IGCSE (9-1) Chemistry notes

Chapter-3
Atomic structure
What is meant by the terms atom and molecule?
Atom is the smallest unit of an element.
A molecule consists of two or more atoms chemically bonded by
covalent bonds.

The structure of an atom

 Elements are made of tiny particles of matter called atoms

 Each atom is made of subatomic particles
called protons, neutrons and electrons
 The nucleus of the atom contains protons and neutrons.

The Mass & Charge of Subatomic Particles Table

What is meant by the terms atomic number, mass number, isotopes and
relative atomic mass (Ar)?
 The number of protons in an atom's nucleus is called its atomic number or
proton number.

The mass number (sometimes known as the nucleon number) counts the
total number of protons and neutrons in the nucleus of the atom.

Isotopes are atoms (of the same element) which have the same atomic
number but different mass numbers. They have the same number of protons
but different numbers of neutrons.

Relative atomic mass is the average mass of an atom.

How to Calculate Relative Atomic Mass?

Chapter-4
The periodic table
How elements are arranged in the periodic table in order of atomic
number, in groups and periods?

Ans: Elements are arranged in order of atomic number and so that elements with
similar chemical properties are in columns, known as groups. Elements in the same
periodic group have the same amount of electrons in their outer shell, which gives
them similar chemical properties. Elements with the same number of shells of
electrons are arranged in rows called periods.

How to deduce the electronic configurations of the first 30 elements from the
in positions in the periodic table?

Ans:- Electrons orbit around the nucleus in a region Known as energy shells.

1st shell - The electrons in the first shell has the least energy
2nd shell-The electrons in the second shell has higher energy than the 1st shell's
electron

Electrons will fill the shells closer to the nucleus before filling any further out. 1st
shell holds 2 electrons, 2nd and 3rd hold 8 electrons.

rule for 1st 20 elements electronic configuration

2,8,8,2
Electronic Configuration of the First 20 Elements

Metals vs Non-metals
Basic oxides (alkaline) Acidic oxides
Good conductor of electricity Poor conductor of electricity

High melting and boiling point Low melting and boiling point

Malleable non-malleable

Identifying metals from periodic table position

A zigzag line separates the metals on the left from the non-metals on the
right.

Why are noble gases unreactive?

They have a full outer shell of electrons so they don’t easily lose, gain or
share electrons.

Chapter- 5 and 6

Mole calculations
Word Equations

 These show the reactants and products of a chemical reaction using their full
chemical names
 The reactants are those substances on the left-hand side of the arrow
 They react with each other and form new substances
 The products are the new substances which are on the right-hand side of the
arrow
 The arrow means the conversion of reactants into products

Example:

Sodium hydroxide + hydrochloric acid ⟶ sodium chloride + water

Representing reactions as equations

 Chemical equations use the chemical symbols of each reactant and product
 When balancing equations, there has to be the same number of atoms of
each element on both side of the equation
 A symbol equation must be balanced to give the correct ratio of reactants
and products.

Example: NaOH (aq) + HCl (aq) ⟶ NaCl (aq) + H2O (l)

How to balance equations

1. Count each type of atom in reactants and products.

2. Place coefficients, as needed, in front of the symbols or formulas to balance
the number of atoms or molecules of the substances on the right or left side.
3. Repeat steps 1 and 2 until the equation is balanced.

Zn + HCI = ZnCl2 + H2 (unbalanced equation)

Zn + 2HCI = ZnCl2 + H2

 Don’t forget to add state symbols when writing balanced equations:

Calculating Relative Mass

 The symbol for the relative formula mass is Mr and it refers to the total
mass of the substance.
 If the substance is molecular you can use the term relative molecular mass.
 To calculate the Mr of a substance, you have to add up the relative atomic
masses of all the atoms present in the formula.

Examples:

Moles: The mole is a unit of the amount of substance

how to carry out calculations involving amount of substance, relative
atomic mass (Ar) and relative formula mass (Mr)?

First of all balance both the sides then do the calculation using this
formula

number of moles = (given) mass (g) /mass of 1 mole (g)

Example: Finding the number of moles in 54g of water,

Relative atomic masses: H = 1, 0 = 16.

1 mol of H20 has a mass of 18g.

Number of moles = (given) mass in g/ mass of 1 mole (g)

No. of moles=54/18 = 3mol.

calculate reacting masses using experimental data and chemical

equations

Example

Calculate the mass of magnesium oxide that can be made by completely

burning 6.0 g of magnesium in oxygen in the following reaction:

2Mg (s) + O2 (g) ⟶ 2 MgO (s)

Calculating Percentage Yield

  Yield is the term used to describe the amount of product you get
from a reaction
 We never get 100% yield in a chemical process for several reasons
 These are:
o Some reactants may be left behind in the equipment
o The reaction may be reversible
o during separation stages such as filtration or distillation
o Products can also be lost during transfer from one container to
another.

Actual & Theoretical Yield

 The actual yield is the recorded amount of product obtained

 The theoretical yield is the amount of product that would be
obtained.

Formula for calculating percentage yield:

Percentage yield= Actual yield/theoretical yield x 100

Investigating the formula of a metal oxide by combustion

Procedures:

 Measure mass of crucible with lid

  Add a ribbon of magnesium into crucible and measure the
total mass with lid
 calculate the mass of the metal by subtracting the mass of
empty crucible
 Strongly heat the crucible over a Bunsen burner
 Lift the lid frequently to allow sufficient air into the crucible for
the magnesium to fully oxidizeS without letting magnesium
oxide escape
 When the reaction is finished, allow the crucible and
contents to cool.
 Measure the mass of crucible and contents
 Calculate the mass of metal oxide by subtracting the mass of
empty crucible.

Mass of metal: mass of crucible + metal-mass of empty crucible/g

Mass of oxygen:
mass of crucible + contents at end of experiment-mass of crucible +
magnesium/g

Magnesium oxygen

Mass a b

Mole a / Ar b / Ar

=x =y
Ratio x : y

Represent the ratio into the form ‘MxOy‘ E.g, MgO

Finding Formulae of Compounds:

Aim:
To determine the formula of hydrated copper sulfate,
CuSO4. xH2O

Method:

 Measure mass of the crucible

 Add a known mass of hydrated salt
 Heat over a Bunsen burner until the blue salt turns
completely white, which indicates that all the water has been
lost
 Record the mass of the crucible and contents
 Avoid overheating the salt as it could decompose and give
you a larger mass change
 Mass of the white anhydrous salt:
Measure mass of white anhydrous salt remaining
  Mass of water:
Subtract mass of the white anhydrous salt remaining from
the mass of known hydrated salt.

Anhydrous salt water

Mass a b

Moles a / Mr b / Mr

=y =x

Ratio 1 : x

Step 3 – Represent the ratio in the form ‘salt.xH2O’

Empirical & Molecular Formulae

Empirical formula: gives the simplest whole number ratio of atoms

of each element in the compound.

Molecular formula: gives the exact numbers of atoms of each element

present in the formula of the compound.

Example:

The empirical formula of X is C4H10S1 and the relative formula mass of

X is 180. What is the molecular formula of X?

