mole and avogadro constant

120

Ch.12 Reacting mass

Prior knowledge

1. Molecules

= Simple molecular structure exist as molecules or atom (Group 0)

2. Atom

‐ consist of proton, neutron and electron

3. Relative molecular mass

‐ Relative molecular mass is used to describe the relative masses of molecules.

4. Formula mass

‐ Formula mass is used to describe the relative masses of elements or compound other than molecules

12.1 The mole, Avogadro constant and molar mass

Mole and Avogadro constant

‐ Atom is very small, e.g. one aluminium foil have billions billions of atom.

- It is inconvenient (不方便) to write down the exact number of atom when we do calculation.

∴A unit is introduced as a counting unit : mole ( symbol: mol)

‐ One mole of ________ contain 6.02 x 1023 of __________

‐ Mole is a unit, just like dozen(打)

e.g One dozen of cake contains 12 pieces of cake

One mole of atoms contains 6.02 x1023 atoms

One mole of water molecules contains 6.02 x 1023 water molecules

Avogadro constant (L) = 6.02 x 1023

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Formula 1:

Number of particles in a substance = mole x L

e.g How many sodium atoms are there in 3.8 mole of sodium atoms?

Ans: 3.8 x L

= 3.8 x 6.02 x 1023

= 2.29 x 1024

∴ there are 2.29 x 1024 atoms in 3.8 moles of sodium atoms.

e.g How many moles of molecules are there in 2.77 x 1023 oxygen molecules?

Anw: (2.77 x 1023 ) / L

= 0.46 mol

∴ there are 0.46 mole of oxygen molecules

e.g How many ions are there in 3 mole of K2SO4 ?

Ans: 1 K2SO4 particle contains 2 K+ ion and 1SO42‐ ( ∵2K+ + 1SO4

2‐ K2SO4)

∴ 1 mole of K2SO4 contains 2 moles of K+ and 1 mole of SO42‐

3 mole of K2SO4 contains 6 moles of K+ and 3 mole of SO42‐

3 mole of K2SO4 contains 9 moles of ions

9 x 6.02 x 1023

= 5.418 x 1024 ion

Practice

Complete the following table

Substance Mole Number

CO2 molecules 1.5

K2CO3 3 Number of formula unit K2CO3:

Number of ion:

Na atom 1.204 × 1024

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Mole and molar mass

Definition: The mass of one mole of a substance is called its molar mass ( g mol‐1)

Molar mass = Sum of relative atomic mass of all the atoms in the substance

=Relative molecular mass or formula mass

Example

Calculate the molar masses of the following compounds

a) SO2 b) sucrose (C6H12O6) c) iron(III) sulphate (Fe2(SO4)3)

Ans: a) Relative atomic mass of S + 2 x Relative atomic mass of O

= 32.1 + (16x2) =64.1

So molar mass of SO2 = 64.1 g mol‐1

b) (6 x relative atomic mass of C) + (12 x relative atomic mass of H) + (6 x relative atomic mass of O)

= 6 x 12 + 12 x 1 + 6 x 16

= 342

c) 2 x 55.8 + 3 x (32.1 + 4 x 16) =399

Practice

Calculate the molar masses of the following substances:

(N = 14, Cu = 63.5, O = 16, H = 1, P = 31)(Refer to table of ion if you forgot)

Substance Molar Mass

nitrogen

copper(II) nitrate

ammonium phosphate

sulphuric acid

chlorine gas

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Formula 2:

Example 1

What is the mass of 3 moles of chlorine molecules?

Answer:

‐ You have mole, molar mass so by the above formula, you can calculate the mass.

Molar mass of chlorine:

35.5 x 2 = 71

Mass of chlorine gas:

3 = mass / 35.5

mass = 213 g

∴ The mass of 3 moles of chlorine is 213g.

Example 2

What is the number of moles of potassium carbonate (K2CO3) present in 62.2 g of the substance?

Answer:

Molar mass of potassium carbonate:

= 138.2

Mole of K2CO3:

= 62.2/138.2

= 0.45 mol

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Practice

1. What is the mass of 8.00 moles of methane (CH4) molecules? (Relative atomic masses: H = 1.0, C = 12.0)

2. What is the number of moles of silver oxide (Ag2O) present in 81.2 g of the substance? (Relative atomic

masses: O = 16.0, Ag = 107.9)

Calculation involving two formulas

Example

How many molecules are there in 81.6 g of ammonia (NH3)?

