TOPIC 1 : EXPERIMENTAL TECHNIQUES

CHEMICAL WASTE

A chemical waste is a product or unwanted material from a chemical reaction or an expired

product which is no longer needed for any experiment in the next activity.

EXAMPLES OF WASTE PRODUCTS FROM CHEMICAL REACTIONS

 Unused chemical e.g. acids, alkalis, solid chemicals etc

 Broken laboratory glassware, sharp objects
 Plastics, waste papers, and rubbers
 Poisonous gases from reactions

SAFE WAYS OF DISPOSING CHEMICAL WASTES

1. Use of normal waste bins.

This way can be used for wastes such as non-recycle plastics, pieces of wood, rubber and dirty
papers.

2. Use of special controlled waste containers.

These are used to dispose sharp objects like scalpel, broken glass ware, sample tubes and items
contaminated with hazardous chemicals. These must be emptied regularly and not allowed to
overflow.

3. Draining using a lot of water.

Wastes which can be drained include harmless soluble inorganic salts, and contaminated and
used acids and alkalis, detergents.

4. Recycling.

Recyclable materials include unbroken glass, packing waste and paper.

5. Incineration.

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In this case the wastes are put into an incinerator and burned. Wastes that can be incinerated
include syringes, needles, all organic solvents, soluble organic wastes, paraffin and mineral oils.

DESIGNING SCIENTIFIC INVESTIGATION

A scientific investigation is the systematic process of trying to find an answer or a solution to a

problem or observation.

COMPONENTS OF A SCIENTIFIC INVESTIGATION

1. Problem identification

Identifying a problem involves asking questions about the natural world. Examples of scientific
questions are:

 What causes rusting?

 Why do plastics not decompose easily?
2. Formulation of a hypothesis

A hypothesis is a guessed answer to a problem. A hypothesis is formulated from the scientist’s

experiences and knowledge.

3. Experimentation

An experiment is a series of investigations intended to accept, modify or reject a hypothesis.

4. Observation and data collection

This involves writing observations correctly, stating the correct units of each measurement.
Scientific data is usually collected using tables.

5. Interpretation of data

Interpreting data means explaining the observations, trends in relation to the aim of the
experiment.

6. Conclusion

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The conclusion is drawn based upon the collected data. It is either a conformation or the
rejection of the hypothesis under investigation. If the hypothesis is correct, it is confirmed and
adopted and if false it is declared null and void hence rejected. When the hypothesis is rejected
another one is formulated and tested.

PURITY OF A SUBSTANCE

A pure substance is a material that has constant composition and consistent properties throughout
the sample. In other words, a pure substance does not contain impurities or any contaminants.

CRITERIA FOR PURITY

Chemists check whether a given substance is pure or not by carrying out a number of tests on the
substance. The three basic tests carried out to check the purity of a substance are:

a. Melting point
b. Boiling point
c. Paper chromatography

CHECKING PURITY OF A SUBSTANCE BY MELTING POINT

A pure substance has a specific melting point. The presence of impurities affects the melting
point in two ways:

 Impurities lower the melting point of the substance. For example, pure ice melts completely
at exactly 0ºC but the ice we make at home from tap water will start to melt before 0ºC.
 Impurities cause the substance to melt over a range of temperatures.

CHECKING PURITY OF A SUBSTANCE BY BOILING POINT

A pure substance has a fixed boiling point. The presence of impurities affects the boiling point in
two ways:

 Impurities raise the boiling point of a substance. The more impurities the substance contains,
the higher its boiling point will be.
 Impurities cause the substance to boil over a range of temperatures.

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CHECKING PURITY OF A SUBSTANCE BY PAPER CHROMATOGRAPHY

Paper chromatography is used to separate mixtures of soluble substances. These are often
coloured substances such as food colourings, inks, dyes or plant pigments.

In chromatography, substances are picked up and carried by a mobile phase which moves
through a stationary phase.

The stationary phase

It is the part of the chromatography which is in solid state e.g. the filter paper.

The mobile phase

It is the part of the chromatography which is in gaseous or liquid state e.g. the solvent.

The different dissolved substances in the mixture are attracted to the two phases in different
proportions. This causes them to travel at different rates through the paper.

Procedure

 Using a pencil, a ‘start line’ is drawn near the bottom of the chromatography paper.
 The mixture to be separated is spotted on the start line.
 The bottom of the chromatography paper is dipped into the solvent and the solvent travels up
the paper.
 The solvent picks up the substance being tested and carries them up the paper.
 The different components in the substance rise to different heights.
 The “solvent front” is marked. The solvent front is the furthest point reached by the solvent
on chromatography paper.

INTERPRETING CHROMATOGRAM

Chromatogram is the pattern formed by substances that have been separated by chromatography.
To determine purity of a substance, the chromatogram is interpreted as follows:

 A pure substance produces only one spot on the chromatogram.

 An impure substance produces two or more spots on the chromatogram.

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RELATIVE FLOW VALUES (Rf)

Relative flow value is the ratio of the distance travelled by the substance (spot) to the distance
travelled by the solvent.

Mathematically;

Relative flow value = Distance travelled by substance

Distance travelled by solvent

Both distances are measured from the centre of the sample spot on the base line.

From this sample chromatogram, the relative flow values for components B, C and D are
calculated as follows:

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 8
1. For component B, Rf = = 0.8
10
6
2. For component C, Rf = = 0.6
10
2
3. For component D, Rf = = 0.2
10

The relative flow value is always less than 1 because the distance moved by spot is always less
than the distance moved by solvent.

INTERPRETTING RELATIVE FLOW VALUES

The relative flow value can be used to identify the purity of a substance.

 A pure substance has only one relative flow value.

 An impure substance has more than one relative flow value.

TESTING FOR WATER, IONS AND GASES

Chemists carry out different chemical tests in order to find out the identity of a compound or a
gas.

TEST FOR IONS

An ion is a charged atom. Ionic compounds consist of two kinds of ions: cations and anions.
The positively charged ions are called cations. The negatively charged ions are called anions.

TESTING AQUEOUS CATIONS

The commonly tested cations are calcium (Ca2+), aluminium (Al3+), zinc (Zn2+), copper (Cu2+),
iron (II), Fe2+ and iron (III), Fe3+.

The aqueous cations can be identified using aqueous (dilute) sodium hydroxide and aqueous
ammonia.

TESTING FOR AQUEOUS CATIONS USING AQUEOUS SODIUM HYDROXIDE

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When aqueous sodium hydroxide solution is added to various salt solutions, most form
precipitates and some soluble complex ions. A precipitate is an insoluble solid that emerges
from a liquid solution.

Procedure

 Add a few drops of aqueous sodium hydroxide to the solution that contains the cation. Note
and record the colour of any precipitate formed.

The table below shows expected colours of precipitates and their solubility in excess dilute
sodium hydroxide.

Test substance Cation Observation after adding aqueous NaOH

3 drops In excess
Copper sulphate Cu2+ Blue precipitate Insoluble
2+
Iron(II) chloride Fe Pale green precipitate Insoluble
Aluminium sulphate Al3+ White precipitate White precipitate
dissolves
Magnesium sulphate Mg2+ White precipitate Insoluble
Calcium sulphate Ca2+ White precipitate Insoluble
Zinc sulphate Zn2+ White precipitate White precipitate
dissolves

TESTING FOR CATIONS USING AQUEOUS AMMONIA

 Add a few drops of aqueous ammonia to the solution containing the cation.
 Note and record the colour of the precipitate formed.

The table below shows the expected colours of precipitates and their solubility in excess dilute
ammonia.

Cation Observation after adding aqueous ammonia (NH3)

3 drops In excess
Calcium (Ca2+) White precipitate Insoluble

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 Aluminium (Al3+) White precipitate Insoluble
Zinc (Zn2+) White precipitate Precipitate dissolves
Copper (Cu2+) Blue precipitate Precipitate dissolves
Iron (II) (Fe2+) Green precipitate Insoluble
Iron (III) (Fe3+) Red brown precipitate Insoluble
2+
Lead (Pb ) White precipitate Insoluble

TESTING FOR ANIONS

The commonly tested anions are sulphates, halides and nitrates.

a. TESTING FOR SULPHATE IONS (SO4 2  )

Add a few drops of dilute hydrochloric acid to the sample, followed by a few drops of barium
chloride solution. A white precipitate forms if sulphate ions are present.

b. TESTING FOR HALIDE IONS

A halide ion is the halogen atom bearing a negative charge. For example, F , Cl , Br  and I  .

Add a few drops of dilute nitric acid to the sample, followed by a few drops of dilute silver
nitrate solution. Observe and record the colour of any precipitate that forms.

The table below gives expected colours of precipitates with acidified silver nitrate.

Halide ion Colour of precipitate

Fluoride ions (F ) No precipitate

Chloride ions (Cl ) White precipitate

Bromide ions (Br  ) Pale yellow precipitate

Iodide ions (I  ) Yellow precipitate

c. TESTING FOR NITRATES (NO3  )

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Nitrate ions can be detected by reducing them to ammonia.

 To a little nitrate solution, add aqueous sodium hydroxide.

 Drop aluminium foil into the solution.
 Warm gently.

If nitrate ions are present, ammonia gas is given off. Ammonia gas turns damp red litmus paper
blue.

TESTING FOR GASES

The commonly tested gases are ammonia, carbon dioxide, hydrogen, chlorine, oxygen and
sulphur dioxide.

a. TESTING FOR AMMONIA (NH3)

Introduce a damp red litmus paper into a jar containing ammonia. The damp red litmus paper
turns blue if the gas is ammonia.

b. TESTING FOR CARBON DIOXIDE (CO2)

Bubble the gas produced through lime water (calcium hydroxide solution). Lime water turns
milky if the gas is carbon dioxide.

c. TESTING FOR HYDROGEN GAS (H2)

Introduce a burning splint at the mouth of a gas jar containing hydrogen. The gas burns with a
‘pop’ sound if its hydrogen.

d. TESTING FOR CHLORINE (Cl2)

Hold a damp blue litmus paper at the mouth of a test tube containing chlorine gas. The damp
blue litmus paper gets bleached if the gas is chlorine.

e. TESTING FOR OXYGEN GAS (O2)

Introduce a glowing splint into a gas jar containing oxygen gas. The glowing splint relights
(bursts into flames).

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f. TESTING FOR SULPHUR DIOXIDE (SO2)

Bubble the gas through acidified potassium dichromate (VII) solution. The acidified potassium
dichromate (VII) changes from purple to colourless if the gas is sulphur dioxide.

TESTING FOR THE PRESENCE OF WATER

The presence of water can be tested using anhydrous copper (II) sulphate or anhydrous cobalt
chloride.

a. ANHYDROUS COPPER (II) SULPHATE (CuSO4) TEST FOR WATER

Anhydrous copper sulphate is white. When water is added to a sample of anhydrous copper
sulphate, it turns blue. Change of colour form white to blue indicates that the added liquid is
water.

b. ANHYDROUS COBALT (II) CHLORIDE (CoCl2) TEST FOR WATER

Anhydrous cobalt (II) chloride is blue in colour. When water is added to a sample of anhydrous
cobalt (II) chloride, it turns pink.

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TOPIC 2 : NITROGEN, SULPHUR AND PHOSPHORUS

NITROGEN

 It is a group V element and is in period 2 of the Periodic Table.

 It has an atomic number is 7.
 Its electron configuration is 2.5
 It has a valency of 3. It can also exhibit a valency of 5.
 It makes about 78% of the air.

SOURCES OF NITROGEN

The following are some of the sources of nitrogen

a. Air (atmosphere)
b. The earth’s crust
c. Lightning
d. Proteins
e. Amino acids
f. Leguminous plants e.g. groundnuts, pigeon peas.
g. Fertilizers e.g. ammonium nitrate (NH4NO3), ammonium phosphate, (NH4)3PO4 and sodium
nitrate (NaNO3).

PHYSICAL PROPERTIES OF NITROGEN

 It is colourless.
 It is odourless.
 It is insoluble in water.
 It is less dense than air.

CHEMICAL PROPERTIES OF NITROGEN

Nitrogen is a diatomic gas that has strong triple covalent bonds (N  N) between the atoms of its
molecules. These bonds require a lot of energy to break. The strong bonds make nitrogen to be
unreactive (inert) under normal conditions.

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However under very high temperatures it can react with other elements such as alkali metals,
alkaline earth metals and hydrogen.

