown in Figure 11.6 is the one whichappears in most chemistry textbooks. The series only includes the common metals. Tellstudents that they should memorize the order of metals in this series.

N12

Note 12The dotted line between lead andcopper divides the common metalsinto 2 groups: those above the linereact with dilute hydrochloric acid togive metal chloride and hydrogen,while those below the line do not.

Metals at the top of the series are the most reactive; those at

the bottom are the least reactive.

Figure 11.6 Metal reactivity series for common metals.

21



(a) Arrange the four metals in decreasing order of reactivity.

(b) Give possible names of the four metals.

(a)

(b)

Experiment Metal

A B C D

Strong heating in air burns with a brick-redflame

forms a black powder burns with a lilacflame

burns with a dazzlingwhite flame

Reaction with coldwater

moderate reaction no reaction violent reaction;burns by itself

no reaction

Reaction with dilutehydrochloric acid

fast reaction no reaction (experiment notperformed)( )

fast reaction

11.4 Chemical equations 11.4

Representing the reaction between magnesium andoxygen

When magnesium burns in air (or oxygen), magnesium oxide

is formed. We can represent this reaction by a word equation:

magnesium + oxygen magnesium oxide

reactants product(reacting substances) (the substance produced)

We may also show the reaction by a diagram (Figure 11.7).

( )

( )

+

( ) ( )

( 11.7)

1 oxygen molecule1

2 magnesium atoms2

2 formula units of magnesium oxide2

Mg Mg

O O

Mg2+

Mg2+

O2–

O2–

Figure 11.7 The formation of magnesium oxide in terms of particles — 2 atoms of Mg react with 1 molecule of O2 to form2 formula units of MgO. (1 formula unit of magnesium oxide consists of 1 magnesium ion, Mg

2+and 1 oxide ion, O

2–.)

Mg O2

MgO ( MgO (Mg2+) (O2–) )

N13

Note 13A cation is smaller than theatom from which it isderived; an anion is largerthan the atom from which itis derived.

A11.2(a) C, A, D, B(b) C: potassium;

A: calcium; D: magnesium; B: copper

22

Part III Metals

We may also represent the reaction by writing a chemical

equation:


reactants product

Equation as the summary of a reaction

A chemical equation (or an equation) is a statement, in

formulae and suitable symbols. It shows the physical states and

relative numbers of particles of the reactants and products in a

chemical reaction.

Let us use the following equation as an example:


1. The reactants involved

These are magnesium (Mg) and oxygen (O2), written on the

left-hand side of the arrow.

2. The products formed

This is magnesium oxide (MgO), written on the right-hand

side of the arrow.

3. Physical states of the substances involved

Mg and MgO are solids, represented by a state symbol (s);

O2 is a gas (g). Other state symbols are: liquid (�) and

aqueous solution (aq).

4. The relative number of particles (atoms, molecules, ions

or formula units)

2 atoms of Mg would react with 1 molecule of O2 to

produce 2 formula units of MgO.

N14

Note 14An equation which is too long to be writtenin one line, for example:

2A + B + 3C 4D + E + 5F + 3Gmay be represented as:

2A + B + 3C 4D + E + 5F + 3G

or2A + B + 3C

4D + E + 5F + 3G,but not as

2A + B + 3C 4D + E + 5F + 3G,

in order to avoid ambiguity.

N15

Note 15Explain to students clearly the differencebetween the state symbols (l) and (aq).

N16

Note 16Another point is: The equation gives therelative number of moles and henceindirectly the relative masses of thevarious substances involved. This pointwill be dealt with in Chapter 12.


( )


1.

(Mg) (O2)

2.

(MgO)

3.

Mg MgO (s)

O2 (g)

( (� )

(aq) )

4.

Mg O2

MgO

23


SO2(g) + 2NaOH(aq) Na2SO3(aq) +

H2O(�)

11.3

Express in words the information provided by the following

equation:

SO2(g) + 2NaOH(aq) Na2SO3(aq) + H2O(�)

Class practice 11.3

N17

A11.31 molecule of sulphur dioxide gas reacts with 2 formula units of aqueous sodium hydroxide, toproduce 1 formula unit of aqueous sodium sulphite and 1 molecule of liquid water.

More about an equation

Balanced equation

Generally, an equation must be balanced with respect to (1) the

number of atoms of any kind and (2) the net ionic charges.

The numbers before the formulae of reactants and products

in a balanced equation are called stoichiometric coefficients.

2Mg(s) + O2(g) 2MgO(s) is a balanced equation. The

stoichiometric coefficients are 2, 1 and 2 respectively. Note that

‘1’ is understood and should be left out — thus O2 is written

instead of 1O2.

(1)

(2)


2 1 2

1

O2 1O2

2CO(g) + O2(g) 2CO2(g)

(a)

(i) (ii)

(b)

(i) (ii)

(c)

11.4

Consider 2CO(g) + O2(g) 2CO2(g).

(a) How many carbon atoms are on the

(i) left-hand side (ii) right-hand side of the arrow?

(b) How many oxygen atoms are on the

(i) left-hand side (ii) right-hand side of the arrow?

(c) Is it a balanced equation?

Class practice 11.4A11.4(a) (i) 2 (ii) 2(b) (i) 4 (ii) 4(c) Yes

One-way reaction and reversible reaction

The single arrow ‘ ’ between the two sides of an equation

indicates that the reaction goes one way only. At the end, all

reactants react to form products.

Note 17Point out to students that it is often simpler torepresent a reaction by a chemical equation thanwritten description (as shown in Q11.3). An equationalso gives information about the composition of thesubstances involved, from their formulae.

24

Part III Metals

The double arrow ‘ ’ is used in some equations, e.g.

N2(g) + 3H2(g) 2NH3(g). The ‘ ’ means that the reaction

is reversible, that is, both forward (left to right) and backward

(right to left) reactions occur at the same time. At the end, both

reactants and products are present.

Information not available from equations

There are some limitations of chemical equations. For example,

the equation 2H2(g) + O2(g) 2H2O(�) does not tell us the

conditions under which hydrogen and oxygen can combine.

Also, it does not tell us whether the reaction is fast or slow.

Steps in writing a chemical equation

The steps in writing an equation are shown below, with an

example for illustration.

hydrogen + oxygen water

In balancing equations, stoichiometric coefficients must be

placed in front of formulae where necessary. The formulae

themselves must not be changed. Thus, for the above reaction,

it would be incorrect to write

H2(g) + O2(g) H2O2(�), ✗

or H2(g) + O(g) H2O(�). ✗

( N2(g) + 3H2(g)

2NH3(g))

( )

( )

2H2(g) + O2(g) 2H2O(�)

+

H2(g) + O2(g) H2O2(�) ✗

H2(g) + O(g) H2O(�) ✗

N18

Note 18Here, the forward reaction is:

N2(g) + 3H2(g) 2NH3(g);the backward reaction is:

2NH3(g) N2(g) + 3H2(g)

It is a common practice to write an equation in which coefficientsare the smallest possible whole numbers. That is why the equation2H2(g) + O2(g) 2H2O(�) is shown here.

Learning tip

2H2(g) + O2(g) 2H2O(�)

25


Writing a chemical equation

Step 1 Find out what the reactants and products are. Write downthe word equation for the reaction.

hydrogen + oxygen water

(After some practice, you will be able to skip thisstep.)

Step 2 Write the equation by replacing names of reactants andproducts with their correct formulae.

H2 + O2 H2O(unbalanced)

Step 3 Balance the equation with respect to atoms of any kind.

To balance the number of oxygen atoms:2 oxygen atoms on the left-hand side, only 1 oxygenatom on the right-hand side, so put ‘2’ before H2O;the equation is still unbalanced

H2 + O2 2H2O

To balance the number of hydrogen atoms:2 hydrogen atoms on the left-hand side, 4 hydrogenatoms on the right-hand side, so put ‘2’ before H2; theequation is now balanced

2H2 + O2 2H2O

Step 4 Write the state symbol after each formula to give thecomplete balanced equation.

2H2(g) + O2(g) 2H2O(�)(complete balanced equation)

Problem-solving strategy

1

(

)

2

H2 + O2 H2O

3

H2O

2

H2 + O2 2H2O

H 2

2

2H2 + O2 2H2O

4

2H2(g) + O2(g) 2H2O(�)

N19

Note 19In public examinations, candidatesusually need not wr ite statesymbols in equations, unless theyare instructed to do so in thequestion.

( )

( )

1.

(a) Cl2O7 (b) (NH4)2Cr2O7

(c) 3Fe2(SO4)3(d) 3Na2CO3 · 10H2O

2.

__________ Pb3O4(s)

__________ PbO(s) +

__________ O2(g)

11.5

1. State the number of atoms of each kind for the given

number of formula units below:

(a) Cl2O7 (b) (NH4)2Cr2O7

(c) 3Fe2(SO4)3 (d) 3Na2CO3 · 10H2O

2. Balance the following by adding suitable stoichiometric

coefficients:

__________ Pb3O4(s) __________ PbO(s) + _________ O2(g)

Class practice 11.5

A11.51. (a) 2 Cl atoms, 7 O atoms

(b) 2 N atoms, 8 H atoms, 2 Cr atoms, 7 O atoms(c) 6 Fe atoms, 9 S atoms, 36 O atoms(d) 6 Na atoms, 3 C atoms, 39 O atoms, 60 H atoms

2. 2Pb3O4(s) 6PbO(s) + O2(g)

N20

Note 20Students may get confused about (1) the subscript after an atomic symbol, e.g. 3Fe 2 (SO 4 )3

(2) the subscript after brackets, e.g. 3Fe2(SO4) 3 and (3) the stoichiometric coefficient in front of aformula, e.g. 3 Fe2(SO4)3. Explain to students how to calculate the number of atoms of each kind.

26

Part III Metals

11.5 Metal reactivity series and the tendencyof metals to form positive ions

11.5

Metals react by losing electrons

N21

Note 21Down a group, atomic size increases, so the ‘pull’ (attraction) ofthe nucleus on the outer shell electrons becomes smaller. Thusthe outermost shell electrons can be lost more easily.

Metals react by losing electrons to form positive ions.

Reactivity and readiness to lose electrons

Reactivity of a metal depends on how readily its atoms lose

electrons.

The readiness of elements to lose electrons decreases across

a period and increases down a group. See Figure 11.8.

11.8

increasing readiness to lose electrons

increasing reactivity of metals

increasingreadiness tolose electrons

increasingreactivity ofmetalsFigure 11.8 Readiness to lose

electrons (and hence reactivity ofmetals) decreases across a periodand increases down a group.

( )

A metal higher in the reactivity series has a higher

reactivity, and its atoms would lose outermost shell

electrons to form cations more easily.

27


11.6 Displacement reactions of metals inaqueous solution

11.6

Copper in silver nitrate solution

When we place copper in silver nitrate solution, the copper

slowly dissolves. Some shiny silver crystals form on the copper

surface. The solution gradually turns pale blue.

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)colourless pale blue

We say that copper displaces the silver metal from the

silver nitrate solution.

Iron in copper(II) sulphate solution

A similar displacement reaction occurs when we place an iron

nail into copper(II) sulphate solution.

Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)silvery blue pale reddish white green brown

On the other hand, if we put a piece of copper into iron(II)

sulphate solution, no reaction occurs.

