HKDSE Chemistry Bridging Programe 1C

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First published July, 2009

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 Summary Chapter 11

11 12 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 Summary

34 35

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

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 10 Occurrence and extraction of metals

10.1 Uses of metals in our daily lives

10.1

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.

)

10.2 Uses related to properties of metals

10.2

Table 10.1 lists the uses of some common metals. 10.1 Uses

Metal Iron

Copper

construction, transport

hard, strong, malleable and ductile; Note 1 cheap

magnet

magnetic

electrical wires

excellent conductor of electricity, very ductile, corrosion resistant

water pipes

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

cooking utensils (b) Aluminium cables are much lighter than copper cables, thus the pylons supporting them can be less sturdily built, with a saving in cost.

overhead power cables

aircraft body

N1

The world uses 9 times more iron than all the other metals put together.

Note 3 (a) Aluminium is more than twice as conductive of electricity as the same mass of copper.

Aluminium

Reasons for use

(c) Aluminium is not used in plumbing because it is difficult to weld aluminium pipes by ordinary welding.

non-poisonous, strong, malleable and ductile, corrosion resistant

excellent conductor of heat, nonpoisonous, strong, malleable, corrosion resistant

very good conductor of electricity, low density, ductile; cheaper than copper

N2 & N3

light but strong, corrosion resistant

1

Part III Metals

Uses

Metal saucepans, kitchen foil

very good conductor of heat, nonpoisonous, corrosion resistant, very malleable

soft drink cans

non-poisonous, low density (hence convenient to carry), corrosion resistant, very malleable ( )

N4

window frames

corrosion resistant, strong

N4

Titanium

supersonic aircraft

light but very strong, very corrosion resistant, malleable

N5

Gold

jewellery, coins

attractive golden yellow colour, extremely corrosion resistant (so always shiny), most malleable and ductile, quite soft when pure

N6

Aluminium

Note 4 (a) The protective oxide layer can be thickened by anodizing. (b) The surface oxide layer can be dyed to give various attractive colours.

Note 5 (a) Titanium weighs only 60% as the same volume of steel. Both are equally strong. (b) At high temperatures, titanium remains strong and its shape is not distor ted. (When speeding through air, a supersonic aircraft becomes heated up by friction with air.)

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

(

)

shield for heat, sunlight and radiation

excellent reflective quality

Mercury

thermometers

liquid at room temperature, expands on heating, does not wet walls of tubes

Silver

jewellery, coins

attractive silvery colour, corrosion resistant, malleable and ductile, moderately soft

electrical and electronic components

best conductor of electricity, malleable and ductile, corrosion resistant

Table 10.1 Uses of some metals and their properties related to the uses. 2

Reasons for use

N7

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


Class practice 10.1

10.1

The following table gives information on some metals (represented by letters A to E).

Metal

Relative price

Density –3

(g cm )

Melting point (°C)

Rank order in thermal conductivity (1 = best) (1

A

B

most expensive

cheapest

(

Rank order in electrical conductivity (1 = best)

)

(1

19.3

1063

3

3

7.9

1540

8

8

C

10.5

961

1

1

D

very expensive

4.5

1933

11

11

9.2

420

6

6

medium

E

)

Other characteristics

)

very expensive

E

A

attractive golden yellow colour, quite soft, chemically inert

hard and strong attractive silvery colour, quite soft, corrosion resistant

strong, very corrosion resistant

poisonous, quite corrosion resistant

Refer to the above information to answer the questions below: (a) Which is the most widely used metal? Why?

(a)

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

(b) (c)

(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.

(d)

(e) Which two metals are the best conductors of electricity? Give a reason why one of these is used much more than the other.

(e)

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.

3

Part III Metals 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.

10.3 Occurrence of metals in nature

10.3

Most metals combine with other elements as compounds, called ores.

(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.

N8

Four important ores found in nature. (a) Bauxite — the main ore of aluminium. It is mostly N9 aluminium oxide

Note 9 Al2O3. Pure aluminium oxide is white. The bauxite ore shown is brown because it contains appreciable amounts of iron(III) oxide.

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

(a) Al2O3 (b)

CuFeS2.

CuFeS2

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

(c)

oxide Fe2O3.

(III) Fe2O3

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

(d)

sulphide PbS.

(II) PbS

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

(

)

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

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

1. 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 to test for hydrogen in the ‘pop’ sound test.

2. 3.

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

(II)

to produce the metal and oxygen. We can use a glowing splint N10 to test for any oxygen evolved. mercury(II) oxide red

silver oxide brownish black

4

mercury + oxygen

(II)

+

silvery

silver + oxygen silvery

+


Heating the metal ore with carbon (carbon reduction) Extraction of iron from its ore — haematite (iron(III)

( )

oxide)

(

(III))

Iron can be extracted from its ore haematite. At high temperatures, carbon can react with iron(III) oxide in N11 haematite. In the process, carbon removes the oxygen from N12

(III)

iron(III) oxide to form iron.

(III)

iron(III) oxide + carbon

iron + carbon dioxide

(III) +

+

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

(

sulphide)

(II))

Extraction can be divided into two stages: The first stage is to heat the ore galena (PbS) in air to form

(PbS)

lead(II) oxide:

(II)

lead(II) sulphide + oxygen

lead(II) oxide + sulphur dioxide

(II) +

(II) +

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

(II)

with carbon and is changed to lead. lead(II) oxide + carbon

lead + carbon dioxide

(II) +

+

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

(

)

(

10.1)

follows: electricity

aluminium oxide

aluminium + oxygen

(from bauxite)

+ (

)

Note 11 There is a competition for oxygen. At high temperatures, carbon holds on to oxygen more tightly than lead does, hence carbon can take oxygen away from lead(II) oxide. Besides carbon, carbon monoxide and hydrogen 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() Note 12 Bunsen flame is not hot enough to give an obvious result in the extraction reaction. However, trace of iron can be detected when a magnet is placed near the resultant reaction mixture.

5

Part III Metals

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

negative (–) electrode (–)

tapping hole molten aluminium

molten aluminium oxide and cryolite

Figure 10.1 Electrolysis of aluminium oxide.

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

Common ores of some metals and their methods of N13 extraction are listed in Table 10.2.

Note 13 Common names of ores need not be memorized.

6

( 950°C

10.2

(2047°C) Na3AlF6)


Most difficult

Metal

Year of discovery

Main ore/source from which metal is obtained

Major metallic compound in the ore/source

/

/

in deposits and sea 1807 A.D. water

KCl

Sodium

common salt in sea 1807 A.D. water; rock salt

NaCl

Calcium

1808 A.D. limestone

CaCO3

1808 A.D. magnesite

MgCO3

Potassium

Magnesium Aluminium

1827 A.D.

Zinc

Usual method of extraction

electrolysis of molten ore

Al2O3 . 2H2O

bauxite zinc blende

ZnS

Step 1:

heat in air

sulphide Ease of extraction

Step 2: 1000 A.D.

Note 14 Copper and mercury are quite unreactive. Therefore in the controlled heating of their sulphides in air, the sulphur par t will be oxidized to sulphur dioxide, leaving the metals behind, e.g. HgS + O2

Note 15 Much silver and some gold are also recovered from by-products in the extraction of other metals.

Lead ancient (B. C.)

Mercury

PbS

copper pyrite

) cinnabar

(

CuFeS2

Gold

(as free element) (

5000

metal

similar to that of Zn heat in

sulphide

metal

N14

displacement from solution

N15

air

Ag2S (

5000 B. C. Easiest

metal

HgS

silver glance (or as free element) ( )

Silver

carbon

Fe2O3

galena

Copper

heat with

oxide haematite

3000

heat with carbon

Hg + SO2

3000 B. C. Iron

oxide

oxide

)

Au

)

mechanical separation (to get rid of impurities) ) (

Table 10.2 Methods of extraction of some metals from common ores. 7

Part III Metals

10.2

Class practice 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

(a)

(b)

(c)

(e) platinum

(f) magnesium

(g) lead

(d) (g)

(e)

(f)

(d) silver

A10.2 (a) Electrolysis.

)

electrolysis

aluminium oxide

aluminium + oxygen

(b) Electrolysis. electrolysis

sodium chloride

sodium + chlorine

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

10.2

shown in Figure 10.2. (c) Heating the metal ore with carbon. heat with carbon

haematite


(d) Displacement from solution containing silver.

silicon 27.7%

(e) Physical separation. (f) Electrolysis.

oxygen 46.6%

electrolysis

magnesium chloride

magnesium + chlorine

(g) Heating the metal ore in air and then with carbon lead(II) sulphide

Lead(II) oxide + sulphur dioxide

5.0%

heat with carbon

Lead(II) oxide

lead + carbon dioxide

all other elements

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

1.5%

magnesium 2.1%

potassium

sodium

calcium

2.8%

3.6%

2.6% Note 16 If all the gold available in the world today were melted to make a cube, the cube would be 15 –18 m on each side. It has been estimated that the total gold resource not yet extracted is only 50% of the existing amount.

Although gold and silver are the earliest metals discovered N16 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. N17 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. 8

aluminium 8.1%

iron

heat in air

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


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

1.

2.

the ease of mining its ore and the cost

2.

3.

the ease of extracting the metal from its ore and the cost of N18 3. Note 18 extraction

)

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

10.1

Example 10.1 Discovery of metals (a) Some information about the extraction methods of four different metals

A

(a)

B

C

D

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

Extraction method

A

Mechanical separation

B

Carbon reduction

C D

Electrolysis Heating the metal ore in air

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

(i)

(ii) Arrange the reactivity of the metals in ascending order.

(ii)

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

(iii)

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

(

)

( )

(b)

Solution (a) (i) A, D, B, C (ii) The extraction methods are indication of the reactivity of metals. Thus, the ascending order of reactivity of the four metals is the same as the order of ease of extraction of these metals. That is, A, D, B, C. (iii) The order of discovery of metals relates directly to the reactivity and hence the ease of extraction of the metals. Therefore, the correct order is A, D, B, C.

(a) (i)

A

D

B

C

(ii)

( A

D

B

) C

(iii) A

D

B

C cont'd

9

Part III Metals

(b) Though aluminium is the most abundant metal, much N19 of it exists in clay (mainly silicates). It is not economical to extract aluminium from clay. In fact, we extract aluminium from bauxite by electrolysis. Much electricity is needed and electricity is expensive.

(b) ( )

On the other hand, we extract iron from haematite by heating haematite with carbon in a blast furnace. The cost is relatively low.

➲ Try Chapter Exercise Q21

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

10.6 Conserving metals

➲

21

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 (4

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

conservation):

)

1.

Reusing metal articles

1.

(reuse)

2.

Replacing metals by other materials

2.

3.

Reducing the use of metal articles

3.

(reduce)

4.

Recycling used metals

4.

(recycle)

(replace)

Reusing metal articles However, this method of conserving metals is still not very N20 popular at present.

Note 20 People are accustomed to the throw-away style of using things. They have yet to learn that world resources are limited and there is an urgent need to conserve materials.

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. 10

PVC (

)


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.