Examples are shown below:

1. Nitrogen reacts with alkali metals at high temperatures

potassium + nitrogen potassium nitride
6K (s) + N2 (g) 2K3N (s)
2. Nitrogen reacts with alkaline earth metals
magnesium + nitrogen magnesium nitride
3Mg (s) + N2 (g) Mg3N2 (s)
3. Nitrogen reacts with hydrogen to produce ammonia
nitrogen + hydrogen  ammonia


N2 (s) +  2NH3 (g)

3H2 (s) 

 ) means that the reaction is reversible.

The double arrow ( 

USES OF NITROGEN

a. It is used in the production of ammonia

b. It is used in food preservation.
c. It is used in oil tankers to prevent fires because of its inertness.
d. Liquid nitrogen is used as a refrigerant that is a very low-temperature coolant.
e. Liquid nitrogen is used to shrink-fit machine parts.

COMPOUNDS OF NITROGEN

The compounds of nitrogen include the following:

 Ammonia (NH3)
 Nitric acid (HNO3)
 Oxides of nitrogen such nitrogen monoxide (NO) and nitrogen dioxide (NO2).

PREPARATION OF AMMONIA

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Ammonia can be prepared by heating any ammonium salt, such as ammonium chloride with an
alkali, such as calcium hydroxide.

calcium + ammonium calcium + water + ammonia

hydroxide chloride chloride
Ca(OH)2 (s) + NH4Cl (s) CaCl2 (s) + H2O (l) + NH3 (g)

Water vapour is removed from the ammonia gas by passing the gas formed through a drying
tower containing calcium oxide.

PROPERTIES OF AMMONIA

a. It is colourless
b. It has a pungent (choking) smell
c. It is very soluble in water.
d. It turns blue litmus paper red. This tells us that ammonia is a basic substance.

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e. It reacts with hydrogen chloride gas to form a white smoke. The smoke is made of tiny
particles of solid ammonium chloride.
NH3 (g) + HCl (g) NH4Cl (g)
f. It reacts with acids to form salts. For example it reacts with nitric acid to form ammonium
nitrate.
NH3 (g) + HNO3 (aq) NH4NO3 (aq)

USES OF AMMONIA

a. It is used in making nitric acid.

b. Used in making plastics.
c. Used in the manufacture of fertilizers.
d. It is used to make hard water soft.
e. It is used in the manufacture of ammonium chloride used in dry cells.
f. Used in making explosives.

PREPARATION OF NITRIC ACID

Nitric acid can is prepared by heating potassium nitrate with concentrated sulphuric acid. The
reaction for the process is:

potassium + sulphuric acid potassium + nitric acid

nitrate hydrogen sulphate

KNO3 (s) + H2SO4 (l) KHSO4 (aq) + HNO3 (aq)

USES OF NITRIC ACID

1. Used for making nitrate fertilizers.

2. Used to manufacture explosives e.g. TNT and dynamite.
3. Used to manufacture plastics.
4. Used in the purification of metals e.g. silver, gold, platinum etc.
5. Used to manufacture dyes and drugs.
6. Used as an oxidizing agent in textile industries.
7. Used in refining gemstones.

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SULPHUR

 It is a group VI and period 3 element of the Periodic Table.

 It has an atomic number of 16.
 Its electron configuration is 2.8.6.

SOURCES OF SULPHUR

 Volcanic regions
 Crude oil
 Metal ores
 Natural gas in form of hydrogen sulphide (H 2S)

PHYSICAL PROPERTIES OF SULPHUR

 It is a brittle yellow substance.

 It is made up of crown shaped molecules, each containing eight atoms, S8.

 It is insoluble in water.
 It is soluble in organic solvents such as methylbenzene and carbon disulphide.
 It has a low melting point.
 It does not conduct heat and electricity.
 It has two allotropes: rhombic and monoclinic sulphur.

CHEMICAL PROPERTIES OF SULPHUR

 It reacts with metals to form sulphides.

Magnesium + Sulphur Magnesium sulphide
Mg (s) + S (s) MgS (s)
 It burns in oxygen to produce sulphur dioxide.

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 S (s) + O2 (g) SO2 (s)

USES OF SULPHUR

Sulphur is used to:

 Produce sulphuric acid.

 Vulcanise rubber. Vulcanisation is the addition of sulphur to rubber in order to strengthen it.
 Make matches, pesticides, drugs and paper.
 Make sulphur concrete.
 Manufacture gun powder.
 Manufacture plastic flowers.

PRODUCTION OF SULPHURIC ACID

Sulphuric acid (H 2SO 4 ) is produced by the contact process. The process has the following
steps:

a. Sulphur is burnt is oxygen to produce sulphur dioxide.

S (s) + O2 (g) SO2 (g)
b. Sulphur dioxide is reacted with oxygen to produce sulphur trioxide.
SO2 (g) + O2 (g) SO3 (g)
c. Sulphur trioxide is mixed with concentrated sulphuric acid to produce oleum.
SO3 (g) + H2SO4 (l) H2S2O7 (l)
d. The oleum is added to water to produce sulphuric acid.
H2S2O7 (l) + H2O (l) 2H2SO4 (l)

USES OF SULPHURIC ACID

Sulphuric acid is used:

 To manufacture inorganic fertilizers such as ammonium sulphate.

 To make paints and dyes.
 In the production of synthetic fibres e.g. nylon.

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 As an acid in car batteries.
 To make soaps and detergents.
 In petroleum refining.
 As dehydrating agent. A dehydrating agent is a substance that removes water from another
substance. The dehydration of sucrose is shown in the equation below.

PHOSPHOROUS

 It is a group V and period 3 element of the Periodic Table.

 Its atomic number is 15.
 Its electron configuration is 2.8.5.
 It has a valency of 3 but it can also exhibit a valency of 5.

SOURCES OF PHOSPHOROUS

 Bones
 Banana peels
 Composite farm manure
 Crab shells
 Eggs
 The earth’s crust in form of phosphates
 Mineral rocks mainly phosphates

PHYSICAL PROPERTIES OF PHOSPHOROUS

 It a yellow solid at room temperature.

 It does conduct heat or electricity.
 It has two allotropes: white and red phosphorous.
 It is insoluble in water.
 It has a melting point of 44ºC and a boiling point of 280ºC.
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CHEMICAL PROPERTIES OF PHOSPHOROUS

 It reacts with oxygen to form oxides such as phosphorous pentaoxide and phosphorous
trioxide.
 It combines easily with halogens.
 It reacts with metals.

USES OF PHOSPHOROUS

Phosphorous is used:

 To manufacture inorganic fertilizers such as ammonium phosphate.

 In the production of phosphoric acid.
 To make matches.
 To make toothpaste, detergents and baking powder.

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TOPIC 3 : CHEMICAL BONDING AND
PROPERTIES OF MATTER
PHYSICAL PROPERTIES OF IONIC COMPOUNDS

 They are made up of ions and not atoms nor molecules.

 They have strong electrostatic forces.
 They are soluble in water.
 They are insoluble in organic solvents like benzene or propanone.
 They have high melting and boiling points.
 They are hard brittle solids at room temperature.
 They conduct electricity when in molten or aqueous state. Molten state means “when melted”
while aqueous state means “when dissolved in water”.

PHYSICAL PROPERTIES OF COVALENT COMPOUNDS

 They are made up of molecules and not ions.

 They low intermolecular forces.
 Most are insoluble in water.
 They are soluble in organic solvents like benzene and propanone.
 They have low melting and boiling points.
 Most are gases or volatile liquids at room temperature.

STRUCTURAL DIFFERENCES BETWEEN IONIC AND COVALENT COMPOUNDS

1. STRUCTURE OF IONIC COMPOUNDS

The ions in the compound are arranged in a regular pattern. Cations and anions are alternated.
The pattern repeats several times. The result is a crystal lattice. The oppositely charged ions are
held by strong ionic bonds, forming a giant structure. An example of a giant structure of
sodium chloride is shown below.

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EFFECTS OF STRUCTURES ON THEIR PHYSICAL PROPERTIES

 The strong electrostatic attraction between the ions in the structure requires a lot of heat to
break down. This causes ionic compounds to have high melting and boiling points.
 In solid state, ionic compounds do not conduct electricity. This is because the ions are locked
in fixed positions crystal lattice and are not able to move freely. However, when in molten
state or in aqueous solution ionic compounds conduct electricity because the crystalline
structure is broken down and the ions are free to move and conduct electric current.

An aqueous solution that conducts electricity is called an electrolyte. For this reason, ionic
compounds are said to be electrolytes.

2. STRUCTURES OF SIMPLE COVALENT COMPOUNDS

In simple covalent compounds, the atoms forming the molecules are held by strong covalent
bonds. However, the molecules are attracted by weak intermolecular force called van der Waal’s
forces or hydrogen bonds.

EFFECTS OF SIMPLE COVALENT STRUCTURES ON THEIR PHYSICAL

PROPERTIES

 The weak intermolecular forces cause simple covalent compounds to have low melting and
boiling points. When heated, they melt quickly or vapourise.
 Since their structure do not contains ions, covalent compounds do not conduct electricity.
They are non-electrolytes.

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DISTINGUISHING IONIC AND COVALENT SOLUTIONS USING CONDUCTIVITY

Materials

2 cells, a bulb (3.8V, 0.3A), 3 connecting wires fitted with crocodile clips, iron nails or graphite
rods, 40 ml of each of the following; sodium chloride solution, sugar solution , dilute sulphuric
acid and ethanol, distilled water and a beaker (100 ml)

a. Set up the apparatus as shown below

b. Pour 40 ml of sodium chloride solution into the beaker

c. Dip the nails into the solution
d. Observe the bulb and record “light” or “no light” in the table of results.
e. Remove the nails from the beaker.
f. Rinse the beaker and the nails with distilled water.
g. Repeat steps b to f using sugar solution, dilute sulphuric acid and ethanol.
Table of results

Liquid Observation
Sodium chloride
Sugar solution
Sulphuric acid
Ethanol

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h. Classify the liquids as ionic or covalent.
HINT
When the bulb gives light, the liquid used is ionic and when the bulb does not give light, the
liquid use is covalent.

COMMON EXAMPLES OF ELECTROLYTES AND NON–ELECTROLYTES

Electrolytes Non–electrolytes
Sodium chloride solution Pure water
Copper (II) sulphate solution Sugar solution
Sodium hydroxide solution Paraffin wax
Hydrochloric acid Ethanol
Sulphuric acid Urea
Ethanoic acid Molten sulphur

PURE COVALENT BONDS AND DATIVE COVALENT BONDS

a. Pure covalent bond

It is a covalent bond whereby the shared pair of electrons are equally donated by the atoms
involved. For example, the bonding in ammonia, NH3 is pure covalent bonding because both
hydrogen and nitrogen contribute an electron to the shared pair.

b. Dative covalent bond

It is a covalent bond whereby only one atom contributes all the shared pair of electrons. A very
good example of dative bonding occurs in the formation of ammonium ion, NH +4 .

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Nitrogen atom in the molecule of ammonia donates a lone pair of electrons to hydrogen ion, H +

POLAR AND NON–POLAR COVALENT BONDS

a. Polar covalent bond

A polar covalent bond is formed when electrons are not shared equally between atoms. This
happens when the atoms have electronegativity differences. Electrons are attracted to the more
electronegative atom. The more electronegative atom acquires a partial negative charge
(symbolized   ) and the less electronegative atom acquires a partial positive charge
(symbolized   ). The bond between two polar molecules is called a dipole.

Molecules that have partial charges are called polar covalent molecules. The existence of
partial charges in some covalent compounds enables them to conduct electricity in aqueous
solutions. A good example of polar covalent compound is hydrogen chloride (HCl).

b. Non–polar covalent bond

A non–polar covalent bond is formed when electrons are equally shared between atoms. This
happens when atoms have the same electronegativity such as in hydrogen (H–H) molecule and a
chlorine (Cl –Cl) molecule. Molecules that have non-polar bonds are called non–polar covalent
molecules. They do not have partial charges as a result they do not conduct electricity.

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TYPES OF INTERMOLECULAR FORCES

There are two types of intermolecular forces; hydrogen bonds and van der Waal’s forces. These
bonds are generally week.

a. Hydrogen bonding

This type of bonding occurs between molecules containing hydrogen and a more electronegative
atom than the hydrogen atom. The attraction between the partially negative atom and the
partially positive atom is called the hydrogen bond.

b. Van der Waals Forces

These are weak attractive or repulsive forces between and outside molecules. Van der Waals
forces are present in most molecules.