(Relative atomic masses: H = 1.0, N = 14.0; Avogadro constant = 6.02 x 1023 mol‐1)

125

Practice

1. Consider 26.1 g of glucose (C6H12O6)

a) What is the number of moles of glucose present?

b) How many glucose molecules are present?

(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0; Avogadro constant = 6.02 x 1023 mol‐1)

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12.2 Percentage by mass of an element in a compound

1. Calculating percentage by mass of an element in the compound

Example:

What is the percentage by mass of Fe in Fe2O3? (Relative atomic mass of Fe = 55.8, O =16)

Answer: sum of relative atomic mass of Fe / molar mass of Fe2O3

= (55.8 x 2)/(55.8 x 2 + 16 x 3) x 100%

= 69.9%

∴ Percentage by mass of Fe In iron(III) oxide = 69.9%

Practice

What is the percentage by mass of oxygen in Na2CO3･10H2O? (09MC Q5)

(Relative atomic masses: H = 1, C = 12, O = 16, Na = 23)

127

2. Calculating the mass of an element in the compound

Example 1

Calculate the mass of copper in 15.0 g of copper(II) sulphate‐5‐water (CuSO4･5H2O).

(Relative atomic masses: H = 1.0, O = 16.0, S = 32.1, Cu = 63.5)

Solution

Formula mass of CuSO4･5H2O

= 63.5 + 32.1 + 16.0 × 4 + 5 × (1.0 × 2 +16.0)

= 249.6

Percentage by mass of Cu in CuSO4･5H2O

= 25.4%

That means for every gram of CuSO4･5H2O, there is 25.4% (or 0.254 gram) of Cu in it.

So mass of Cu in 15.0 g of CuSO4･5H2O

= 15.0 g × 25.4%

= 3.81 g

Practice

1. Calculate the mass of potassium in 7.91 g of potassium dichromate (K2Cr2O7).

(Relative atomic masses: O = 16.0, K = 39.1, Cr = 52.0)

(Hints: first calculate the percentage by mass, then % x total mass)

128

2. An ore contains 80% of zinc sulphide by mass. Assuming that the other components in this ore do not

contain zinc, what mass of the ore is required to extract 0.70 g of zinc? (11MC Q8)

(Relative atomic masses : S = 32.1, Zn = 65.4)

129

12.3 Chemical formulae of compounds

‐ In this section, we are going to learn three types of chemical formulae.

1. Empirical formula

2. Molecular formula

3. Structural formula

1. 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.

2. 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.

3. 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

130

12.4 Empirical formulae and molecular formulae derived from experimental data

‐ In the past, chemists don’t know the ratio of each element in a compound

e.g we have a copper oxide, is it copper(I) oxide (Cu2O) or copper(II) oxide (CuO)?

‐ So experiment is carried to determine the formula

‐ The formula found in experiment is called empirical formula (實驗式)

Example of finding empirical formula

Question type 1: Simple calculation (06Q37)(13Q13)

Example 1: Determining the empirical formula of magnesium oxide

‐ In this set‐up, we put pure magnesium in the crucible

‐ We weight the

1. crucible +lid,

2. crucible + lid with magnesium ,

3. crucible + lid + magnesium oxide

‐ By simple calculation, we get the mass of magnesium oxide,

and the mass of oxygen.

‐ Then follow the calculation below.

‐ We get a mole ratio: Mg : O =1 :1

∴ The empirical formula is MgO

131

Example 2: Determining the empirical formula of red copper oxide

Item Mass(g)

Combustion tube 18.100

Combustion tube + oxide of copper 18.701

Combustion tube + copper 18.579

Problem‐solving strategy

132

Practice

Hydrogen is passed over heated oxide of metal M to reduce the oxide to metal M. The following results are

obtained:

Mass of empty combustion tube = 52.2 g

Mass of combustion tube + oxide (before experiment) = 68.1 g

Mass of combustion tube + metal M (after experiment) = 65 g

What is the empirical formula of the oxide of metal M? (Relative atomic masses: O = 16.0, M = 63.5)

M O

Mass

Mole

Mole ratio

133

Question type 2: Determining from the percentage by mass (08Q31)(10Q33) (11Q38) (12Q3)

Example:

Determining the empirical formula of a compound from the percentage by mass

Compound Y was found to contain iron and oxygen only. Experiments showed that it contains 70% iron and

30% oxygen by mass. Calculate the empirical formula of Y.