Note 21 The demand for metals is ever increasing due to increasing world population and rising living standards. For example, in 1900, about 15 000 tonnes of aluminium were produced; today, the amount is about 30 000 000 tonnes a year, an increase of 2000 times.

1.

Metal resources are limited. However, demand for metals N21 keeps on increasing. Recycling is an effective way of conserving metals.

2.

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

3.

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 N22 and extraction from ores. 4.

Recycling promotes public awareness of resource scarcity.

Key terms

4.

Note 22 Mining 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.

Page 1. bauxite

4

2. copper pyrite

4

3. extraction

4

4. galena

4

5. haematite

4

6. recycling

10

7. replacing

10

8. reusing

10 11

Part III Metals

Summary 10.1 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.

properties prices 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, in nature. Most metals exist as

gold

and in ores.

compounds

platinum

) occur in elemental form

10.4 Extraction of metals from their ores ores

4.

Metals can be obtained from

5.

Metals can be extracted from their ores by alone (a) heating the metal ore carbon (b) heating the metal ore with electrolysis (c)

extraction

by a process called

.

10.5 Discovery of metals 6.

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

7.

availability 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: Reusing (a) metal articles Replacing (b) metals by other materials Reducing (c) the use of metal articles Recycling (d) used metals

10.

12

Recycling

conservation

.

metals means melting down used metals and using them again.

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 N1 factors: 1.

Note 1 Ask students why it is unfair and misleading to compare the reactivity of magnesium and iron in the following way:

The lowest temperature at which the reaction starts Add a magnesium ribbon to water at 25°C; add iron powder to dilute hydrochloric acid at 80°C; observe which metal gives out gas bubbles faster.

2.

The rate (speed) of reaction

3.

The amount of heat energy given out during reaction

11.2 Comparing reactivity of common metals Reactions of metals with air

Note 2 (a) A little calcium nitride Ca3N2 (pale yellow) is also formed. This point need not be mentioned to students.

1. 2. 3.

11.2

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

(b) The equations provided in Section 11.2 only appear in the Teacher’s 11.1. Edition. After learning equations, students should, in Activity, turn all the word equations in Section 11.2 into chemical equations.

Metal

Conditions for reaction

Potassium

gentle heating

Observation It burns vigorously with a lilac (pale purple) flame to produce an orange powder.

Word equation potassium + oxygen potassium superoxide (KO2) (KO2) K(s) + O2(g)

Sodium

B

gentle heating

U

It burns vigorously with a golden yellow flame to produce a white powder.

R

KO2(s)

sodium + oxygen sodium oxide (Na2O) (Na2O) 4Na(s) + O2(g)

2Na2O(s)

N

Calcium

strong heating Note 3 A little magnesium nitride Mg3N2 (pale yellow) is also formed. This

Magnesium

strong heating point need not be mentioned to students.

It burns quite vigorously with a brick-red flame to produce a white powder.

calcium + oxygen calcium oxide (CaO) (CaO) 2Ca(s) + O2(g)

It burns with a very bright white flame to produce a white powder.

2CaO(s)

magnesium + oxygen magnesium oxide (MgO) (MgO) 2Mg(s) + O2(g)

N2

N3

2MgO(s)

13

Part III Metals

Metal Metal

Conditions for Conditions for reaction reaction

Aluminium

strong heating

Observation Observation Aluminium powder burns to give out much heat; a white powder forms.

Word equation equation Word aluminium + oxygen aluminium oxide (Al2O3) (Al2O3) 4Al(s) + 3O2(g)

2Al2O3(s)

B

Zinc

U

strong heating

R N

Zinc powder burns to give out some heat; a powder (yellow when hot, white when cold) forms.

zinc + oxygen

2Zn(s) + O2(g)

zinc oxide (ZnO) (ZnO)

N4

2ZnO(s)

( ) Iron

strong heating

Iron powder burns with yellow showery sparks to produce a black solid.

iron + oxygen iron(II) iron(III) oxide (Fe3O4) (II III) (Fe3O4) 3Fe(s) + 2O2(g)

Lead

D O

strong heating

N O T

Copper

B U R N

Mercury

Silver

Platinum

Gold

N O R E A C T I O N

It melts to silvery balls; a powder (orange when hot, yellow when cold) is seen on the surface.

Fe3O4(s)

lead + oxygen oxide (PbO)

lead(II) (II) (PbO)

2Pb(s) + O2(g)

2PbO(s)

( ) very strong heating

Its surface turns black.

copper + oxygen 2Cu(s) + O2(g) copper(II) oxide (CuO) 2CuO(s) (II) (CuO)

very strong heating

A red powder forms on the surface.

mercury + oxygen 2Hg() + O2(g) 2HgO(s) mercury(II) oxide (HgO) (II) (HgO)

—

No apparent change even on very strong heating.

—

—


—

—


—

Table 11.1 Reactions of some common metals with air. Note 4 Zinc oxide has a certain structure (which is white) at room temperature; it has a different structure (which is yellow) at high temperatures. The different colours of lead(II) oxide at different temperatures can also be explained by difference in structure.

14

N4


Figure 11.1 summarizes the reactivity of common metals with air (oxygen).

11 . 1 (

Metal

Reaction with air

Potassium

Reactivity of metals with air decreases

, Na

Calcium

, Ca

Magnesium

, Mg

Aluminium

, Al

Zinc

, Zn

Iron

, Fe

do not burn

, Cu

Mercury Silver

, Hg , Ag

Platinum Gold

burn

, Pb

Copper

Figure 11.1 The reactivity of common metals towards air (oxygen). ( )

,K

Sodium

Lead

)

no reaction

, Pt

, Au

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 N5 quickly on the water surface with a hissing sound, burning with a lilac flame. potassium + water 2K(s) + 2H2O()

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

potassium hydroxide solution + hydrogen

+

+

2KOH(aq) + H2(g)

The resultant solution is alkaline because potassium hydroxide is formed. It will turn red litmus blue. 15

Part III Metals

Action of sodium on water Sodium reacts with water in a similar way as potassium, but less vigorously.

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

When we add a small piece of sodium to water, it moves N6 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 2Na(s) + 2H2O()

+

sodium hydroxide solution + hydrogen 2NaOH(aq) + H2(g)

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

11.2

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.

hydrogen gas

hydrogen gas

water inverted funnel

water inverted funnel

Figure 11.2 The reaction of calcium with cold water.

calcium calcium

A milky suspension is produced as the white calcium hydroxide formed is only slightly soluble in water. calcium + water Ca(s) + 2H2O()

calcium hydroxide + hydrogen Ca(OH)2(s) + H2(g)

+ 16

+


Action of magnesium, aluminium, zinc and iron on steam 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

Mg(s) + 2H2O()

Mg(OH)2(s) + H2(g)

11.3

With steam, however, the reaction is vigorous. We can carry out the experiment using a set-up as shown in Figure 11.3. magnesium ribbon

wet sand

delivery tube

hydrogen water

heat

Figure 11.3 The reaction of heated magnesium with steam. Note 7 The aluminium oxide layer is so thin that the metal still looks shiny and silvery white.

trough Note 8 We can remove the protective oxide layer on aluminium by rubbing it with a solution of mercury(II) chloride and hydrochloric acid.

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 Mg(s) + H2O(g)

magnesium oxide + hydrogen

+

+

+

+

MgO(s) + H2(g)

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. Zn(s) + H2O(g)

zinc + steam iron + steam 3Fe(s) + 4H2O(g)

ZnO(s) + H2(g)

zinc oxide + hydrogen

iron(II) iron(III) oxide + hydrogen

+

(II

III) +

Fe3O4(s) + 4H2(g)

Aluminium does not seem to react with steam. This is because aluminium metal is usually covered with a very thin N7 layer of aluminium oxide. The oxide layer protects the metal N8 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

11.4

the reactivity of common metals towards water. 17

Part III Metals

Reaction with water

Metal Potassium

Note 9 Point out to students that heated magnesium, aluminium, zinc and iron react with steam to form metal oxides, not hydroxides.

Reactivity of metals towards water decreases

Sodium

, Na

Calcium

, Ca

Magnesium

, Mg

Aluminium

, Al

Zinc

, Zn

Iron

, Fe

Lead

, Cu

Mercury Silver

, Hg

heated metals react with steam

metal + water metal hydroxide + hydrogen

metal + steam metal oxide + hydrogen

heated metals do not react with water or steam

—

, Ag

Platinum Gold

metals react with cold water

, Pb

Copper

Figure 11.4 The reactivity of common metals towards liquid water/steam.

,K

Equation

, Pt

, Au

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

11.1

Class practice 11.1 Rubidium (Rb) reacts with water to form hydrogen and a compound. Is this compound rubidium oxide or rubidium hydroxide? Why?

(Rb)

Reactions of metals with dilute hydrochloric acid and dilute sulphuric acid 11.5 Figure 11.5 shows the reactivity of common metals towards N10 dilute hydrochloric acid or dilute sulphuric acid.

Learning tip Potassium or sodium reacts with dilute acids explosively, so NEVER attempt to perform this experiment in a school laboratory. Note 10 Dilute sulphuric acid forms insoluble sulphates (such as CaSO4, PbSO4) which would stop the reaction after a short while; dilute nitric acid is not a typical acid towards metals (as it has oxidizing properties).

18

N9


Potassium

Reactivity of metals towards dilute hydrochloric /sulphuric acid decreases

, Na

Calcium

, Ca

Magnesium

, Mg

Aluminium

, Al

Zinc

, Zn

Iron

, Fe

, Hg , Ag

Platinum Gold

reacts with acid, more slowly down the series

metal + hydrochloric acid metal chloride + hydrogen or metal + sulphuric acid metal sulphate + hydrogen

, Cu

Mercury Silver

explosive reaction

very slow reaction

, Pb

Copper Figure 11.5 The reactivity of common metals towards dilute hydrochloric acid/dilute sulphuric acid.

,K

Sodium

Lead

Equation

Reaction with dilute acid

Metal

no reaction

—

, Pt

, Au

Reactive metals react with dilute hydrochloric acid and sulphuric acid to give salts and hydrogen.

Check your concept

✘ ✔

Zinc reacts with both concentrated and dilute sulphuric acid to give out hydrogen gas.

✘

Zinc reacts with concentrated sulphuric acid to give out sulphur dioxide rather than hydrogen gas. (Details will be discussed in Chapter 31.)

✔

( )

19

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 N11 11.6).

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

Potassium

Note 12 The dotted line between lead and copper divides the common metals into 2 groups: those above the line react with dilute hydrochloric acid to give metal chloride and hydrogen, while those below the line do not.

,K

Sodium

, Na

Calcium

, Ca

Magnesium

, Mg

Aluminium

, Al

Zinc

, Zn

Iron

, Fe

Lead

most reactive

decreasing reactivity

N12

, Cu , Hg

Mercury

, Ag

Platinum Gold

11.6)

, Pb

Copper

Silver

(

, Pt

, Au

least reactive

Figure 11.6 Metal reactivity series for common metals.

Metals at the top of the series are the most reactive; those at the bottom are the least reactive.