EFFECTS OF INTERMOLECULAR FORCES ON PHYSICAL PROPERTIES

 Water has a higher boiling point when compared to many covalent compounds because of the
presence of hydrogen bonds between its molecules.
 Surface tension of water is great due to hydrogen bonding among water molecules.
 The presence of hydrogen bonds between water molecules gives it its liquid state.
 Van der Waals force significantly to higher boiling points for some elements such as
halogens.

ALLOTROPY

This is the existence of an element in more than one form in the same state. The different forms
of the same element which exist in the same physical state are called allotropes. Oxygen has two
allotropes; oxygen and ozone. Sulphur has two allotropes; rhombic and monoclinic. Carbon has
two allotropes; graphite and diamond.

STRUCTURE OF GRAPHITE

In graphite, each carbon atom forms covalent bonds to three others. This gives rings of six
atoms. The rings form flat sheets that can slide over each other easily.

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The sheets are held together by weak forces.

PHYSICAL PROPERTIES OF GRAPHITE

 It is soft and slippery.

 Dark grey in colour
 It conducts electricity. This is because each carbon atom has four valence electrons but only
forms three bonds. So, the fourth electron is free to move through the graphite, carrying
charge.

USES OF GRAPHITE

 Used as a lubricant for engines and locks.

 It is mixed with clay to make pencil lead.
 Used as electrode in electric circuits.
 Used for connecting brushes in generators.

STRUCTURE OF DIAMOND

In diamond, each carbon atom forms four covalent bonds to four others. The four carbon atoms
are arranged in a tetrahedral structure. This makes the structure very strong and rigid, which
melts at 3550ºC.

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PHYSICAL PROPERTIES OF DIAMOND

 It is the hardest substance on earth ever known.

 It is a colourless substance.
 Its sparkles in light when cut.
 It does not conduct electricity.

USES OF DIAMOND

 Used for making jewellery e.g. necklace, rings, and earrings. This is because of its attractive
appearance.
 Used to make drilling equipment and cutting tools because it is very hard.

SILICON DIOXIDE (SILICA)

Silicon dioxide (SiO2) occurs naturally as quartz, the main mineral in sand. Like diamond,
silicon dioxide forms a giant structure shown below.

In silicon dioxide, each silicon atom forms four covalent bonds with oxygen atoms and each
oxygen atom bonds covalently to two silicon atoms. The result is a very hard substance with a
melting point 1710ºC. The table below compares the properties of diamond and silicon dioxide.

Property Diamond Silicon dioxide

Hardness Hardest substance on earth Very hard substance
Conductivity Does not conduct electricity Does not conduct electricity
Melting point Very high melting point, 3550ºC High melting point, 1710ºC

PROPTERIES OF METALS

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 They have high melting and boiling points. This is due to the strong metallic bonds between
the positive metal ions and the mobile sea of electrons.
 They have high densities because the atoms are very closely packed.
 They are ductile i.e. they can be drawn out into thin wires.
 They are malleable i.e. they can be hammered into different shapes.
 They are sonorous i.e. they produce a ringing sound.
 They are shiny.
 They are good conductors of heat. This is because they have free electrons which transfer the
heat through the metal structure.
 They are good conductors of electricity due to the free electrons within the metal structure.
These mobile electrons carry charge when a voltage is applied across the metal.

USES OF METALS IN RELATION TO THEIR PROPERTIES

Property Uses
Good electrical conductors Used in electrical wiring and in appliances
such as TVs, radios, computers
Good thermal conductors Used to make cooking utensils e.g. pots
Malleable Used to make shaped objects e.g. car bodies,
kitchen ware etc.
Ductile Use to make connecting wires e.g. copper wire
Sonorous Used to make bells
Shiny Used to make jewellery e.g. earrings, bungles
and necklaces.

ALLOYS

An alloy is a mixture of two or more elements one whereby atleast one of them is a metal.

EXAMPLES OF ALLOYS

a. Stainless steel. It is a combination of iron (metal) and chromium (metal).

b. Brass. It is a mixture copper (metal) and zinc (metal).

Page | 27
c. Bronze. It is a combination of copper (metal) and tin (metal).

PROPERTIES AND USES OF ALLOYS

Alloy Composition Properties Used to make

Brass 70% copper  Harder than pure copper  Kitchenware
30% zinc  Gold coloured  Musical instruments
 Screw
 Radiators
Stainless steel 74% iron  Shiny  Cutlery
18% chromium  Strong  Surgical instruments
8% carbon  Doesn’t rust  Kitchen sinks
Steel 99% iron  Hard and strong  Bridges and bridges
1% carbon  Body of cars and railway
tracks
Bronze 90% copper  Hard and strong  Statues and monuments
10% tin  Has shiny surface  Medals, swords and
 Doesn’t rust easily artistic materials

Page | 28
TOPIC 4 : STOICHIOMETRY
CHEMICAL FORMULAE OF COMPOUNDS

The formula of a compound is made up from the symbols of the elements present and numbers to
show the ration in which the different atoms are present.

STEPS TO BE FOLLOWED IN WRITING CHEMICAL FORMULAE

In order to write a correct chemical formula of a compound, the following steps can be used.

 Write down the correct symbol of the elements.

 Write down the valency of each element.
 Exchange the valencies of the elements.
 Write down the exchanged valencies as subscripts in the formula.

Examples

Write down the formula of each of the following compounds given the valencies of each
element.

a. sodium sulphate
Solution
Symbol Na SO4
Valency 1 2
1 2
Exchange the valencies Na SO4
The correct formulas is Na2SO4

b. Magnesium chloride
Solution
Symbol Mg Cl
Valency 2 1
Formula MgCl2

Page | 29
c. Calcium hydroxide
Solution
Symbol Ca OH
Valency 2 1
Formula Ca(OH)2
d. Magnesium oxide
Solution
Symbol Mg O
Valency 2 2
Formula Mg2O2 which is often simplified to
MgO

CHEMICAL EQUATIONS

A chemical equation is a shorthand way representation of what happens in a chemical reaction.

Example

Magnesium reacts with oxygen to form magnesium oxide. Write a chemical equation for this
reaction.

SOLUTION

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

IMPORTANT FACTS ABOUT CHEMICAL EQUATIONS

 Reactants and products are separated by an arrow.

Magnesium + Oxygen Magnesium oxide
(Reactants) (Product)

Reactants are always written on the left hand side while products are written on the right hand
side of the arrow.

 The plus (+) sign in chemistry when used on the left hand side of the arrow means “reacts
with”.

Page | 30
 The arrow means “to form” the products on the right.
 A double arrow (  ) means that the reaction is reversible.

BALANCING CHEMICAL EQUATIONS

Balancing a chemical equation means ensuring that there is an equal number of atoms of each
element on either side of the equation.

GUIDELINES FOR BALANCING CHEMICAL EQUATIONS

1. Write down the correct formulae for each substance.

2. Balance the atoms on both sides of the equation by multiplication of molecules of reactants
or products.
3. Do not change the formula of a substance or element in order to balance the equation.
4. In the brackets, after each substance write down its physical state.
The following are the short forms for different physical states which are commonly used in
chemical equations.
Physical state State symbol
solid state (s)
liquid state (l)
gaseous state (g)
aqueous solution (aq)

METHODS OF BALANCING CHEMICAL EQUATIONS

There are two methods of balancing chemical equations: trial and error method and the
systematic method. In this book we will use the systematic method to balance equations.

Example 1

Balance the equation C3H8 (g) + O2 (g) CO2 (g) + H2O (l)

Solution

Let the balanced equation be

aC3H8 (g) + bO2 (g) cCO2 (g) + dH2O (l)

Page | 31
Using C atoms only

3a = c …………....................................................................................[1]

Using H atoms only

8a = 2d ………………………………………………….…………….[2]

Using O atoms only

2b = 2c + d ……………………………………………..……….….[3]

Solving the equations

Let a = 1. Then use this assumption to solve for all the equations using substitution
method.

c = 3a = 3 (1) = 3

2d = 8a = 8(1)

8
d = = 4
2

2b = 2(3) + 4

10
b = = 5
2

The balanced equation is therefore

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g)

Example 2:

Balance the equation H2 (g) + N2 (g) NH3 (g)

Solution

Let the balanced equation be

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aH2 (g) + bN2 (g) cNH3 (g)

Using H atoms only

2a = 3c…………………………………………………..[1]

Using N atoms only

2b = c…………………………………….…………….[2]

2 1
Let a = 1, then c = and b = .
3 3

The balanced equation becomes

1 2
H2 (g) + N2 (g) NH 3 (g)
3 3

The coefficients must be whole numbers; therefore the whole equation is multiplied by 3 to
remove the fractions:

1 2
3H2 (g) + 3× N 2 (g) 3× NH 3 (g)
3 3

Finally the balanced equation is

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

RELATIVE FORMULA MASS (R.F.M) OF A COMPOUND

The relative formula mass (RFM) of a compound is the sum of the relative atomic masses
(RAM) of atoms in the numbers shown in the formula. Both relative atomic and formula masses
have the same units. The unit is the atomic mass unit (amu).

CALCULATING RELATIVE FORMULA MASS OF COMPOUNDS

To calculate the relative formula mass of a compound, add the relative atomic masses for all the
atoms present in that compound.

Page | 33
Example 1

Work out the relative formula mass (R.F.M) of ethanol (CH3CH2OH).

(RAM of C =12, H = 1 and O = 16).

Solution

R.F.M of CH3CH2OH = (12  2) + (1  6) + (16  1)

= 24 + 6 + 16

= 46 amu

Example 2

Work out the relative formula mass (R.F.M) of aluminium sulphate, Al2(SO4)3.

(RAM of Al = 27, S = 32, and O = 16)

Solution

RFM of Al2(SO4)3 = (27  2) + (32  3) + (16  4  3)

= 54 + 96 + 192

= 342 amu

THE MOLE

The mole refers to an amount of a substance containing 6.023  1023 particles. The mole is a unit
of measurement just like “pair”, “dozen”, “ream” among others. The number 6.023  1023 is
called Avogadro’s constant.

The particles in moles could be atoms, molecules, ions and electrons. For example

1 mole of hydrogen atoms = 6.023  1023 H atoms.

1 mole of potassium atoms = 6.023  1023 K atoms

Page | 34
1 mole of oxygen molecules = 6.023  1023 O2 molecules

1 mole of copper ions = 6.023  1023 Cu2+ ions

1 mole of electrons = 6.023  1023 electrons

RELATIONSHIP BETWEEN MOLE AND MASS OF A SUBSTANCE

1 mole of the particles of any substance is equivalent to the relative atomic mass (RAM) of the
atom in the molecule of the substance expressed in grams.

For example, the relative atomic mass of sodium (Na) is 23 amu. Therefore, 1 mole of sodium
atoms weighs 23g. Similarly, 23g of sodium atoms contains 6.023  1023 sodium atoms.

MOLAR MASS

It is the mass of one mole of a substance. It is simply the relative atomic mass (RAM) of an atom
or the relative formula mass (RFM) of a compound expressed in grams. The unit of molar mass
is the gram per mol. It is denoted as g/mol or gmol-1 .

MOLE CALCULATIONS

a. CONVERTING MASS INTO MOLES

Number of moles = Mass of substance

Molar mass

Example 1
Work out the number of moles of aluminium (Al) present in 108g of the substance.

(RAM of Al = 27)

Solution

108g
Number of moles =
27 g/ mol

= 4 moles

Page | 35
Example 2

Calculate the number of moles present in 19.6g of sulphuric acid (H2SO4). (RAM of H = 1, S =
32 and O = 16).

Solution

Molar mass of H2SO4 = (1  2)  (32  1)  (16  4)

= 98g/mol

19.6 g
Number of moles =
98g/ mol

= 0.2 mole

b. CONVERTING MOLES INTO MASS

Mass of a substance = Number of moles  molar mass

Example 1

How many grams are there in 2 moles of iron (Fe)? (RAM of Fe = 56).

Solution

Mass of iron = 2 moles  56g/mol

= 112g.

Example 2

Calculate the mass of 0.25 mole of ethanol (CH3CH2OH). (RAM of C = 12, H =1 and O = 16).