Relative atomic masses: carbon : 12 hydrogen : 1 sulfur : 32

Step 1 – Calculate the relative formula mass of the empirical formula

(C x 4) + (H x 10) + (S x 1) = (12 x 4) + (1 x 10) + (32 x 1) = 90

Step 2 – Divide the relative formula mass of X by the mass of the

empirical formula

180 / 90 = 2

Step 3 – Multiply each number of elements by 2

(C4 x 2) + (H10 x 2) + (S1 x 2) = (C8) + (H20) + (S2)

Molecular Formula of X = C8H20S2.

Calculating Concentrations of Solutions

 A solid substance that dissolves in a liquid is called a solute, the
liquid is called a solvent and the two when mixed together form
a solution
 Concentration simply means the amount of solute there is in a
specific volume of the solvent
 The greater the amount of solute in a given volume then
the greater the concentration

FORMULA:
Concentration (in mol/dm3)= mole/volume

Calculating volumes of Gases

To find the volume
Volume = Moles x Molar Volume

To find the moles

Moles = Volume ÷ Molar Volume

Examples:
Calculate the volume (in cm3) of 0.01 g of hydrogen at rtp (A,: H = 1).
1 mol H2 has a mass of 2g:
number of moles= mass / mass of 1 mol
0.01 g of hydrogen is 0.01/2 mol = 0.005mol.

Because we want the volume in cm3 we use the molar volume as

24000cm3: volume = number of moles x molar volume
0.005mol of hydrogen occupies 0.005 x 24000 = 120cm3

Practical: know how to determine the formula of a metal oxide by

combustion (e.g. magnesium oxide) or by reduction (e.g. copper (II)
oxide)

Procedure:
• Weigh a ceramic dish.
• Put about 3g of copper oxide in the ceramic dish and weigh the
dish again.
• Place the ceramic dish in a tube as shown.
• Pass hydrogen gas over the copper oxide.
•Burn the excess hydrogen, which comes out of the small hole in the
boiling tube.
• Heat the copper oxide strongly until the reaction is finished (pink
brown copper metal will be seen).
The mass decreases at the end, because the hydrogen combines with
the oxygen from the copper oxide to form water. The oxygen is
removed from the copper oxide and only copper is left in the dish at
the end of the experiment. Because oxygen is removed from the
copper oxide we say that the copper oxide has been reduced.
Calculation
Step-1
Calculate the mass of copper oxide, by subtracting the mass of the
dish from the mass of the dish + copper oxide
Step-2
Work out the mass of copper remaining at the end by subtracting
the mass of the dish from the mass of the dish + copper at the end
Step-3
Calculate the mass of oxygen by subtracting the mass of copper at
the end from the mass of copper oxide.

Copper oxygen
Mass a b
Mole a / Ar b / Ar
=x =x
Ratio x : x
Represent the ratio into the form ‘CuxOx ‘ E.g, CuO

Chapter-7
bIonic bonding

How ions are formed by electron loss or gain?

An ion is an electrically charged atom or group of atoms formed by
the loss or gain of electrons. This loss or gain of electrons takes place to
obtain a full outer shell of electrons.When they lose electrons, they
become positively charged and are named cations. When they gain
electrons, they are negatively charged and are named anions.
charges of these ions: • metals in Groups 1, 2 and 3 • non-metals in Groups 5,
6 and 7 • Ag+, Cu2+, Fe2+, Fe3+, Pb2+, Zn2+ • hydrogen (H+), hydroxide
(OH–), ammonium (NH4+), carbonate (CO32–), nitrate (NO3- ), sulfate
(SO42–
Formulae for compounds formed between the ions listed above

Ionic Bonds: Dot & Cross Diagrams

Magnesium Oxide Dot & Cross Diagram

 Magnesium is a group 2 metal so will lose two outer electrons to another

atom to have a full outer shell of electrons
 A positive ion with the charge +2 is formed
 Oxygen is a group 6 non-metal so will need to gain two electrons to have a
full outer shell of electrons
 Two electrons will be transferred from the outer shell of the magnesium
atom to the outer shell of the oxygen atom
 Oxygen atom will gain two electrons to form a negative ion with charge -2

Formula of ionic compound: MgO

ionic bonding in terms of electrostatic attractions

The positive and negative charges are held together by the strong electrostatic
forces of attraction between oppositely charged ions. This is what holds ionic
compounds together.

why compounds with giant ionic lattices have high melting and boiling points?

There are many strong ionic bonds which has strong force of attraction between
the oppositely charged ions, and it requires a very high energy to be broken
down.

ionic compounds do not conduct electricity when solid, but do conduct

electricity when molten and in aqueous solution

when the ionic compounds are solid,the ions in the compound are tightly packed
and when it is molten the ions move freely.
CHAPTER-8
Chapter 9
Metallic bonding
What is metallic bonding?
It is the strong force of attraction between an array of positive ions and
sea of delocalized electrons.

Factors affecting the strength of metallic bonding

Size of positive ions- greater size means weaker metallic bond.

Charge of positive ion-higher charge means stronger force of

attraction.

Diagram showing metallic lattice structure with delocalized

electrons

Explaining the Properties of Metals

The link between metallic bonding and the properties of
metals

Metals have high melting and boiling points

o There are many strong metallic bonds in giant metallic

structures.
o A lot of heat energy is needed to break these bonds.

Metals conduct electricity

Metals have a sea of delocalized electrons and they are

free to move throughout the structure.

Metals are malleable and ductile

o Metals have Layers of positive ions which

can slide over eash other when a force is applied.

Chapter 10
Electrolysis
why do not covalent compounds conduct electricity?

Most covalent compounds don’t conduct electricity as they have no

delocalized electrons which can carry electricity.

Why ionic compounds conduct electricity only when molten or in

aqueous solution?
When the ionic compounds are solid, the ions in the compound are
tightly packed and when it is molten the ions move freely.

Anion and cation

Positive ions within the electrolyte move towards the negatively charged
electrode which is the cathode.

Negative ions within the electrolyte move towards the positively charged
electrode which is the anode.

Experiments to investigate electrolysis, using inert electrodes, of

molten compounds (including lead(II) bromide) and aqueous
solutions (including sodium chloride, dilute sulfuric acid and
copper(II) sulfate) and to predict the products.
Diagram showing the electrolysis of lead(II) bromide

Method:

 Add lead(II) bromide into beaker and heat so that it turns molten,
which will allow the ions to be free to move and conduct electric
charges.
 Add two graphite rods as the electrodes and connect this to a
battery.
 Turn on the battery and allow electrolysis to take place.
 Negative bromide ions will move to the positive electrode (anode)
and lose two electrons to form bromine molecules. There is
bubbling at the anode as the bromine gas is given off.
 Positive lead ions move to the negative electrode (cathode) and
gain electrons to form lead metal which is deposited on the bottom
of the electrode.
Electrode Products:
Anode: Bromine gas
Cathode: Lead metal.

2 rules for electrolyzing aqueous solutions

Positive ions: The least reactive element will be discharged

Negative ions: If there are Cl-,Br-,I- we will get Cl2,Br2,I2 otherwise OH-
will be discharged to give H2O or O2 gas

For aqueous NaCl solution:- Na+,H+,Cl-,OH-

Anode- 2Cl -2e- Cl2

Cathode- 2H+ + 2e- H2

If the question says dilute solution of NaCL.

There will be a large amount of of Na+,Cl-,H+,OH-

Anode- 2Cl- – 2e- = Cl

4OH- -4e- =2H2O+02

Cathode- 2H+ + 2e- =H2.