20


Class practice 11.2 A11.2 (a) C, A, D, B (b) C: potassium; A: calcium; D: magnesium; B: copper

11.2 Metal

Experiment

A

B

C

D

Strong heating in air

burns with a brick-red flame

forms a black powder

burns with a lilac flame

burns with a dazzling white flame

Reaction with cold water

moderate reaction

no reaction

violent reaction; burns by itself

no reaction

Reaction with dilute hydrochloric acid

fast reaction

no reaction

(experiment not performed)

fast reaction

(

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

)

(a)

(b) Give possible names of the four metals. (b)

11.4 Chemical equations

11.4

Representing the reaction between magnesium and oxygen (

)

When magnesium burns in air (or oxygen), magnesium oxide is formed. We can represent this reaction by a word equation: magnesium + oxygen reactants (reacting substances)

(

)

magnesium oxide

+

product (the substance produced)

(

)

(

)

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

O O Note 13 A cation is smaller than the atom from which it is derived; an anion is larger than the atom from which it is derived.

Mg

(

11.7)

1 oxygen molecule 1

Mg

2 magnesium atoms 2

Mg

2+

2–

O

2+

Mg

2–

O

N13

2 formula units of magnesium oxide 2

Figure 11.7 The formation of magnesium oxide in terms of particles — 2 atoms of Mg react with 1 molecule of O22–to form 2+ 2 formula units of MgO. (1 formula unit of magnesium oxide consists of 1 magnesium ion, Mg and 1 oxide ion, O .) Mg O2 2+ 2– MgO ( MgO (Mg ) (O ) )

21

Part III Metals

We may also represent the reaction by writing a chemical equation: 2Mg(s) + O2(g)

2MgO(s)

reactants

product

2Mg(s) + O2(g)

2MgO(s)

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: 2Mg(s) + O2(g) 1.

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

The reactants involved These are magnesium (Mg) and oxygen (O2), written on the N14

1.

left-hand side of the arrow. 2.

(Mg)

2.

side of the arrow.

(MgO)

Physical states of the substances involved Mg and MgO are solids, represented by a state symbol (s);

3.

O 2 is a gas (g). Other state symbols are: liquid () and N15

Mg

MgO

aqueous solution (aq). 4.

(O 2 )

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

3.

2MgO(s)

(s)

O2 (

The relative number of particles (atoms, molecules, ions N16

( ) (aq)

or formula units) 2 atoms of Mg would react with 1 molecule of O 2 to

(g) )

4.

produce 2 formula units of MgO.

22

Note 14 An equation which is too long to be written in one line, for example: 2A + B + 3C 4D + E + 5F + 3G may be represented as: 2A + B + 3C 4D + E + 5F + 3G or 2A + B + 3C 4D + E + 5F + 3G, but not as 2A + B + 3C 4D + E + 5F + 3G, in order to avoid ambiguity.

Note 15 Explain to students clearly the difference between the state symbols (l) and (aq). Note 16 Another point is: The equation gives the relative number of moles and hence indirectly the relative masses of the various substances involved. This point will be dealt with in Chapter 12.

Mg

O2 MgO


11.3

Class practice 11.3 Express in words the information provided by the following

N17

equation: SO2(g) + 2NaOH(aq)

SO2(g) + 2NaOH(aq)

Na2SO3(aq) + H2O()

H2O()

A11.3 1 molecule of sulphur dioxide gas reacts with 2 formula units of aqueous sodium hydroxide, to produce 1 formula unit of aqueous sodium sulphite and 1 molecule of liquid water.

More about an equation Balanced equation

Na 2SO 3(aq) +

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

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

(1)

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

(2)

The numbers before the formulae of reactants and products in a balanced equation are called stoichiometric coefficients. 2Mg(s) + O 2 (g)

2MgO(s) is a balanced equation. The

2Mg(s) + O2(g)

2MgO(s)

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

2

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

Consider 2CO(g) + O2(g)

2CO2(g).

(a) How many carbon atoms are on the (i) left-hand side

1O2

11.4

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

(ii) right-hand side of the arrow?

2CO(g) + O2(g)

(ii) right-hand side of the arrow?

(c) Is it a balanced equation?

2CO2(g)

(a)

(b) How many oxygen atoms are on the (i) left-hand side

2

1 O2

Class practice 11.4

1

(i)

(ii)

(i)

(ii)

(b)

(c)

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.

23

Part III Metals

The double arrow ‘ N2(g) + 3H2(g)

’ is used in some equations, e.g.

2NH3(g). The ‘

(

’ means that the reaction N18

N 2 (g) + 3H 2 (g)

2NH 3 (g))

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

2H2(g) + O2(g)

2H2O()

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

Note 18 Here, the forward reaction is: N2(g) + 3H2(g) 2NH3(g); the backward reaction is: 2NH3(g) N2(g) + 3H2(g)

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) or

H2(g) + O(g)

H2O2(), ✗ H2O(). ✗

H2(g) + O2(g) H2(g) + O(g)

H2O2() H2O()

Learning tip It is a common practice to write an equation in which coefficients are the smallest possible whole numbers. That is why the equation 2H2(g) + O2(g) 2H2O() is shown here.

24

2H2(g) + O2(g)

2H2O()

✗ ✗


Problem-solving strategy Writing a chemical equation Step 1 Find out what the reactants and products are. Write down the word equation for the reaction. hydrogen + oxygen

1

water

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

( )

Step 2 Write the equation by replacing names of reactants and products with their correct formulae. H2 + O2

2

H2O

H2 + O2

(unbalanced)

(

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

H 2O

2H2O

2

To balance the number of hydrogen atoms: 2 hydrogen atoms on the left-hand side, 4 hydrogen atoms on the right-hand side, so put ‘2’ before H2; the equation is now balanced 2H2 + O2

)

3

To balance the number of oxygen atoms: 2 oxygen atoms on the left-hand side, only 1 oxygen atom on the right-hand side, so put ‘2’ before H2O; the equation is still unbalanced H2 + O2

H2O

H2 + O2

2H2O

2H2O

H2

Step 4 Write the state symbol after each formula to give the complete balanced equation. 2H2(g) + O2(g)

2H2O()

2 N19

2H2 + O2

(complete balanced equation)

2H2O

4 Note 19 In public examinations, candidates usually need not write state symbols in equations, unless they are instructed to do so in the question.

A11.5 1. (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)

2H2(g) + O2(g) (

2.

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 Balance the following by adding suitable stoichiometric coefficients:

__________ Pb3O4(s)

)

11.5

Class practice 11.5 1.

2H2O()

1. (a) Cl2O7 (b) (NH4)2Cr2O7 (c) 3Fe2(SO4)3 (d) 3Na2CO3 · 10H2O

N20

2.

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

Note 20 Students 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 a formula, e.g. 3 Fe2(SO4)3. Explain to students how to calculate the number of atoms of each kind.

__________ PbO(s) + __________ O2(g)

25

Part III Metals

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

11.5

Metals react by losing electrons 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. Note 21 Down a group, atomic size increases, so the ‘pull’ (attraction) of the nucleus on the outer shell electrons becomes smaller. Thus the outermost shell electrons can be lost more easily.

N21

11.8

increasing readiness to lose electrons

increasing reactivity of metals

Figure 11.8 Readiness to lose electrons (and hence reactivity of metals) decreases across a period and increases down a group. (

increasing readiness to lose electrons

increasing reactivity of metals

)

A metal higher in the reactivity series has a higher reactivity, and its atoms would lose outermost shell electrons to form cations more easily.

26


11.6 Displacement reactions of metals in aqueous 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) colourless

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

Cu(s) + 2AgNO3(aq)


pale blue

We say that copper displaces the silver metal from the silver nitrate solution.

Iron in copper(II) sulphate solution

(II)

A similar displacement reaction occurs when we place an iron nail into copper(II) sulphate solution. Fe(s) + CuSO4(aq) silvery white

blue

N22

FeSO4(aq) + Cu(s) pale green

(II)

reddish brown

N23

Fe(s) + CuSO4(aq)

FeSO4(aq) + Cu(s)

(II)

On the other hand, if we put a piece of copper into iron(II) sulphate solution, no reaction occurs.

A metal (M1) higher in the reactivity series will displace any metal (M 2 ) lower in the series from the solution of a

(M 1 ) (M 2 )

compound of M2.

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. Note 22 Suppose a student is asked to describe a simple experiment to compare the reactivity of copper and iron. The displacement reaction here can be quoted. However, it would be wrong to quote the example of setting up an electrolytic cell. With an external electric current applied, the relative reactivity of the two metals cannot be compared.

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

27

Part III Metals

Example 11.1 Predicting chemical reactions based on the metal reactivity series Predict, with reasoning, whether a reaction takes place in each 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.

11.1

(a) (b)

(II)

(c)

State what type of reaction (if any) takes place and write an appropriate equation for the reaction. Solution (a) No reaction occurs. Zinc is lower than magnesium in the metal reactivity series.

(a)

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

(b)

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

(II)

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

No reaction occurs. All metals lower than copper in the metal reactivity series would have no reaction with dilute sulphuric acid.


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

➲

11.7 Ionic equations

19

11.7

Representing some reactions by ionic equations An ionic equation is an equation involving ions in aqueous solution. Only those ions formed or changed during the reaction are included.

Consider the reaction between copper metal and aqueous silver nitrate solution. The equation for the reaction is: Cu(s) + 2AgNO3(aq)


Cu(s) + 2AgNO3(aq)


28


But the ionic equation +

Cu(s) + 2Ag (aq)

2+

+

Cu (aq) + 2Ag(s)

2+

Cu(s) + 2Ag (aq)

Cu (aq) + 2Ag(s)

shows more clearly what has happened. Note that the nitrate –

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

–

(NO3 )

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

Problem-solving strategy Writing an ionic equation Step 1 Write the full balanced equation for the reaction. Cu(s) + 2AgNO3(aq)

1


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

Step 2 Rewrite the equation:

• Rewrite the formulae of electrolytes in aqueous •

2

solution as separate formulae of cations and anions Keep other formulae unchanged

•

AgNO 3 and Cu(NO 3) 2 are electrolytes in aqueous solution. +

Cu(NO 3) 2

AgNO 3

•

–

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

–

+

Cu(s) + 2Ag (aq) + 2NO3 (aq) –

2+

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

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

–

Cu(s) + 2Ag (aq) + 2NO3 (aq) – 2+ Cu (aq) + 2NO3 (aq) + 2Ag(s) Step 4 Check that the ionic charge is balanced in the ionic equation. +

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

2+

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

3 –

+

Cu(s) + 2Ag (aq) + 2NO3 (aq) –

2+

Cu (aq) + 2NO3 (aq) + 2Ag(s) 4 +

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

2+

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

29

Part III Metals

Example 11.2 Writing ionic equations

11.2

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)

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

Solution (a) Rewrite the equation: 2+

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

2–

Pb(s) + Cu (aq) + SO4 (aq)

2–

2+

Pb (aq) + SO4 (aq) + Cu(s)

(a)

Cancel out the spectator ions on both sides of the equation: 2+

2–


2–

2+


2–

2+

2–

2+



2–

2+


The ionic charge is balanced on both sides of the equation. Thus, the ionic equation is: 2+

2+

Pb(s) + Cu (aq)

Pb (aq) + Cu(s)

net charge = +2

net charge = +2 2+

–

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

–

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

2+

+

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

= +2

(b) –

2+

Zn (aq) + 2NO3 (aq) + 2Ag(s) –

+

Zn(s) + 2Ag (aq) + 2NO3 (aq) –

2+

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

2+

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

+

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

30

–

+

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

Zn (aq) + – 2NO3 (aq) + 2Ag(s)

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

Pb (aq) + Cu(s)

= +2

2+

Zn (aq) + – 2NO3 (aq) + 2Ag(s)

Cancel out the spectator ions on both sides of the equation: +

2+

Pb(s) + Cu (aq)

(b) Rewrite the equation: +

2–

2+


2+

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


Example 11.3 Balancing the ionic equation

11.3

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

yCl2(g) + 6OH (aq)

–

–

zCl (aq) + ClO3 (aq) + 3H2O()

y

z

–

Solution Since there are 6 oxygen atoms on LHS and 6 oxygen atoms – on RHS, the stoichiometric coefficient for ClO3 must be 1. An ionic equation must be balanced with respect to ionic charges.