Solution

Molar mass of ethanol = (12  2)  (1 6)  (16  1) g/mol = 46g/mol

Mass of ethanol = 0.25 mole  46g/mol

= 11.5g

Page | 36
c. USING AVOGADRO’S CONSTANT TO FIND THE NUMBER OF PARTICLES IN A
GIVEN SUBSTANCE

Number of particles = Avogadro’s constant  Number of moles

The Avogadro’s constant is denoted as L = 6.023  1023 particles.

Example 1

Find the number of atoms in 3 moles of iron (Fe). (L = 6.023  1023 particles)

Solution

Number of Fe atoms = 3 moles  6.023  1023 atoms

= 1.8069  1024 atoms.

Example 2

If 54g of aluminium were collected from aluminium ore, how many atoms of aluminium were
collected? (RAM of Al = 27, L = 6.023  1023 particles).

Solution

54 g
Number of moles of Al =
27 g/ mol

= 2 moles

Number of atoms of Al = 2 moles  6.023  1023 atoms

= 1.2046  1024 atoms.

Example 3

How many ions are there in 13g of potassium ions (K+). (RAM of K = 39, L = 6.023  1023
particles).

Solution

Page | 37
 13g
Number of moles of K+ ions =
39 g/ mol

= 0.333 mole

Number of K+ ions = 0.333 mole  6.023  1023 ions

= 2.005659  1023 ions.

MOLAR GAS VOLUME

Molar gas volume is the volume occupied by 1 mole of a gas.

It is 24 dm3 room temperature and pressure (r.t.p). Room temperature and pressure are 25⁰C
(298K) and 1 atmosphere respectively. It is 22.4 dm3 at standard temperature and pressure (s.t.p).
Standard temperature and pressure are 0⁰C (273K) and 1 atmosphere respectively.

The molar volume is true for any kind of a gas. It does not matter whether the gas exists as atoms
or molecules, or whether its atoms are large or small.

CALCULATIONS INVOLVING MOLAR VOLUME

At r.t.p:

Number of moles of a gas = Volume of gas

24 dm3
Volume of gas = Number of moles  24 dm3

At s.t.p:

Number of mole of a gas = Volume of gas

22.4 dm3
Volume of gas = Number of moles  22.4 dm3

Example 1

Work out the number of moles of carbon dioxide (CO2) contained in 60 dm3 of carbon dioxide
measured at room temperature and pressure (r.t.p).

Page | 38
Solution

Number of moles = Volume of carbon dioxide

24 dm3
60 dm3
=
24 dm3

= 2.5 moles

Example 2

How many moles are there in 24 cm3 of hydrogen (H2) at room temperature and pressure?

Solution

First, convert 24 cm3 into cubic decimeters.

24
24 cm3 = dm3
1000

= 0.024 dm3

Number of moles = Volume of hydrogen

24 dm3
0.024 dm3
=
24 dm3

= 0.001 mole

Example 3

What volume (in cm3) does 0.25 mole of oxygen occupy at standard temperature and pressure?

Solution

Volume of gas = Number of moles  22.4 dm3

= 0.25  22.4 dm3

= 5. 6 dm3

Page | 39
Volume in cm3 = 5.6  1000

= 5600 cm3

Example 4

What volume does 22 g of carbon dioxide occupy at room temperature and pressure? (RAM of C
= 12 and H =1).

Solution

Number of moles = Mass of compound

Molar mass
22 g
=
[(12×1) + (16× 2)]gmol-1

22g
=
44 gmol-1

= 0.5 mole

Volume of gas = Number of moles  24 dm3

= 0.5  24 dm3

= 12 dm3

WATER OF CRYSTALLIZATION

Water of crystallization refers to the water present in crystals of some salt compounds. Water of
crystallization is also called water of hydration.

Salts containing water of crystallization are called hydrates or hydrated compounds. Some of
the hydrates are shown in the table below.

Name of salt Formula Number of molecules of water of

crystallization
Copper (II) sulphate pentahydrate CuSO4.5H2O 5

Page | 40
 Magnesium sulphate MgSO4.7H2O 7
heptahydrate
Sodium carbonate decahydrate Na2CO3.10H2O 10
Cobalt (II) chloride hexahydrate CoCl2.6H2O 6

When hydrated salts are heated, they lose their water of crystallization. The resulting crystal is
called anhydrous, meaning without water.

DETERMINING THE PERCENTAGE OF WATER IN MOLECULAR AND

HYDRATED IONIC COMPOUNDS

The percentage of water in any molecular or hydrated ionic compound can be determined both
experimentally and theoretically.

 Experimentally, the percentage of water in a hydrate is found by comparing the mass of

water driven off, usually by heating, to the total mass of the compound.
 Theoretically, the percentage of water is found by dividing the water of crystallization
relative formula mass by the relative formula mass of the hydrate.

Example 1

Calculate the percentage of water crystallization in sodium carbonate decahydrate

(Na2CO3.10H2O). RAM of Na = 23, C = 12, O = 16 and H = 1).

Solution

Relative formula mass of water = (10  18)

= 180 amu

Relative formula mass of Na2CO3.10H2O = (23  2) + (12  1) + (16  3) + 180

= 46 + 12 + 48 + 180

= 286 amu

Page | 41
 Percentage of water = mass of water crystallization
mass of compound
180
=  100%
286

= 62. 937%

Example 2

On heating 1.250 g hydrated barium chloride, 1.060 g of anhydrous barium chloride remained.
Calculate the percentage of water of crystallization in the compound.

Solution

Total mass of hydrate = 1.250 g

Mass of anhydrous salt = 1.060 g

Mass water of crystallization = mass of hydrate – mass of anhydrous

= 1.250 g – 1.060 g

= 0.19 g

Percentage of water of crystallization = mass of water

mass of hydrate
0.19
=  100
1.250

= 15. 2%

Experiment

Aim

To determine the percentage of water in hydrated copper sulphate

Materials

Page | 42
Hydrated copper sulphate, evaporating basin (tin), burner, triple beam balance, tripod stand and
wire gauze.

Procedure

1. Weigh the tin on the triple beam balance and record its mass in the appropriate space in the
table of results.
2. Add the crystals of hydrated copper sulphate until the reading increases by approximately 5g.
3. Record the mass of the tin plus the hydrated copper sulphate.
4. Find the mass of the hydrated copper sulphate.
5. Heat the tin gently until the hydrated copper sulphate turns into a white powder.
6. Weigh the tin plus white powder (anhydrous copper sulphate) and record the results in the
appropriate space in the table.
7. Work out the mass of the white powder.
8. Find the mass of water lost.
9. Calculate the percentage of water in the hydrated copper sulphate.

Table of results

Item Mass (g)

Tin
Tin + hydrated copper sulphate
Hydrated copper sulphate
Tin + anhydrous copper sulphate
Anhydrous copper sulphate
Water lost

 Mass of hydrated copper sulphate = mass of tin + hydrated copper sulphate – mass of tin
 Mass of anhydrous copper sulphate = mass of tin + anhydrous copper sulphate – mass of
tin
 Mass of water lost = mass of hydrated copper sulphate – mass of white
powder

Page | 43
EMPIRICAL AND MOLECULAR FORMULAE OF A COMPOUND

1. EMPIRICAL FORMULA

Empirical formula is the formula that contains the lowest number of atoms that make up a
compound.

For example in the molecular formula of glucose, C6H12O6 the ratio of atoms is 6:12:6. This can
be simplified to 1:2:1. Therefore, the empirical formula of glucose would be written as CH2O.

CALCULATING EMPIRICAL FORMULA

The empirical formula of any compound can be worked out from:

 mass composition of a compound

 percentage composition by mass of a compound

GENERAL GUIDELINES FOR CALCULATING EMPIRICAL FORMULA

a. Convert mass of each element into moles.

b. Convert the moles into simplest mole ratio.
c. Write the empirical formula using simplest mole ratio.

Example 1

A compound contains 52.2% carbon (C), 13.0% hydrogen (H), and 34.8% oxygen (O). Work out
its empirical formula.

(RAM of C =12, H =1 and O = 16)

SOLUTION

Since percentage is usually out 100, then assume you have 100 g of the substance. Then convert
the percentage of each element into mass in grams. Thus

Carbon = 52.2 g, H = 13.0 g and 34.8 g.

Page | 44
Element Mass (g) Number of moles Simplest mole ratio
Carbon (C) 52.2 52.2 4.35
= 4.35 =2
12 2.175
Hydrogen (H) 13.0 13.0 13
= 13 =6
1 2.175
Oxygen (O) 34.8 34.8 2.175
= 2.175 =1
16 2.175

The empirical formula of the compound is: C2H6O

Example 2

A compound was found to contain 3.2 g of copper, 0.6 g of carbon and 2.4 g of oxygen. Find its
empirical formula. (RAM of Cu = 64, C = 12 and O =16)

Solution

Element Mass (g) Number of moles Simplest mole ratio

Copper (Cu) 3.2 3.2 0.05
= 0.05 =1
64 0.05
Carbon (C) 0.6 0.6 0.05
= 0.05 =1
12 0.05
Oxygen (O) 2.4 2.4 0.15
= 0.15 =3
16 0.05

The empirical formula of the compound is: CuCO3

2. MOLECULAR FORMULA

A molecular formula is a formula that shows the exact or actual number of atoms of each
element in one molecule of a compound.

CALCULATING MOLECULAR FORMULA

Page | 45
a. Identify or work out the empirical formula
b. Identify the molar mass of the compound
c. Calculate the empirical formula mass
d. Find the empirical formula units using the formula:
Empirical formula units = Molar mass
Empirical formula mass
e. Multiply the subscripts in the empirical formula. The result is the required molecular
formula.

Example 3

Calculate the molecular formula of a compound if its empirical formula is CH2O and has a molar
mass of 180 g/mol. (RAM of C =12, H = 1 and O = 16).

Solution

Empirical formula = CH2O

Empirical formula mass = (12  1) + (1  2) + (16  1)

= 30 g/mol

Molar mass of compound = 180 g/mol

180 g/ mol
Empirical formula units =
30 g/ mol

= 6

Molecular formula = 6(CH2O)

= C6H12O6

CONCENTRATION OF SOLUTIONS

The concentration of a solution refers to the amount of solute dissolved in a specific volume of a
solvent.

Page | 46
WAYS OF EXPRESSING CONCENTRATION OF A SOLUTION

There are three ways of expressing concentration of a solution:

1. In moles per unit volume

Concentration = Number of moles of solute
Volume of solution

The units of concentration in this case are mol/dm3, mol/cm3 and mol/litre

2. In mass per unit volume

Concentration = Mass of solute
Volume of solution

The units of concentration in this case are g/dm3, g/cm3 and g/litre

3. As a percentage
Concentration = Mass of solute  100%
Mass of solution

CALCULATING CONCENTRATION OF SOLUTIONS

Example 1

Calculate the concentration of a solution containing 4 moles of sodium hydroxide (NaOH) in 2

dm3 of water.

Solution

Concentration = Number of moles

Volume of solution
= 4 moles
2 dm3
= 2mol/dm3

Example 2

Calculate the concentration of sodium carbonate (NaCO3) solution containing 53 g of the salt
dissolved in 2 dm3.

Page | 47
Solution

Number of moles = Mass

Molar mass
53g
=
106 g/ mol
= 0.5 mole

0.5 mole
Concentration =
2 dm3

= 0.25 mol/dm3

Example 3

12 g of sodium chloride (NaCl) is dissolved in 100g of water. Calculate the concentration of the
solution.

Solution

Mass of solute = 12 g

Mass of solution = 12 g + 100 g

= 112 g

Concentration = mass of solute  100%

mass of solution
12 g
= 100%
112 g
= 0.1% NaCl

MOLARITY

Molarity is defined as the number of moles of a solute per unit volume of the solution. The unit
of molarity is the mol per cubic decimeter. It is also called the molar, abbreviated (M).