(II)

(II)

Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

(II)

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)

N22

Note 22Suppose a student is asked to describe a simpleexperiment to compare the reactivity of copperand iron. The displacement reaction here can bequoted. However, it would be wrong to quote theexample of setting up an electrolytic cell. With anexternal electric current applied, the relativereactivity of the two metals cannot be compared.

N23

Note 23Remind students to write the appropr iateequations in examinations. If the question asksfor ‘a balanced chemical equation’, either the fullequation or ionic equation (if applicable) can beput down. But if it asks for ‘an ionic equation’,only the ionic equation will be accepted.

We may use the reactivity series to explain this rule. A

metal higher in the reactivity series is more reactive, so its

atoms lose electrons more readily to form cations. The cations

of the less reactive metal would accept these electrons, forming

back the atoms of the less reactive metal.

A metal (M1) higher in the reactivity series will displace any

metal (M2) lower in the series from the solution of a

compound of M2.

(M1)

(M2)

28

Part III Metals

Example 11.1Predicting chemical reactions based on the metalreactivity series

Predict, with reasoning, whether a reaction takes place ineach of the following:

(a) Zinc is added to magnesium chloride solution.

(b) Magnesium is added to lead(II) nitrate solution.

(c) Silver is added to dilute sulphuric acid.

State what type of reaction (if any) takes place and write anappropriate equation for the reaction.

Solution

(a) No reaction occurs. Zinc is lower than magnesium inthe metal reactivity series.

(b) Displacement reaction occurs. Magnesium is higherthan lead in the metal reactivity series. It can displacelead from lead(II) nitrate solution.

Mg(s) + Pb(NO3)2(aq) Mg(NO3)2(aq) + Pb(s)

(c) No reaction occurs. All metals lower than copper in themetal reactivity series would have no reaction withdilute sulphuric acid.


11.1

(a)

(b) (II)

(c)

(a)

(b)

(II)

Mg(s) + Pb(NO3)2(aq) Mg(NO3)2(aq) + Pb(s)

(c)

➲ 19

11.7 Ionic equations 11.7

Representing some reactions by ionic equations

Consider the reaction between copper metal and aqueous

silver nitrate solution. The equation for the reaction is:


An ionic equation is an equation involving ions in aqueous

solution. Only those ions formed or changed during the

reaction are included.

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) +

2Ag(s)

29


But the ionic equation

Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)

shows more clearly what has happened. Note that the nitrate

ions (NO3–) remain unchanged in the reaction. These ions, which

do not actually take part in a reaction, are called spectator ions.

They do not appear in ionic equations.

Steps in writing an ionic equation

Steps in writing an ionic equation are summarized below: N24

Note 24Many students find it difficult to write ionic equations. The onlyexample given here is a metal displacement reaction. Ionicequations of other reaction types (e.g. neutralization, precipitation,metal/acid reaction) are given in Chapter 18 of Book 2.


(NO3–)

Writing an ionic equation

Step 1 Write the full balanced equation for the reaction.


Step 2 Rewrite the equation:

• Rewrite the formulae of electrolytes in aqueoussolution as separate formulae of cations andanions

• Keep other formulae unchanged

AgNO3 and Cu(NO3)2 are electrolytes in aqueoussolution.

Cu(s) + 2Ag+(aq) + 2NO3–(aq)

Cu2+(aq) + 2NO3–(aq) + 2Ag(s)

Step 3 Cancel out the spectator ions on both sides of the equation.

Cu(s) + 2Ag+(aq) + 2NO3–(aq)

Cu2+(aq) + 2NO3–(aq) + 2Ag(s)

Step 4 Check that the ionic charge is balanced in the ionicequation.

Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)net charge = +2 net charge = +2


1

Cu(s) + 2AgNO3(aq)

Cu(NO3)2(aq) + 2Ag(s)

2

• Cu(NO3)2 AgNO3

•

Cu(s) + 2Ag+(aq) + 2NO3–(aq)

Cu2+(aq) + 2NO3–(aq) + 2Ag(s)

3

Cu(s) + 2Ag+(aq) + 2NO3–(aq)

Cu2+(aq) + 2NO3–(aq) + 2Ag(s)

4


= +2 = +2

30

Part III Metals

Example 11.2Writing ionic equations

Write the ionic equations for the following reactions.

(a) Pb(s) + CuSO4(aq) PbSO4(aq) + Cu(s)

(b) Zn(s) + 2AgNO3(aq) Zn(NO3)2(aq) + 2Ag(s)

Solution

(a) Rewrite the equation:

Pb(s) + Cu2+(aq) + SO42–(aq) Pb2+(aq) + SO4

2–(aq) +Cu(s)

Cancel out the spectator ions on both sides of theequation:

Pb(s) + Cu2+(aq) + SO42–(aq) Pb2+(aq) + SO4

2–(aq) +Cu(s)

The ionic charge is balanced on both sides of theequation. Thus, the ionic equation is:

Pb(s) + Cu2+(aq) Pb2+(aq) + Cu(s)net charge = +2 net charge = +2

(b) Rewrite the equation:

Zn(s) + 2Ag+(aq) + 2NO3–(aq) Zn2+(aq) +

2NO3–(aq) + 2Ag(s)

Cancel out the spectator ions on both sides of theequation:

Zn(s) + 2Ag+(aq) + 2NO3–(aq) Zn2+(aq) +

2NO3–(aq) + 2Ag(s)

The ionic charge is balanced on both sides of theequation. Thus, the ionic equation is:

Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)net charge = +2 net charge = +2

11.2

(a) Pb(s) + CuSO4(aq)PbSO4(aq) + Cu(s)

(b) Zn(s) + 2AgNO3(aq)Zn(NO3)2(aq) + 2Ag(s)

(a)

Pb(s) + Cu2+(aq) + SO42–(aq)

Pb2+(aq) + SO42–(aq) + Cu(s)

Pb(s) + Cu2+(aq) + SO42–(aq)

Pb2+(aq) + SO42–(aq) + Cu(s)

Pb(s) + Cu2+(aq) Pb2+(aq) + Cu(s)= +2 = +2

(b)

Zn(s) + 2Ag+(aq) + 2NO3–(aq)

Zn2+(aq) + 2NO3–(aq) + 2Ag(s)

Zn(s) + 2Ag+(aq) + 2NO3–(aq)

Zn2+(aq) + 2NO3–(aq) + 2Ag(s)

Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)= +2 = +2

31


Example 11.3Balancing the ionic equation

Find the values of y and z in the ionic equation:

yCl2(g) + 6OH–(aq) zCl–(aq) + ClO3–(aq) + 3H2O(�)

Solution

Since there are 6 oxygen atoms on LHS and 6 oxygen atomson RHS, the stoichiometric coefficient for ClO3

– must be 1.An ionic equation must be balanced with respect to ioniccharges.

Net charge of reactants = –6 = net charge of products∴ –6 = z(–1) + (–1)∴ z = 5

An ionic equation must also be balanced with respect tonumber of atoms of any kind. Consider chlorine atoms,

number of Cl atoms on right-hand side = 5 + 1 = 6∴ number of Cl atoms on left-hand side = 6∴ y = 3


11.3

y z

yCl2(g) + 6OH–(aq)zCl–(aq) + ClO3

–(aq) + 3H2O(�)

6ClO3

– 1

= –6 =

–6 = z(–1) + (–1) z = 5

Cl = 5 + 1 = 6 Cl = 6

y = 3

➲ 18

(a) Mg(s) + 2AgNO3(aq)

Mg(NO3)2(aq) + 2Ag(s)

(b) 3Mg(s) + 2Al(NO3)3(aq)

3Mg(NO3)2(aq) + 2Al(s)

11.6

Write the ionic equations for the following equations.

(a) Mg(s) + 2AgNO3(aq) Mg(NO3)2(aq) + 2Ag(s)

(b) 3Mg(s) + 2Al(NO3)3(aq) 3Mg(NO3)2(aq) + 2Al(s)

Class practice 11.6

A11.6(a) Mg(s) + 2Ag

+(aq) Mg

2+(aq) + 2Ag(s)

(b) 3Mg(s) + 2Al3+

(aq) 3Mg2+

(aq) + 2Al(s)

11.8 Extraction of metals from their ores 11.8

The process of getting a metal from its ore is called extraction.

32

Part III Metals

We can draw two general relationships from the table:

• The lower the position of a metal in the reactivity series,

the earlier it was first discovered. (This is because less

reactive metals form less stable compounds, from which

the metals can be extracted more easily.)

• The method used to extract a metal from its ore depends

on the position of the metal in the reactivity series.

•

10.2

•

(

)

Potassium , K

Sodium , Na

Calcium , Ca

Magnesium , Mg

Aluminium , Al

Zinc , Zn

Iron , Fe

Lead , Pb

Copper , Cu

Mercury , Hg

Silver , Ag

Platinum , Pt

Gold , Au

most reactive

Reactivity

least reactive

latest

Order ofdiscovery

earliest

33


Example 11.4Relating the extraction method of a metal to its position in the metal reactivity series

Some information about the extraction methods of fivedifferent metals A, B, C, D and E are listed as follows:

A B C D E

Arrange the positions of metals in the metal reactivity seriesbased on the above information, from the lowest to thehighest. Explain your answer.

Solution

Mechanical separation is the easiest way to extract a metalfrom its ore. Thus, metal A must be the least reactive (i.e. thelowest in the metal reactivity series).

Metal D should be in a position higher than that of A, butlower than that of E as it can be displaced out from itssolution by E.

Metal C should be in the highest position among the fivemetals since it is the most difficult to be extracted.

Metal B should be in a position higher than that of E since itis extracted by carbon reduction.

∴ The positions of five metals in the metal reactivity seriesshould be:

A < D < E < B < C

A( )

D A EE

C

B E

∴

A < D < E < B < C

11.4

Metal

A

B

C

D

E

Extraction method

Mechanical separation

Carbon reduction

Electrolysis

Displacement from solution by metal E E D

Heating the metal ore in air

34

Part III Metals

1. (a) (

(III))

(b)

2.

(a) (III) + +

(b) +

11.7

1. (a) Is it possible to reduce haematite (containing iron(III)

oxide) by heating it with magnesium? Why?

(b) Is this method used in industry to extract iron from

haematite? Why?

2. Write balanced chemical equations for the following

reactions:

(a) iron(III) oxide + carbon iron + carbon dioxide

(b) silver oxide silver + oxygen

Class practice 11.7

A11.71. (a) Yes. Magnesium is more reactive than iron.

(b) No. Magnesium is much more expensive than iron.2. (a) 2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g)

(b) 2AgO(s) 2Ag(s) + O2(g)


1. balanced equation 23

Page

3. displacement reaction 27

5. one-way reaction 23

6. reactivity 13

7. reversible reaction ( ) 23

8. spectator ion 29

9. stoichiometric coefficient 23

2. chemical equation 22

4. ionic equation 28

35


SS uu mm mm aa rr yy11.1 Different reactivities of metals

1. is the readiness to react.

11.2 Comparing reactivity of common metals

2. The reactivity of metals can be found by comparing their reactions with air, water and diluteacids. Refer to p.13 – 19 for the results of the reactions.