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

6 ClO3

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

–

1

= –6 = –6 = z(–1) + (–1) z=5

An ionic equation must also be balanced with respect to number 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

➲ Try Chapter Exercise Q18 Cl Cl

=5+1=6 =6

y=3

➲ A11.6 + (a) Mg(s) + 2Ag (aq) 3+ (b) 3Mg(s) + 2Al (aq)

18

2+

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

11.6

Class practice 11.6 Write the ionic equations for the following equations. (a) Mg(s) + 2AgNO3(aq) (b) 3Mg(s) + 2Al(NO3)3(aq)

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

(a) Mg(s) + 2AgNO3(aq) Mg(NO3)2(aq) + 2Ag(s) (b) 3Mg(s) + 2Al(NO3)3(aq) 3Mg(NO3)2(aq) + 2Al(s)


11.8

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

31

Part III Metals

We can draw two general relationships from the table:

10.2

• 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.)

latest

Order of discovery

)

Sodium

, Na

Calcium

, Ca , Mg

Aluminium

, Al

Zinc

, Zn

Iron

, Fe

, Cu , Hg

Mercury Silver

, Ag

Platinum Gold

Reactivity

, Pb

Copper

, Pt

on the position of the metal in the reactivity series.

least reactive

, Au

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

32

most reactive

Magnesium

Lead

earliest

,K

Potassium

•


11.4

Example 11.4 Relating the extraction method of a metal to its position in the metal reactivity series Some information about the extraction methods of five different metals A, B, C, D and E are listed as follows:

Metal

Mechanical separation

B

Carbon reduction

C

Electrolysis

E

B

C

D

E

Extraction method

A

D

A

Displacement from solution by metal E E

D

Heating the metal ore in air

Arrange the positions of metals in the metal reactivity series based on the above information, from the lowest to the highest. Explain your answer. Solution Mechanical separation is the easiest way to extract a metal from its ore. Thus, metal A must be the least reactive (i.e. the lowest in the metal reactivity series).

A (

)

Metal D should be in a position higher than that of A, but lower than that of E as it can be displaced out from its solution by E.

D

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

C

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

B

∴ The positions of five metals in the metal reactivity series should be: A
A E

E

E

∴ A
33

Part III Metals

11.7


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

1.

(a)

oxide) by heating it with magnesium? Why? (b) Is this method used in industry to extract iron from haematite? Why? 2.

(b)

Write balanced chemical equations for the following reactions:

2.

(a) iron(III) oxide + carbon (b) silver oxide


silver + oxygen

(a)

(b) A11.7 1. (a) (b) 2. (a) (b)

( (III))

(III) +

+

+

Yes. Magnesium is more reactive than iron. No. Magnesium is much more expensive than iron. 4Fe(s) + 3CO2(g) 2Fe2O3(s) + 3C(s) 2AgO(s) 2Ag(s) + O2(g)

Key terms Page 1. balanced equation

23

2. chemical equation

22

3. displacement reaction

27

4. ionic equation

28

5. one-way reaction

23

6. reactivity

13

7. reversible reaction

34

(

)

23

8. spectator ion

29

9. stoichiometric coefficient

23


Summary 11.1 Different reactivities of metals 1.

Reactivity

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 dilute acids. Refer to p.13 – 19 for the results of the reactions.

11.3 The metal reactivity series 3.

metal reactivity series The order of reactivity.

4.

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

is a series of common metals arranged in decreasing

Reaction of metal with Metal

K Na

Appearance of metal dull (stored under paraffin oil)

air burns violently, forming oxide (Example 1)

Ca Mg Al generally dull

Zn

metal + water reacts explosively, forming metal metal chloride and hydroxide hydrogen (Example 6) + hydrogen no reaction (Example 4)

reacts with metal + steam decreasing metal oxide vigour, forming + hydrogen oxide (Example 2) (Example 5)

Fe Pb Cu Hg generally shiny

a layer of oxide formed on surface (Example 3)

dilute hydrochloric acid

water/steam

reacts with decreasing vigour: metal + hydrochloric acid metal chloride + hydrogen reduced with increasing ease: (Example 7) metal oxide + carbon metal + carbon dioxide (Example 8)

no reaction no reaction

Ag Au

Reduction of oxide with carbon (1500°C)

no reaction

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()

Example 5:

Zn(s) + H2O(g)

Example 6:

2K(s) + 2HCl(aq)

Displacement reactions not applicable these 3 metals react with water in aqueous solution to give H2 gas

a metal displaces any other metal lower in the series from a solution of its compound (Example 10)

decomposed to metal and oxygen by heating alone (Example 9)

2NaOH(aq) + H2(g) ZnO(s) + H2(g) 2KCl(aq) + H2(g) (NEVER attempt this experiment!) 35

Part III Metals

Example 7:

Fe(s) + 2HCl(aq)

Example 8:

2PbO(s) + C(s)

Example 9:

2HgO(s)

Example 10:

Mg(s) + 2AgNO3(aq)

Ionically:

FeCl2(aq) + H2(g) 2Pb(s) + CO2(g)

2Hg() + O2(g) +

Mg(s) + 2Ag (aq)

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

Mg (aq) + 2Ag(s)

11.4 Chemical equations 5.

chemical reaction A (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.

6.

A

7.

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

reversible

reaction is represented by a double arrow ‘

’.

11.5 Metal reactivity series and the tendency of metals to form positive ions 8.

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

11.6 Displacement reactions of metals in aqueous solution 9.

A metal (M1)

higher

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. lower

11.7 Ionic equations 10.

ionic equation An is an equation involving ions in aqueous solution; only those ions formed or changed during the reaction are included.

11.

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

as


36

position

12.

The ease of extracting a metal from its ores is related to the reactivity series.

of the metal in the

13.

lower The a metal in the reactivity series, the earlier it was first discovered. This is less because less reactive metals form stable compounds, from which the metals can be extracted more easily.

Chapter 12 Reacting masses

12.1 The mole concept in general

12.1

The Avogadro constant Learning tip Every pure substance has a formula. The simplest unit of a substance is its formula unit. In the case of a molecular compound, a formula unit is in fact a molecule. In ionic compounds, there are no molecules. Thus, a formula unit of magnesium chloride is MgCl2, which consists of 1 2+ – Mg ion and 2 Cl ions.

MgCl2

1

2+

Mg

2

Cl

–

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

Chemists use a very special unit, called mole (abbreviation: mol) to show the number of formula units. One mole contains a N1 23

large number of particles, 6.02 10 . This number is called 23

–1

Avogadro constant (L). Thus L = 6.02 10 mol .

( mol) 6.02 10

N2

(L) 23

To count by weighing

10 mol

23

L = 6.02

–1

We can count indirectly — by weighing. Note 2 (a) Determination of L using different methods leads to values which are very close to each other. 23 23 The most recent values lie between 6.022 24 10 and 6.022 80 10 . By approximation, 23 –1 we take L = 6.02 10 mol . (b) At one time, L was referred to as the Avogadro number. Now the term Avogadro constant is used as its value is constant.

Example 12.1 Counting substance by weighing

12.1

Suggest an indirect way of counting 24

(b) 3.01 10 copper atoms.

(a) 50 000 $1 coins

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

(a) 50 000 24 (b) 3.01 10 ( g g)

(a) Weigh out 7.08 50 000 = 354 000 g or 354 kg of $1 coins. –22

(b) Weigh out 1.05 10 copper metal.

24

(3.01 10 ) = 316 g of pure

= 7.08 –22 = 1.05 10

(a)

7.08 50 000 = 354 000 g 354 kg

(b)

1.05 10 316 g

–22

24

(3.01 10 ) = 37

Part III Metals

12.1

Class practice 12.1 Given that 1 mole of silver has a mass of 108 g, suggest an indirect way of counting 1.204 10

Defining mole

23

1

A12.1 1 mole of silver contains 6.02 23 10 silver atoms. ∴ number of moles of silver atoms = number of atoms/L

108 g 1.204 10

silver atoms.

23

23

= 1.204 1023 6.02 10 = 0.200 mol Molar mass of silver (Ag) –1 = 108 g mol

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

(

formula is the amount containing the same number of

)

12.0 g

-12

formula units as the number of atoms in exactly 12.0 g of carbon-12.

Mole and mass

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

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

One mole of a substance (or species) has a mass corresponding N3

1

(

)

to its formula mass expressed in gram unit. The molar mass of a substance is the mass in grams of one –1

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

(

N4

)

(

–1

g mol )

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

For example, the molar mass of carbon dioxide is 44.0 g –1

–1

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

44.0 g mol

–1

18.0 g mol

–1

(H2O) –1

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

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

–1

44.0, not 44.0 g mol .

–1

g

(b)

)

(

44.0

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.

mass

no. of moles (mol) mass (g) = –1 molar mass (g mol )

NUMBER OF MOLES

no. of moles (mol) no. of formula units = –1 Avogadro constant (mol )

(mol) =

38

(mol) (g) –1

(g mol )

=

–1

(mol )

number of formula units


12.2


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

1.

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

(a)

Fe2(SO4)3

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

(c) What is the mass of 0.2000 mole of Fe2(SO4)3? 2.

(c) 0.2000

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

A12.2 1. (a) 55.8 2 + (32.1 + 16.0 4) 3 = 399.9 –1 (b) 399.9 g mol (c) 399.9 0.2000 = 79.98 g

2. Molar mass of silver (Ag) –1 = 108 g mol Number of moles of silver atoms mass (g) = –1 molar mass (g mol )

Important relationships — a summary

=

27.0 mol 108

Fe2(SO4)3

2.

27.0 g

(1)

(

= 0.250 mol Number of silver atoms = number of moles L 23 = 0.250 (6.02 10 ) 23 = 1.51 10

Based on definition of mole, we can get the following important relationships: (1) Mass of 1 mole of a substance or species (g)

formula mass expressed in

=

(2) Number of moles (mol) = (3) Number of moles (mol) =

gram unit mass (g)

N5

–1

(2)

molar mass (g mol )

(mol) (g)

=

number of formula units

–1

(g mol )

–1

Avogadro constant (mol )

(3)

(mol) =

–1

molar mass (g mol )

–1

(mol )

–1

Avogadro constant (mol ) (4) Note 5 There are 3 terms in each of the relationships (2), (3) and (4). By knowing any two of them, the third one can be calculated.