INVESTIGATING THE ELECTROLYSIS OF AQUEOUS

SOLUTIONS
 Set up the apparatus as shown.

. • Pour concentrated sodium chloride solution into the glass tube.

• Place a test-tube containing sodium chloride solution over each

electrode. The test-tubes must not completely cover the electrodes or
ions will be unable to flow and there will be no current.

• Connect the battery to the electrodes.

• The experiment should be done in a fume cupboard or well ventilated

room because chlorine gas is poisonous.

We can see if something is happening by looking for bubbles of gas or

a metal forming at the electrodes. Any gases can be tested.

In this experiment we see bubbles of gas at both electrodes. When

the gases are tested we find that hydrogen forms at the negative
electrode and chlorine forms at the positive electrode.

Electrolysis of aqueous sodium chloride

 In the electrolysis of aqueous sodium chloride the half

equation at the negative electrode (cathode) is:
 2H+ + 2e– ⟶ H2 Reduction

 At the positive electrode (anode) chlorine gas is

produced by the discharge of chloride ions:

2Cl– – 2e– ⟶ Cl2 Oxidation

Experiment: Electrolysis of dilute sulfuric acid

 In the electrolysis of dilute sulfuric acid the half equation at

the negative electrode (cathode) is:

2H+ + 2e– ⟶ H2 Reduction

 At the positive electrode (anode) oxygen gas is produced by

the discharge of water molecules:

2H2O – 2e– ⟶ O2 + 2H+ Oxidation

when we do this experiment the amount of hydrogen we obtain is

more than twice as much as the oxygen. This is because oxygen
is more soluble in water than hydrogen.

Electrolysis of aqueous copper(II) sulfate

 In the electrolysis of aqueous copper(II)sulfate the half equation at

the negative electrode (cathode) is:

Cu2+ + 2e– ⟶ Cu Reduction

 At the positive electrode (anode) oxygen gas is produced by the

discharge of water molecules:

2H2O – 2e– ⟶ O2 + 2H+ Oxidation

If the electrolysis is continued for a long time the copper (II) ions
will all be used up, and so the color of the solution will fade from
blue to colorless. Copper ions and hydroxide ions are discharged at
the electrodes. Hydrogen ions from the water aren't being
discharged and neither are the sulfate ions. The solution turns into
dilute sulfuric acid (H2S04). The electrolysis will then continue as
for dilute sulfuric acid

Inorganic chemistry
Chapter 11
The Alkali metals

 The group 1 metals are known as the alkali metals

Because they form alkaline solutions when they react with

water

 The group 1 metals are lithium, sodium, potassium, rubidium,

caesium and francium and they are found in the first column
of the periodic table
 The alkali metals have similar chemical properties because
they each have one electron in their outermost shell
 Some of these properties are:
o They are all soft metals which can easily be cut with a

knife
o They have relatively low densities and low melting points

o They are very reactive (they only need to lose one electron

to become stable).
REACTIONS WITH WATER

All these metals react in the same way with water to produce a
metal hydroxide and hydrogen:

Alkali metal + water -+ alkali metal hydroxide + hydrogen 2M +

2H20 -+ 2MOH + H2

The main difference between the reactions is how quickly they

happen. As you go down the group, the metals become more reactive
and the reactions occur more rapidly.

With sodium

observations

• The sodium floats.

. • There is fizzing because hydrogen gas is produced.

. • The piece of sodium gets smaller and eventually disappears. The

sodium is used up in the reaction.

With lithium

The reaction is very similar to sodium's reaction, except that it is

slower

With potassium

Potassium's reaction is faster than sodium's. It burns with a lilac

flame. Potassium often spits around and explodes.

With rubidium and caesium

These react even more violently than potassium, and the reaction can
be explosive. Rubidium hydroxide and cesium hydroxide are formed.

Reactions with Oxygen

The alkali metals react with oxygen in the air and forms metal
oxides, which is why the alkali metals tarnish when it is exposed to
the air

The metal tarnish more rapidly as you go down the group.

If we heat each of the metals in the air using a Bunsen burner, we get
a much more vigorous reaction and it is more difficult to see which
metal is most reactive because all the reactions are so rapid. That’s
why we identify metals by their flame color.

Lithium burns with a red flame to form lithium oxide.

Sodium burns with a yellow flame to form sodium oxide.

Potassium burns with a lilac flame to form potassium oxide.

Predicting Properties in Group 1

 Following these trends, we can say that:

o Rubidium, caesium and francium will react even more
vigorously with air and water than the first three alkali metals
 Of the alkali metals, lithium is the least reactive (as it is at the
top of group 1) and francium would be the most reactive (as it’s
at the bottom of group 1)
 Using the information given in the trends we would predict
that rubidium:
o would be a soft solid
o is more dense than potassium
o has a lower melting point than potassium
Group 1: Reactivity & Electronic Configurations

 The reactivity of the group 1 metals increases as you go down

the group
 As you go down group 1, the number of shells of electrons
increases by 1
o This means that the outermost electron gets further away
from the nucleus, so there are weaker forces of
attraction between the outermost electron and the nucleus
o Less energy is required to break the force of attraction as
it gets weaker, so the outer electron is lost more easily
o So, the alkali metals get more reactive as you go down the
group.

Chapter 12
Group 7 (Halogens)

Physical Properties

 The elements in group 7 are known as the halogens

o These are fluorine, chlorine, bromine, iodine and astatine
 These elements are non-metals that are poisonous
 All halogens have similar reactions as they each have seven
electrons in their outermost shell
 Halogens are diatomic because they form a covalent bond.
Predicting Properties in Group 7

 Chlorine, bromine and iodine react with metals and non-metals to

form compounds

Metal Halides

 The halogens react with some metals to form ionic

compounds which are metal halide salts
 E.g., sodium is a group 1 metal:
o 2 Na + Cl2 → 2 NaCl
 Calcium is a group 2 metal:
o Ca + Br2 → CaBr2
 The halogens decrease in reactivity moving down the group, but
they still form halide salts with some metals including iron
 The rate of reaction is slower for halogens which are further down
the group such as bromine and iodine.

Non-metal Halides

The halogens react with non-metals to form simple molecular covalent

structures

For example, the halogens react with hydrogen to form hydrogen halides
(e.g., hydrogen chloride)
Reactivity decreases down the group, so iodine reacts less vigorously
with hydrogen than chlorine

Displacement Reactions

A halogen displacement reaction occurs when a more reactive halogen

displaces a less reactive halogen.

Chlorine with Bromides & Iodides

If we add chlorine solution to colorless potassium bromide or potassium

iodide solution a displacement reaction occurs:

The solution becomes orange as bromine is formed or

The solution becomes brown as iodine is formed

Chlorine is above bromine and iodine in group 7 so it is more reactive

Chlorine will displace bromine or iodine from an aqueous solution of the

metal halide:

Cl2 + 2KBr → 2KCl + Br2

chlorine + potassium bromide → potassium chloride + bromine

Cl2 + 2KI → 2KCl + I2

chlorine + potassium iodide → potassium chloride + iodine

Bromine with Iodides

Bromine is above iodine in group 7 so it is more reactive

Bromine will displace iodine from an aqueous solution of the metal

iodide

bromine + potassium iodide → potassium bromide + iodine

Br2 + 2KI → 2KBr + I2

Why the reactivity of group 7 decreases as you go down the group?