11.3 The metal reactivity series

3. The is a series of common metals arranged in decreasingorder of reactivity.

4. The following table summarizes the appearances and reactions of metals in the reactivity series.

Example 1: 4Na(s) + O2(g) 2Na2O(s)

Example 2: 2Ca(s) + O2(g) 2CaO(s)

Example 3: 2Cu(s) + O2(g) 2CuO(s)

Example 4: 2Na(s) + 2H2O(�) 2NaOH(aq) + H2(g)

Example 5: Zn(s) + H2O(g) ZnO(s) + H2(g)

Example 6: 2K(s) + 2HCl(aq) 2KCl(aq) + H2(g) (NEVER attempt this experiment!)

a metal displacesany other metallower in the seriesfrom a solution ofits compound(Example 10)

Metal Appearanceof metal

Reduction ofoxide with

carbon (1500°C)

Displacementreactionsair water/steam dilute hydrochloric

acid

K dull (storedunder paraffinoil)

Reaction of metal with

burns violently,forming oxide(Example 1)

reacts explosively, formingmetal chloride andhydrogen (Example 6)Na

Ca

Mg

generally dullAl

Zn

Fe

Pb

reacts withdecreasingvigour, formingoxide (Example 2)

Cu

Hg

Ag

Au

generally shiny no reaction

no reaction

a layer of oxideformed onsurface(Example 3)

reacts with decreasingvigour: metal +hydrochloric acidmetal chloride + hydrogen(Example 7)

metal + water metalhydroxide+ hydrogen

(Example 4)

no reaction

metal + steam metal oxide+ hydrogen

(Example 5)reduced withincreasing ease:metal oxide + carbon

metal + carbondioxide

(Example 8)

no reaction

decomposed tometal and oxygen byheating alone(Example 9)

not applicable � these3 metals react withwater in aqueoussolution to give H2 gas

Reactivity

metal reactivity series

36

Part III Metals

Example 7: Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g)

Example 8: 2PbO(s) + C(s) 2Pb(s) + CO2(g)

Example 9: 2HgO(s) 2Hg(�) + O2(g)

Example 10: Mg(s) + 2AgNO3(aq) Mg(NO3)2(aq) + 2Ag(s) or

Ionically: Mg(s) + 2Ag+(aq) Mg2+(aq) + 2Ag(s)

11.4 Chemical equations

5. A (or an equation) is a statement, in formulae and suitablesymbols. It shows the physical states and relative numbers of particles of the reactants andproducts in a chemical reaction.

6. A reaction is represented by a double arrow ‘ ’.

7. The steps in writing a chemical equation are shown in ‘Problem-solving strategy’ on p.25.

11.5 Metal reactivity series and the tendency of metals to form positive ions

8. Metals react by electrons to form cations. Different metals have differentreactivities because they have different tendencies to lose electrons. Atoms of a reactive metal loseelectrons readily.

11.6 Displacement reactions of metals in aqueous solution

9. A metal (M1) in the reactivity series will displace any one (M2)in the series from the solution of a compound of M2. This is because a more

reactive metal loses electrons more easily.

11.7 Ionic equations

10. An is an equation involving ions in aqueous solution; onlythose ions formed or changed during the reaction are included.

11. An ionic equation must be balanced with respect to the aswell as the number of atoms. (Refer to the ‘Problem-solving strategy’ on p.29.)

11.8 Extraction of metals from their ores

12. The ease of extracting a metal from its ores is related to the of the metal in thereactivity series.

13. The a metal in the reactivity series, the earlier it was first discovered. This isbecause less reactive metals form stable compounds, from which the metals canbe extracted more easily.

less

lower

position

ionic charges

ionic equation

lower

higher

losing

reversible

chemical reaction

37

Chapter 12 Reacting masses

The Avogadro constant

Chemists use a very special unit, called mole (abbreviation:

mol) to show the number of formula units. One mole contains a

large number of particles, 6.02 � 1023. This number is called

Avogadro constant (L). Thus L = 6.02 � 1023 mol–1.

To count by weighing

We can count indirectly — by weighing.

(

mol)

6.02 � 1023

(L) L = 6.02 �

1023 mol–1

12.1 The mole concept in general 12.1

Every pure substance has a formula. The simplest unit of asubstance is its formula unit.

In the case of a molecular compound, a formula unit is in fact amolecule. In ionic compounds, there are no molecules. Thus, aformula unit of magnesium chloride is MgCl2, which consists of 1Mg

2+ion and 2 Cl

–ions.

Learning tip

MgCl2 1 Mg2+

2 Cl–

N1

Note 2(a) Determination of L using different methods leads to values which are very close to each other.

The most recent values lie between 6.022 24 � 1023

and 6.022 80 � 1023

. By approximation,we take L = 6.02 � 10

23mol

–1.

(b) At one time, L was referred to as the Avogadro number. Now the term Avogadro constant isused as its value is constant.

Note 1The word mole is derived from the Latin word meaning a collection or pile.

N2

Example 12.1Counting substance by weighing

Suggest an indirect way of counting

(a) 50 000 $1 coins (b) 3.01 � 1024 copper atoms.

(Given that the mass of a $1 coin = 7.08 g; the mass of onecopper atom = 1.05 � 10–22 g)

Solution

(a) Weigh out 7.08 � 50 000 = 354 000 g or 354 kg of $1coins.

(b) Weigh out 1.05 � 10–22� (3.01 � 1024) = 316 g of pure

copper metal.

12.1

(a) 50 000(b) 3.01 � 1024

( = 7.08g = 1.05 � 10–22

g)

(a) 7.08 � 50 000 = 354 000 g354 kg

(b) 1.05 � 10–22� (3.01 � 1024) =

316 g

38

Part III Metals

1 108 g

1.204 � 1023

12.1

Given that 1 mole of silver has a mass of 108 g, suggest an

indirect way of counting 1.204 � 1023

silver atoms.

Class practice 12.1

Defining mole

One mole (mol) of a substance (or species) represented by a

formula is the amount containing the same number of

formula units as the number of atoms in exactly 12.0 g of

carbon-12.

(

) 12.0 g -12

Mole and mass

One mole of a substance (or species) has a mass corresponding

to its formula mass expressed in gram unit.

For example, the molar mass of carbon dioxide is 44.0 g

mol–1; that of water (H2O) is 18.0 g mol–1.

Note: (a) The unit of molar mass is g mol–1. (b) The relative

molecular mass (or formula mass) of carbon dioxide is just

44.0, not 44.0 g mol–1.

We should notice that the ‘mole’ can relate the ‘mass’ of a

substance to the ‘number of formula units’ contained in the

substance.

The molar mass of a substance is the mass in grams of one

mole of the substance. (Unit of molar mass: g mol–1) N4

Note 4Remind students that the abbreviation of mole is mol (not m); that of gram is g (not gm).

massNUMBER OF

MOLESnumber of

formula units

no. of moles (mol)

=no. of formula units

Avogadro constant (mol–1

)

no. of moles (mol)

=mass (g)

molar mass (g mol–1

)

1 ( )

44.0 g mol–1 (H2O)

18.0 g mol–1

(a) g

mol–1 (b) (

) 44.0 44.0 g mol–1

( ) (

g mol–1)

(mol)

=(mol

–1)

(mol)

=(g)

(g mol–1

)

A12.11 mole of silver contains 6.02 �10

23silver atoms.

∴ number of moles of silver atoms= number of atoms/L

=

= 0.200 molMolar mass of silver (Ag) = 108 g mol

–1

1.204 � 1023

6.02 � 1023

∴ mass of pure silver metal to be weighed out (for counting indirectly)= number of moles � molar mass= 0.200 � 108 g= 21.6 g

N3

Note 3‘Species’ has a broader sense than substance. Sodium chloride is aspecies and also a substance, but chloride ion is a species, not asubstance. Chloride ions alone cannot be isolated.

39


1. (a) Fe2(SO4)3

(b) F e 2 ( S O 4 ) 3

(c) 0.2000 Fe2(SO4)3

2. 27.0 g

12.2

1. (a) Calculate the formula mass of Fe2(SO4)3.

(b) What is the molar mass of Fe2(SO4)3?

(c) What is the mass of 0.2000 mole of Fe2(SO4)3?

2. Calculate the number of atoms in 27.0 g of pure silver.

Class practice 12.2

A12.21. (a) 55.8 � 2 + (32.1 + 16.0 �

4) � 3= 399.9

(b) 399.9 g mol–1

(c) 399.9 � 0.2000 = 79.98 g

2. Molar mass of silver (Ag) = 108 g mol

–1

Number of moles of si lveratoms

=

= mol

= 0.250 molNumber of silver atoms= number of moles � L= 0.250 � (6.02 � 10

23)

= 1.51 � 1023

27.0108

mass (g)

molar mass (g mol–1

)

Important relationships — a summary

Based on definition of mole, we can get the following

important relationships:

(1) Mass of 1 mole of a=

formula mass expressed in

substance or species (g) gram unit

(2) Number of moles (mol) =

(3) Number of moles (mol) =

(4) Mass of 1 formula unit (g)

=molar mass (g mol–1)

Avogadro constant (mol–1)

number of formula units


mass (g)

molar mass (g mol–1)N5

Note 5There are 3 terms in each of therelationships (2), (3) and (4). By knowingany two of them, the third one can becalculated.

(1) ( ) (g)

=

(2) (mol)

=

(3) (mol)

=

(4) (g)

=(g mol–1)

(mol–1)

(mol–1)

(g)

(g mol–1)

Example 12.2Calculations involving mole concepts

12.2

A beaker contains 44.44 g of calcium chloride CaCl2.Calculate

(a) the formula mass of CaCl2

(b) the molar mass of CaCl2

(c) the number of moles of CaCl2 in the beaker

(d) the number of formula units of CaCl2 in the beaker

(e) the number of Cl– ions in the beaker.

cont'd

44.44 g CaCl2

(a) CaCl2

(b) CaCl2

(c) CaCl2

(d) CaCl2

(e) Cl–

40

Part III Metals

Solution

(a) Formula mass of CaCl2 = 40.1 + 35.5 � 2 = 111.1

(b) Molar mass of CaCl2 = 111.1 g mol–1

(c) Number of moles of CaCl2 =

= mol

= 0.4000 mol

(d) Number of formula units of CaCl2

= number of moles (mol) � Avogadro constant (mol–1)

= 0.4000 � 6.02 � 1023

= 2.41 � 1023

(e) Since 1 formula unit of CaCl2 contains 2 Cl– ions,

number of Cl– ions = 2.41 � 1023� 2

= 4.82 � 1023

44.44111.1

mass (g)

molar mass (g mol–1)

(a) CaCl2 = 40.1 + 35.5 � 2 = 111.1

(b) CaCl2 = 111.1 g mol–1

(c) CaCl2

=

= mol

= 0.4000 mol

(d) CaCl2

= (mol) �(mol–1)

= 0.4000 � 6.02 � 1023

= 2.41 � 1023

(e) CaCl2

Cl–

Cl– = 2.41 � 1023� 2

= 4.82 � 1023

44.44

111.1

(g)

(g mol–1)

N6

Note 6(a) Remind students again that formula mass

has no unit, while molar mass has theunit of g mol

–1.

(b) Tell students not to use ‘· ’ as themultiplication sign ‘�’. Thus it is 35.5 �2, not 35.5 · 2.

Note 7(a) Remind students to write the ‘subject’ of

an expression clearly. Thus it is ‘Molarmass of CaCl2 = 111.1 g mol

–1’, not

‘CaCl2 = 111.1 g mol–1

’.(b) Do not write ‘gm’ for ‘g’, ‘m’ or ‘M’ for

‘mol’.