(Relative atomic masses: O = 16.0, Fe = 55.8)

Assume that there are 100 g of Y. Then there are 70 g of iron and 30 g of oxygen.

Fe O

Mass 70 30

Moles 70/55.8

= 1.25

30/16

= 1.88

Mole ratio 1.25/1.25

= 1

1.88/1.25

= 1.5

‐ So the empirical formula of Y is Fe2O3.

Practice

In an oxide of metal M, the mass percentage of M is 55.0%. What is the chemical formula of this oxide?

(Relative atomic masses: 0 = 16.0, M = 39.1) (12Q3)

134

Question type 3: Calculation of mass of each element before finding empirical formula (09Q33)

An oxide of metal M reacts completely with carbon to give 7.37 g of metal M and 4.36 g of carbon dioxide.

What is the empirical formula of the oxide? (Relative atomic mass of M=55.8)

Answer

‐ To get the empirical formula, you need the mass of metal and oxygen

‐ We cannot get the mass of oxygen directly, so we need to first calculate the percentage by mass

(質量百分比) of oxygen in CO2

∴the empirical formula of the oxide is M2O3.

Practice

Complete reduction of 41.1 g of an oxide of metal X by hydrogen gas produces metal X and 4.28 g of water.

What is the empirical formula of the oxide? (Relative atomic masses: H = 1.0, O = 16.0, X = 207.2)

135

Prior knowledge

1. Molecular formula and empirical formula

‐ Empirical formula of a covalent compound shows the simplest whole number ratio of atoms in the

compound may not be same as the actual molecular formula

Determination of molecular formulae

Question type 4: Determining molecular formula using empirical formula and relative molecular mass

Example

A compound X contains 85.7% carbon and 14.3% hydrogen. Its relative molecular mass is 56.0.

Find a) its empirical formula; and

b) its molecular formula.

Answer

Suppose we have 100.0 g of compound X, containing 85.7 g of carbon and 14.3 g of hydrogen.

∴ the empirical formula of compound X is CH2

b) Let (CH2)n be the molecular formula of compound X.

Relative molecular mass of X = n(12.0 + 2 x 1.0) = 14n

∴14n = 56.0

n = 4

∴ the molecular formula of compound X is (CH2)4 or C4H8.

136

Practice

A chemical analysis of caffeine indicates its composition to be 49.5% carbon, 5.2% hydrogen, 16.5% oxygen

and 28.9% nitrogen by mass.

a) Calculate the empirical formula of caffeine.

b) The relative molecular mass of caffeine is 194.0.

What is its molecular formula? (Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0)

137

Question type 5: Determining the number of water of crystallization in a hydrated salt (03Q11, 14Q5)

‐ Some ionic compound would form a hydrated ionic compound (crystal)

water molecules are included in the crystal lattice

e.g formula of hydrated copper(II) sulphate is CuSO4•5H2O

Example

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.

(Relative atomic masses: H = 1.0, O = 16.0, S = 32.1, Cu = 63.5)

Answer:

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

= 2.01 g

Since n should be a whole number, a reasonable value of n would be 5.

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

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

(2) The anhydrous salt has absorbed some moisture from the atmosphere during weighing.

138

Practice

1. A sample of MgSO4.xH2O(s) of mass 123.2 g contains 63.0 g of water of crystallisation. What is the value of

x ? (Relative atomic masses: H = 1.0, 0 = 16.0, Mg = 24.3, S = 32. 1)(03Q11)

*2. Hydrated salt X •nH2O contains 51.16% of water by mass. Given that the molar mass of X is 120.3 g,

what is n? (Relative atomic masses: H = 1.0, 0 = 16.0)(14Q5)

139

12.5 Calculations involving chemical equation

‐ Consider the equation representing the reaction between hydrogen and oxygen

2 H2(g) + O2(g) 2 H2O

Mole ratio: 2 : 1 : 2

‐If we know the mole/no. /mass of one of the chemical reagent in the equation, we can calculate the rest.

Example 1:

2 H2(g) + O2(g) 2 H2O(l)

Given the mole of hydrogen gas are 4, what is the mole of water after all the hydrogen are consumed?