All group 7 elements have 7 electrons in their outer shell. All of them
need to gain 1 electron to become stable. Down the group 7 the atomic
size increases so the distance between nucleus and outer shell electron
increases. The force of attraction between nucleus and outer shell
electron decreases.

Chapter 13
Gases in the atmosphere
 Composition of Air

how to determine the percentage by volume of oxygen in air using

phosphorous?

The initial level of water is marked on the side of the bell jar with a
waterproof pen or a sticker.
• The bung is removed from the bell jar and the phosphorus is touched
with a hot metal wire in order to ignite it.

• The bung is quickly put back into the bell jar.

• The phosphorus burns, the bell jar becomes filled with a white smoke
and the level of water rises inside the bell jar. The smoke eventually
clears as the phosphorus oxide dissolves in the water.

• When the level of water inside the bell jar stops rising, the final level
is marked.

• By making careful measurements of water levels before and after the

experiment you can determine the percentage of oxygen in the air.
how to determine the percentage by volume of oxygen in air using
experiments involving the reactions of metals (e.g. iron) with air?

Combustion
 Carbon Dioxide from Thermal Decomposition

 Carbon dioxide can be made by the heating of copper (ii) carbonate.

 The thermal decomposition of copper(II)carbonate occurs readily on
heating
 Copper(II) carbonate is a green powder and slowly becomes dark as
black copper(II) oxide is produced
 The carbon dioxide given off can be tested by passing the gas
through limewater and looking for it to turn milky
 The equation for the reaction is

CuCO3 (s) → CuO (s)+ CO2 (g)

copper(II) carbonate → copper(II) oxide + carbon dioxide

Greenhouse gases

Carbon dioxide, nitrous oxide and water vapor are greenhouse gases.

Greenhouse gases absorb the infra-red radiation reradiated from the

Earth’s surface. That’s how the gases increases the global temperature
of the Earth. This effect is known as global warming.
Effects of global warming

 Melting of polar ice caps

 Sea level rise
 Flooding in many areas of the world
 Loss of habitat as the forest areas will be drowned by water
 Changes in rainfall pattern.

Sources of greenhouse gases

 By burning fossil fuels

 Burning of car fuel
 Factories
 Overgrazing

Chapter 14
Reactivity series
Metals Reaction with water reaction with acids

Potassium Most reactive metals Reacts violently Reacts violently

Sodium Reacts quickly Reacts violently

Calcium Reacts less strongly Reacts vigorously

Magnesium Reacts vigorously

Aluminum Reacts vigorously

Carbon increasing reactivity

Zinc moderately reactive

Iron reacts less vigorously

Lead

Hydrogen

Copper least reactive

Silver Unreactive

Gold

Metal Displacement Reactions

 The reactivity of metals decreases going down the

reactivity series.
 This means that a more reactive metal will displace a less
reactive metal from its compounds
 Two examples are:
o Reacting a metal with a metal oxide (by heating)

o Reacting a metal with an aqueous solution of a metal

compound
 For example copper(II) oxide can be reduced by heating it
with zinc.
 The reducing agent in the reaction is zinc:

Zn + CuO → ZnO + Cu

zinc + copper(II) oxide → zinc oxide + copper

Displacement reactions between metals & aqueous solutions

of metal salts
 The reactivity between two metals can be compared
using displacement reactions in salt solutions of one of
the metals
 Displacement reactions occur when the solid metal is
more reactive than the metal that is in the compound.
Reactivity series

Rusting of Iron
 Rusting is a chemical reaction between iron, water and oxygen
to form hydrated iron(III)oxide
 Oxygen and water must be present for rusting to occur.
 Rusting is a redox process
Formula of rusting is Fe2O3.xH20
Preventing Iron Rusting
Barrier Methods
 Rust can be prevented by coating iron with barriers such as
painting and oil that prevents the iron from coming into contact
with water and oxygen.
 However, if the coatings are washed away or scratched, the iron
is once again exposed to water and oxygen and will rust.
Galvanising
In galvanizing, iron is coated with zinc. The coating prevents air and water
reaching the iron and so no rusting takes place. When the coating is
scratched, even then the rusting doesn’t occur as zinc is more reactive
than iron.
Sacrificial protection
In sacrificial protection a highly reactive metal is attached to iron the
metal layer prevents iron to react with water and air. The highly reactive
metal corrodes instead of iron.
Oxidation- It is the gain of oxygen and loss of electrons
Reduction- It is the gain of electrons and loss of oxygen
Reducing agent- it removes oxygen
Oxidizing agent-it adds oxygen
Redox reaction- when both oxidation and reduction takes place
investigate reactions between dilute hydrochloric and sulfuric acids and
metals (e.g. magnesium, zinc and iron
Set up four test-tubes and put about 2 cm3 of dilute hydrochloric acid
into each one.
• Put a small piece of magnesium, zinc, iron or copper into each test-tube
and observe any reaction that occurs.
• If there is fizzing, collect or trap the gas and test with a lighted splint - a
squeaky pop indicates the presence of hydrogen gas.

• Repeat the experiments with dilute sulfuric acid.

 Results

Magnesium-Reacts vigorously with lots of fizzing. The gas

produced gave a squeaky pop with a lighted splint

Zinc- Steady reaction. Fizzing. Enough gas eventually collected

to produce a squeaky pop with a lighted splint. A

Iron -Slow fizzing. Very little gas was collected in the time
available.

Chapter 15
Extraction and uses of metals
Sources of Metals

 The Earth’s crust contains metals and metal compounds

such as gold, copper, iron oxide and aluminum oxide
 Rocks from which metals are obtained are called ores
 They have to be extracted from their ores through
processes such as electrolysis, using a blast furnace or by
reacting with more reactive material
 In many cases the ore is an oxide of the metal, therefore
the extraction of these metals is a reduction process
.Common examples of oxide ores
are iron and aluminum ores which are
called hematite and bauxite.
 Unreactive metals do not have to be extracted chemically
as they are often found as the uncombined element
 Extracting of metals

Why metals below carbon aren’t aren’t extracted by electrolysis?

For electrolysis, electricity is needed which is expensive that’s carbon

reduction process is used to extract the metals below carbon because carbon
is cheaper than electricity.

Using Metals

 The uses of aluminum, copper and steel are summarized in these tables:

Uses of Aluminium

Uses of Copper
 Uses of steel

Alloys
 Alloys are mixtures of metals, where the metals are mixed together
physically but are not chemically combined.
 Alloys have properties that can be very different to the metals they
contain, for example they can have greater strength, hardness or
resistance to corrosion.
 Alloys contain atoms of different sizes, which distorts the regular
arrangements of atoms
 This makes it more difficult for the layers to slide over each other,
so they are usually much harder than the pure metal.