N7

Example 12.3Calculations involving mole concepts

12.3

Find the mass of (a) 1 Na atom (b) 1 H2O molecule (c) 1 formula unit of NaCl (d) 1 SO4

2– ion.

Solution

One mole of a substance (or species) corresponds to itsmolar mass and contains the Avogadro constant of formulaunits.

∴ mass of 1 formula unit =

(a) Mass of 1 Na atom = g

= 3.82 � 10–23 g

23.0

6.02 � 1023



cont'd

( ) (a) Na(b) H2O(c) NaCl(d) SO4

2–

( )

∴

=

(a) Na

= g

= 3.82 � 10–23 g

23.0

6.02 � 1023

(g mol–1)

(mol–1)

N8

Note 8It is a good practice to write a unit foreach separate line (if applicable).

41


(b) Mass of 1 H2O molecule = g

= 2.99 � 10–23 g

(c) Mass of 1 formula unit of NaCl = g

= 9.72 � 10–23 g

(d) Mass of 1 SO42– ion = g

= 1.60 � 10–22 g

32.1 + 16.0 � 4

6.02 � 1023

23.0 + 35.5

6.02 � 1023

1.0 � 2 + 16.0

6.02 � 1023

(b) H2O

= g

= 2.99 � 10–23 g

(c) NaCl

= g

= 9.72 � 10–23 g

(d) SO42–

= g

= 1.60 � 10–22 g

32.1 + 16.0 � 4

6.02 � 1023

23.0 + 35.5

6.02 � 1023

1.0 � 2 + 16.0

6.02 � 1023


1. Calculate the mass of

(a) 0.200 mol of chlorine atoms.

(b) 0.200 mol of chlorine molecules.

(c) chlorine which contains the same number of molecules as

there are in 1.20 mol of water.

2. Complete the following table.

1.

(a) 0.200

(b) 0.200

(c) 1.20

2.

Substance Molar mass

(g mol–1

)Mass (g)

No. of moles(mol)

Number of molecules/formula unit

Sodium hydroxide(a)0.250

Helium (b) 0.20

Sulphur dioxide(c) 3.01 � 1024

3.01 � 1023Compound X

X(d)

23.0

A12.31. (a) 0.200 � 35.5 = 7.10 g

(b) 0.200 � (35.5 � 2) = 14.2 g(c) 1.20 mol of chlorine (Cl2) contains the same

number of molecules as 1.20 mol of water (H2O).Mass of chlorine = 1.20 � (35.5 � 2) g = 85.2 g

2. (a) Molar mass of sodium hydroxide (NaOH)= 23.0 + 16.0 + 1.0 g mol

–1

= 40.0 g mol–1

Mass of NaOH = 0.250 � 40.0 = 10.0 g

Number of formula units of NaOH= 0.250 � 6.02 � 10

23

= 1.51 � 1023

(b) Since helium is monoatomic, molar mass ofHe = 4.0 g mol

–1.

Number of moles of He molecules = mol

= 0.050 mol

Number of He molecules = 0.050 � 6.02 � 1023

= 3.0 � 1022

0.204.0

(d) Number of moles of X

= mol

= 0.500 molMolar mass of X

= g mol–1

= 46.0 g mol–1

23.00.500

3.01 � 1023

6.02 � 1023

(c) Molar mass of sulphur dioxide (SO2)= 32.1 + 16.0 � 2 g mol

–1

= 64.1 g mol–1

Number of moles of SO2

= mol

= 5.00 mol

3.01 � 1024

6.02 � 1023

Mass of SO2

= 5.00 � 64.1 g= 321 g

42

Part III Metals

Composition from formulae

From the formula of a compound, we can work out the

percentage by mass of each element in the compound. In

general,

12.2 Percentage by mass of an element in acompound

12.2

Fraction by mass of element A in a compound

= relative atomic mass of A � no. of atoms of A in formulaformula mass of the compound

Percentage by mass of element A in a compound

= � 100%

relative atomic mass of A � no. of atoms of Ain formula

formula mass of the compound

A

=

A �

A

A

= � 100%

A �

A

Example 12.4Calculating the percentage by mass of an element in acompound

Calculate the percentage by mass of copper in copper(II)sulphate-5-water, CuSO4 · 5H2O.

Solution

Formula mass of CuSO4 · 5H2OCu S O4 5H2O

= 63.5 + 32.1 + 16.0 � 4 + 5 � (1.0 � 2 + 16.0)= 249.6

% by mass of Cu

= � 100%

= � 100%

= 25.4%

Self-Test 12.4

Calculate the percentage by mass of sulphur, oxygen andhydrogen in copper(II) sulphate-5-water, CuSO4 · 5H2O.

63.5249.6

relative atomic mass of Cu � no. of Cu atoms in formula

formula mass of CuSO4 · 5H2O

12.4

(II) CuSO4 · 5H2O

CuSO4 · 5H2O

Cu S O4 5H2O= 63.5 + 32.1 + 16.0 � 4 + 5 � (1.0 � 2

+ 16.0)= 249.6

Cu

= � 100%

= � 100%

= 25.4%

(II) CuSO4 · 5H2O

63.5249.6

Cu �

CuCuSO4 · 5H2O

Self-Test 12.4% by mass of S

= � 100% = 12.9%

% by mass of O

= � 100%

= 57.7%

% by mass of H

= � 100%

= 4%

1.0 � 10249.6

16.0 � 4 + 16.0 � 5249.6

32.1249.6

43


Example 12.5Calculating the mass of water of crystallization in ahydrated salt

Find the mass of water of crystallization in 15.0 g ofcopper(II) sulphate-5-water.

Solution

Mass of water of crystallization

= mass of CuSO4 · 5H2O � fraction by mass of water inCuSO4 · 5H2O

= mass of CuSO4 · 5H2O �

= 15.0 g �

= 5.41 g

Self-Test 12.5

Find the mass of water of crystallization in 20 g of FeCl2 ·4H2O.

5 � (1.0 � 2 + 16.0)

249.6

5 � formula mass of H2O

formula mass of CuSO4 · 5H2O

12.5

15.0 g (II)

= CuSO4 · 5H2O �

CuSO4 · 5H2O

= CuSO4 · 5H2O �

= 15.0 g �

= 5.41 g

20 g FeCl2 · 4H2O

5 � (1.0 � 2 + 16.0)249.6

5 � H2OCuSO4 · 5H2O

Relative atomic mass from formulae

Relative atomic mass of an element may be calculated from

formula of its compound and percentage mass of the element

in the compound.

Example 12.6Calculating relative atomic mass of an element

12.6

The chloride of a metal M has the formula MCl3 andcontains 34.4% by mass of M. Find the relative atomic massof M.

Solution

Let the relative atomic mass of M be a.

Fraction by mass of M in MCl3

=

=

∴ a = 55.8

Hence the relative atomic mass of M is 55.8.

a

a + 35.5 � 3

34.4

100

relative atomic mass of M � 1

formula mass of MCl3

cont'd

M MCl 3

M 34.4%M

M a

MCl3 M

=

=

∴ a = 55.8

M 55.8

a

a + 35.5 � 3

34.4

100

M � 1MCl3

Self-Test 12.5Mass of water crystallization

= 20 g �

= 7.24 g

4 � (1.0 � 2 + 16.0)(55.8 + 35.5 � 2) + 4 � (1.0 � 2 + 16.0)

44

Part III Metals

Self-Test 12.6

The bromide of a metal X has the formula XBr2 and contains25.6% by mass of X. Find the relative atomic mass of X.

X XBr2

X 25.6% X

1. 100 g (NaNO3)

2. 4.6 g

(Na2CO3 · 10H2O)

3. MO M

79.87% M

4. 26.88 g MCl

5.68 g M

12.4

1. What is the mass of nitrogen present in the sample of

sodium nitrate (NaNO3) which contains 100 g of sodium?

2. What is the mass of water of crystallization present in the

sample of sodium carbonate-10-water (Na2CO3 · 10H2O)

which contains 4.6 g of sodium?

3. A metal oxide MO contains 79.87% by mass of the metal

M. Find the relative atomic mass of M.

4. 26.88 g of a metal chloride MCl contains 5.68 g of chlorine.

Find the relative atomic mass of the metal M.

Class practice 12.4

4. % by mass of Cl in MCl = � 100% = 21.13%

∴ % by mass of M in MCl = 100% – 21.13%= 78.87%

Let the relative atomic mass of M be a.Fraction by mass of M in MCl

=

=

∴ a = 132.5

aa + 35.5

78.87100

relative atomic mass of M � 1formula mass of MCl

5.6826.88

3. Let the relative atomic mass of M be a.Fraction by mass of M in MO

=

=

∴ a = 63.5

aa + 16.0

79.87100

relative atomic mass of M � 1formula mass of MO

2. Number of moles of Na = mol = 0.2 mol

Since number of moles of Na : number of moles ofNa2CO3 · 10H2O = 2 : 1∴ number of moles of Na2CO3 · 10H2O = 0.1 molMass of Na2CO3 · 10H2O= 0.1 � (23.0 � 2 + 12.0 + 16.0 � 3 + 10 � (1.0

� 2 + 16.0)) g = 28.6 g

% by mass of H2O in Na2CO3 · 10H2O

=

= 62.9%Mass of H2O= 28.6 g � 62.9%= 17.99 g

4.623.0

A12.41. Number of moles of Na

= mol = 4.35 mol

Since number of moles of Na : number of moles of NaNO3 = 1 : 1

10023

∴ number of moles of NaNO3 = 4.35 molMass of NaNO3

= 4.35 � (23.0 + 14.0 + 16.0 � 3) g = 369.75 g% by mass of N in NaNO3

= � 100% = 16.5%

Mass of N = 369.75 g � 16.5% = 61.0 g

14.023.0 + 14.0 + 16.0 � 3

10 � (1.0 � 2 + 16.0)(23.0 � 2 + 12.0 + 16.0 � 3 + 10 � (1.0 � 2 + 16.0))

Empirical formula

The empirical formula of a compound is the formula which

shows the simplest whole number ratio of the atoms or ions

present. It is applicable to all compounds.

Ionic formula

The ionic formula of an ionic compound is the formula which

shows the simplest whole number ratio of the ions present, and

also the charges carried by them.

Molecular formula

The molecular formula of a substance shows the actual number

of each kind of atoms in one molecule of the substance. It is only

applicable to molecular compounds and elements consisting of

molecules.

12.3 Chemical formulae of compounds 12.3

Self-Test 12.6Let the relative atomic mass of X be a.

Fraction by mass of X in XBr2 =

=

a = 55.0

aa + 79.9 � 2

25.6100

relative atomic mass X � 1formula mass of XBr2

� 100%

N9

Note 9Tell students not to write ionic formulae when writingchemical equations. For example, the reaction betweenmagnesium and silver nitrate should be written as

Mg(s) + 2AgNO3(aq) Mg(NO3)2(aq) + 2Ag(s)

but notMg(s) + 2Ag

+NO3

–(aq) Mg

2+(NO3

–)2(aq) + 2Ag(s)

45


Structural formula

The structural formula of a molecular substance is the formula

which shows how the constituent atoms are joined up in one

molecule of the substance.

The following flow chart (Figure 12.1) shows the sequence

and methods by which the various types of formulae are

found.