Answer:

As mole ratio of H2: H2O = 2:2 = 1:1

So the mole of water = 4 mol

Example 2:

2 H2(g) + O2(g) 2 H2O(l)

Given the mole of water produced are 3, what is the mole of oxygen consumed?

Answer:

As mole ratio of O2: H2O = 1:2

So the mole of oxygen consumed = 1.5 mol

Prior knowledge:

Formula 1:

No. of _____ = mole x L

L = 6.02 x 1023

Formula 2:

Mole = mass / molar mass

140

Example 3:

Calculate the mass of magnesium oxide produced when 2.43 g of magnesium burns completely in air.

Step 1:

Write down the balanced chemical equation for the reaction.

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

Step 2:

Convert the mass(es) of the given substance(s) into number of moles

Molar mass of Mg = 24.3

Number of moles of Mg:

Step 3:

Use the balanced chemical equation to calculate the number of moles of the substance asked in the

question.

As mole ratio of Mg: MgO = 2:2 ( =1:1)

So number of moles of MgO = 0.1 mol

Step 4:

Convert the number of moles of that substance into mass

Molar mass of MgO

= (24.3 + 16.0)

= 40.3

Mass of MgO produced

= 0.100 × 40.3

= 4.03 g

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Practice

1. Calculate the mass of sodium hydroxide produced when 12 g of sodium reacts completely with water.

Step 1:

Write down the balanced chemical equation for the reaction.

Step 2:

Convert the mass(es) of the given substance(s) into number of moles

Step 3:

Use the balanced chemical equation to calculate the number of moles of the substance asked in the

question.

Step 4:

Convert the number of moles of that substance into mass

142

2. Magnesium reacts with lead(II) oxide according to the following equation:

Mg(s) + PbO(s) MgO(s) + Pb(s)

Calculate the mass of magnesium required to react completely with 10.55 g of lead(II) oxide.

(Relative atomic masses: O = 16.0, Mg = 24.3, Pb = 207.2)

143

Limiting reactant (限量反應物)

‐ In chemical reaction, the reactant that is totally consumed when the reaction completes is called the

limiting reactant

‐ The other reactants in the reaction are in excess.

e.g 2H2 + O2 2H2O

500mole 0.4mole

H2 is in excess, O2 is the limiting reagent

Example

Titanium is prepared by the reaction of titanium(IV) chloride with molten magnesium. The reaction can be

represented by the equation:

TiCl4(g) + 2Mg(l) Ti(s) + 2MgCl2(l)

What mass of titanium can be produced by heating 64.6 g of titanium(IV) chloride with 19.2 g of

magnesium?

(Relative atomic masses: Mg = 24.3, Cl = 35.5, Ti = 47.9)

Ans:

144

Practice

1. When heated, lithium reacts with nitrogen to form lithium nitride as shown in the equation below:

6Li(s) + N2(g) 2Li3N(s)

8.28 g of lithium were heated with 10.6 g of nitrogen. Calculate the mass of lithium nitride formed.

(Relative atomic masses: Li = 6.9, N = 14.0)

(Hints: 1. Calculate the mole of reactant, 2. Determine which one is excess / limiting reagent

3. Calculate the mole of product by using the mole of limiting reagent)

2. Refer to the following chemical equation:

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

N moles of Fe2O3 are allowed to react with 2N moles of CO under suitable conditions until the reaction stops.

How many moles of Fe are formed?

A. N

B. 2N

C. 2/3 N

D. 4/3 N

145

Theoretical yield and percentage yield

‐ Theoretical yield (理論產量)(假設100%的reactant變成product)

e.g. 2H2 + O2 2H2O

‐ Actual yield (實際產量). The actual yield of a reaction is often less than the theoretical yield because:

• the reaction is incomplete.

• impurities are present in the reactants.

• side reactions occur in which unwanted side products are produced.

The efficiency of a chemical reaction can be expressed by the percentage yield:

Example

In an experiment, 15.9 g of copper(II) oxide was heated with 0.60 g of hydrogen :

CuO(s) + H2(g) Cu(s) + H2O(l)

(a) Calculate the theoretical yield of copper.

(b) Given the percentage yield of copper is 82%. Calculate the actual yield of copper.