(g) –1 (g mol )

=

Example 12.2 Calculations involving mole concepts

–1

(mol )

12.2

A beaker contains 44.44 g of calcium chloride CaCl 2 . Calculate (a) the formula mass of CaCl2

CaCl 2

(b) CaCl2

the number of moles of CaCl2 in the beaker

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

44.44 g (a) CaCl2

(b) the molar mass of CaCl2 (c)

(g)

=

(4) Mass of 1 formula unit (g) =

)

–

the number of Cl ions in the beaker.

(c)

CaCl2

(d)

CaCl2

(e)

Cl

–

cont'd

39

Part III Metals

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

N6

–1

(b) Molar mass of CaCl2 = 111.1 g mol (c)

= 40.1 + 35.5 2 = 111.1

(a) CaCl2

N7

mass (g)

–1

(b) CaCl2

Number of moles of CaCl2 = –1 molar mass (g mol ) 44.44 = mol 111.1 = 0.4000 mol

(c)

= 111.1 g mol

CaCl2 (g)

=

–1

(g mol )

(d) Number of formula units of CaCl2 –1

= number of moles (mol) Avogadro constant (mol ) = 0.4000 6.02 10 23

= 2.41 10 (e)

111.1

mol

= 0.4000 mol –

(d) CaCl2 = (mol) –1 (mol ) 23 = 0.4000 6.02 10 23 = 2.41 10

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

23

number of Cl ions = 2.41 10 2 23

= 4.82 10

Note 6 (a) Remind students again that formula mass has no unit, while molar mass has the –1 unit of g mol . (b) Tell students not to use ‘·’ as the multiplication 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 ‘Molar –1 mass of CaCl 2 = 111.1 g mol ’, not –1 ‘CaCl2 = 111.1 g mol ’. (b) Do not write ‘gm’ for ‘g’, ‘m’ or ‘M’ for ‘mol’.

(e) Cl

Find the mass of (a) 1 Na atom (c) 1 formula unit of NaCl

CaCl 2

–

23

= 2.41 10 2 23

= 4.82 10

12.3 (

(b) 1 H2O molecule 2– (d) 1 SO4 ion.

(a) (b) (c) (d)

Solution One mole of a substance (or species) corresponds to its molar mass and contains the Avogadro constant of formula units. –1 molar mass (g mol ) ∴ mass of 1 formula unit = –1 Avogadro constant (mol ) (a) Mass of 1 Na atom =

–

Cl

Example 12.3 Calculations involving mole concepts

23.0 23

6.02 10

–23

= 3.82 10

)

Na H2O NaCl 2– SO4

(

)

N8

∴

g

–1

(g mol )

=

–1

(mol )

g cont'd

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

40

44.44

=

23

(a) =

Na 23.0 6.02 10

= 3.82 10

23

–23

g g


(b) Mass of 1 H2O molecule = 1.0 2 + 16.0 g 23 6.02 10 –23

= 2.99 10 (c)

23

6.02 10

–23

= 9.72 10 (d) Mass of 1 SO4

2–

–22

=

Number of formula units of NaOH 23 = 0.250 6.02 10 23 = 1.51 10 (b) Since helium is monoatomic, molar mass of –1 He = 4.0 g mol . 0.20 Number of moles of He molecules = mol 4.0 = 0.050 mol Number of He molecules = 0.050 6.02 10 22 = 3.0 10

2.

SO4 =

g

23

–23

g

2–

32.1 + 16.0 4 23

6.02 10

= 1.60 10

–22

g

g

23

1. (a) 0.200

(b) 0.200 mol of chlorine molecules.

(b) 0.200

(c) chlorine which contains the same number of molecules as there are in 1.20 mol of water.

(c)

Molar mass Substance

–1

Mass

(g)

(g mol )

(a)

Sodium hydroxide

(b)

Helium

(c)

Sulphur dioxide

(d)

Compound X X

1.20

2.

Complete the following table.

24

g

6.02 10

(a) 0.200 mol of chlorine atoms.

= 3.01 1023 mol 6.02 10 = 5.00 mol

g

23

12.3

Calculate the mass of

(c) Molar mass of sulphur dioxide (SO2) –1 = 32.1 + 16.0 2 g mol –1 = 64.1 g mol Number of moles of SO2

–23

23.0 + 35.5

= 9.72 10

g


6.02 10

NaCl

(c)

g

(d) A12.3 1. (a) 0.200 35.5 = 7.10 g (b) 0.200 (35.5 2) = 14.2 g (c) 1.20 mol of chlorine (Cl 2) 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) –1 = 23.0 + 16.0 + 1.0 g mol –1 = 40.0 g mol Mass of NaOH = 0.250 40.0 = 10.0 g

1.0 2 + 16.0

= 2.99 10

g

4 g ion = 32.1 + 16.0 23 6.02 10 = 1.60 10

H2O =

g

23.0 + 35.5

Mass of 1 formula unit of NaCl =

(b)

No. of moles (mol)

Number of molecules/ formula unit

0.250 0.20

Mass of SO2 = 5.00 64.1 g = 321 g

23.0

3.01 10

24

3.01 10

23

(d) Number of moles of X 23 = 3.01 1023 mol 6.02 10 = 0.500 mol Molar mass of X 23.0 –1 = g mol 0.500 –1

= 46.0 g mol

41

Part III Metals

12.2 Percentage by mass of an element in a compound

12.2

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, Fraction by mass of element A in a compound =

A

relative atomic mass of A no. of atoms of A in formula formula mass of the compound

A A

=

Percentage by mass of element A in a compound relative atomic mass of A no. of atoms of A in formula 100% = formula mass of the compound

A

A A

=

Example 12.4 Calculating the percentage by mass of an element in a compound

12.4 (II) CuSO4 · 5H2O

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

CuSO4 · 5H2O

Formula mass of CuSO4 · 5H2O Cu S O4 5H2O = 63.5 + 32.1 + 16.0 4 + 5 (1.0 2 + 16.0) = 249.6 % by mass of Cu relative atomic mass of Cu no. of Cu atoms in formula = 100% formula mass of CuSO4 · 5H2O = 63.5 100% 249.6 = 25.4% Self-Test 12.4

Self-Test 12.4 % by mass of S 32.1 100% = 12.9% = 249.6 % by mass of O 16.0 4 + 16.0 5 100% = 249.6 = 57.7%

% by mass of H 1.0 10 100% = 249.6 = 4%

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

42

100%

Cu S O4 5H2O = 63.5 + 32.1 + 16.0 4 + 5 (1.0 2 + 16.0) = 249.6 Cu Cu = =

Cu CuSO4 · 5H2O

100%

63.5 100% 249.6

= 25.4%

(II) CuSO4 · 5H2O


Example 12.5 Calculating the mass of water of crystallization in a hydrated salt

12.5 15.0 g

(II)

Find the mass of water of crystallization in 15.0 g of copper(II) sulphate-5-water. Solution Mass of water of crystallization = mass of CuSO4 · 5H2O fraction by mass of water in CuSO4 · 5H2O 5 formula mass of H2O = mass of CuSO4 · 5H2O formula mass of CuSO4 · 5H2O = 15.0 g

= CuSO4 · 5H2O CuSO4 · 5H2O

= CuSO4 · 5H2O

5 H2O CuSO4 · 5H2O

5 (1.0 2 + 16.0) 249.6

= 15.0 g

Self-Test 12.5 Mass of water crystallization 4 (1.0 2 + 16.0) = 20 g (55.8 + 35.5 2) + 4 (1.0 2 + 16.0) = 7.24 g

= 5.41 g Self-Test 12.5

5 (1.0 2 + 16.0) 249.6

= 5.41 g

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

20 g

FeCl 2 · 4H 2 O

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 12.6 Calculating relative atomic mass of an element

12.6

The chloride of a metal M has the formula MCl 3 and contains 34.4% by mass of M. Find the relative atomic mass of M.

M M

MCl3 34.4%

M

Solution Let the relative atomic mass of M be a.

M

Fraction by mass of M in MCl3 relative atomic mass of M 1 = formula mass of MCl3 34.4 100 ∴

=

MCl3 =

a

a M 1

M MCl3 34.4

a + 35.5 3

100

a = 55.8

∴

Hence the relative atomic mass of M is 55.8. cont'd

=

a a + 35.5 3

a = 55.8 M

55.8 43

Part III Metals

Self-Test 12.6

Self-Test 12.6 Let the relative atomic mass of X be a. relative atomic mass X 1 Fraction by mass of X in XBr2 = formula mass of XBr2 25.6 a = 100 a + 79.9 2 a = 55.0

X

The bromide of a metal X has the formula XBr2 and contains 25.6% by mass of X. Find the relative atomic mass of X.

XBr 2

X

25.6%

A12.4 1. Number of moles of Na ∴ number of moles of NaNO3 = 4.35 mol 100 = mol = 4.35 mol Mass of NaNO3 23 Since number of moles of Na : number of moles of NaNO3 = 1 : 1 = 4.35 (23.0 + 14.0 + 16.0 3) g = 369.75 g % by mass of N in NaNO3 14.0 100% = 16.5% = Class practice 12.4 23.0 + 14.0 + 16.0 3

X

12.4

Mass of N = 369.75 g 16.5% = 61.0 g

1.

What is the mass of nitrogen present in the sample of sodium nitrate (NaNO3) which contains 100 g of sodium?

1.

2.

What is the mass of water of crystallization present in the sample of sodium carbonate-10-water (Na2CO3 · 10H2O)

2.

100 g

(NaNO3)

4.6 g (Na2CO3 · 10H2O)

which contains 4.6 g of sodium? 3. 4.

A metal oxide MO contains 79.87% by mass of the metal M. Find the relative atomic mass of M.

3.

26.88 g of a metal chloride MCl contains 5.68 g of chlorine. Find the relative atomic mass of the metal M.

4.

% by mass of H2O in Na2CO3 · 10H2O

4.6 mol = 0.2 mol 23.0 Since number of moles of Na : number of moles of Na2CO3 · 10H2O = 2 : 1 ∴ number of moles of Na2CO3 · 10H2O = 0.1 mol Mass of Na2CO3 · 10H2O = 0.1 (23.0 2 + 12.0 + 16.0 3 + 10 (1.0 2 + 16.0)) g = 28.6 g

2. Number of moles of Na =

=

10 (1.0 2 + 16.0) 100% (23.0 2 + 12.0 + 16.0 3 + 10 (1.0 2 + 16.0))

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

MO

M

79.87%

M

26.88 g 5.68 g

MCl M

3. Let the relative atomic mass of M be a. Fraction by mass of M in MO relative atomic mass of M 1 = formula mass of MO

12.3 Chemical formulae of compounds

12.3

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

Note 9 Tell students not to write ionic formulae when writing chemical equations. For example, the reaction between magnesium and silver nitrate should be written as Mg(s) + 2AgNO3(aq)

N9

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

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.

but not – + Mg(s) + 2Ag NO3 (aq)

2+

–

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

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. 44

5.68 100% = 21.13% 26.88 ∴ % by mass of M in MCl = 100% – 21.13% = 78.87%

4. % by mass of Cl in MCl =

Let the relative atomic mass of M be a. Fraction by mass of M in MCl relative atomic mass of M 1 = formula mass of MCl

78.87 a = 100 a + 35.5 ∴ a = 132.5

79.87 a = 100 a + 16.0 ∴ a = 63.5


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.