( 12.1)

for ionic compounds

Example magnesium chloride(an ionic compound)

( )

Compound underinvestigation

Example ethanoic acid

(a molecular compound)

( )

qualitative analysis

Mg, ClElements present in the

compound C, H, O

quantitative analysis (to find composition by mass)( )

MgCl2Empirical formula CH2O

for molecular compounds

finding the charges of ions present determination of relative molecular mass

Mg2+(Cl–)2

Ionicformula

Molecularformula C2H4O2

study of properties of compound

Structuralformula CC

H

O HH

O

H

Figure 12.1 Determination of variousformulae, with examples.

Some examples of empirical, ionic, molecular and

structural formulae of a few substances are given in Table 12.1.12 .1

46

Part III Metals

SiO2Quartz

(K+)2Cr2O72–K2Cr2O7

Potassium dichromate

Fe2+SO42– . 7H2OFeSO11H14

Iron(II) sulphate-7-water

(II)

NH4+Cl–NH4ClAmmonium chloride

C2H6OMethoxymethane

C2H6O C2H6O

C2H6O

Ethanol

C3H6Propene

C2H4CH2

CH2

Ethene

N�N

O=C=O

N2– –

–

–

–

–

–

Nitrogen

Structural formulaMolecularformulaIonic formulaEmpirical

formulaSubstance

Carbon dioxide CO2 CO2

C HCH

H H

C CC

H

H H

H H H

C OC

H

H H

H H

H

O CC

H

H H

H H

H

– –

– –

– –

–– –

Table 12.1 The different formulae of some substances.

We should note that the empirical and molecular formulae

of a compound may be the same (e.g. carbon dioxide) or

different (e.g. ethene). The molecular formula is the empirical

formula multiplied by some whole number (1, 2, 3, etc.).

Even different compounds may have the same empirical

formula and same molecular formula (e.g. ethanol and

methoxymethane). But they have different structural formulae.

( ) (

)

( 1 2 3 )

( )

12.2

47


-1-

(a) (b)

12.5

But-1-ene has the structural formula .

Write down its (a) molecular formula and (b) empirical formula.

Class practice 12.5

C CC

H

H H

H H H

C

H

H

A12.5(a) C4H8

(b) CH2

C CC

H

H H

H H H

C

H

H

12.4 Determination of empirical formulae 12.4

The empirical formula of a compound can be calculated from

its composition by mass. The composition of a compound has

to be determined by experiment.

Example 1

Finding the empirical formula of copper oxide

To determine the empirical formula of copper oxide, we have

to find the ratio by mass of copper and oxygen in the

compound. We can pass town gas over a known mass of heated

copper oxide. The hydrogen and carbon monoxide in the town

gas remove oxygen from the copper oxide; this leaves reddish

brown copper whose mass is found. A set-up for doing the

experiment is shown in Figure 12.2.

town gassupply

copper oxide

heat

hole in test tube

excess town gasburns here

Figure 12.2 To find the empirical formula of copper oxide by passing town gas over the heatedcompound.

48

Part III Metals

The empirical formula is worked out by changing the

composition by mass to the simplest whole number mole ratio,

as illustrated below:

Specimen results

Mass of test tube = 18.100 g

Mass of test tube + copper oxide = 18.701 g

Mass of test tube + copper = 18.579 g

Mass of copper in oxide = (18.579 – 18.100) g

= 0.479 g

Mass of oxygen in oxide = (18.701 – 18.579) g

= 0.122 g

Calculations

∴ the empirical formula of copper oxide is CuO.

Example 2

Finding the empirical formula of magnesium oxide

A known mass of magnesium is heated strongly in a crucible

(also of known mass) until it catches fire (see Figure 12.3). The

crucible lid is carefully lifted up slightly from time to time. This

lets in air to react with magnesium.

= 18.100 g

+ = 18.701 g

+ = 18.579 g

= (18.579 – 18.100) g

= 0.479 g

= (18.701 – 18.579) g

= 0.122 g

Number of moles of atoms (mol)

( =mass in g

) molar mass

(mol)

0.1220.479

0.479= 0.007 54

63.5

Masses (in g) ( g )

Cu O

0.122= 0.007 63

16.0

Relative number of moles(divided by the smallest number) 0.007 54

= 10.007 54

0.007 63= 1.01

0.007 54� 1

∴ CuO

( )

(

12.3)

49


magnesium coil crucible

rocksil

tripod

pipeclaytriangle

heat verystrongly

Figure 12.3 To find the empirical formula ofmagnesium oxide by heating magnesium in air.

From the experimental results, the empirical formula of

magnesium oxide can be worked out to be MgO. Try the

experiment yourselves.MgO

1.

1.200 g

1.173 g 0.240 g

2. CxHy

1.000 g

0.857 g x y

12.6

1. 1.200 g of a compound containing only carbon, hydrogen

and oxygen gave 1.173 g of carbon dioxide and 0.240 g of

water on complete combustion. Find the empirical formula

of the compound.

2. A compound has the empirical formula CxHy. On analysis,

1.000 g of the compound is found to contain 0.857 g of

carbon. Find the values of x and y.

Class practice 12.6

12.5 Determination of molecular formulae 12.5

Molecular formula may be determined from empirical formula

and relative molecular mass. This is because molecular formula

is a whole number multiple of empirical formula.

Example 12.7Determining empirical formula and molecular formulausing percentage by mass

Compound X was found to contain carbon and hydrogenonly. Experiments showed that it contained 80% carbon and20% hydrogen by mass. If its relative molecular mass was30.0, find its empirical formula and molecular formula.

XX

80% 20%30.0

12.7

cont'd

∴ empirical formula of the compound is CHO2.

2.

∴ empirical formula of the compound is CH2.

A12.61. Let CxHyOz be the empirical formula of the compound.

Number of moles of CO2 = mol = 0.0267 mol

∴ number of moles of C = 0.0267 mol

Number of moles of H2O = mol = 0.0133 mol

∴ number of moles H = 0.0133 � 2 mol = 0.0266 molMass of C in the compound = 0.0267 � 12.0 g = 0.3204 gMass of H in the compound = 0.0266 � 1.0 g = 0.0266 g∴ mass of O in the compound = (1.200 – 0.3204 – 0.0266) g = 0.849 g

0.24(1.0 � 2 + 16.0)

1.17344.0

∴ number of moles of O in the compound = mol

= 0.053 mol

0.84916.0

Masses (in g)

Number of moles(mol)

Relative numberof moles

0.3204

C

0.0267

� 1

0.02670.0266

0.0266

H

0.0266

= 1

0.02660.0266

0.849

O

0.053

� 2

0.0530.0266

Masses (in g)

Number ofmoles (mol)


0.857

C

= 0.07142

0.85712.0

= 1

0.071420.07142

0.143

H

= 0.143

0.1431.0

� 2

0.1430.07142

50

Part III Metals

Solution

Assume there were 100 g of X, then there would be 80 g ofcarbon and 20 g of hydrogen.

100 g X 80 g20 g

∴ the empirical formula of X is CH3.

Note: Experimental errors would probably result in a smalldifference from whole numbers. In calculating relativenumber of moles, it is an accepted practice to ‘round off’these values to the nearest whole numbers. However, caremust be taken in doing so, e.g. 2.98 can be taken to be 3, but2.8 is usually not taken as 3.

Let its molecular formula be (CH3)n, where n is the wholenumber.

Relative molecular mass of (CH3)n = 30.0

n(12.0 + 1.0 � 3) = 30.0

15.0n = 30.0

∴ n = 2

∴ molecular formula of X is C2H6.

X CH3

2.983 2 . 8

3

(CH3)n

n

(CH3)n = 30.0

n(12.0 + 1.0 � 3) = 30.0

15.0n = 30.0

∴ n = 2

X C2H6

80 = 6.712.0


( =mass in g

) molar mass

(mol)

( =(g)

)

80 20

C H

Masses (in g) ( g )

20 = 201.0

6.7 = 16.7

Relative number of moles(divided by the smallest number)

( )

20 = 2.98 � 36.7

Example 12.8Determining empirical formula and molecular formulausing masses of combustion products

A compound Y containing only carbon, hydrogen andoxygen burned completely in air to form carbon dioxide andwater as the only products. 2.43 g of Y gave 3.96 g of carbondioxide and 1.35 g of water. Find the empirical formula of Y.If its relative molecular mass was 160, find also its molecularformula.

Y

2.43 g Y3.96 g 1.35 g

Y Y160

12.8

cont'd

Note 10Refer to ‘Supplementary information: Combustion analysis oforganic compounds’ in the Teacher’s Guide.

N10

51


Solution

Since all the C in CO2 and H in H2O came from thecompound,

mass of C in the compound = 3.96 g �

= 1.08 g;

mass of H in the compound = 1.35 g �

= 0.15 g

The rest of mass of the compound must come from oxygen.

∴ mass of O in compound = (2.43 – 1.08 – 0.15) g = 1.20 g

Now go on to find the empirical formula as follows:

1.0 � 2

1.0 � 2 + 16.0

12.0

12.0 + 16.0 � 2

CO2 C H2O H

C

= 3.96 g � = 1.08 g

H

= 1.35 g � = 0.15 g

O

= (2.43 – 1.08 – 0.15) g = 1.20 g

1.0 � 2

1.0 � 2 + 16.0

12.0

12.0 + 16.0 � 2

Note: We should not take 1.2 to be equal to 1, because wecannot allow for so large an experimental error.

∴ empirical formula of compound is C6H10O5.

Let its molecular formula be (C6H10O5)n, where n is awhole number.

Relative molecular mass of (C6H10O5)n = 160n(12.0 � 6 + 1.0 � 10 + 16.0 � 5) = 160

162n = 160n = 0.988

� 1

∴ molecular formula of compound Y is C6H10O5.

1.2 1

∴ C6H10O5

(C6H10O5)n

n

(C6H10O5)n = 160n(12.0 � 6 + 1.0 � 10 + 16.0 � 5) = 160

162n = 160n = 0.988

� 1

Y C6H10O5

1.20

O

1.20 = 0.07516.0

0.075 = 10.075

1.08 = 0.09012.0


( =mass in g

) molar mass

(mol)

( =(g)

)

1.08 0.15

C H

Masses (in g) ( g )

0.15 = 0.151.0

0.090 = 1.20.075


( )

0.15 = 20.075

1 � 5 = 51.2 � 5 = 6

(multiplied by the smallest possible whole number

(5 here) to turn all values into whole numbers)

( ( )

)

2 � 5 = 10

52

Part III Metals

Example 12.9Determining molecular formula using empirical formulaand relative molecular mass

A compound has an empirical formula CH2 and a relativemolecular mass of 42. Find its molecular formula.

Solution

Let the molecular formula of the compound be (CH2)n,where n is a whole number.

Relative molecular mass of (CH2)n = 42

n(12.0 + 1.0 � 2) = 42

n = 3

∴ the molecular formula is (CH2)3, i.e. C3H6.

12.9

C H 2

42

(CH 2) n

n

(CH2)n = 42

n(12.0 + 1.0 � 2) = 42

n = 3

∴ (CH 2) 3

C3H6

0.81 g 1.32 g

0.45 g

320

12.7

A compound containing only carbon, hydrogen and oxygen gave

the following results on analysis: 0.81 g of the substance gave

1.32 g of carbon dioxide and 0.45 g of water on complete

combustion. Find the empirical formula of the compound. If the

relative molecular mass was 320, find also its molecular formula.