(Relative atomic masses: H = 1.0, O = 16.0, Cu = 63.5)

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

Step 2: Molar mass of CuO = (63.5 + 16.0)

= 79.5

Step 3:

As the mole ratio of CuO : H2 = 1 : 1 , 0.20 mol of CuO would react with 0.20 mol of H2.

Since 0.30 mol of H2 is heated, H2 is in excess.

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

From the equation, mole ratio of CuO : Cu = 1 : 1.

Number of moles of Cu formed = 0.20 mol

So theoretical yield of Cu = 0.20 × 63.5 = 12.7 g

(b) Actual yield of Cu = theoretical yield (g) × percentage yield (%)

= 12.7 g × 82%

= 10.4 g

146

Practice

Methanol (CH3OH) can be produced from carbon monoxide and hydrogen according to the following

equation: CO(g) + 2H2(g) CH3OH(g) (Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)

(a) Calculate the theoretical yield of methanol when 430 g hydrogen reacts with excess carbon monoxide.

(b) Given the percentage yield of methanol is 45%, calculate the actual yield of methanol.

147

Ch.13 Corrosion of metals and their protection

13.1 What is corrosion?

Corrosion: The deterioration (變壞) of a metal due to its reaction with oxygen, moisture or other

substances in the environment

13.2 Corrosion of iron : Rusting

‐ The corrosion of iron would have a special term to describe : rusting

‐ Rust (銹) is the reddish brown solid that forms on the surface of the iron

‐ Its chemical name is hydrated iron(III) oxide ( Fe2O3∙x H2O)

‐ In hydrated iron(III) oxide, the number of water molecules which combine with one formula unit of Fe2O3

varies.

∴ We write x H2O.

‐ Rust is easy to detach the surface of iron, so fresh iron surface is then exposed in the air keeping rusting

until the whole iron is disintegrated.

‐ Rusting is a slow process. Usually takes several days for rusting to occur.

Condition necessary for rusting to occur

∴ Rusting need oxygen and water

148

What happens during rusting?

1. During rusting, iron atoms first lose electrons to form iron(II) ions.

Fe(s) Fe2+(aq) + 2e‐ (09Q20)

2. The electrons then combine with oxygen and water to form hydroxide ions.

1/2 O2(g) + H2O(l) + 2e‐ 2OH‐(aq)

3. The precipitate further reacts with dissolved oxygen and water, forming iron(III) hydroxide, Fe(OH)3(s).

4Fe(OH)2(s) + O2(aq) + 2H2O(l) 4Fe(OH)3(s)

The iron(II) ions and hydroxide ions precipitate as iron(II) hydroxide. Iron(II) hydroxide then reacts with

dissolved oxygen to form rust, which is hydrated iron(III) oxide (Fe2O3•xH2O).

Overall equation to represent rusting:

4Fe(s) + 3O2(g) + 2xH2O(l) 2Fe2O3•xH2O(s)

13.3 What factors can speed up the rusting process?

1. Presence of dissolved ionic substances

∵ rusting involve the electron transfer process. The presence of mobile ions would increase the electricity

conductivity. This also increases the speed of rusting.

∴ Sea water would speed up the rusting process.

2. Presence of acidic pollutants

3. Higher temperature

∵ at a higher temperature, chemical reaction becomes faster

4. Scratched or bent the surface. (刮掉/彎曲的表面)

‐ Electron lose easier on the scratching/bending surface so the rusting process speed up.

5. Attach (接觸) to a less reactive metal.

‐ e.g When Fe attach Cu/Ag/Pb ….. the rusting process speed up.

‐ If Fe attach to a more reactive metal( Zn, Al,Mg), rusting would not occur until those more reactive metal

are completely corroded.

149

13.4 To observe rusting using a rust indicator

‐ Rust indicator: a mixture of potassium hexacyanoferrate(III) ( K3Fe(CN)6) and phenolphthalein

‐ K3Fe(CN)6 gives a blue color in the presence of iron(II) ions = presence of rusting

∵Rusting involve

Fe(s) Fe2+(aq) + 2e‐

13.5 How to prevent rusting?

Using Protection coatings

1. Coating iron with paint, oil, grease or plastic

Painting

‐ cheapest, simplest, but when the paint is scratched, the metal would exposed to air rusting start

∴ not suitable for the moving parts of machines

‐ Coating plastic is more expensive than painting.