Page | 48
1 mol/dm3 = 1 Molar

Molarity = Number of moles

Volume (dm3)

But, number of moles = mass

Molar mass

Then, molarity = mass

molar mass  volume

PROBLEM SOLVING ON MOLARITY

Example 1

Calculate the molarity of a solution of sodium hydroxide (NaOH), which was made by
dissolving 10 g of solid sodium hydroxide in 250 cm3 of water. (RAM of Na = 23,O = 16 and H
=1)

Solution

Number of moles = mass of NaoH

molar mass of NaOH  volume

250
Volume = dm3 = 0.25 dm3
1000

Molar mass of NaOH = (23×1)+(16×1)+(1×1)g/mol

= 40 g/mol

10g
Molarity =
40g/mol×0.25dm3

= 1M

Example 2

Page | 49
Calculate the mass of potassium hydroxide (KOH), which needs to be used to prepare 500 cm3 of
a 2M solution in water. (RAM of K = 39, O =16 and H =1).

Solution

Number of moles = concentration of KOH  volume

500
= 2 mol/dm3  dm3
1000

= 2 mol/dm3  0.5 dm3

= 1 mole

Mass of KOH = Number of moles  molar mass

= 1 mole  [(39  1) + (16  1) + (1  1)] g/mol

= 1 mole  56 g/mol

= 56 g

STANDARD SOLUTIONS

A standard solution is a solution whose concentration is known.

METHODS OF PREPARING STANDARD SOLUTIONS

There are two methods of preparing standards solutions; dissolution of solutes and dilution of
stock solutions.

1. DISSOLUTION OF SOLID SOLUTES

In this method, a solid solute of a measured mass is dissolved in a solution of a known volume.
To prepare a standard solution by this method, one needs to know:

 the concentration of the solution to be prepared.

 the volume of the solution to prepared.

STEPS FOR PREPARING STANDARD SOLUTION BY SOLUTE DISSOLUTION

Page | 50
a. Calculate the mass of the solute to be dissolved.
The formula used is: mass of solute = molarity  molar mass  volume of
solution
b. Weigh the calculated mass of the solute using a balance.
c. Dissolve the solute completely in a beaker.
d. Transfer the solution quantitatively into a volumetric flask of the required volume.
e. Add water into the volumetric flask up to the mark.
f. Put a stopper and shake to mix thoroughly.

NB: Always keep prepared solutions in bottles and label them. Do not keep them in the
volumetric flask.

Example

Describe how you would prepare 250 ml of 2M potassium carbonate (K2CO3) solution using
solid potassium carbonate. (RAM of K = 39, C = 12 and O =16).

Solution

 Firstly, the mass of potassium carbonate to be dissolved is calculated using the formula:

Mass of K2CO3 = molarity  molar mass  volume

= 2M  [(39×2)+(12×1)+(16×3)]g/mol  0.25 dm3

= 2M  138 g/mol  0.25 dm3

= 69 g

 69g of solid potassium carbonate is weighed using a balance and placed in a beaker.
 Using distilled water, the solid potassium carbonate is dissolved quantitatively.
 Then, the potassium carbonate solution is transferred into a 250 ml volumetric flask. The
beaker is rinsed with distilled water several times, adding the solution into the volumetric
flask.
 Distilled water is added into the flaks up to the mark.

Page | 51
 The flask is covered on top and shaken to mix thoroughly. The concentration of the solution
will be 2M potassium carbonate.

2. DILUTION OF STOCK SOLUTIONS

Dilution is the process of making a highly concentrated solution less concentrated. The stock
solution is the concentrated solution that will be diluted before using it. In order to prepare a
standard solution by dilution, one needs to know:

 the volume of the solution to be prepared.

 the concentration of the stock solution.
 the concentration of new solution

DILUTION FORMULA

When a solution is diluted, the number of moles of solute in the solution does not change. This
means that the number of moles of solute before dilution is equal to the number of moles solute
in the solution after dilution.

Since number of moles = concentration  volume

Letting C1 = initial concentration

V1 = initial volume

C2 = final concentration (concentration of diluted solution)

V2 = final volume (new volume of solution)

The dilution law is given as:

C1V1 = C2V2

STEPS FOR PREPARING STANDARD SOLUTION BY DILUTION

1. Calculate the volume of the stock solution to be transferred.

C 2 × V2
The formula is given as: V1 =
C1

Page | 52
2. Measure the calculated volume using a measuring cylinder of pipette.
3. Transfer it into a volumetric flask of the required volume.
4. Rinse the measuring cylinder with distilled water and transfer the water into the volumetric
flask.
5. Add water into the flask up to the mark. Put a stopper and shake to mix.

Example 1

60 cm3 of the solution whose concentration is 15 g/cm3 was diluted with distilled water by
raising its volume to 60 cm3. Calculate the concentration of the new solution.

SOLUTION

By dilution formula:

C1V1 = C2V2

15 g/cm3  60 cm3 = C2  80 cm3

15g/ cm3× 60 cm 3
C2 =
80 cm3

= 11.25 g/cm3

Example 2

Calculate the amount of water that must be added to 5 cm3 of 2M hydrochloric acid to dilute it to
0.1M.

Solution

By dilution formula

C1V1 = C2V2

2M  5 cm3 = 0.1 M  V2

2 M×5cm3
V2 =
0.1M

Page | 53
 = 100 cm3

Water added = V2 – V1

= 100 cm3 – 5 cm3

= 95 cm3

ACID – BASE TITRATION

Titration is the gradual addition of one solution to another until the reaction between them is
complete.

FACTS ABOUT ACID – BASE TITRATION

 The ultimate goal of a titration is to determine the concentration of unknown solution. The
process of determining the exact concentration (molarity) of a solution is called
standardization.
 During the titration, one solution (titrant) is titrated against or added to another solution
(analyte) until the reaction between the components in the solutions is complete.

TITRANT

This is the solution of a known concentration which is added to another solution to determine the
concentration of the unknown solution.

ANAYLTE

It is the solution of unknown concentration.

END POINT

It is the balanced point whereby an acid is completely neutralized by a base. The end point is
shown by an indicator.

INDICATOR

Page | 54
An indicator is a dye that has a different colour in acidic or basic solutions. Examples of
indicators are used during a titration are phenolphthalein and Methyl orange. When the end point
is reached, there is a change in colour of the indicator.

APPARATUS USED IN A TITRATION

 Clamp and clamp stands

 Burette or syringe
 Dropper bottle
 Measuring cylinder
 Funnel
 Conical flask or beaker
 White tile

STEPS TO BE FOLLOWED WHEN CARRYING OUT A TITRATION

1. The apparatus is set up as shown below.

2. The titrant is added into the burette.

3. A specific volume of the solution to be titrated is poured into a conical flask.

Page | 55
4. The indicator is added into the conical flask.
5. The titrant is slowly added to the solution being titrated until there is a colour change.
6. The volume of the titrant added from the burette is recorded.

CALCULATING CONCENTRATION OF UNKNOWN SOLUTION

If the titrant and analyte have a mole ratio of 1:1 in the balanced equation, the concentration of
the unknown solution is calculated using the formula:

C a× Va = C b × Vb
where Ca = concentration of the acid used
Va = volume of the acid used (cm3)
Cb = concentration of the alkali used
Vb = volume of the alkali used (cm3)

On the other hand if the titrant and analyte have a mole ratio of greater than 1:1 in the balanced
equation (e.g. 2:1), the concentration of the unknown solution is calculated using the formula:

C a × Va C b × Vb
=
Na Nb

where

Na = number of moles of the acid shown in the chemical equation

Nb = number of moles of the alkali shown in the chemical equation

PROBLEM SOLVING ON TITRATION

Example 1

You are provided with a burette, a funnel, a measuring cylinder, a beaker, clamp and clamp
stand, 0.1M sulphuric acid (0.1M H2SO4), sodium hydroxide solution of unknown concentration,
phenolphthalein indicator and a white tile or paper.

Page | 56
a. Set up the apparatus as shown below.

b. Fill the burette with sulphuric acid (H2SO4) up to the zero mark.
c. Pour 15 cm3 of sodium hydroxide into a beaker.
d. Add three drops of phenolphthalein indicator to the sodium hydroxide.
e. Slowly, add the sulphuric acid from the burette to sodium hydroxide, shaking the beaker all
the time, until the pink colour disappears.
f. Note and record the volume of sulphuric acid used.
Initial volume of sulphuric Final volume of sulphuric Volume of sulphuric acid
acid (cm3) acid (cm3) used (cm3)

g. Empty the beaker and rinse it with distilled water.

h. Repeat steps c to f.
i. Calculate the average volume of sulphuric acid used.

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j. Write a balanced equation for the reaction.
k. Calculate the concentration of hydrochloric acid.

Example 2

In a titration, 11.6 cm3 of 3.0M of sulphuric acid (H2SO4) was required to neutralize 25 cm3 of
sodium hydroxide (NaOH) of unknown concentration.

a. Identify each of the following in the titration; titrant, analyte

Solution
The titrant is sulphuric acid. The analyte is sodium hydroxide.
b. Write down a balanced equation for the reaction.
Solution
2NaOH (aq) + H2SO4 (aq) Na2SO4 (aq) + 2H2O (l)
c. Calculate the concentration of sodium hydroxide.
Solution
The titrant and analyte have a mole ratio of 2:1. Therefore concentration of sulphuric acid is
calculated using the formula:
C a × Va C b × Vb
=
Na Nb
C a × Va N a
Cb = ×
Na Vb

3.0 M×11.6 cm3 2

= ×
1 25cm3
= 2.784M

The concentration of sodium hydroxide is 2.784M.

LIMITING AND EXCESS REAGENT

The limiting reagent is the reactant that is completely used up in a reaction while the other is
still available. The quantity of the limiting reagent controls the amount of product formed by the
reaction. The excess reagent is the reactant that remains after the chemical reaction has reached
equilibrium.

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DETERMINING LIMITING AND EXCESS REAGENTS IN A CHEMICAL REACTION

To determine the limiting reagent and the excess reagent in a chemical reaction, there are four
important guidelines.

1. Ensure the chemical equation is balanced.

2. Convert given masses into moles.
3. Calculate the number of moles of product produced by each reactant.
4. The reactant that produces the smaller number of moles of product is the limiting reagent.

Example 1

Magnesium reacts with oxygen to form magnesium oxide according to the equation:

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

If 6g of magnesium reacts with 2 g of oxygen determine the:

a. limiting reagent and excess reagent

b. mass of magnesium oxide formed. (RAM of Mg = 24 and O = 16)

Solution

a. Number of moles of magnesium = mass of magnesium

molar mass
6g
=
24 g/ mol
= 0.25 mole

Number of moles of oxygen = mass of oxygen

molar mass
2g
=
32 g/ mol
= 0.0625 mole

To find the number of moles of product formed by each reactant, we use the balanced equation.

i. From the balanced equation

2 moles Mg = 2 moles MgO

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 0.25 mole Mg = 0.25 mole MgO

ii. From the balanced equation

1 mole O2 = 2 moles MgO
0.0625 mole O2 = y

y = 0.0625  2

= 0.125 mole MgO

Comparing the number of moles of product each reactant produces, it can be seen that O2 gives
out the smaller number.

Hence, O2 is the limiting reagent while Mg is the excess reagent.

b. The quantity of the product formed is controlled by the limiting reagent.

Thus
0.0625 mole of O2 = 0.125 mole of MgO
Mass of MgO = Number of moles  molar mass
= 0.125 mole  40 g/mol
= 5g

THEORETICAL AND PERCENTAGE YIELD OF A CHEMICAL REACTION

The yield is the amount of product obtained form a chemical reaction. There are two kinds of
yield: theoretical and actual.

Theoretical yield the amount of a product obtained from the balanced equation without doing an
experiment while actual yield is the amount of a product obtained by experiment.

CALCULATING PERCENTAGE YIELD OF A REACTION

The percentage yield of a chemical reaction is calculated using the formula:

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Percentage yield = Actual yield  100
Theoretical yield

Example 1:

2.34 g of aluminium reacts with excess copper (II) sulphate solution to produce 3.89 of copper
according to the equation:

2Al (s) + 3CuSO4 (aq) Al2(SO4)3 (aq) + 3Cu (s)

Determine the theoretical, actual and percentage yield of copper.

(RAM of Cu = 64 and Al = 27)

Solution

a. To find the theoretical yield

From the balanced equation
2 moles Al = 3 moles Cu
2  27g Al = 3  64g Cu
54g Al = 192g Cu
2.34g Al = y
2.34  192
y =
54
= 8.32g
The theoretical yield of copper is 8.32g;

b. The actual yield of copper is 3.89g

c. Percentage yield = Actual yield  100

Theoretical yield
3.89 g
= ×100
8.32 g
= 46. 75%

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CHAPTER 6 : HEATS OF REACTION
TYPES OF REACTIONS IN RELATION TO HEAT CHANGES

There are two types of chemical reactions in relation to heat changes. These are:

 Exothermic reactions
 Endothermic reactions
1. EXOTHERMIC REACTIONS

An exothermic reaction is a chemical reaction that releases heat energy from the system to the
surroundings. The surroundings can be the container or the container. When the heat is given out,
the reaction mixture and its surroundings get hotter.