Class practice 12.7

Example 12.10Determining the number of water of crystallization

5.60 g of hydrated copper(II) sulphate CuSO4 · nH2O washeated in a crucible to drive off the water of crystallization.The white residue was anhydrous copper(II) sulphate,which was found to have a mass of 3.59 g.(a) Deduce a reasonable value for n.(b) Explain why the answer you gave in (a) differs a bit

from the value actually calculated.

5.60 g (II) (CuSO4 ·nH2O)

(II) 3.59 g

(a) n

(b) (a)

12.10

cont'd

Now go on to find the empirical formula as follows:

A12.7Since all the C in CO2 and H in H2O came from the compound,

mass of C in the compound = 1.32 g � = 0.36 g

mass of H in the compound = = 0.05 g0.45 g � 1.0 � 2

1.0 � 2 + 6.0

12.012.0 + 16.0 � 2 The rest of the compound must be oxygen.

∴ Mass of O in compound = (0.81 – 0.36 – 0.05) g = 0.40 g

∴ empirical formula of compound is C6H10O5

Relative number ofmoles (divided by thesmallest number)Multiplied by thesmallest possiblewhole number (5here) to turn all valuesinto whole number

1.2 � 5 = 6 2 � 5 = 10 1 � 5 = 5

C

= 1.20.030.025

H

= 20.050.025

O

= 10.0250.025

Masses (in g)

Number of moles ofatoms (mol)

(= )

0.36C

= 0.030.3612

0.05H

= 0.050.051.0

0.40O

= 0.0250.4016mass in g

molar mass

Let its molecular formula be (C6H10O5)n, where n is a whole number.Relative molecular mass of (C6H10O5)n = 320n(12.0 � 6 + 1.0 � 10 + 16.0 � 5 ) = 320

162n = 320n = 2

∴ molecular formula of compound is C12H20O10.

53


Solution

(a) Mass of water of crystallization = (5.60 – 3.59) g = 2.01 g

Formula mass of CuSO4 = 63.5 + 32.1 + 16.0 � 4= 159.6

Relative molecular mass of H2O = 1.0 � 2 + 16.0= 18.0

n should be a whole number. A reasonable value of nwould therefore be 5.

(b) The experimental value of n (4.98) is lower than 5. Thismight be due to two reasons:

(1) Not all water of crystallization has been removedin the heating process.

(2) Weighing had been delayed, so that theanhydrous salt had absorbed some moisture fromthe atmosphere.

Self-Test 12.10

When 9.99 g of hydrated iron(II) chloride (FeCl2 · nH2O) washeated in a crucible, 5.41 g of anhydrous iron(II) chloridewas left. Find the value of n.


(a) = (5.60 – 3.59) g = 2.01 g

CuSO4 = 63.5 + 32.1 +16.0 � 4

= 159.6

H2O = 1.0 � 2 + 16.0= 18.0

n5

(b) n (4.98) 5

(1)

(2)

9.99 g (II) (FeCl2 ·nH2O) 5.41 g

(II) n

➲ 21

3.59 = 0.0225159.6

Number of moles (mol)

( =mass in g

) molar mass

(mol)

( =(g)

)

3.59 2.01

CuSO4 H2O

Masses (in g) ( g )

2.01 = 0.11218.0

0.0225 = 10.0225


( )

0.112 = 4.980.0225

12.6 Calculations based on equations 12.6

Calculations from equations — reacting masses

The theoretical amounts of substances used up or produced in

a reaction can be calculated from its balanced equation as

shown below:

—

Formula mass of FeCl2 = 55.8 + 35.5 � 2 = 126.8

Self-Test 12.10Mass of water of crystallization = (9.99 – 5.41) g = 4.58 g

Mass (in g)

Number of moles(mol)


5.41

FeCl2

= 0.04275.41126.8

= 10.04270.0427

4.58

H2O

= 0.25444.5818.0

≈ 60.25440.0427

∴ n = 6

54

Part III Metals

Calculating the reacting masses

Step 1 Write down the balanced equation for the reaction.


Step 2 Convert the amounts of given substances into molequantities.

(Assume that the mass of Mg is 2.43 g.)

Molar mass of Mg = 24.3 g mol–1

Number of moles of Mg =

= mol

= 0.100 mol

Step 3 Calculate the mole quantities of the required substancesusing the ratio, as given by the stoichiometric coefficientsof the equation.Since oxygen is in excess, all the Mg is changed toMgO. (Mg is called the limiting reactant in this case,as it is all used up. Obviously, amounts of productsformed depend on the amount of the limitingreactant only.)

From the equation, mole ratio of Mg : MgO = 2 : 2 = 1 : 1

Hence number of moles of MgO formed = 0.100 mol

Step 4 Change the mole quantities of the required substances backinto mass as required by the question.

Molar mass of MgO = (24.3 + 16.0) g mol–1

= 40.3 g mol–1

� mass of MgO formed = 0.100 � 40.3 g= 4.03 g

The steps taken are illustrated below:

2.4324.3

mass (g)



12Mg(s) + O2(g) 2MgO(s)

2

( Mg 2.43 g )

Mg = 24.3 g mol–1

Mg

=

= mol

= 0.100 mol

3

MgMg

MgO Mg

()

Mg MgO= 2 : 2 = 1 1

MgO =0.100 mol

4

MgO= (24.3 + 16.0) g mol–1

= 40.3 g mol–1

� MgO= 0.100 � 40.3 g= 4.03 g

2.4324.3

(g)

(g mol–1)

by ratio

Requiredinformation of B

Giveninformation of A

Number ofmoles of A

Number ofmoles of B

B

A A

B

55


Study the following examples.

Example 12.11Calculations based on equation (with limiting reactant)

Calculate the mass of magnesium oxide formed when 2.43 gof magnesium are burnt with 1.28 g of oxygen.

Solution

Step 1: 2Mg(s) + O2(g) 2MgO(s)

Step 2: Molar mass of Mg = 24.3 g mol–1

Number of moles of Mg = mol

= 0.100 mol

Molar mass of O2 = 32.0 g mol–1

Number of moles of O2 = mol

= 0.0400 mol

Step 3: From the equation, mole ratio of Mg : O2 = 2 : 1.

∴ 0.0400 mol of O2 would react with 0.0400 � 2 =0.0800 mol of Mg

Since 0.100 mol of Mg is used, Mg is in excess.

O2 is the limiting reactant in this case, as it is all usedup.

From the equation, mole ratio of O2 : MgO = 1 : 2,

∴ number of moles of MgO = 0.0400 � 2 mol

= 0.0800 mol

∴ mass of MgO formed = 0.0800 � 40.3 g

= 3.22 g

1.2832.0

2.4324.3

12.11

2.43 g 1.28 g

1 2Mg(s) + O2(g) 2MgO(s)

2 Mg = 24.3 g mol–1

Mg = mol

= 0.100 mol

O2 = 32.0 g mol–1

O2 = mol

= 0.0400 mol

3 M g O 2

= 2 1

∴ 0.0400 mol O2 0.0400 � 2= 0.0800 mol Mg

0.100 mol MgMg

O2

O2

O2 MgO= 1 2

∴ MgO= 0.0400 � 2 mol= 0.0800 mol

∴ MgO= 0.0800 � 40.3 g= 3.22 g

1.2832.0

2.4324.3

Example 12.12

Calculations based on equation (complete reaction)

Calculate the mass (in kg) of copper produced by thecomplete reaction of 1.59 kg of copper(II) oxide in thefollowing reaction:

CuO(s) + H2(g) Cu(s) + H2O(�)

1.59 kg (II) ( kg

)

CuO(s) + H2(g) Cu(s) + H2O(�)

12.12

cont'd

N11

Note 11This example illustrates what to do if the mass givenis not in g unit. Some students may just divide 1.59by 79.5 to calculate number of moles of CuO.

56

Part III Metals

Solution

The only substances involved in calculations here are CuOand Cu.


CuOCu

➲ 22

First method

Step 1: CuO(s) + H2(g) Cu(s) + H2O(�)1 CuO(s) + H2(g) Cu(s) + H2O(�)

Step 2: Mass of CuO = 1.59 kg= 1.59 � 1000 g = 1590 g

Molar mass of CuO = (63.5 + 16.0) g mol–1

= 79.5 g mol–1

Number of moles of CuO = mol = 20.0 mol

2 CuO = 1.59 kg

= 1.59 � 1000 g = 1590 g

CuO = (63.5 + 16.0) g mol–1

= 79.5 g mol–1

CuO = mol = 20.0 mol

Step 3: From equation, mole ratio of CuO : Cu = 1 : 1. ∴ number of moles of Cu = 20.0 mol

mass of Cu produced = 20.0 � 63.5 g= 1270 g = 1.27 kg

3 CuO Cu = 1 1

∴ Cu = 20.0 molCu = 20.0 � 63.5 g

= 1270 g = 1.27 kg

159079.5

159079.5

Second method

Since 1 mole of CuO produces 1 mole of Cu,

∴ 79.5 g of CuO produces 63.5 g of Cu, or

79.5 kg of CuO produces 63.5 kg of Cu.

∴ 1.59 kg of CuO produces (63.5 � )

= 1.27 kg of Cu.

1 mol CuO 1 mol Cu

∴ 79.5 g CuO 63.5 g Cu

79.5 kg CuO 63.5 kg Cu

∴ 1.59 kg CuO (63.5 � )

= 1.27 kg Cu

1.5979.5

1.5979.5

Example 12.13

5.91 g of iron was dissolved in excess dilute hydrochloricacid to give a solution containing Fe2+ ions. The solution wasthen boiled with concentrated nitric acid to oxidize all Fe2+

ions into Fe3+ ions. Excess sodium hydroxide solution wasadded to precipitate all Fe3+ ions as iron(III) hydroxide,Fe(OH)3. The precipitate was filtered off, washed, dried andfinally heated to convert all into iron(III) oxide, Fe2O3.

5.91 gFe2+

Fe2+ Fe3+

F e 3 + ( I I I )Fe(OH)3

Fe(OH)3

(III) Fe2O3

12.13

cont'd

Calculations based on equation (actual yield andtheoretical yield)

57


(a) Calculate the theoretical mass of iron(III) oxideobtained (the theoretical yield).

(b) The mass of iron(III) oxide actually obtained fromexperiment (actual yield) was 7.95 g. Compare this withthe theoretical mass in (a) and give two possiblereasons for the difference.

Solution

(a) Step 1: The whole process may be represented by asequence of steps:

Fe Fe2+ Fe3+ Fe(OH)3 Fe2O3

5.91 g ? g

Step 2: To get the answer, one method is to writebalanced equations for each of the reactions.From these, we can calculate masses of Fe2+, Fe3+

and Fe(OH)3 in turn, and finally that of Fe2O3.

A much simpler method is to write the overallexpression representing the mole ratio of thegiven substance (Fe) and the required substance(Fe2O3), without writing any equations:

2Fe Fe2O3 (the ‘2’ before Fe is added tobalance number of Fe atoms)

Thus mole ratio of Fe : Fe2O3 = 2 : 1.

Molar mass of Fe = 55.8 g mol–1

Molar mass of Fe2O3 = 55.8 � 2 + 16.0 � 3 = 159.6 g mol–1

Number of moles of Fe used = mol

= 0.106 mol

Step 3: Number of moles of Fe2O3 formed = mol

= 0.0530 mol

∴ theoretical mass of Fe2O3 formed = 0.0530 �

159.6 g

= 8.46 g

(b) The actual yield (7.95 g) is smaller than the theoreticalyield (8.46 g).