(

12.1)

The following flow chart (Figure 12.1) shows the sequence and methods by which the various types of formulae are found. Example magnesium chloride (an ionic compound) (

Compound under investigation

Example ethanoic acid (a molecular compound) (

)

)

qualitative analysis

Elements present in the compound

Mg, Cl

C, H, O

quantitative analysis (to find composition by mass) ( ) Empirical formula

MgCl2 for ionic compounds

finding the charges of ions present

2+

–

Mg (Cl )2

Ionic formula

CH2O for molecular compounds

determination of relative molecular mass

Molecular formula

C2H4O2

study of properties of compound

Figure 12.1 Determination of various formulae, with examples.

Structural formula

H

H

O

C

C

O

H

H

Some examples of empirical, ionic, molecular and

12.1

structural formulae of a few substances are given in Table 12.1. 45

Part III Metals

Empirical formula

Ionic formula

Molecular formula

Structural formula

–

–

N2

NN

Carbon dioxide

CO2

–

CO2

O=C=O

Ethene

CH2

–

C2H4

Substance Nitrogen

H H H C

C

H

H H H

Propene

CH2

C3H6

–

H C

C

C

H

H H H

Ethanol

C2H6O

C2H6O

–

H C

C

O H

H H

Methoxymethane

Ammonium chloride Iron(II) sulphate-7water (II) Potassium dichromate Quartz

C2H6O

C2H6O

–

+

–

NH4Cl

NH4 Cl

FeSO11H14

2– 2+ Fe SO4 . 7H2O

K2Cr2O7

+

(K )2Cr2O7

SiO2

2–

–

H

H

H C

O C

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.).

(

) 1

2

3

(

)

Even different compounds may have the same empirical formula and same molecular formula (e.g. ethanol and methoxymethane). But they have different structural formulae.

46

(

)


Class practice 12.5

But-1-ene has the structural formula H

H

H

H

H

C

C

C

C

H

H

12.5

A12.5 (a) C4H8 (b) CH2

-1-

H . H

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

H

H

H

H

C

C

C

C

H

H

H

(a)

12.4 Determination of empirical formulae

(b)

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

12.2

experiment is shown in Figure 12.2.

hole in test tube copper oxide town gas supply

excess town gas burns here

heat

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

47

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

= 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

= (18.579 – 18.100) g

= 0.479 g

= 0.479 g

= (18.701 – 18.579) g

= (18.701 – 18.579) g

= 0.122 g

= 0.122 g

Mass of oxygen in oxide

= 18.701 g = 18.579 g

Calculations

Masses (in g)

(

g

)

Cu

O

0.479

0.122

Number of moles of atoms (mol) (=

mass in g ) molar mass

0.479

(mol)

63.5

0.122

= 0.007 54

16.0

Relative number of moles (divided by the smallest number)

0.007 54 0.007 54

∴ the empirical formula of copper oxide is CuO.

= 0.007 63

0.007 63 =1

0.007 54

∴

= 1.01 1

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.

48

)

12.3)

Chapter 12 Reacting masses A12.6 1. Let CxHyOz be the empirical formula of the compound. crucible 1.173 Number of moles of CO2 = mol = 0.0267 mol 44.0 ∴ number of moles of C = 0.0267 mol pipeclay 0.24 Number of moles of H2O = mol = 0.0133 mol triangle (1.0 2 + 16.0) ∴ number of moles H = 0.0133 2 mol = 0.0266 mol Mass of C in the compound = 0.0267 12.0 g = 0.3204 g Mass 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

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

magnesium coil rocksil H

O

Masses (in g)

0.3204

0.0266

0.849

Number of moles (mol)

0.0267

0.0266

0.053

Relative number of moles

0.0267 0.0266 1

0.0266 0.0266 =1

0.053 0.0266 2

heat very strongly

tripod

C

∴ empirical formula of the compound is CHO2.

0.849 mol 16.0 = 0.053 mol

∴ number of moles of O in the compound =

From the experimental results, the empirical formula of magnesium oxide can be worked out to be MgO. Try the experiment yourselves.


2.

C

H

Masses (in g)

0.857

0.143

Number of moles (mol)

0.857 12.0 = 0.07142

0.143 1.0 = 0.143

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

MgO

12.6 1. 1.200 g 1.173 g

0.240 g

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. Relative number of moles

0.07142 0.07142 =1

0.143 0.07142 2

12.5 Determination of molecular formulae

2.

C xH y 1.000 g 0.857 g

x

y

∴ empirical formula of the compound is CH2.

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 12.7 Determining empirical formula and molecular formula using percentage by mass Compound X was found to contain carbon and hydrogen only. Experiments showed that it contained 80% carbon and 20% hydrogen by mass. If its relative molecular mass was 30.0, find its empirical formula and molecular formula.

12.7

X 80% 30.0

X 20%

cont'd

49

Part III Metals

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

Masses (in g)

(

g

100 g

X

20 g C

H

80

20

80 = 6.7 12.0

20 = 20 1.0

6.7 =1 6.7

20 = 2.98 3 6.7

)

80 g



(mol) (=

(g)

)

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

)

∴ the empirical formula of X is CH3. Note: Experimental errors would probably result in a small difference from whole numbers. In calculating relative number of moles, it is an accepted practice to ‘round off’ these values to the nearest whole numbers. However, care must be taken in doing so, e.g. 2.98 can be taken to be 3, but 2.8 is usually not taken as 3. Let its molecular formula be (CH3)n, where n is the whole number. Relative molecular mass of (CH3)n = 30.0 n(12.0 + 1.0 3) = 30.0

X

2.98 3

2.8

3 (CH3)n n (CH3)n

= 30.0 n(12.0 + 1.0 3) = 30.0

15.0n = 30.0 ∴

n=2

15.0n = 30.0 ∴

n=2

∴ molecular formula of X is C2H6.

Example 12.8 Determining empirical formula and molecular formula using masses of combustion products A compound Y containing only carbon, hydrogen and oxygen burned completely in air to form carbon dioxide and water as the only products. 2.43 g of Y gave 3.96 g of carbon dioxide and 1.35 g of water. Find the empirical formula of Y. If its relative molecular mass was 160, find also its molecular Note 10 formula. N10 Refer to ‘Supplementary information: Combustion analysis of organic compounds’ in the Teacher’s Guide.

50

CH3

X

C2H6

12.8

Y 2.43 g 1.35 g Y

3.96 g Y 160

Y

cont'd


Solution Since all the C in CO 2 and H in H 2 O came from the compound, 12.0 mass of C in the compound = 3.96 g 12.0 + 16.0 2 = 1.08 g; 1.0 2 mass of H in the compound = 1.35 g 1.0 2 + 16.0 = 0.15 g

CO 2

C

= 3.96 g

12.0

H = 1.35 g

1.0 2

)

= 0.15 g

1.0 2 + 16.0 O

Now go on to find the empirical formula as follows:

g

= 1.08 g

12.0 + 16.0 2

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

(

H

C

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

Masses (in g)

H 2O

= (2.43 – 1.08 – 0.15) g = 1.20 g

C

H

O

1.08

0.15

1.20

1.08 = 0.090 12.0

0.15 = 0.15 1.0

1.20 = 0.075 16.0

0.090 = 1.2 0.075

0.15 =2 0.075

0.075 =1 0.075

1.2 5 = 6

2 5 = 10

15=5



(mol) (=

(g)

)


)

(multiplied by the smallest possible whole number (5 here) to turn all values into whole numbers) ( ( ) )

Note: We should not take 1.2 to be equal to 1, because we cannot allow for so large an experimental error. ∴ empirical formula of compound is C6H10O5. Let its molecular formula be (C6H10O5)n, where n is a whole number. Relative molecular mass of (C6H10O5)n = 160 n(12.0 6 + 1.0 10 + 16.0 5) = 160 162n = 160 n = 0.988 1 ∴ molecular formula of compound Y is C6H10O5.

1.2 ∴

1 C6H10O5 (C 6 H 10 O 5 ) n

n (C6H10O5)n = 160 n(12.0 6 + 1.0 10 + 16.0 5) = 160 162n = 160 n = 0.988 1 Y

C6H10O5 51

Part III Metals

Example 12.9 Determining molecular formula using empirical formula and relative molecular mass

12.9

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

CH2 42

Solution Let the molecular formula of the compound be (CH 2) n, where n is a whole number.

( C H 2) n n

Relative molecular mass of (CH2)n = 42

(CH2)n

n(12.0 + 1.0 2) = 42

= 42 n(12.0 + 1.0 2) = 42

n=3

n=3

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

∴

( C H 2) 3 C3H6

A12.7 Since all the C in CO2 and H in H2O came from the compound, 12.0 = 0.36 g mass of C in the compound = 1.32 g 12.0 + 16.0 2 mass of H in the compound =

0.45 g 1.0 2 = 0.05 g 1.0 2 + 6.0

The rest of the compound must be oxygen. ∴ Mass of O in compound = (0.81 – 0.36 – 0.05) g = 0.40 g

12.7

Class practice 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. Now go on to find the empirical formula as follows: H C 0.05 Masses (in g) 0.36 Number of moles of atoms (mol) mass in g (= ) molar mass

0.36 = 0.03 12

0.05 = 0.05 1.0

0.81 g 0.45 g 320

O 0.40 0.40 = 0.025 16

12.10

Example 12.10 Determining the number of water of crystallization 5.60 g of hydrated copper(II) sulphate CuSO4 · nH2O was heated 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. C

52

Relative number of 0.03 = 1.2 moles (divided by the 0.025 smallest number) Multiplied by the smallest possible whole number (5 1.2 5 = 6 here) to turn all values into whole number

1.32 g

H

O

0.05 =2 0.025

0.025 =1 0.025

2 5 = 10

15=5

5.60 g nH 2 O)

(II) (CuSO 4 ·

(II) (a) (b)

3.59 g n (a) cont'd

∴ empirical formula of compound is C6H10O5

Let its molecular formula be (C6H10O5)n, where n is a whole number. Relative molecular mass of (C6H10O5)n = 320 n(12.0 6 + 1.0 10 + 16.0 5 ) = 320 162n = 320 n =2 ∴ molecular formula of compound is C12H20O10.