Chapter 16
Acids, alkalis and titration
Color Indicators

 Litmus is not suitable for titrations as the color change is not sharp
and it goes through a purple transition color in neutral solutions
making it difficult to determine an endpoint
 Litmus is very useful as an indicator paper and comes in red and
blue versions, for dipping into solutions or testing gases.
The pH Scale
1. The pH scale goes from 0 – 14 (extremely acidic substances can have
values of below 0)
2. All acids have pH values of below 7, all alkalis have pH values of above
7
3. The lower the pH then the more acidic the solution is
4. The higher the pH then the more alkaline the solution is
5. A solution of pH 7 is described as being neutral

Universal indicator
Acids & Alkalis
 When acids are added to water, they form positively
charged hydrogen ions (H+)
 The presence of H+ ions is what makes a solution acidic
 When alkalis are added to water, they form negative hydroxide
ions (OH–)
 The presence of the OH– ions is what makes the aqueous solution
an alkali.
Neutralization
A neutralization reaction occurs when an acid reacts with an alkali
When these substances react together in a neutralization reaction, the
H+ ions react with the OH– ions to produce water.
Acid-Alkali Titrations
 Take a known volume of acid into a conical flask using a pipette.
 Add a suitable indicator into the conical flask (methyl
orange/phenolphthalein).
 Don’t use litmus as it doesn’t give any sharp color change.
  Take alkali into burette. Note its initial volume.
 Pour down alkali into the conical flask until the indicator changes
color.
 Note the final burette reading (volume).Calculate how much alkali
was needed to neutralize the acid.

Chapter 17
Acid,bases and salt preparation
Solubility Rules
 All Na+,k+,NH4+,NO3- salts are soluble.
 All Cl- salts are solube (except Agcl,PbCl2)
 All SO4-2 salts are soluble (except BaSO4,PbSO4)
 All CO3-2 and OH- salts are insoluble (except Na+,K+,NH4+)

Acids, Bases & Protons

Bronsted lowry theory
Acid is a proton donor and base is a proton acceptor
Arrhenius theory
An acid is a substance that dissolves in water and releases hydrogen
ions.
A base is a substance that dissolve in water and releases hydroxide
ions.
Reactions of Acids
Reactions of acids with metals
 Only metals above hydrogen in the reactivity series will react with
dilute acids
  The more reactive the metal then the more vigorous the reaction
will be
 Metals that are placed high on the reactivity series such as
potassium and sodium are very dangerous and
react explosively with acids
 When acids react with metals they form a salt and hydrogen gas:
 The equation is:
metal + acid ⟶ salt + hydrogen

Reaction of acids with bases

 When an acid reacts with a base, a neutralization reaction occurs
 In all acid-base neutralization reactions, a salt and water are
produced:
acid + base ⟶ salt + water
Reactions of Acids with Metal Carbonates
 Acids will react with metal carbonates to form
metal salt, carbon dioxide and water

Bases
What makes a base act like a base?
 Bases are substances which can neutralise an acid.Some are
dissolved in water. They are called alkalis.
 A base is insoluble in water but alkalis are soluble in water.
 Bases are usually oxides, hydroxides or carbonates of metals.
Preparing soluble salts
1. Add excess metal solution to sulphuric acid (to remove all the acids)
2. Stir the mixture (to ensure the acid and the metal solution comes in
contact and the reaction is complete)
3. Filter off the excess metal solution
4. Heat the metal sulphate solution in an evaporating basin.
5. Use a glass rod to check whether crystal forms on cooling
(taking a small drop of solution on a glass rod cools the solution &
may get stuck in the rod. The salt crystalizes as the solubility of salt
when in cold is less than in hot)
6. .Stop heating & leave the evaporating basin to cool. This allows
crystals to form.
7. Filter off the crystals from the uncrystalised solution
8. Dry the crystals using blotting paper/ in an oven/ leave to dry....

Why we didn't directly remove all the water from the solution by
heating?
Answer: because crystals contain water in the form of water of
crystallization. Removing all water by heating will give you metal
sulphate powder rather than crystals.

Preparing soluble salts by titration

1. Add the same volume of Acid and alkali without indicator (as
indicator would contaminate the salt solution)
2. Heat metal sulphate solution in an evaporating basin
3. Use a glass rod to check whether crystal forms on cooling
(taking a small drop of solution on a glass rod cools the solution &
may get stuck in the rod. The salt crystalizes as the solubility of salt
when in cold is less than in hot)
4.Stop heating & leave the evaporating basin to cool. This allows
crystals to form.
5.Filter off the crystals from the uncrystalised solution
6. Dry the crystals using blotting paper/ in an oven/ leave to dry....

Why do we need a different method?

As sodium, potassium and ammonium salts are soluble so adding
excess would result the sodium/potassium/NH4 to dissolve in
water/solution and there would be nothing to filter off.
How does titration help?
Titration gives you the exact amount/volume of acid and alkali that
would neutralize each other.
Making insoluble salts using precipitation method
1) Add water to both the reactant solids to make their solution
2) Mix the two solution
3) Stir the mixture
4) Filter off the solid precipitate from the mixture using a filter
paper.
5) Wash the solid precipitate using distilled water to remove excess
ions.(Tap water contains impurities)
6) Leave the solid to dry using a blotting paper.

practical: prepare a sample of pure, dry hydrated copper(II) sulfate

crystals starting from copper(II) oxide
Reaction: CuO(S) + H2SO4(aq) = CuSO4(aq) + H20 (l)
1. Add excess Copper(ii) oxide to sulphuric acid (to remove all the
acids)
2. Stir the mixture (to ensure the acid and the metal solution comes in
contact and the reaction is complete)
3. Filter off the excess Copper oxide
4. Heat the copper sulphate solution in an evaporating basin.
5. Use a glass rod to check whether crystal forms on cooling
 (taking a small drop of solution on a glass rod cools the solution &
may get stuck in the rod. The salt crystalizes as the solubility of salt
when in cold is less than in hot)
6. .Stop heating & leave the evaporating basin to cool. This allows
crystals to form.
7. Filter off the crystals from the uncrystalised solution
8. Dry the crystals using blotting paper/ in an oven/ leave to dry.
practical: prepare a sample of pure, dry lead(II) sulfate
Pb(NO3)2 (aq) + K2SO4 (aq) → PbSO4 (s) + 2KNO3 (aq)
1) Add water to lead(ii)nitrate and potassium sulphate to make their
solution
2) Mix the two solution
3) Stir the mixture
4) Filter off the solid lead(II) sulfate precipitate from the mixture
using a filter paper.
5) Wash the lead(II) sulfate precipitate using distilled water to
remove excess ions.(Tap water contains impurities).
6) Leave the solid to dry using a blotting paper.
Chapter 18
Chemical tests
Test for the purity of the water
Test: Measure the boiling point.
Result: pure water boils at 100 C

Physical chemistry
Chapter 19
Energetics
calorimetric experiments for combustion

Measure 100 cm3 of cold water using a measuring cylinder and transfer
the water to a copper can.
• Take the initial temperature of the water.
• measure the mass of a spirit-burner which contains ethanol with its
lid on it. Keep the mouth of the spirit burner closed by a lid to prevent
the alcohol from evaporating.
• Light the wick to heat the water. Stop heating when you have a
reasonable temperature rise of water (say, about 40.0°C). put the lid
back on it to stop the fire
• Stir the water thoroughly and measure the maximum temperature of
the water.
• Weigh the spirit-burner again with its lid on.
calorimetric experiments for reactions such as combustion,
displacement
Place a polystyrene cup in a 250 cm3 glass beaker.
• Transfer 50 cm3 of 0.200 mol/dm3 copper {II) sulfate solution into the
polystyrene cup using a measuring cylinder.
• Weigh 1.20g of zinc using a weighing boat on a balance.
• Record the initial temperature of the copper (II) sulfate solution.
• Add the zinc.
• Stir the solution as quickly as possible.
• Record the maximum temperature reached.
Initial temperature of copper (II) sulfate solution/°C=17
Maximum temperature of copper (II) sulfate solution/°C= 27.3
Q = mc T = 50 x 4.18 x (27.3 - 17.0)
= 2152.7 J or 2.1527kJ
Number of moles (n) of zinc added = mass (m)/relative atomic mass (A,)
1.20 /65 = 0.0185mol
Number of moles (n) of copper (II) sulfate added = volume (V) x
concentration (C)
= 0.050 X 0.200
=0.0100mol
Molar enthalpy change of reaction = heat energy change (Q)/ number
of moles of copper sulfate reacted (n)
= 2· 1527/0.01
= 215kJ/mol