Possible reasons are:

(1) Iron used might be impure.

(2) There was loss of materials during the variousexperimental processes, e.g. filtration.

0.1062

5.9155.8

(a) (III) ( )

(b) (III) ( ) 7.95 g (a)

(a) 1

Fe Fe2+ Fe3+ Fe(OH)3 Fe2O3

5.91 g ? g

2

F e 2 + F e 3 +

Fe(OH)3

Fe2O3

(Fe) (Fe2O3)

2Fe Fe2O3 (Fe 2F e

)

Fe Fe2O3 =2 1

Fe = 55.8 g mol–1

Fe2O3 = 55.8 � 2+ 16.0 � 3 = 159.6 g mol–1

Fe

= mol = 0.106 mol

3 Fe2O3

= mol = 0.0530 mol

∴ Fe2O3

= 0.0530 � 159.6 g = 8.46 g

(b) (7.95 g) (8.46g)

(1)(2) (

)

0.1062

5.9155.8

58

Part III Metals

1.5 g

(a)

(b)

(c)

12.8

A student performed the following experiment to obtain calcium

hydroxide from calcium metal. 1.50 g of calcium granules was

dissolved in large amount of water. The precipitate of calcium

hydroxide was filtered off, washed and dried.

(a) Write down the equation for the reaction of calcium with

water.

(b) Calculate the theoretical mass of calcium hydroxide

obtained.

(c) The mass of calcium hydroxide obtained from the

experiment was much less than the theoretical value.

Explain why there is such difference.

Class practice 12.8


1. actual yield 56

Page

3. composition by mass 47

4. empirical formula 44

7. limiting reactant 54

8. molar mass 38

9. mole 37

10. molecular formula 44

11. percentage by mass 42

12. structural formula 45

13. theoretical yield 56

5. fraction by mass 42

2. Avogadro constant 37

6. ionic formula 44

A12.8(a) Ca(s) + 2H2O(�) Ca(OH)2(aq) + H2(g)(b) Mole ratio of Ca : Ca(OH)2 = 1 : 1

Molar mass of Ca = 40.1 g mol–1

Molar mass of Ca(OH)2 = 40.1 + (16.0 + 1.0) � 2 g mol–1

= 74.1 g mol–1

Number of moles of Ca used = mol

= 0.0374 mol

Number of moles of Ca(OH)2 formed = 0.0374 mol∴ theoretical mass of Ca(OH)2 formed

= 0.0374 � 74.1 g= 2.77 g

(c) (1) Calcium used might be impure.(2) There was loss of material during the various

experimental processes, e.g. filtration.1.5040.1

59


SS uu mm mm aa rr yy12.1 The mole concept in general

1. Chemists use (abbreviation: mol) to show the number of formula units.

2. The is the number of atoms in exactly 12 g of carbon-12. It isequal to 6.02 � 1023 mol–1.

3. One of a substance (or species) represented by a formula is the amountcontaining 6.02 � 1023 formula units.

4. The of a substance (or species) is the mass in grams of onemole of it. The unit of molar mass is g mol–1.

5. Important relationships involving moles:

• Mass of 1 mole of a substance (or species) = formula mass expressed in gram unit

• Number of moles (mol) =

• Number of moles (mol) =

• Mass of 1 formula unit (g) =

12.2 Percentage by mass of an element in a compound

6. The percentage by mass of an element in a compound can be found by the equation:

Percentage by mass of element A in a compound

= � 100%

12.3 Chemical formulae of compounds

7. are part of language of chemistry. They give informationabout the substances concerned and are not just short-hand representation. Some commonchemical formulae include formula, formula,

formula and formula.

______________________________ mass of A � no. of atoms of A in formula

formula mass of the compound

________________________________ (g mol–1)


number of _______________________________


______________ (g)


structural

relative atomic

molar mass

formula units

mass

molar mass

mole

Avogadro constant (L)

mole

ionic

molecular

empirical

Chemical formulae

60

Part III Metals

12.4 Determination of empirical formulae

8. formula of a compound is the formula which shows the simplest wholenumber ratio of the atoms or ions present.

9. The empirical formula of a compound can be calculated from its. The composition of a compound has to be determined by

experiment.

12.5 Determination of molecular formulae

10. may be determined from empirical formula and relativemolecular mass. This is because molecular formula is a whole number multiple of empiricalformula.

12.6 Calculations based on equations

11. The amounts of substances used up or produced in a reaction can be calculatedfrom its balanced equation.

theoretical

Molecular formula

composition by mass

Empirical

61

Chapter 13 Corrosion of metals and their protection

13.1 Corrosion of metals 13.1

Most metals corrode. Generally, a metal higher in the metal

reactivity series will corrode more rapidly.

Rusting refers to the corrosion of iron. As iron is the most

widely used metal, rusting is the most common type of

corrosion.

N1

Note 1We can say ‘corrosion of copper’ but not ‘rusting of copper’.

The reaction of a metal with air, water or other substances

in the surroundings, leading to gradual deterioration of the

metal, is called corrosion.

✘ Corrosion of copper is also called rusting.

✔ Rusting refers to the corrosion of iron only. We can say‘corrosion of copper’, but not ‘rusting of copper’.

Check your concept

✘

✔

13.2 Rusting 13.2

Conditions for rusting

We can use the set-up shown in Figure 13.1 to find whether

both water and air are involved in rusting. We have to leave the

test tubes to stand for several days.

13.1

Anhydrous calcium chloride is a drying agent. It removes water(moisture) from the air.

Learning tip

( )

62

Part III Metals

anhydrous calcium chloride

cotton wool

iron nail boiled distilledwater

oil layer

distilled water

no rusting rusting occurs

Tube 11

Tube 22

Tube 33Figure 13.1 Iron rusts only in the presence of

both water and air.

( )

For rusting to occur, two things must be present: water and

oxygen.

1. 13.1 1

2

3

2.

13.1

1. Refer to Figure 13.1. Explain why Tube 1 has no water, Tube

2 has no air, while Tube 3 has both water and air.

2. Iron rusts on the Earth. Will a piece of iron rust on the

Moon?

Class practice 13.1

A13.11. In Tube 1, moisture in air has been absorbed by anhydrous

calcium chloride. In Tube 2, dissolved air in water has beendriven out by boiling. Besides, the oil layer on top preventsair from dissolving in water again.

Iron nail is immersed in distilled water in Tube 3. Distilled water containsdissolved air.

2. No; there is no air on the Moon. (In March 1998, NASA revealed that there was strong evidence for theexistence of a large quantity of ice at the poles of the Moon. Thus the old ideathat there was no water on the Moon might have to be changed.)

Simple chemistry of rusting

Rusting is a slow chemical process. In the first stage of rusting,

some iron atoms lose electrons to form Fe2+(aq) ions.

Fe(s) Fe2+(aq) + 2e–

Then a series of reactions follow. The overall reaction can

be represented by an equation:

4Fe(s) + 3O2(g) + 2nH2O(�) 2Fe2O3 · nH2O(s)rust (reddish brown)

Rust is in fact hydrated iron(III) oxide (Fe2O3 · nH2O),

where n is a variable number.

N2

Note 2The theory of rusting is beyondthe scope of the HKDSE syllabus.

Fe2+(aq)

Fe(s) Fe2+(aq) + 2e–

( )

(III) (Fe2O3 ·

nH2O) n

4Fe(s) + 3O2(g) + 2nH2O(�) 2Fe2O3 · nH2O(s)

63


Example 13.1Classifying chemical reactions in terms of their rates

Different chemical reactions occur at different rates.Reactions can be roughly classified into three typesdepending on rate:

(1) Instantaneous reactions

(2) Reactions with a moderate rate

(3) Slow reactions

Give two examples of each type.

Solution

(1) Instantaneous reactions: explosions (Figure 13.2),precipitations.

(2) Reactions with a moderate rate: addition of magnesiumto dilute hydrochloric acid; addition of calcium towater.

(3) Slow reactions: rusting of iron; corrosion of stonework(Figure 13.3).

(1)

(2)

(3)

(1) ( 13.2)

(2)

(3)( 13.3)

13.1

Figure 13.2 An explosion is a very fast reaction. Figure 13.3 Corrosion of stonework is a slow reaction.

13.3 Factors that speed up rusting 13.3

Presence of acidic solutions or soluble salts

Acidic solutions increase the speed of rusting.

Soluble salts (e.g. sodium chloride) also speed up rusting. ( )

N3 &N4

Note 3Acids speed up rusting because they(1) promote the formation of Fe

2+(aq):

Fe(s) Fe2+

(aq) + 2e–

(2) increase the conductivity of solution.

Note 4Alkaline solutions (e.g. 0.1 M NaOH) will notspeed up rusting. On the contrary, they willinhibit rusting by shifting the position ofequilibrium of the following cathodic reactionto the left:

O2(g) + 2H2O(�) + 4e–

4OH–(aq)

64

Part III Metals

High temperature

An increase in temperature always increases the rate of

chemical reactions, including rusting.

Other factors

Other factors that speed up rusting include the presence of

• a less reactive metal (such as copper or silver) in contact

with iron.

• uneven or sharply pointed regions in the iron piece.

• (

)

•

13.4 To observe rusting using rust indicator 13.4

An iron nail is placed in a warm gel containing a rust indicator

(which contains potassium hexacyanoferrate(III) K3[Fe(CN)6])

(Figure 13.4). At the very start of rusting, iron forms Fe2+(aq)

ions. The rust indicator is used as a sensitive test for Fe2+(aq)

ions, forming a blue colour. Thus, the appearance of a blue

colour indicates rusting.

(

(III) )

( 13.4)

Fe2+(aq)

Fe2+(aq)

N5 &N6

tip

head

shank

Petri dish

gel containing rust indicator solution

nail

Figure 13.4 Detecting rusting by a rust indicatorsolution. Blue patches around the iron nail appearin a few minutes.

Example 13.2Using rust indicator to investigate rusting

Refer to the experimental results shown in Figure 13.5.There are three Petri dishes, each containing a gel with arust indicator solution.

Dish 1 contains a single iron nail.

Dish 2 contains an iron nail partly wrapped with amagnesium ribbon.

Dish 3 contains an iron nail partly wrapped with a copperstrip.

13.5

1

2

3

13.2

cont'd

Note 5There is no need to mention the principlebehind rust indicator. The rust indicatorshould be treated as any indicator (such asmethyl orange or phenolphthalein) and itsonly use is to show where rusting occurs.

Note 6The rust indicator contains:(a) Potassium hexacyanoferrate(III)

(K3[Fe(CN)6]), which reacts withFe

2+(aq) ions to form a deep blue

complex.

(b) Sodium chloride, which increaseselectrical conductivity of solution, sothat corrosion (causing the colour tochange to blue) occurs more quickly.

(c) Agar, which makes the warm solutionset on cooling to form a gel. Diffusionof blue patches formed around the nailcan thus be slowed down for easierobservation.

65


(a) What do you observe in each case?

(b) Explain the observed results.

(c) From your answers in (a) and (b), suggest a method toprotect iron from rusting.

(a)

(b)

(c) (a) (b)

Solution

(a) Dish 1: A blue colour appears, mainly around the headand tip of the nail.

Dish 2: Gas bubbles appear around magnesium. Noblue colour appears.