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

(a)

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

CuSO4

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

Masses (in g) ( g Number of moles (mol) (=

= 1.0 2 + 16.0 = 18.0

CuSO4

H2O

3.59

2.01

3.59 = 0.0225 159.6

2.01 = 0.112 18.0

0.0225 =1 0.0225

0.112 = 4.98 0.0225

)

mass in g ) molar mass (g)

= 63.5 + 32.1 + 16.0 4 = 159.6

H2O

(mol) (=

= (5.60 – 3.59) g = 2.01 g

)


)

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

n 5

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

(b)

(1)

Not all water of crystallization has been removed 12.10 in the heating process. Self-Test Mass of water of crystallization = (9.99 – 5.41) g = 4.58 g

(2)

Weighing had been delayed, so that the anhydrous salt had absorbed some moisture from the atmosphere. Formula mass of FeCl2 = 55.8 + 35.5 2 = 126.8

Self-Test 12.10

Mass (in g)

FeCl2

H2O

5.41

4.58

When 9.99 g of hydrated iron(II) chloride (FeCl2 · nH2O) was heated in a crucible, 5.41 g of anhydrous iron(II) chloride was left. Find the value of n. Number of moles 5.41 4.58


(mol) Relative number of moles

126.8

= 0.0427

0.0427 =1 0.0427

18.0

n

5

(1) (2)

9.99 g nH2O)

(II) (FeCl 2 · 5.41 g (II)

= 0.2544

0.2544 ≈6 0.0427

(4.98)

➲

n 21

∴ n=6

12.6 Calculations based on equations Calculations from equations — reacting masses

12.6 —

The theoretical amounts of substances used up or produced in a reaction can be calculated from its balanced equation as shown below:

53

Part III Metals

Problem-solving strategy Calculating the reacting masses Step 1 Write down the balanced equation for the reaction. 2Mg(s) + O2(g)

1 2Mg(s) + O2(g)

2MgO(s)

Step 2 Convert the amounts of given substances into mole quantities.

2 (

(Assume that the mass of Mg is 2.43 g.) Molar mass of Mg = 24.3 g mol

–1

Number of moles of Mg =

Mg

mass (g)

–1

(g mol )

2.43 = 24.3 mol = 0.100 mol 3

Mg Mg MgO

Mg

(

Hence number of moles of MgO formed = 0.100 mol Step 4 Change the mole quantities of the required substances back into mass as required by the question.

) MgO

Mg =2:2=1 1 MgO 0.100 mol

–1

–1

mass of MgO formed = 0.100 40.3 g = 4.03 g

=

4

The steps taken are illustrated below:

MgO –1 = (24.3 + 16.0) g mol –1 = 40.3 g mol MgO = 0.100 40.3 g = 4.03 g

Number of moles of A by ratio Required information of B A

A

B

54

–1

(g)

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

Number of moles of B

= 24.3 g mol

=

Step 3 Calculate the mole quantities of the required substances using the ratio, as given by the stoichiometric coefficients of the equation. Since oxygen is in excess, all the Mg is changed to MgO. (Mg is called the limiting reactant in this case, as it is all used up. Obviously, amounts of products formed depend on the amount of the limiting reactant only.)

Given information of A

)

Mg

–1

molar mass (g mol ) 2.43 = mol 24.3 = 0.100 mol

= 40.3 g mol

2.43 g

Mg

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

2MgO(s)

B


Study the following examples. Example 12.11 Calculations based on equation (with limiting reactant)

12.11

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

2.43 g

1.28 g

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

2MgO(s)

Step 2: Molar mass of Mg = 24.3 g mol Number of moles of Mg =

–1

2.43 mol 24.3

1

2Mg(s) + O2(g)

2

Mg

= 24.3 g mol

Number of moles of O2 =

–1

O2

1.28 mol 32.0

= 32.0 g mol

= 0.0400 mol 3

Mg =2

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

–1

= 1.28 mol 32.0 = 0.0400 mol

O2

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

–1

= 2.43 mol 24.3 = 0.100 mol

Mg

= 0.100 mol Molar mass of O2 = 32.0 g mol

2MgO(s)

O2

1

∴ 0.0400 mol O2 0.0400 2 = 0.0800 mol Mg

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

0.100 mol Mg Mg

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

O2 O2

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

O2

∴ number of moles of MgO = 0.0400 2 mol

=1

= 0.0800 mol

MgO

2

∴ MgO = 0.0400 2 mol = 0.0800 mol

∴ mass of MgO formed = 0.0800 40.3 g = 3.22 g

∴

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

MgO = 0.0800 40.3 g = 3.22 g

12.12

Example 12.12 Calculations based on equation (complete reaction) Calculate the mass (in kg) of copper produced by the complete reaction of 1.59 kg of copper(II) oxide in the following reaction: CuO(s) + H2(g)

Cu(s) + H2O()

1.59 kg

N11

(II) (

kg

) CuO(s) + H2(g)

Cu(s) + H2O() cont'd

55

Part III Metals

Solution CuO

The only substances involved in calculations here are CuO and Cu.

Cu

First method Step 1: CuO(s) + H2(g) 1 CuO(s) + H2(g)

Second method

Cu(s) + H2O() Cu(s) + H2O()

Since 1 mole of CuO produces 1 mole of Cu, ∴

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

79.5 kg of CuO produces 63.5 kg of Cu.

Molar mass of CuO = (63.5 + 16.0) g mol –1 = 79.5 g mol Number of moles of CuO = 2 CuO

∴

–1

1 mol CuO ∴

= 1.59 1000 g = 1590 g = (63.5 + 16.0) g mol –1 = 79.5 g mol

∴

=

CuO

3 ∴

Cu Cu

Cu

1 mol Cu

79.5 g CuO

63.5 g Cu

79.5 kg CuO

63.5 kg Cu

1.59 kg CuO

1.59 (63.5 79.5 )

–1

1590 mol = 20.0 mol 79.5 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 CuO

1.59 1.59 kg of CuO produces (63.5 79.5 ) = 1.27 kg of Cu.

1590 mol = 20.0 mol 79.5

= 1.59 kg

CuO

79.5 g of CuO produces 63.5 g of Cu, or

=1

= 1.27 kg Cu

1

= 20.0 mol = 20.0 63.5 g = 1270 g = 1.27 kg


Example 12.13 Calculations based on equation (actual yield and theoretical yield) 5.91 g of iron was dissolved in excess dilute hydrochloric 2+ acid to give a solution containing Fe ions. The solution was 2+ then boiled with concentrated nitric acid to oxidize all Fe 3+ ions into Fe ions. Excess sodium hydroxide solution was 3+ added to precipitate all Fe ions as iron(III) hydroxide, Fe(OH)3. The precipitate was filtered off, washed, dried and finally heated to convert all into iron(III) oxide, Fe2O3.

➲

22

12.13

5.91 g 2+ Fe Fe Fe

2+

Fe

3+

3+

(III)

Fe(OH)3 Fe(OH)3 (III) Fe2O3 cont'd

56


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

(a)

(b) The mass of iron(III) oxide actually obtained from experiment (actual yield) was 7.95 g. Compare this with the theoretical mass in (a) and give two possible reasons for the difference.

(b)

(III) (

)

(

(III) 7.95 g

)

(a)

Solution (a) Step 1: The whole process may be represented by a sequence of steps: Fe 5.91 g

Fe

2+

Fe

3+

(a)

1 2+

Fe(OH)3

Fe2O3 ?g

3+

Fe Fe 5.91 g

Fe

Fe(OH)3

Fe2O3 ?g

2

Step 2: To get the answer, one method is to write balanced equations for each of the reactions. 2+ 3+ From these, we can calculate masses of Fe , Fe and Fe(OH)3 in turn, and finally that of Fe2O3.

Fe

2+

Fe

3+

Fe(OH) 3 Fe2O3

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

(Fe)

2Fe Fe 2O 3 (the ‘2’ before Fe is added to balance number of Fe atoms)

(Fe 2 O 3 )

Thus mole ratio of Fe : Fe2O3 = 2 : 1. Molar mass of Fe = 55.8 g mol

2Fe

–1

Fe2O3 (Fe Fe

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

) Fe

Number of moles of Fe used = 5.91 mol 55.8

2

1 = 55.8 g mol

–1

Fe 5.91 = 55.8 mol = 0.106 mol Fe2O3 3 0.106 = mol = 0.0530 mol 2 Fe2O3 ∴ = 0.0530 159.6 g = 8.46 g

∴ theoretical mass of Fe2O3 formed = 0.0530 159.6 g = 8.46 g (b) The actual yield (7.95 g) is smaller than the theoretical yield (8.46 g).

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

=

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

0.106 mol Step 3: Number of moles of Fe2O3 formed = 2 = 0.0530 mol

(1) Iron used might be impure.

Fe 2 O 3

Fe

= 0.106 mol

Possible reasons are:

2

(b)

(7.95 g)

(8.46

g) (1) (2)

( ) 57

Part III Metals

12.8

Class practice 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

1.5 g

hydroxide was filtered off, washed and dried. (a) Write down the equation for the reaction of calcium with water.

(a) (b)

(b) Calculate the theoretical mass of calcium hydroxide obtained.

(c)

(c) The mass of calcium hydroxide obtained from the experiment was much less than the theoretical value. Explain why there is such difference. A12.8 (a) Ca(s) + 2H2O() Ca(OH)2(aq) + H2(g) (b) Mole ratio of Ca : Ca(OH)2 = 1 : 1 –1 Molar mass of Ca = 40.1 g mol –1 –1 Molar mass of Ca(OH)2 = 40.1 + (16.0 + 1.0) 2 g mol = 74.1 g mol 1.50 Number of moles of Ca used = mol 40.1 = 0.0374 mol

Key terms

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.

Page

58

1. actual yield

56

2. Avogadro constant

37

3. composition by mass

47

4. empirical formula

44

5. fraction by mass

42

6. ionic 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


Summary 12.1 The mole concept in general mole

1.

Chemists use

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

2.

Avogadro constant (L) The 23 –1 equal to 6.02 10 mol .

3.

mole One of a substance (or species) represented by a formula is the amount 23 containing 6.02 10 formula units.

4.

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

5.

Important relationships involving moles:

is the number of atoms in exactly 12 g of carbon-12. It is

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

mass ______________ (g) –1

molar mass (g mol ) formula units number of _______________________________ –1

Avogadro constant (mol ) –1

• Mass of 1 formula unit (g) =

molar mass ________________________________ (g mol ) –1

Avogadro constant (mol )

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 =

relative atomic ______________________________ mass of A no. of atoms of A in formula

formula mass of the compound

100%

12.3 Chemical formulae of compounds 7.

Chemical formulae

are part of language of chemistry. They give information about the substances concerned and are not just short-hand representation. Some common empirical ionic chemical formulae include formula, formula, molecular structural formula and formula.

59

Part III Metals

12.4 Determination of empirical formulae Empirical

8.

formula of a compound is the formula which shows the simplest whole number ratio of the atoms or ions present.

9.

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.

12.5 Determination of molecular formulae 10.

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.

12.6 Calculations based on equations 11.

60

theoretical The amounts of substances used up or produced in a reaction can be calculated from its balanced equation.

Chapter 13 Corrosion of metals and their protection

13.1 Corrosion of metals

13.1

The reaction of a metal with air, water or other substances in the surroundings, leading to gradual deterioration of the metal, is called corrosion.