calorimetry experiments for dissolving reactions

• Place a polystyrene cup in a 250 cm3 glass beaker.
• Transfer 100cm3of water into the polystyrene cup using a measuring
cylinder.
• Record the initial temperature of the water.
• Weigh 5.20g of ammonium chloride on a balance.
• Add the ammonium chloride to water and stir the solution vigorously
until all the ammonium chloride has dissolved.
• Record the minimum temperature.
Initial temperature of water/°C=18.3
Minimum temperature of salt solution=15.1
Q = mcT = 100 x 4.18 x (18.3-15.1) = 1337.6J
= 1.3376kJ
The relative formula mass of ammonium chloride (NH4CI) is 14 + 4 X 1 +
35.5 = 53.5.
Number of moles (n) of ammonium chloride dissolved n = mass/
relative formula mass
5.20/53.5=0.0972 mol
Molar enthalpy change of solution = heat energy change/ number of
moles of ammonium chloride dissolved (n)
1.3376 / 0.0972 = 13.kJ/mol.

Calorimetric experiments for neutralization reactions

• Place a polystyrene cup in a 250 cm3 glass beaker.
• Transfer 25 cm3 of 2.00 mol/dm3 potassium hydroxide into the
Polystyrene cup using a measuring cylinder.
• Record the initial temperature.
• Fill a burette with 50.00cm3 of dilute hydrochloric acid.
• Use the burette to add 5.00cm3 of dilute hydrochloric acid to the
Potassium hydroxide.
• Stir vigorously and record the maximum temperature reached.
• Don’t add the acid in one go, Continue adding 5.00cm3 portions of
dilute hydrochloric acid at a time, stir and record the maximum
temperature each time, Until a total volume of 50.00cm3 has been
added.
Calculations- use the same approach as experiment 2
Chapter 20
Rates of reaction
AN INVESTIGATION OF THE RATE OF REACTION BETWEEN MARBLE
CHIPS AND DILUTE HYDROCHLORIC ACID
CaC03 (s) + 2HCl(aq) .... CaCl2(aq) + H20(1) + C02 (g)
Procedure
• Use a measuring cylinder to measure 25cm3 of 2.00mol/dm3 dilute
Hydrochloric acid.
• Add 5.00g of large marble chips to a conical flask and place a piece
of cotton wool at the opening of the flask. The cotton wool prevents
the acid to spray out and lets the carbon dioxide to escape
After the end of the reaction some marble chips will be left over
because all the acids will be used up
• Place everything on a balance and reset it to zero.
• Add the acid to the marble chips and record the reading on the
balance every 30 seconds.
The mass goes down because the carbon dioxide escapes through the
cotton wool.
Why the rate of reaction is highest at the beginning and falls as the
reaction proceeds?
Answer:- At the beginning of the reaction, there are lots of reactant
particles, they collide with each other more often that’s why the rate of
reaction is high. As the reaction proceeds the reactant particles get
used up, they collide with each other less often, and the rate of
reaction decreases.

Using different reaction conditions to see how it affects the rate of

reaction
Using smaller marble chips:
As we can see from the graph, the curve for the smaller marble chips is
much steeper than the one with the bigger marble chips, which means
that the rate of reaction is higher for the smaller marble chips used.
Because using smaller marble chips increases the surface area, which
means that more of the solid is exposed to the acid particles. The
amount of carbon dioxide produced is the same because the acid is
limited.
Effect of concentration of a solution on the rate of reaction
Method:
 Measure 50 cm3 of sodium thiosulfate solution into a flask
 Measure 5 cm3 of dilute hydrochloric acid into a measuring
cylinder
 Draw a cross on a piece of paper and put it underneath the flask
 Add the acid into the flask and immediately start the stopwatch
 Look down at the cross from above and stop the stopwatch when
the cross can no longer be seen
  Repeat using different concentrations of sodium thiosulfate
solution (mix different volumes of sodium thiosulfate solution
with water to dilute it)
Result:
With an increase in the concentration of a solution, the rate of reaction
will increase. This is because there will be more reactant particles in a
given volume, which allows more frequent and successful collisions and
increases the rate of reaction.
INVESTIGATE THE EFFECT OF DIFFERENT SOLIDS ON THE CATALYTIC
DECOMPOSITION OF HYDROGEN PEROXIDE SOLUTION
Procedure:
• Measure 100 cm3 of 2 vol hydrogen peroxide and transfer to a 250
cm3
conical flask.
• Weigh out 0.20g of manganese(IV) oxide on a balance.
• Add the manganese(IV) oxide to the hydrogen peroxide and quickly
replace the bung with the gas syringe attached on it.
• Record the amount of oxygen produced every 20 seconds for
3 minutes and plot a graph of volume of oxygen versus time.
• Repeat the reaction with 0.20 g of lead(IV) oxide and copper(II) oxide
but keep everything else the same.
Result: manganese(IV) oxide is a very effective
catalyst for the decomposition of hydrogen peroxide. In comparison,
lead(IV) oxide is less effective as the rate of the reaction is much slower
and copper(II) oxide does not act as a catalyst for this reaction at all.

Explaining Rates
The following factors influence the rate of reaction:
 Increasing concentration
 Increasing temperature
 Increase the surface area of a solid reactant
 Use of a catalyst
Concentration of a Solution
Explanation: Increasing the concentration of a solution will increase the
rate of reaction .This is because there will be more reactant particles in
a given volume, allowing more frequent and successful collisions per
second
Temperature
Explanation: Increase in the temperature, the rate of reaction will
increase
This is because the particles will have more kinetic energy than the
required activation energy, more frequent and successful collisions will
occur, which increases the rate of reaction.

Surface area of a solid

Explanation: With an increase in the surface area of a solid reactant,
the rate of reaction will increase. This is because more surface area of
the particles will be exposed to the other reactant, that’s why more
successful and frequent collision occurs.
pressure of a gas- when the pressure is increased,the particles are
brought together and more frequent and successful collisions take
place which increases the rate of reaction.