Examples of exothermic reactions are:

 Combustion
 Neutralization reactions
 Rusting
 Dehydration of sucrose by sulphuric acid
 Reaction between sodium hydroxide and water

2. ENDOTHERMIC REACTIONS

An endothermic reaction is a chemical reaction that takes in heat energy from the surroundings.
When the heat energy is absorbed from the surroundings, the reaction container or vessel feels
colder.

Examples of endothermic reactions are:

 Photosynthesis
 Dissolution of salts in water e.g. Ammonium nitrate
 Evaporation of liquid water
 Melting of ice cubes

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INVESTIGATING TEMPERATURE CHANGES INVOLVED IN EXOTHERMIC AND
ENDOTHERIMC REACTIONS AND PROCESSES

Experiment

Aim
To describe the temperature changes involved in exothermic and endothermic reactions.
Materials
2 test tubes in a rack, a measuring cylinder, thermometer, spatula, tap water, ammonium chloride
and sodium hydroxide pellets.
Procedure
1. Pour 5 cm3 of water into each test tube.
2. Measure the initial temperature of water in each test tube and record the results in the
appropriate spaces in the table.
3. Add half spatula of ammonium chloride (NH4Cl) in one test tube and shake gently.
4. Measure the temperature of ammonium chloride solution and record the results in the table.
5. Repeat steps c and d using sodium hydroxide (NaOH) pellets.

Table of results

Solution Initial temperature Final temperature Temperature change

(°C) (°C) (Final temperature –
Initial temperature) (°C)
Ammonium chloride
(NH4Cl)
Sodium hydroxide
(NaOH)

6. State whether the change in each case is exothermic or endothermic.

Observations
a. When ammonium chloride is dissolved in water, the temperature of the solution is lower
when compared to the initial temperature of the water. In addition, the test tube felt cold
when touched.

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b. When sodium hydroxide is dissolved in water, the temperature of the solution is higher when
compared to the initial temperature of the water. In addition, the test tube felt hot when
touched.
Interpretation of results
A decrease in temperature means that heat energy was absorbed during the reaction while
increase in temperature means that heat energy was released into the surroundings during the
reaction. The dissolving of ammonium chloride in water is therefore endothermic while the
dissolving of sodium hydroxide in water is exothermic.
Conclusion
In endothermic reactions, temperature decreases while in exothermic reactions, temperature
increases.
Let Ti = Initial temperature
Tf = Final temperature
T = Change in temperature
In endothermic reactions,  T is negative since the final temperature is lower than the initial
temperature.
In exothermic reactions,  T is positive since the final temperature is greater than the initial
temperature.

IDENTIFYING EXOTHERMIC AND ENDOTHERMIC REACTIONS FROM

THERMO-CHEMICAL EQUATIONS
A thermo-chemical equation is a balanced equation that includes physical states of all reactants
and products and the enthalpy change.
Enthalpy
Enthalpy is the energy stored in the bonds of a substance. It is given the symbol H and is
measured in kilojoules (kJ).
Enthalpy change
It is the change in energy going from reactants to products. It is shown as ∆H (‘delta H’). ∆H is
also called the heat of reaction.
For an exothermic reaction, enthalpy change is negative while for an endothermic reaction it is
positive.

Page | 64
EXAMPLES OF THERMO-CHEMICAL EQUATIONS

Identify whether each of the following is endothermic or exothermic basing on enthalpy change.

a. NaOH (aq) + HCl (aq) NaCl (aq) + H2O(l),  H  57 kJ/ mol

Solution

The reaction is exothermic since enthalpy change has a negative sign.

b. H2O (l) 2H2 (g) + O2 (g),  H  +575 kJ/ mol

Solution

The reaction is endothermic since enthalpy change has a positive sign.

ENERGY LEVEL DIAGRAMS

An energy level diagram is a graph showing energy changes involving reactants and products
during a chemical reaction.

ENERGY LEVEL DIAGRAM FOR EXOTHERMIC REACTIONS

In exothermic reactions, energy level decreases as the reaction proceeds. For this reason, the
reactants are shown at a higher level than products on the diagram.

The arrow point points downwards. It shows that heat energy is lost or given out.

Page | 65
ENERGY LEVEL DIAGRAMS FOR ENDOTHERMIC REACTIONS

In endothermic reactions, energy level increases as the reaction proceeds. Therefore, reactants
are shown below products in the diagram.

The arrow points upwards. It shows that heat energy is taken in or gained.

WORKED EXAMPLES

Draw energy level diagrams for each of the following reactions:

a. H2 (g) + Cl2 (g) 2HCl (aq),  H  92 kJ/ mol

b. NH4NO3 (s) + H2O (l) NH 4 + (aq) + NO3 (aq) ,  H  +28.1kJ/ mol

Solution

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BOND ENERGIES

Bond energy is the amount of energy required to break one mole of chemical bonds in a
molecular element or compound. Different chemical bonds have different bond energies. The
unit of bond energy is the kilojoule per mole (kJ/mol). Examples of bond energies are shown
below.

Chemical bond Bond energy (kJ/mol)

H–H 436
O=O 498
O–H 464
C–H 413
C–C 346
Cl–Cl 242
C=O 805

ENERGY CHANGES INVOLVED BOND BREAKING AND BOND FORMATION

During any chemical reaction, old bonds are broken. At the same time, new bonds are formed.
To break a bond, requires energy. This energy is needed to separate the atoms or ions against the
attractive forces. Therefore, bond breaking is endothermic.

On the other hand, bond formation gives out heat energy. Therefore bond formation is
exothermic.

CALCULATING OVERALL ENERGY CHANGES OF REACTIONS USING BOND

ENERGIES

The overall energy change is the difference between the energy required to break existing bonds
and the energy given off when new bonds are made.

H = energy required to break bonds – energy given out when bonds are made

Page | 67
Example

Methane burns in excess oxygen to produce carbon dioxide and water according to the equation:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)

a. Calculate the overall enthalpy change for the reaction.

b. State whether the reaction is endothermic or exothermic based on enthalpy change.
[Hint: use the bond energies date given in the table above]

Solution

a. Breaking 4 C–H bonds in methane requires

4  413 kJ = 1652kJ

Breaking 2 O=O bonds in oxygen requires

2  498 kJ = 996kJ

Total energy required to break bonds = (1652 + 996)kJ

= 2648kJ

Making 2 C=O bonds in carbon dioxide gives out

2×805 kJ = 1610kJ

Making 4 O–H bonds in water gives out

4× 464 kJ = 1856kJ

Total energy given out = (1610 + 1856)kJ

= 3466kJ

Overall energy change = 2648 – 3466

H = 818 kJ

b. Basing on enthalpy change the reaction is exothermic.

Page | 68
CHAPTER 6 : ALKANOLS
Alkanols are a family of organic compounds that belong to the class of oxycarbons. They contain
oxygen (O), carbon (C) and hydrogen (H) atoms in their compounds. Alkanols are also called
alcohols.

FUNCTIONAL GROUP OF ALKANOLS

The functional group of Alkanols is –OH, the hydroxyl group.

NAMING ALKANOLS

Alkanols are named by replacing the –e in the parent alkane having a similar number of carbon
atoms with –anol.

Number of carbon atoms Name of alkanol

1 Methanol
2 Ethanol
3 Propanol
4 Butanol
5 Pentanol
6 Hexanol
7 Heptanol
8 Octanol
9 Nonanol
10 Decanol

GENERAL FORMULA OF ALKANOLS

The general formula of alkanols is CnH2n+1OH. The “n” stands for the number of carbon atoms
in the molecule. To find the formula of any alkanol given the number of carbon atoms, simply
substitute ‘n’ in the general formula, with the value given, and simplify using rules of ordinary
algebra.

Page | 69
Example:

Work out the formula of an alkanol with each of the following number of carbon atoms (a) 2
carbon atoms, (b) 4 carbon atoms (c) 10 carbon atoms

SOLUTION

a. If n = 2, then C 2 H (2×2)+1OH = C2H5OH

b. If n = 4, then C4 H (2×4)+1OH = C4H9OH

c. If n = 10 then C10 H (2×10)+1OH = C10H21OH

MOLECULAR AND CONDENSED FORMULAE OF ALKANOLS

A molecular formula is a formula shows the actual number of atoms of each element in a
molecule. For alkanols, the functional group is also shown in the molecular formula.

A Condensed formula is a formula which shows how atoms are attached to each other in the
structure of the molecule.

The table below shows the molecular and condensed formulae of the first ten alkanols.

Name of alkanol Molecular formula Condensed formula

Methanol CH3OH CH3OH

Ethanol C2H5OH CH3CH 2 OH

Propanol C3H7OH CH3 (CH 2 ) 2OH

Butanol C4H9OH CH3 (CH 2 )3OH

Pentanol C5H11OH CH3 (CH 2 ) 4OH

Hexanol C6H13OH CH3 (CH 2 )5OH

Heptanol C7H15OH CH3 (CH 2 )6 OH

Octanol C8H17OH CH3 (CH 2 )7 OH

Nonanol C9H19OH CH3 (CH 2 )8 OH

Decanol C10H21OH CH3(CH2)9OH

Page | 70
STRUCTURAL FORMULAE OF ALKANOLS

A structural formula is a formula which shows how atoms are bonded to each other in a
molecule. The structural formulae of the first ten alkanols are shown below.

Molecular formula Structural formula

CH3OH

C2H5OH

C3H7OH

C4H9OH

C5H11OH

C6H13OH

C7H15OH

C8H17OH

Page | 71
 C9H19OH

C10H21OH

SKELETAL FORMULAE OF ALKANOLS

A skeletal formula is a diagrammatic representation of a molecule in which lines represent bonds

between atoms. The lines are drawn in zigzag fashion. Carbon atoms are implied when two
bonds meet and no atom listed except for functional groups. The skeletal formulae of the first ten
alkanols are shown below.

Molecular formula Skeletal formula

CH3OH

C2H5OH

C3H7OH

C4H9OH

C5H11OH

C6H13OH

C7H15OH

C8H17OH

C9H19OH

C10H21OH

Page | 72
CLASSIFICATION OF ALKANOLS

Alkanols can be classified as primary, secondary or tertiary. The classification is based on the
location of the hydroxyl group.

1. PRIMARY ALKANOL

It is an alkanol in which the hydroxyl group is attached to a carbon atom which is bonded to only
one other carbon atom. For example

2. SECODARY ALKANOL

It is an alkanol in which the hydroxyl group is bonded to a carbon atom which is bonded to two
other carbon atoms. Examples are shown below

3. TERTIARY ALKANOL

It is an alkanol in which the hydroxyl group is bonded to a carbon atom which is bonded to other
three carbon atoms. An example is shown below.

Page | 73
PREPARATION OF ETHANOL

Ethanol can be prepared using both indigenous methods and industrial technology.

1. INDIGENOUS METHOD OF PREPARING ETHANOL

The indigenous name for ethanol is kachaso. The process for preparing ethanol using indigenous
ways is as follows:

 Maize bran (madeya) is mixed with sugar solution or sugarcane or juices of fruits such as
mangoes or masuku.
 The mixture is kept for about 3 – 5 days for fermentation to take place.
 When fermentation is complete, the mixture is distilled in order to obtain ethanol using the
apparatus shown below.

 When the mixture is heated, ethanol boils faster than water because its boiling point is lower.

Page | 74
 The gaseous ethanol rises up in the pot and passes through the delivery tube.
 The cold water in the condenser cools and condenses the gaseous ethanol.
 Liquid ethanol is finally collected in the receiving bottle.
2. INDUSTRIAL METHODS OF PREPARARING ETHANOL

In industries, ethanol can be prepared by fermentation of sugars by yeast and by hydration of

ethene.

a. PREPARATION OF ETHANOL BY FERMENTATION OF SUGAR BY YEAST

Fermentation is a chemical process whereby sugars are converted to alcohol and carbon dioxide
by a catalyst.