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 N1 widely used metal, rusting is the most common type of corrosion.

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

Check your concept

✘ ✔

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’.

✔

13.2 Rusting

13.2

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

13.1

both water and air are involved in rusting. We have to leave the test tubes to stand for several days.

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

(

)

61

Part III Metals

anhydrous calcium chloride

cotton wool

oil layer

iron nail

boiled distilled water

Tube 1 1

Figure 13.1 Iron rusts only in the presence of both water and air. ( )

distilled water

Tube 3 3

Tube 2 2 no rusting

rusting occurs

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

Class practice 13.1 1. 2.

A13.1 1. In Tube 1, moisture in air has been absorbed by anhydrous calcium chloride. In Tube 2, dissolved air in water has been driven out by boiling. Besides, the oil layer on top prevents air from dissolving in water again.

13.1

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.

1.

1 2

3

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

13.1

Iron nail is immersed in distilled water in Tube 3. Distilled water contains dissolved air. 2. No; there is no air on the Moon. (In March 1998, NASA revealed that there was strong evidence for the existence of a large quantity of ice at the poles of the Moon. Thus the old idea that there was no water on the Moon might have to be changed.)

2.

Simple chemistry of rusting Rusting is a slow chemical process. In the first stage of rusting, N2 2+

some iron atoms lose electrons to form Fe (aq) ions. Fe(s)

2+

–

Fe (aq) + 2e

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

2+

Fe (aq) Fe(s)

Then a series of reactions follow. The overall reaction can

2+

–

Fe (aq) + 2e

be represented by an equation: 4Fe(s) + 3O2(g) + 2nH2O()

2Fe2O3 · nH2O(s) rust (reddish brown)

4Fe(s) + 3O2(g) + 2nH2O()

(

Rust is in fact hydrated iron(III) oxide (Fe 2O 3 · nH 2O),

)

(III) (Fe 2 O 3 ·

where n is a variable number. 62

2Fe2O3 · nH2O(s)

nH2O)

n


13.1

Example 13.1 Classifying chemical reactions in terms of their rates Different chemical reactions occur at different rates. Reactions can be roughly classified into three types depending on rate:

(1)

(1) Instantaneous reactions

(2)

(2) Reactions with a moderate rate

(3)

(3) Slow reactions Give two examples of each type. Solution

(1)

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

13.2)

(2)

(2) Reactions with a moderate rate: addition of magnesium to dilute hydrochloric acid; addition of calcium to water.

(3) (

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

Figure 13.2 An explosion is a very fast reaction.

(

13.3)

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. Note 3 Acids speed up rusting because they 2+ (1) promote the formation of Fe (aq): 2+ – Fe(s) Fe (aq) + 2e (2) increase the conductivity of solution.

Note 4 Alkaline solutions (e.g. 0.1 M NaOH) will not speed up rusting. On the contrary, they will inhibit rusting by shifting the position of equilibrium of the following cathodic reaction to the left: – – O2(g) + 2H2O() + 4e 4OH (aq)

N3 & N4

(

)

63

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 N5 & (which contains potassium hexacyanoferrate(III) K3[Fe(CN)6])

(

N6

(III)

2+

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

(

2+

13.4)

2+

ions. The rust indicator is used as a sensitive test for Fe (aq) ions, forming a blue colour. Thus, the appearance of a blue

)

Fe (aq) 2+

Fe (aq)

colour indicates rusting. Note 5 There is no need to mention the principle behind rust indicator. The rust indicator should be treated as any indicator (such as methyl orange or phenolphthalein) and its only use is to show where rusting occurs. Note 6 The rust indicator contains: (a) Potassium hexacyanoferrate(III) (K 3 [Fe(CN) 6 ]), which reacts with 2+ Fe (aq) ions to form a deep blue complex.

(b) Sodium chloride, which increases electrical conductivity of solution, so that corrosion (causing the colour to change to blue) occurs more quickly. (c) Agar, which makes the warm solution set on cooling to form a gel. Diffusion of blue patches formed around the nail can thus be slowed down for easier observation.

gel containing rust indicator solution

head

shank

nail

Figure 13.4 Detecting rusting by a rust indicator solution. Blue patches around the iron nail appear in a few minutes.

Petri dish tip

13.2

Example 13.2 Using rust indicator to investigate rusting

64

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

13.5

Dish 1 contains a single iron nail.

1

Dish 2 contains an iron nail partly wrapped with a magnesium ribbon.

2

Dish 3 contains an iron nail partly wrapped with a copper strip.

3 cont'd


(a) What do you observe in each case?

(a)

(b) Explain the observed results.

(b)

(c)

(c)

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

iron nail only

(a)

iron nail wrapped with a magnesium ribbon

(b)

iron nail wrapped with a copper strip

Figure 13.5 Investigating rusting with a rust indicator.

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

(a)

1

Dish 2: Gas bubbles appear around magnesium. No blue colour appears.

2

Dish 3: A blue colour appears, mainly around the head and tip of the nail. The blue patches are larger than those observed in dish 1.

3

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

1 (b)

1

(The blue patches are formed mainly around the head and tip of the nail. This is because these regions are sharply pointed; iron there 2+ can change into Fe (aq) ions more easily.) Dish 2: A blue colour does not appear, indicating that the iron nail does not rust. This is because magnesium, being more reactive, loses electrons more easily than iron. Because of this, iron is prevented from losing electrons. Iron 2+ therefore cannot form Fe (aq) ions and does not rust. (Magnesium reacts with hot water to form hydrogen. This accounts for the appearance of gas bubbles around magnesium.)

( 2+

Fe (aq)

)

2

2+

Fe (aq) ( ) cont'd

65

Part III Metals

Dish 3: A blue colour appears, showing that the iron nail rusts. The iron nail wrapped with a copper strip rusts more quickly than the iron nail alone. This is because copper is less reactive than iron, causing iron to lose electrons more easily. This speeds up the rusting of the iron nail. (c)

Join iron to a more reactive metal (e.g. magnesium or zinc).


3

(c)

( )

➲

13.5 Protecting iron from rusting

15

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.

(

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

Coating with paint, plastic, oil or grease As soon as the paint cracks or peels off, rusting occurs.

Objects unlikely to be scratched can be coated with paint. For N7 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 N8 scratched.) Oil or grease is used for protection instead.

Coating with another metal

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

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 N9 iron. Some roofs and buckets are made from galvanized iron.

66

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

1.

)


2.

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

2.

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. Learning tip Tin is a less reactive metal than iron.

3.

Electroplating: Electroplating is a process in which a thin

(

3.

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.

Learning tip The negative terminal of a car battery is called the cathode. So this kind of electrical protection is known as cathodic protection. More details about the meaning of ‘cathode’ will be discussed in Chapter 31.

Sacrificial protection During rusting, iron (Fe(s)) loses electrons to form iron(II) ions 2+

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

(Fe(s)) (II)

2+

(Fe (aq))

metal will lose electrons in preference to iron. This would 2+

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

(Fe(s))

2+

Fe (aq) 67

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. N10 13.6) Galvanizing is not used in making food cans because zinc ions are toxic. 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.

zinc coating broken surface

zinc

Figure 13.6 Sacrificial protection of iron by zinc.

iron

oxygen and water cannot reach iron, so no rusting occurs

oxygen reacts with zinc instead of iron — no rusting

iron

Check your concept

✘ ✔

Tin-plating is a kind of sacrificial protection.

✘

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

✔

Using alloys of iron (0.15–1.5%)

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.

68

(

)


Class practice 13.2

13.2

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

Stainless steel does not rust, yet it is seldom used to make large objects. Why?

Table 13.1 summarizes the advantages and disadvantages

13.1

of different methods of rust prevention with some examples.

Method of rust prevention (a) Painting

(b) Coating with plastic

Simple chemistry

the added layer prevents the iron object from contact with air and water

(c) Oiling or greasing

(d) Tin-plating

Advantages

Disadvantages

Examples

• economical

• fall off easily

window frames, car bodies

• lasts long • looks good

• more expensive than painting

coat hangers

• does not fall off like paint

• not ‘once and for moving parts of machines all’

• has lubricating effect

• dirt would stick to oil or grease

• tin is corrosion resistant

• when the tin coating is damaged, rusting will occur more quickly than iron alone

‘tin’ cans for storing food

• tin ions are not poisonous

(e) Galvanizing

• in case the zinc coating is damaged, the iron is still protected

• zinc ions are poisonous

galvanized iron plate used in construction, buckets

(f) Chromiumplating

• has a beautiful shiny appearance

• expensive

bathroom fittings, car bumpers

• convenient

• not applicable to car bodies, many objects pipelines, water/fuel tanks

(g) Cathodic protection

the negative terminal of an electric source is connected to the iron object, supplying electrons to prevent it from rusting

69

Part III Metals

Method of rust prevention (h) Sacrificial protection

Simple chemistry a more reactive metal (e.g. magnesium, zinc) in contact is ‘sacrificed’ to form ions; this would prevent iron from forming iron(II) ions

Advantages

Disadvantages

Examples

• an effective way of • the ‘sacrificed’ metal needs protection replacement from time to time

zinc blocks attached to the hull of a ship

• has a beautiful appearance

cutlery, scissors

(II) (i) Alloying

iron is alloyed with chromium, nickel and manganese which are corrosion resistant

• expensive

• a very effective way of protection

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 N11 13.7). This coating protects the aluminium underneath from

(

13.7)

further attack by air and water. Note 11 We can remove the aluminium oxide layer by rubbing it with cotton wool soaked in mercury(II) chloride solution and dilute hydrochloric acid. The treated aluminium then corrodes unhindered.

Figure 13.7 Explaining the corrosion resistance of aluminium. 70

surface attacked by oxygen in the air

aluminium

thin protective layer of aluminium oxide

aluminium


Thickening protective layer on aluminium The protective oxide layer on aluminium is very thin. The oxide layer can be thickened and strengthened by an N12 electrolysis process called anodizing (anodization). After anodizing, aluminium becomes even more corrosion resistant. Also, anodized aluminium can be dyed easily to give attractive colours.

Note 12 –6 (a) Untreated aluminium has an oxide layer about 10 cm 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.

Key terms Page 1. alloying

68

2. anodizing (anodization)

71

3. cathodic protection

67

4. corrosion

61

5. electroplating

67

6. galvanized

66

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

71

Part III Metals

Summary 13.1 Corrosion of metals Corrosion

1.

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

2.

In general, a metal

higher

in the metal reactivity series corrodes faster.

13.2 Rusting 3.

Rusting

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

13.3 Factors that speed up rusting 4.

Factors that speed up rusting include: acidic • Presence of solutions or soluble salts High • temperature less • 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 colour where rusting occurs.

rust indicator

. It shows a blue

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 different methods of rust prevention.

13.6 Socio-economic implications of rusting 7.

Rusting causes enormous socio-economic problems and extensive corrosion protection has to be developed.

13.7 Corrosion resistance of aluminium 8. 9.

72

Aluminium is resistant to corrosion because it has a protective Anodizing

oxide

layer.

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

HKDSE Chemistry Bridging Programe 1C

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