Catalysts
 Catalysts are substances which speed up the rate of a reaction
without themselves being consumed in the reaction at the end.
The mass of a catalyst at the beginning and end of a reaction is
the same. Catalysts work by providing an alternative route for the
reaction, involving a lower activation energy.
Activation energy
It is the minimum energy required to start a reaction.
When temperature is increased particles gain kinetic energy and moves
faster. More particles have energy greater than or equal to the
activation energy and therefore more frequent and successful collisions
take place.
Chapter 21
Reversible reaction and equilibria
 Reversible reaction is one that can go both ways, in other words,
reactants react to form to form products and products react to
form reactants. When writing chemical equations for reversible
reactions, two opposing arrows are used to indicate the forward
and reverse reactions occurring at the same time.
Thermal Decomposition of Ammonium Chloride
 Heating ammonium chloride produces ammonia and hydrogen
chloride gases:
NH4Cl (s) → NH3 (g) + HCl (g)
 As the hot gases cool down they recombine to form solid
ammonium chloride
NH3 (g) + HCl (g) → NH4Cl (s)
 So, the reversible reaction is represented like this:
NH4Cl (s) ⇌ NH3 (g) + HCl (g)
DEHYDRATION OF COPPER(II) SULFATE CRYSTALS
If we heat blue copper(II) sulfate crystals, the blue crystals turn to a
white powder and water is removed. Heating causes the crystals to lose
their water of crystallisation and to form white anhydrous copper(II)
sulfate..
CuS04.5H20(s)=CuSO4(s) + 5H20(l)
If we add water to the white solid, it turns blue again; it also becomes
very warm.
CuS04(s) + 5H2O(I)= CuS04.5H2O(s).
Dynamic equilibrium means the reactions are still continuing, but the
rate of the forward reaction is equal to the rate of the reverse reaction
and the total amounts or concentrations of reactants and products are
now constant.It only occurs in a closed system.

Features of dynamic equilibrium

 Forward and backward reactions occur at equal rates.
 Concentration of reactants and products remains constant
Effect of catalyst on equilibrium position
The presence of a catalyst does not affect the position of equilibrium
but it does increase the rate at which equilibrium is reached. This is
because the catalyst increases the rate of both the forward and
backward reactions by the same amount by providing an alternative
pathway requiring lower activation energy. As a result, the
concentration of reactants and products is still the same at equilibrium
as it would be without the catalyst.
temperature
increasing temperature: the position of equilibrium shifts in the
endothermic direction.
Decreasing temperature: the position of equilibrium shifts in the
exothermic direction.
Example;
2N02(g) ⇌ N204 (g)
Pressure
If you increase the pressure, the position of equilibrium will shift to
reduce it again by producing fewer gaseous molecules. In other words,
the position of equilibrium will move to the right and the reaction will
produce more dinitrogen tetroxide. If you lower the pressure, the
position of equilibrium will shift to increase it again by producing more
gaseous molecules. Therefore the position of equilibrium shifts to the
left.

Organic chemistry
Chapter 22
Introduction to organic chemistry
Definition of a Hydrocarbon
A compound that contains only hydrogen and carbon atoms
 Organic compounds can be represented in a number of ways:
o Emperical Formulae
o Molecular Formulae
o General Formulae
o Structural Formulae
o displayed Formulae
 The empirical formula shows the simplest possible ratio of the
atoms in a molecule
o For example: Hydrogen peroxide is H2O2 but the empirical
formula is HO
 The molecular formula shows the actual number of atoms in a
molecule
 o For example:

The general formula shows a ratio of atoms in a group of compounds in

terms of ‘n’ .
o For example, the general formula of a molecule that belong
to the alkane family is CnH2n+2
 The displayed formula shows the structural arrangement of all the
atoms and bonds in a molecule. Each line represents a pair of shared
electrons in a covalent bond.
 For example:

 In a structural formulae enough information is shown to make the

structure clear, but most of the actual covalent bonds are omitted
 Identical groups can be bracketed together
 Side groups are also shown using brackets
 Straight chain alkanes are shown as follows:
 Three important terms to know in this topic are homologous
series, functional group and isomerism
Homologous Series
This is a series of organic compounds that have similar features and
chemical properties due to them having the same functional group
 All members of a homologous series have:
o The same general formula
o Same functional group
o Similar chemical properties
o The difference in the molecular formula between one member
and the next is CH2
Functional Group
 Functional group: A group of atoms bonded in a specific
arrangement .Some examples are shown here
Isomerism
Isomers are compounds that have the same molecular formula
but different displayed formulae.
Naming Organic Compound

Rules for naming compounds

 When there is more than one carbon atom where a functional group
can be located it is important to distinguish exactly which carbon the
functional group is.
 Each carbon is numbered and these numbers are used to describe
where the functional group is.
 When 2 functional groups are present di- is used as a prefix to the
second part of the name.
Classifying Organic Reactions
Substitution
 A substitution reaction takes place when one functional group is
replaced by another
Example: Methane reacts with bromine under ultraviolet light
CH4 + Br2 → CH3Br + HBr
Methane + Bromine → Bromomethane + Hydrogen Bromide
Addition
 An addition reaction takes place when two or more molecules
combine to form a larger molecule with no other products
C2 H4 + Br2 → C2H4Br2
Ethene + Bromine → Dibromoethane
  In a combustion reaction, an organic substance reacts with
oxygen to form carbon dioxide and water if incomplete
combustion carbon monoxide or soot and water.
o If there is an unlimited supply of air / oxygen, the products are
carbon dioxide and water:
CH4 + 2O2 → CO2 + 2H2O
 This is complete combustion
o If there is a limited supply of air / oxygen, the products are
carbon monoxide and water:
CH4 + ½O2 → CO + 2H2O
 This is incomplete combustion
Chapter 23
Crude oil
What is crude oil made up of?
Crude oil is made up of small, medium and large chain hydrocarbons.
Fractional distillation in the oil refinery
Crude oil is heated and entered into the fractionating column. The
temperature inside the column is higher at the bottom and falls as
moved up. Different fractions of crude oil have different boiling point.
The crude oil fraction boils and rises up the column where they are
cooled and condensed at different heights. Different fractions are
collected at different heights. Lighter fractions are collected at the top
and heavier at bottom.
Structure of a fractionating column

Properties of the main fractions of crude oil

 Viscosity: High viscosity liquids are thick and flow less easily. If
the number of carbon atoms increases, the attraction between
the hydrocarbon molecules also increases which makes the liquid
more viscous with the increasing length of the hydrocarbon chain.
 Color: As carbon chain length increases the colour of the liquid
gets darker as it gets thicker and more viscous
  Melting point/boiling point: As the molecules get larger, the
intermolecular force of attraction becomes greater. So more heat
is needed to separate the molecules. With increasing molecular
size there is an increase in boiling point.
 Volatility: Volatility is the ability of a substance to vaporise. With
increasing molecular size hydrocarbon liquids become less
volatile. This is because the attraction between the molecules
increases with increasing molecular size.
Combustion Products
The burning of fossil fuels releases the gases carbon dioxide, carbon
monoxide, oxides of nitrogen and oxides of sulfur.

Complete versus Incomplete Combustion

Complete combustion occurs when there is sufficient oxygen. After the
end of the reaction water and carbon dioxide are the products formed
For example, the combustion equation for propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
Incomplete Combustion
Incomplete combustion occurs when there is insufficient
oxygen to burn.The products of these reactions are soot, carbon
monoxide and water
Example
2CH4 + 3O2→ 2CO + 4H2O
CH4 + O2→ C + 2H2O
Dangers of Carbon Monoxide
Carbon monoxide is a toxic,poisonous gas Carbon monoxide binds with
haemoglobin and prevents the oxygen carrying capacity of the blood.
Describe how long-chain alkanes are converted to alkenes and shorter-
chain alkanes by catalytic cracking?
The fuel oil fraction is heated to give a gas and then passed over a
catalyst of silicon dioxide (also called silica) and aluminium oxide (also
called alumina) at about 600-700 ' C. Cracking can also be carried out at
higher temperatures without a catalyst (thermal cracking). Cracking is
just an example of thermal decomposition: a big molecule splitting into
smaller ones on heating. Most of the hydrocarbons found in crude oil
have single bonds between the carbon atoms. During the cracking
process, C-C single bonds are broken and new C=C double bonds are
formed.
Why is cracking necessary?
Cracking produces smaller fractions which are more useful. Smaller
fractions burn more easily and efficiently. Smaller fractions have
greater demand than larger ones. For example there are more cars
than Lorries so, greater demand for petrol than diesel oil. Cracking also
produces alkenes which can be used to make plastics such as ethane to
polythene.