Oil, Grease (油)

‐ can be used to protect the moving parts of machine which cannot be painted

‐ serve as lubricant too

2. Tin‐plating

‐ Iron coat with tin

‐ As tin is less reactive than Fe, if the tin coat is scratched, rusting start.

‐ Tin ions is not‐toxic, so we use this method to make food cans.

3. Zinc‐plating

‐ Iron can be coated with a layer of zinc. Iron coated with zinc is called galvanized iron and this method is

called zinc‐plating

‐ When the zinc coating is damaged, zinc ‘sacrifices’ itself to protect the iron.

This is because zinc is more reactive than iron. Thus, zinc corrodes instead of iron.

‐ galvanized iron is not suitable for making food cans because zinc ions are toxic

150

4. Electroplating

‐ plate a very thin layer of metal on iron by an electrical process called electroplating, e.g Cr, Ni

‐ This plating is not easily detached and plating Cr makes the iron article more attractive

‐ Expensive method

5. Using sacrificial metals

‐ When a more reactive metal is attached to iron, the more reactive metal corrodes instead of iron

‐ The more reactive metal ‘sacrifices’ itself to prevent iron from rusting.

‐ This method is called sacrificial protection.(犧牲性保護)

‐ We often use Zn and Mg to do sacrificial protection

‐ We must connect iron and Zn/Mg directly or with a metal wire. Otherwise, the protection will not work.

6. Cathodic protection (陰極保護)

‐ Connect iron to the negative terminal of a battery while conductor (e.g graphite) to the positive terminal

‐ provide electron to Fe, prevent the formation of Fe2+

‐ Car body is protected from rusting by connecting it to the negative terminal of the car battery

Using corrosion resisting alloy (合金) of iron

‐ e.g Stainless steel : alloy made up of Fe, Cr, Ni

151

Practice

1. Suggest the principle of the rust prevention method used in each of the following cases:

A: Can’t contact with water/air

B: sacrificial protection

Method A/B?

galvanized iron buckets

greasing the steel gear system of a bicycle

tin‐plating iron food cans

painting an iron gate

attaching magnesium pieces to the hull of a ship

chromium‐plated car bumpers

*2. The diagram below shows three iron nails of the same size and shape each immersed in a liquid. (14Q3)

Which of the following arrangements represents the ascending order of rate of corrosion of the iron nails?

A. Z<Y<X

B. Y<Z<X

C. Z<X<Y

D. X<Z<Y

3. Tin plating is used to prevent iron cans from rusting because (16Q5)

A. tin provides sacrificial protection to iron.

B. tin layer prevents iron from exposure to air.

C. tin is higher than iron in the metal reactivity series.

D. tin and iron form an alloy which does not corrode.

152

13.7 Corrosion resistance of aluminium

‐ Aluminium reacts with oxygen in the air, an even coating of aluminium oxide forms.

‐ This oxide layer sticks firmly to the surface of the metal and is impermeable to oxygen and water.

‐ It protects the metal beneath from further attack.

‐ Rusting of iron is destructive but oxidation of the aluminium surface is a useful process.

Aluminium Anodization (陽極電鍍)

‐ an electrolytic process used to increase the thickness of the metal oxide layer on the surface of a metal

part.

‐ Aluminium is anodized to increase its corrosion resistance and surface hardness and to allow dyeing more

easily. e.g. the dye on milk bottle caps

153

Experimental set‐up for aluminium anodization

‐ The aluminium object to be anodized is connected to the positive terminal

‐ An aluminium sheet is rolled into cylindrical shape and connected to the negative terminal

‐ The electrolyte is dilute sulphuric acid

‐ During the process, oxygen produced at the anode reacts with the aluminium object to increase the

thickness of the oxide layer

‐ Overall equation

2Al(s) + 3H2O(l) Al2O3(s) + 3H2(g)

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Practice (08 Q3)

Four iron‐made objects are placed separately in get with rust indicator solution containing

potassium hexacyanoferrate(III), and allowed to stand in air for some time.

Complete the following table by writing down the observation and giving the relevant explanation for each of

the cases. (5 marks)

Case Observation Explanation

Iron‐made object fully plated

with zinc


with tin


with zinc , but part of the zinc

scratched to expose the iron

underneath


with tin, but part of the tin

scratched to expose the iron

underneath

mole and avogadro constant

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