Dish 3: A blue colour appears, mainly around the headand tip of the nail. The blue patches are largerthan those observed in dish 1.

(b) Dish 1: A blue colour appears, showing that the ironnail rusts.

(The blue patches are formed mainly aroundthe head and tip of the nail. This is becausethese regions are sharply pointed; iron therecan change into Fe2+(aq) ions more easily.)

Dish 2: A blue colour does not appear, indicating thatthe iron nail does not rust. This is becausemagnesium, being more reactive, loseselectrons more easily than iron. Because of this,iron is prevented from losing electrons. Irontherefore cannot form Fe2+(aq) ions and doesnot rust.

(Magnesium reacts with hot water to formhydrogen. This accounts for the appearance ofgas bubbles around magnesium.)

(a) 1

2

3

1

(b) 1

(

Fe2+(aq) )

2

Fe2+(aq)

()

Figure 13.5 Investigating rusting with a rust indicator.

iron nailonly

iron nailwrapped witha magnesiumribbon

iron nailwrapped witha copper strip

cont'd

66

Part III Metals

Dish 3: A blue colour appears, showing that the ironnail rusts. The iron nail wrapped with a copperstrip rusts more quickly than the iron nailalone. This is because copper is less reactivethan iron, causing iron to lose electrons moreeasily. This speeds up the rusting of the ironnail.

(c) Join iron to a more reactive metal (e.g. magnesium orzinc).


3

(c) ()

➲ 15

13.5 Protecting iron from rusting 13.5

Applying a protective layer

Both water and air are necessary for rusting to occur. Any

method which can keep out one or both of them from iron will

prevent rusting.

Coating with paint, plastic, oil or grease

Objects unlikely to be scratched can be coated with paint. For

example, bridges, ships, car bodies and other large objects are

painted.

A plastic layer looks better and lasts longer. However, it is

more expensive than paint. The underbody of a car, draining

racks, coat hangers and paper clips can be coated with plastic.

The moving parts of machines and woodworking tools are

not painted. (The paint would surely fall off if it is frequently

scratched.) Oil or grease is used for protection instead.

Coating with another metal

Iron can be coated with a thin layer of corrosion-resistant metal

to avoid direct contact with air and oxygen. There are three

common ways to achieve this goal:

1. Galvanizing: Iron coated with zinc is called galvanized

iron. Some roofs and buckets are made from galvanized

iron.

( )

1.

N7

Note 7Paint coatings are not completely impervious to air and water. However,they do protect the underlying iron by slowing down the movement of ionsessential for the electrochemical process of rusting.

As soon as the paint cracks or peels off, rusting occurs.

N8

Note 8This method is not once and for all, because the oilor grease soon rubs off. Thus machines have to beoiled (or greased) from time to time. Besides, dirtwould stick to oil or grease, making a mess.

N9

Note 9Zinc and tin are resistant to corrosion because they form a protective oxide layer on thesurface.

67


2. Tin-plating: Tin-plate is iron coated with tin. Tin is an

unreactive metal. It can protect iron from air and oxygen

and hence prevent rusting. Tin-plating is commonly used

in making food cans since tin and tin ions are not

poisonous.

2.

3. Electroplating: Electroplating is a process in which a thin

layer of metal is plated on an object, for example iron

object. Chromium is a common metal to be electroplated

on iron. Iron coated with electroplated chromium has a

beautiful shiny appearance, but is quite expensive.

Examples are bathroom fittings, car bumpers and motor

cycle parts.

Cathodic protection

Electricity can also prevent rusting. For example, the negative

terminal of a car battery is always connected to the car body.

This supplies electrons to the iron body, preventing it from

losing electrons. Cathodic protection is often used to protect

water/fuel pipelines and storage tanks, ships, offshore oil

platforms and onshore oil well casings.

3. (

)

Tin is a less reactive metal than iron.

Learning tip

The negative terminal of a car battery is called the cathode. So thiskind of electrical protection is known as cathodic protection. Moredetails about the meaning of ‘cathode’ will be discussed in Chapter31.

Learning tip

Sacrificial protection

During rusting, iron (Fe(s)) loses electrons to form iron(II) ions

(Fe2+(aq)). If we connect iron to a more reactive metal, that

metal will lose electrons in preference to iron. This would

prevent Fe(s) from forming Fe2+(aq) ions.

(Fe(s))

(II) (Fe2+(aq))

(Fe(s)) Fe2+(aq)

68

Part III Metals

Galvanizing (zinc-plating) provides a good example. When

the zinc coating is undamaged, the iron is protected from

rusting. In case the coating is partly damaged, the exposed iron

is still protected. Zinc, being more reactive than iron, will form

zinc ions. Thus, iron will not rust but zinc corrodes instead —

zinc is ‘sacrificed’ to ‘save’ iron (Figure 13.6). This kind of

protection against rusting is called sacrificial protection.

Galvanizing is not used in making food cans because zinc ions

are toxic.

oxygen and water cannotreach iron, so no rustingoccurs

zinc

iron

broken surface

zinc coating

oxygen reacts with zinc instead ofiron — no rustingiron Figure 13.6 Sacrificial protection of

iron by zinc.

N10

Note 10(a) Zinc offers sacrificial protection even when the zinc coating is damaged. Thus galvanized iron can be used

for making objects that are often scratched or knocked about during use (e.g. buckets).(b) About one-third of all the zinc produced in the world is used for galvanizing iron.

(

13.6)

✘ Tin-plating is a kind of sacrificial protection.

✔ Tin is resistant to corrosion. As long as the tin coating remainscompletely undamaged, the underlying iron does not rust.However, if the coating is damaged (even only partly), iron willrust more quickly than it does without tin-plating. This isbecause tin is less reactive than iron.

Check your concept

✘

✔

Using alloys of iron

Steel is produced from iron by the addition of the right amount

of carbon (0.15–1.5%). To fight against corrosion, steel can be

alloyed with other metals (such as chromium, nickel and

manganese) to produce stainless steel.

(0.15–1.5%)

( )

69


13.2

Stainless steel does not rust, yet it is seldom used to make large

objects. Why?

Class practice 13.2 A13.2Stainless steel is too expensive to be used in large objects.

Table 13.1 summarizes the advantages and disadvantages

of different methods of rust prevention with some examples.13 .1

• lasts long • looks good

• more expensivethan painting

coat hangers

• does not fall off likepaint

• has lubricatingeffect

(c) Oiling orgreasing

• not ‘once and forall’

• dirt would stick tooil or grease

moving parts ofmachines

(b) Coatingwith plastic

Method of rustprevention Disadvantages Examples Simple chemistry

• economical(a) Painting

Advantages

the added layerprevents the ironobject from contactwith air and water

• fall off easily window frames,car bodies

• tin is corrosionresistant

• tin ions are notpoisonous

(d) Tin-plating • when the tincoating is damaged,rusting will occurmore quickly thaniron alone

‘tin’ cans forstoring food

• convenient(g) Cathodicprotection

the negative terminalof an electric source isconnected to the ironobject, supplyingelectrons to prevent itfrom rusting

• not applicable tomany objects

car bodies,pipelines,water/fuel tanks

• in case the zinccoating is damaged,the iron is stillprotected

(e) Galvanizing • zinc ions arepoisonous

galvanized ironplate used inconstruction,buckets

• has a beautifulshiny appearance

(f) Chromium-plating

• expensive bathroom fittings,car bumpers

70

Part III Metals

Method of rustprevention Disadvantages Examples Simple chemistry Advantages

• an effective way ofprotection

• has a beautifulappearance

• a very effective wayof protection

• expensive

(h) Sacrificialprotection

(i) Alloying

• the ‘sacrificed’metal needsreplacement fromtime to time

zinc blocksattached to thehull of a ship

cutlery, scissors

a more reactive metal(e.g. magnesium,zinc) in contact is‘sacrificed’ to formions; this wouldprevent iron fromforming iron(II) ions

(II)

iron is alloyed withchromium, nickel andmanganese which arecorrosion resistant

Table 13.1 Different methods of rust prevention.

13.6 Socio-economic implications of rusting 13.6

Rusting causes damages to buildings and enormous economic

loss. This is the reason why extensive corrosion protection has

to be developed.

13.7 Corrosion resistance of aluminium 13.7

Protective oxide layer on aluminium

Although aluminium is higher than iron in the reactivity series,

it is resistant to corrosion. This is because aluminium forms a

thin but tough oxide layer which seals the metal surface (Figure

13.7). This coating protects the aluminium underneath from

further attack by air and water.( 13.7)

Figure 13.7 Explaining the corrosionresistance of aluminium.

surface attacked by oxygen in the air thin protective layer of aluminium oxide

aluminium aluminium

N11

Note 11We can remove the aluminium oxide layer by rubbingit with cotton wool soaked in mercury(II) chloridesolution and dilute hydrochloric acid. The treatedaluminium then corrodes unhindered.

71


Thickening protective layer on aluminium

The protective oxide layer on aluminium is very thin. The

oxide layer can be thickened and strengthened by an

electrolysis process called anodizing (anodization). After

anodizing, aluminium becomes even more corrosion resistant.

Also, anodized aluminium can be dyed easily to give attractive

colours.

N12

Note 12(a) Untreated aluminium has an oxide layer about 10

–6cm thick. The oxide film, after

anodizing, usually varies from 0.0005 cm to 0.0025 cm thick.(b) Remind students that the aluminium oxide layer can be thickened by anodization

(electrolysis), not by further reaction with air.(c) Tell students that it is wrong to say anodization can improve the strength of aluminium

such that it can be used to make aircraft bodies.


1. alloying 68

Page

3. cathodic protection 67

4. corrosion 61

7. potassium hexacyanoferrate(III) (III) 64

8. protective layer 66

9. rust indicator 64

10. rusting 61

11. sacrificial protection 67

12. tin-plating 67

5. electroplating 67

2. anodizing (anodization) 71

6. galvanized 66

72

Part III Metals

SS uu mm mm aa rr yy13.1 Corrosion of metals

1. is the gradual deterioration of a metal due to reaction with air, water or othersubstances in the surroundings.

2. In general, a metal in the metal reactivity series corrodes faster.

13.2 Rusting

3. is the corrosion of iron. Rusting requires the exposure of iron to both water andair. Rust is in fact hydrated iron(III) oxide, Fe2O3 · nH2O.

13.3 Factors that speed up rusting

4. Factors that speed up rusting include:

• Presence of solutions or soluble salts

• temperature

• A reactive metal in contact with iron

• Uneven or sharply pointed regions in the iron piece

13.4 To observe rusting using rust indicator

5. We can observe rusting conveniently using a . It shows a bluecolour where rusting occurs.

13.5 Protecting iron from rusting

6. To prevent rusting, we can make use of a suitable method. Refer to Table 13.1 on p.69 for differentmethods of rust prevention.

13.6 Socio-economic implications of rusting

7. Rusting causes enormous socio-economic problems and extensive corrosion protection has to bedeveloped.

13.7 Corrosion resistance of aluminium

8. Aluminium is resistant to corrosion because it has a protective layer.

9. is an electrolysis process used to thicken the aluminium oxide layer onaluminium. Anodized aluminium is even more corrosion-resistant, and can be easily dyed to giveattractive colours.

Corrosion

higher

Rusting

acidic

High

less

rust indicator

oxide

Anodizing

HKDSE Chemistry Bridging Programe 1C

Documents

Transcript of HKDSE Chemistry Bridging Programe 1C