Chapter 24
Alkanes
Alkanes are a group of saturated hydrocarbons
The term saturated means that they only have single carbon-carbon
bonds, there are no double bonds
The general formula of the alkanes is CnH2n+2.Alkanes give
substitution reaction.
Halogens & Alkanes
In a substitution reaction, one atom is replaced with another atom
Alkanes undergo a substitution reaction with halogens in the presence
of ultraviolet radiation
In the presence of ultraviolet (UV) radiation, methane reacts with
bromine in a substitution reaction

The equation for the reaction is

CH4 + Br2 → CH3Br + HBr

methane + bromine → bromomethane + hydrogen bromide

Chlorine will also react with alkanes to form chloromethane

CH4 + Cl2 → CH3Cl + HCl

methane + chlorine → chloromethane + hydrogen chloride

The products belong to a family called halogenoalkanes .

Chapter 25
Alkenes
All alkenes contain a carbon to carbon double bond, C=C.
The general formula of alkene is CnH2n.Alkenes are unsaturated,
because they have carbon to carbon double bond. A carbon-carbon
double can break and form a single bond, allowing more atoms to
attach to the carbon atoms.
Bromination of Ethene
The reaction between bromine and ethene is an example of an addition
reaction.
Testing for Alkenes
Test: Add bromine water
Result with alkanes: No change of the color of the orange solution
Result with alkenes: the orange solution turns colorless.

Chapter 26
Alcohols
'Alcohol' is just one member of a large family (homologous series) of
similar compounds. All Alcohols contain the functional group - OH
covalently bonded to a carbon chain.

how to draw structural and displayed formulae for methanol, ethanol,

propanol (propan-1-ol only) and butanol (butan-1-ol only), and name
each compound?
Combustion
 Alcohols undergo combustion to form carbon dioxide and water
 The complete combustion of ethanol is as follows:
C2H50H(I) + 302(g) = 2CO2 (g) + 3H20 (l)

ETHANOL IS OXIDISED BY HIE AIR IN THE PRESENCE OF MICROBES

(MICROBIAL OXIDATION)
A bottle of wine left open to the air turns sour. The ethanol in the wine
is oxidized by air with the help of microorganisms such as bacteria or
yeast to form ethanoic acid, CH3COOH.
ethanol + oxygen → ethanoic acid + water
Making ethanol by fermentation
Yeast is added to a sugar solution and left in the warm (about 30 °C} for
several days in the absence of air. The absence of air and the
temperature are both important. In the presence of air, enzymes in the
yeast produce carbon dioxide and water instead of ethanol. The
enzymes are protein catalysts and if the temperature is increased
above 40°C they lose their structure and don't work any longer. The
proteins are said to be denatured. At a lower temperature, the reaction
is too slow. Somewhere between 30 and 40°C is the optimum
temperature for the reaction.
Enzymes in the yeast convert glucose into ethanol and carbon dioxide
in a large number of steps. The overall equation for that conversion is
C6H12O6 → 2C2H5OH + 2CO2

Glucose Ethanol carbondioxide

Starting materials: glucose (from sugar cane)
Temperature: 30°c
Catalyst: enzymes in yeast
Other conditions: anaerobic.

MAKING ETHANOL BY THE HYDRATION OF ETHENE

Ethanol is also made by reacting ethene with steam, a process
known as hydration.
CH2=CH2(g) + H20(g)= CH3CH20H(g)
Starting materials: : ethene and steam
Temperature: 300°c
Pressure: 60-70 atmospheres
Catalyst: phosphoric acid.

ETHANOL CAN BE OXIDISED BY HEATING WITH POTASSIUM

DICHROMATE(VI) IN DILUTE SULFURIC ACID
Equation:

Starting materials: ethanol and potassium dichromate (VI) in dilute

sulfuric acid
Conditions: heat under reflux
Color change: color changes from orange to green.
Cr is reduced from oxidation number +6 to +3.

Chapter-27
The Carboxylic Acids
Acids such as ethanoic acid are known as carboxylic acids and
they all contain the functional group –COOH.
Reactions of Carboxylic Acids
The carboxylic acids behave like other acids
They react with metals to form a salt and hydrogen and with
carbonates to form a salt, water and carbon dioxide gas
They take part in neutralization reactions to produce salt and
water
Ethanoic acid (also called acetic acid) is the acid used to make
vinegar.
The salts formed by the reaction of carboxylic acids all have the
name, –anoate at the end.
So methanoic acid forms a salt called methanoate, ethanoic acid
forms a salt called ethanoate etc.
 In the reaction with metals, a metal salt and hydrogen gas
are produced
For example in the reaction of ethanoic acid with magnesium, the
salt magnesium ethanoate is formed:
2CH3COOH + Mg → (CH3COO)2Mg + H2

 In the reaction with hydroxides a salt and water are formed

in a neutralization reaction
For example in reaction with potassium hydroxide the salt
potassium propanoate is formed by reaction with propanoic acid:
CH3CH2COOH + KOH → (CH3CH2COO)K + H2O

 In the reaction with carbonates a metal salt, water and

carbon dioxide gas are produced
For example in reaction with potassium carbonate the salt
potassium butanoate is formed by reaction with butanoic acid:
2CH3CH2CH2COOH + K2CO3 → (CH3CH2CH2COO)2K + H2O +
CO2

Chapter 28
The Esters
Making Esters
Alcohols and carboxylic acids react to make esters in
esterification reactions
Esters are compounds with the functional group R-COO-R
Esters are sweet-smelling liquids used in food flavorings and
perfumes
Ethanoic acid will react with ethanol in the presence of
concentrated sulfuric acid (catalyst) to form ethyl ethanoate:
CH3COOH + C2H5OH → CH3COOC2H5 + H2O

Diagram showing the formation of ethyl ethanoate

Naming Esters
An ester is made from an alcohol and carboxylic acid
The first part of the name indicates the length of the carbon chain
in the alcohol, and it ends with the letters ‘- yl’
The second part of the name also indicates the length of the
carbon chain in the carboxylic acid, but it ends with the letters ‘-
oate’
e.g. the ester formed from pentanol and butanoic acid is called
pentyl butanoate.
Practical: Preparation of Ethyl Ethanoate
Procedure:
 Put 1 cm3 of ethanoic acid and 1 cm3 of ethanol into a boiling
tube.
• add a few drops of concentrated sulfuric acid.
• Place the boiling tube in a beaker of hot water at about 80
degree celsius for 5 minutes.
• Allow the contents of the tube to cool.
• After it is cooled, pour the mixture into a beaker which contains
0.5 mol/dm3 of sodium carbonate solution. Some bubbles of gas
can be seen as the excess acid reacts with the metal carbonate.
The reason for doing this is to be able to smell the ester better.
• A layer of ester, ethyl ethanoate will separate and float on top of
the water. The acid and alcohol will mostly dissolve in the water,
but the ester won't.
THE END