 During fermentation of sugars, sugar (glucose) solution is mixed with yeast.

 The mixture is kept at room temperature for about 3 – 4 days in an apparatus shown below.

 Yeast contains an enzyme called zymase. This enzyme speeds up the decomposition of sugar
into ethanol and water.
 The lime water turns milk, indicating the production of carbon dioxide.

The equation for the reaction is:

yeast
glucose ethanol + carbon dioxide

Page | 75
 yeast
C6H12O6 (aq) 2C2H5OH9 (aq) + 2CO2 (g)

Ethanol produced by this process has of low alcohol content. To obtain a higher percent of
ethanol, the mixture separated by fractional distillation.

b. FORMATION OF ETHANOL BY HYDRATION OF ETHENE

Ethanol is prepared by hydrating ethene (reacting it with steam) in the presence of a catalyst such
as phosphoric acid (H3PO4).

H PO
C2H4 (g) + H2O (l) 3 4 C2H5OH (l)
The reaction has two main characteristics:

 It is exothermic in nature. Therefore, it carried out at a relatively low temperature of about

300ºC.
 It is reversible. Ethanol can be dehydrated back into ethene.
Conc. H SO
C2H5OH (l) 2 4 C2H4 (l) + H2O (l)

PHYSICAL PROPERTIES OF ALKANOLS

1. Alkanols are soluble in water. The solubility of alkanols decreases with an increase in
molecular mass. This is so because the proportion of the –OH group becomes smaller when
the size of the molecule gets bigger.
2. The melting and boiling points of alkanols increase as the molecular size increases. This is so
because, as the molecular size increases, the intermolecular forces also increase. Hence more
energy is needed to overcome.
Alkanols have higher melting and boiling points than hydrocarbons of the same number of
carbon atoms. This is because; alkanols form hydrogen bonds between their hydroxyl groups.
Hydrogen bonds are stronger than van dear Waals’s forces in hydrocarbons. More energy is
therefore required to break the hydrogen bonds in alkanols.
3. Density of alkanols increases as the molecular mass increases. This is because of the increase
in intermolecular forces making them to be very close to each other. This reduces the volume
occupied by the molecule hence increasing density.

Page | 76
4. The viscosity (resistance to flow) of alkanols increases with an increase in molecular size.
This is because the strength of the intermolecular forces increases, making the molecules to
stick together.
5. Alkanols are volatile liquids. Volatility of alkanols decreases as the molecular size increases.
This is because of the increases in the strength of intermolecular forces. Hence more heat is
needed to the molecules.

CHEMICAL PROPERTIES/REACTIONS OF ALKANOLS

1. Alkanols burn in oxygen to produce carbon dioxide and water. For example,

C2H5OH (l) + 3O2 (l) 2CO2 (g) + 3H2O (l)

2. Alkanols react with alkali metals to form alkanoxides and hydrogen gas. For example

Ethanol + Sodium Sodium ethoxide + hydrogen gas

2C2H5OH (l) + 2Na (s) 2C2H5ONa (aq) + H2 (g)

3. Alkanols can be oxidized to alkanoic acids. Oxidation is the addition of oxygen to a

substance. The oxidation of alkanols can be done in two ways:
a. Using oxidizing agents such as acidified potassium dichromate (VI) or acidified potassium
permanganate (VII)

acidified potassium dichromate (VI)

Ethanol + Oxygen Ethanoic acid

C2H5OH (l) + [O] K 2Cr2O7 (aq) 2CH3COOH (aq)

b. Oxidation by atmospheric oxygen. For example, if ethanol is exposed to the air can be
oxidized to and become Ethanoic acid. An example is wine turning sour if left open to the
atmosphere for too long is due to the presence of Ethanoic acid in the wine.
4. Alkanols can be dehydrated by concentrated sulphuric acid to produce alkenes. For example:

Conc. H SO
C2H5OH (l) 2 4 C2H4 (l) + H2O (l)

Concentrated sulphuric acid is use as the dehydrating agent.

Page | 77
5. Alkanols react with alkanoic acids to form alkanoates (esters) and water. The reaction of
alkanoic acids with alkanols to form esters is called esterification.

USES OF ETHANOL

1. Used as a solvent.
2. Used in the manufactures of varnishes, ink, glues, and paints.
3. Used as a fuel in cars. It can be blended with gasoline (petrol) or instead of petrol.
4. Used in alcoholic drinks e.g. wines, wines and spirits.
5. Used in the manufactures of deodorants, perfumes because it evaporates easily.
6. Used as an antiseptic in specified concentrations.
7. Moderate consumption of alcohol has numerous health benefits such as greater protection
heart diseases, decrease common infections and improvement in factors that influence blood
clotting.

DANGERS OF EXCESSIVE CONSUMPTION OF ALCOHOLIC DRINKS

1. Addiction (alcoholism) which can destroy physical and emotional heath.

2. Causes of some cancers such as liver cancer, mouth cancer, pharynx cancer, esophagus
cancer and pancreatic cancer.
3. Causes cardiac problems such as high blood pressure, heart failure and stroke.
4. Causes miscarriages in pregnant women.
5. Can influence a person to commit suicide or suffer serious injury.
6. Influences a person to indulge in risky sexual behaviours leading to unwanted pregnancies,
HIV and AIDS, and other sexually transmitted infections (STIs).
7. Loss of employment leading to poverty.
8. Causes liver cirrhosis.

Page | 78
TOPIC 7 : ALKANALS AND ALKANONOES
Alkanals and alkanones are two separate homologous series that belong to the oxycarbons class
of organic compounds. Alkanals are also called aldehydes while alkanones are also called
ketones.

FUNCTIONAL GROUP OF ALKANALS AND ALKANOES

The functional group of alkanals and alkanones is the same and it is called the carbonyl group,
C=O.

NOMECLATURE OF ALKANALS

The name of the alkanal is derived from the parent alkane with the same number of carbon
atoms. The –e ending from the parent alkane is replaced by –al.

Number of carbon atoms Name of alkanal

1 Methanal
2 Ethanal
3 Propanal
4 Butanal
5 Pentanal

GENERAL FORMULA OF ALKANALS

The general formula of alkanals is Cn H 2n+1CHO . This formula can be used to work the formula
of any alkanal given the value of n.

Example: Work out the formula of the alkanal given the following values of n.

a. n = 0

Page | 79
b. n = 1
c. n = 4

Solution:

a. If n = 0, H (2×0)+1CHO = HCHO

b. If n = 1, C1H (2×1) CHO = CH3CHO

c. If n = 4, C4 H (2×4)+1CHO = C4H9CHO

MOLECULAR AND CONDENSED FORMULAE OF ALKANALS

The table below shows the molecular and condensed formulae of the first five alkanals.

Name of alkanal Molecular formula

Methanal HCHO
Ethanal CH3CHO
Propanal C2H5CHO
Butanal C3H7CHO
Pentanal C4H9CHO

STRUCTURAL FORMULA OF ALKANALS

In alkanals, at least one hydrogen atom is bonded to the carbonyl group. The other group is an
alkyl group. The general structure of alkanals is shown below.

Where R represents an alkyl group such as CH3  , CH3CH 2  , CH3CH 2CH 2  etc, except in
methanal.

The structures of the first five alkanals are shown in the table below.

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 Name of alkanal Structure of alkanal

Methanal

Ethanal

Propanal

Butanal

Pentanal

NOMENCLATURE OF ALKANONES

Alkanones take their name from the parent alkane with similar number of carbon atoms. The
ending –e is removed and replaced with –one. The smallest alkanone is propanone.

Number of carbon atoms Name of alkanone

3 Propanone
4 Butanone
5 Pentanone
6 Hexanone
7 Heptanone

GENERAL FORMULA OF ALKANONES

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The general formula of alkanones is Cn H 2n O . The formula can be used to work out the
molecular formula of an alkanone given the value of n. The molecular formulae of the first five
alkanones are shown in the table below.

Name of alkanone Molecular formula

Propanone C3 H 6 O

Butanone C4 H 8O

Pentanone C5 H10O

Hexanone C6 H12 O

Heptanone C7 H14 O

STRUCTURES OF ALKANONES

In alkanones, the carbonyl group is bonded to two alkyl groups which may be similar or
different.

Where R and R' represent the alkyl groups.

Name of alkanone Structure of alkanone

Propanone

Butanone

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 Pentan-2-one or
2-Pentanone

Hexan-2-one or
2-Hexanone

Heptan-2-one or
2-Heptanone

SOURCES OF ALKANALS AND ALKANONES

Oxidation of alkanols is the primary source of alkanals and alkanones. The oxidizing agents
include acidified potassium permanganate (VII) or acidified potassium dichromate (VI)

a. OXIDATION OF PRIMARY ALKANOLS

Alkanals are produced by oxidation of primary alkanols. For example:

b. OXIDATION OF SECONDARY ALKANOLS

Alkanones are produced by the oxidation of secondary alkanols. For example:

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PHYSICAL PROPERTIES OF ALKANALS

 All alkanals are liquids except methanal which is a gas.

 They are colourless and they have a characteristic smell.
 The solubility of alkanals in water decreases moving along the homologous series. Smaller
alkanals are more soluble than larger alkanals.
 The melting and boiling points of alkanals increases with an increase in relative molecular
mass.

CHEMICAL PROPERTIES OF ALKANALS

 They burn in air to form carbon dioxide and water.

e.g. 2CH 3CHO (l) + 5O 2 (g) 4CO 2 (g) + 4H 2O (l)

 They are easily oxidized to alkanoic acids.

 They can be reduced to primary alkanols. Reduction is the addition of hydrogen to a
substance.
 They are neutral to litmus paper.
 They undergo addition polymerizations.

PHYSICAL PROPERTIES OF ALKANONES

 All alkanones are liquids.

 They are polar because of the carbonyl group.
 Their solubility in water decreases moving along the homologous series.
 Their melting and boiling points increase with increasing molecular mass.

CHEMICAL PROPOERTIES OF ALKANONES

 They burn in air to form carbon dioxide and water.

 They are not easily oxidized due to the absence of hydrogen on the carbonyl group.
 They are neutral to litmus paper.
 They can be reduced to secondary alkanols.

DISTINGUSIHING ALKANALS AND ALKANONES USING CHEMICAL TESTS

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Three main tests are carried out to help identify whether a given unknown substance is an alkanal
or alkanone. The tests are the Fehling’s test, the Brady’s test and the Tollen’s test (silver mirror
test).

a. The Fehling’s test

Fehling’s test is used to confirm the fact that alkanals are easily oxidized to alkanoic acids.

Procedure

To 15 drops of Fehling’s solution in a clean test tube, add about 15 drops of the test liquid. Heat
the contents of the tube gently for about 5 minutes.

Result

 If the blue Fehling’s solution turns to a red precipitate, then the test substance is either an
alkanal or a sugar.
 If the blue colour of Fehling’s solution does not disappear, then the test liquid is an alkanone.
b. The Brady’s test

The Brady’s test is used to detect the presence of the carbonyl group, –C=O, in a compound. The
test uses a reagent called 2,4 Dinitrophenylhydrazine (2,4 –DNPH), which is yellow in colour.

Procedure

 Add 2 cm3 of the test liquid to 2 cm3 of 2,4 DNPH reagent. Shake well.

Result

If an orange precipitate forms, then the added liquid could be either be an alkanal or an alkanone.
This is so because both alkanals and alkanones contain the carbonyl group as their functional
group.

c. The Tollen’s test

The Tollen’s reagent is a mixture of excess ammonia solution and silver nitrate solution. The
Tollen’s reagent oxidizes an alkanal to alkanoic acid while the silver ions present the reagent are
reduced to silver atoms, forming a silver mirror on the inside of the test tube.

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Procedure

 To 3 cm3 of Tollen’s reagent, add 2-3 drops of the test liquid.

Result

 If a silver mirror forms inside the test tube, the unknown liquid is an alkanal.
 If a silver mirror does not form, then the test liquid is an alkanone.

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TOPIC 8 : ALKANOIC ACIDS
Alkanoic acids are also called carboxylic acids.

FUNCTIONAL GROUP OF ALKANOIC ACIDS

The functional group of alkanoic acids is the carboxyl group and it is represented as –COOH.

NOMENCLATURE OF ALKANOIC ACIDS

To name an alkanoic acid, the –e from the parent alkane is replaced by –oic acid. The table
below shows the names of the first ten alkanoic acids.

Number of carbon atoms Name of alkanoic acid

1 Methanoic acid
2 Ethanoic acid
3 Propanoic acid
4 Butanoic acid
5 Pentanoic acid
6 Hexanoic acid
7 Heptanoic acid
8 Octanoic acid
9 Nonaoic acid
10 Decanoic acid

GENERAL FORMULA OF ALKANOIC ACIDS

The general formula of alkanoic acids is CnH2n+1COOH.

Example

What is the molecular formula of an alkanoic acid when n = 6?

Solution: If n = 6 then C6 H (2×6)+1COOH = C6 H13COOH .

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MOLECULAR AND CONDENSED FORMULAE OF ALKANOIC ACIDS

The table below shows the molecular and condensed formulae of the first ten alkanoic acids.

Name of alkanoic acid Molecular formula Condensed formula

Methanoic acid HCOOH HCOOH
Ethanoic acid CH3COOH CH3COOH

Propanoic acid C2 H5COOH CH3CH 2 COOH

Butanoic acid C3H 7 COOH CH3 (CH 2 ) 2COOH

Pentanoic acid C4 H9 COOH CH3 (CH 2 )3COOH

Hexanoic acid C5 H11COOH CH3 (CH 2 ) 4COOH

Heptanoic acid C6 H13COOH CH3 (CH 2 )5COOH

Octanoic acid C7 H15COOH CH3 (CH 2 )6COOH

Nonaoic acid C8 H17 COOH CH3 (CH 2 )7 COOH

Decanoic acid C9 H19COOH CH3 (CH 2 )8COOH

STRUCTURAL FORMULAE OF ALKANOIC ACIDS

The table shows the structures of the first ten alkanoic acids.

Molecular formula Structural formula

HCOOH

CH3COOH

C2 H5COOH

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 C3H 7 COOH

C4 H9 COOH

C5 H11COOH

C6 H13COOH

C7 H15COOH

C8 H19COOH

C9 H19 COOH

SKELETAL FORMULAE OF ALKANOIC ACIDS

Name of alkanoic Condensed Skeletal formula

acid formula
Methanoic acid HCOOH HCOOH

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 Ethanoic acid CH3COOH

Propanoic acid CH3CH 2 COOH

Butanoic acid CH3 (CH 2 ) 2COOH

Pentanoic acid CH3 (CH 2 )3COOH

Hexanoic acid CH3 (CH 2 ) 4COOH

Heptanoic acid CH3 (CH 2 )5COOH

Octanoic acid CH3 (CH 2 )6COOH

Nonaoic acid CH3 (CH 2 )7 COOH

Decanoic acid CH3 (CH 2 )8COOH

NATURAL SOURCES OF ALKANOIC ACIDS

Some of the natural sources of alkanoic acids are:

 Citrus fruits (oranges, lemons) e.g. citric acid.

 Sour milk e.g. lactic acid.
 Vinegar e.g. acetic acid/ethanoic acid.
 Ant, bee and nettle stings e.g. methanoic acid.
 Human sweat e.g. butanoic acid.

PREPRATION OF ALKANOIC ACIDS

Alkanoic acids can be prepared by the oxidation of corresponding alkanols. The oxidising agents
are acidified potassium permanganate (VII) or acidified potassium dichromate (VI).

The oxidation is done in two steps:

a. First, the alkanol is oxidized to an alkanal. For example

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b. Then, the alkanal is further oxidized to an alkanoic acid.

PHYSICAL PROPERTIES OF ALKANOIC ACIDS

a. They are liquids at room temperature.

b. Smaller alkanoic acids are soluble in water; the larger ones are insoluble in water. Solubility
of alkanoic acids in water decrease moving along the series due to an increasing proportion
of the hydrocarbon part of the molecule.
c. The melting points and boiling points of alkanoic acids increase with the increase in
molecular size. There are two reasons for this:
 As the size of the molecule gets bigger, the van der Waal’s forces are increased.
 There is formation of hydrogen bonds between alkanoic acid molecules because the –OH
group of the carboxyl group is more polarized due to the presence of an electron
withdrawing group, –C=O.
d. Viscosity of alkanoic acids increases with an increase in molecular mass. This is due to
increasing stronger and more hydrogen bonds in the acid molecules.

CHEMICAL PROPERTIES OF ALKANOIC ACIDS

 They react with metals to form salts and hydrogen.

Example:
Magnesium + Ethanoic acid Magnesium ethanoate + Hydrogen

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 Mg (s) + 2CH 3COOH (aq) (CH3COO) 2 Mg (aq) + H2

 They react with alkalis to produce salt and water. This reaction is called neutralization
reaction.
Example:
Sodium hydroxide + Ethanoic acid Sodium ethanoate + Water
NaOH (aq) + CH3COOH (aq) CH3COONa (aq) + H 2O (l)

 They react with carbonates to form salts, carbon dioxide and water.
Example:
Calcium + Ethanoic acid Calcium + Carbon + Water
carbonate ethanoate dioxide
CaCO3 (s) + CH3COOH (aq) (CH3COO) 2Ca (aq) + CO 2 (g) + H 2O (l)

 They react with alkanols to produce alkanoates (esters). This reaction is called esterification.
Example:

EFECT OF ALKANOIC ACIDS ON ACID–BASE INDICATORS

Alkanoic acids show acidic behaviours by affecting acid–base indicators:

 They turn blue litmus paper red.

 They have pH values of less than 7.
 They are colourless in phenolphthalein.

ELECTRICAL CONDUCTIVITY OF ALKANOIC ACIDS

Alkanoic acids conduct electricity when in aqueous state. This is because, when dissolved in
water, they ionize. The ions enable the alkanoic acid to conduct electricity.

For example, the ionization of ethanoic acid is shown according to the equation:

CH3COOH (l) + H 2O (l)  CH3COO  (aq) + H 3O + (aq)

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USES OF ALKANOIC ACIDS

 Used as food preservatives e.g. ethanoic acid

 Used as solvents e.g. ethanoic acid
 Used in the production of esters.
 Used in manufacture of medicines such as aspirin.

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TOPIC 9 : ALKANOATES
Alkanoates are organic compounds which are formed when alkanoic acids and alkanols react in
the presence of a catalyst. They are also called esters.

FUNCTIONAL GROUP OF ALKANOATES

The functional group of alkanoates is given as –COO–

NOMENCLATURE OF ALKANOATES

The name of an alkanoate has two parts: the acid part and the alkanol part.

 The alkyl part of the alkanol comes first.

 The acid part comes second but the –oic ending is replaced by –oate.

Example

Work out the names of alkanoates formed when the following substances react:

a. Ethanoic acid and methanol

b. Butanoic acid and ethanol

Solution:

a. The alkyl part of methanol is “methyl”. The acid part ethanoic changes to ethanoate.
The name of the alkanoate is methyl ethanoate.
b. The alky part of ethanol is “ethyl”. The acid part butanoic changes to butanoate.
The name of the alkanoate is ethyl butanoate

DRAWING STRUCTURES OF ALKANOATES FORMED FROM GIVEN ALKANOIC

ACID AND ALKANOL

To draw structures of alkanoates formed from given alkanoic acid and alkanol, use the following
simple steps.

 Write down the structures of the alkanoic acid and the alkanol separately.

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 Take away the ending –H from the acid and the –OH from the alkanol. The result is an
alkanoate and an alky part respectively.
 Combine the two structures, starting with the acid part and then the alkanol part.

Example:

Draw and name the structure of the alkanoate that will be formed when:

a. Methanol reacts with methanoic acid.

b. Propanol reacts with Propanoic acid.

Solution:

a. The structures of methanol and methanoic acid are:

Taking away the ending –H from methanoic acid and –OH from methanol we get:

Combining the methanoate structure and the methyl structure, we get:

b. The structures of propanol and propanoic acid are:

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The structure of the alkanoate formed is shown as:

DEDUCING THE REACTANTS OF ESTERIFICATION GIVEN THE STRUCTURE OF

THE PRODUCT OF THE REACTION

Example

Deduce the names of the reactants of esterification given each of the following structures of
products:

Solution:

a. The alkanoate part contains 4 carbon atoms. The reacting alkanoic acid was butanoic acid.
The alkyl part contains 3 carbon atoms. This means the reacting alkanol was propanol. The
reactants of this reaction were: butanoic acid and propanol
b. The alkanoate apart has 2 carbon atoms. The reacting acid was ethanoic acid. The alkyl part
contains 4 carbon atoms. The reacting alkanol was butanol.
Hence the reactants of this reaction were: ethanoic acid and butanol.

SOURCES OF ALKANOATES

1. Natural sources which include:

 Fruits and flowers
 Fats and oils
2. Alkanoates can be prepared synthetically by condensation reaction of alkanoic acids.

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PROPERTIES OF ALKANOATES

a. They have pleasant fruits smells.

For example
Alkanoate Fruit smell
Ethyl butanoate Pineapple
Pentyl ethanoate Banana

b. They are volatile liquids.

c. Small chain alkanoates are fairly soluble in water. Their solubility decreases with increasing
chain length.
d. They have lower melting and boiling points than the alkanoic acids and alkanols of similar
molecular mass. This is because alkanoates have no free –OH groups so they do no form
hydrogen bonds.
e. They react with sodium hydroxide in the presence of water (i.e. they are hydrolysed) to
produce soap. This reaction is called saponification.

To make soap, a fat (natural ester) is boiled with a strong alkaline solution such as sodium
hydroxide (NaOH). It is hydrolysed to give glycerol and soap (sodium octadecanoate).

The equation for the reaction is:

USES OF ALKANOATES

 Used for flavouring e.g. flavours in sweets ad biscuits.

 Used as fragrances in perfumes.
 Used as solvents in glues since they evaporate easily.

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TOPIC 10 : IDENTIFICATION OF UNKNOWN COMPOUNDS

1. DEDUCING FAMILIES AND FORMULAE OF UNKNOWN ORGANIC

COMPOUNDS

The families and formulae of unknown organic compound can be deduced given relevant
information such as structural formulae, general formulae, and products of chemical reactions,
physical and chemical properties.

Example

An organic compound was found to contain 64.8% carbon, 13.6% hydrogen and 21.6% oxygen
by mass.

a. Calculate the empirical formula of the compound.

b. Deduce the family to which the compound could belong.

(RAM of C = 12, H = 1 and O = 16)

Solution

a. The empirical formula of the compound is calculated as follows:

Element Mass (g) Number of moles Simplest mole ratio

Carbon (C) 64.8 64.8 5.4
= 5.4 =4
12 1.35
Hydrogen (H) 13.6 13.6 13.6
= 13.6 = 10
1 1.35
Oxygen (O) 21.6 21.6 1.35
=1.35 =1
16 1.35

The empirical formula is C4 H10 O

b. The empirical formula can be worked out exactly using the general formula of alkanols,
which is Cn H 2n+1OH . For example C4 H10 O can be written as C4 H 9OH .

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Therefore the compound could be an alkanol.

2. DISTINGUISHING ORGANIC COMPOUNDS BASING ON PROPERTIES

To distinguish organic compounds basing on properties, a flow diagram is used. The flow
diagram is uses chemical tests in order to separated unknown organic compounds. The tests
derive from the physical and chemical properties specific to a given homologous series. Some of
the tests are the solubility test, the bromine test, the acid test, the Fehling’s test (copper mirror
test), the Brady’s test and the Tollen’s test (silver mirror test

IDENTIFICATION TEST TABLE

CHEMICAL TEST
RESULT Solubility Bromine NaOH + 2,4 DNPH Fehling’s Tollen’s
in water solution phenolphthal solution solution
ein
Alkanes Insoluble Red/brown Pink No colour No colour No colour
change change change
Alkenes Insoluble Colourless Pink No colour No colour No colour
change change change
Alkanols Soluble Red/brown Pink No colour No colour No colour
change change change
Alkanoic Soluble Red/brown Colourless No colour No colour No colour
acids change change change
Alkanals Soluble No colour No colour An orange Red A silver
change change precipitate precipitate mirror
forms forms forms
Alkanones Soluble No colour No colour An orange No colour No silver
change change precipitate change mirror
forms formed

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Example

Construct a flow diagram that could be used to identify acetic acid, ethanol, hexene and

hexane, using tests that make use of distilled water, bromine solution, sodium

hydroxide solution and phenolphthalein indicator.

Solution

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