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TABLE OF CONTENTS
Unit1: Chemical bonding
Unit2: Trends in properties of elements in the periodic table
Unit3: Water pollution
Unit4: Effective ways of waste management
Unit5: Categories of chemical reactions
Unit6: Preparation of salts and identification of ions
Unit7: The mole concept and gas laws
Unit8: Preparation and classification of oxides
Unit9: Electrolysis and non-electrolytes
Unit10: Properties of organic compounds and uses of
alkanes
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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UNIT1: CHEMICAL BONDING
1. Stability of atoms
An atom is most stable when its outermost energy level is completely filled with electrons
(i.e. 2electrons for helium or 8 electrons for others).
Therefore, atoms of noble gases are stable because their outermost energy levels are
completely filled up with electrons. They include helium (He), Neon (Ne), Argon (Ar),
Krypton (Kr), Xenon (Xe) and Radon (Rn).
Noble gas
Symbol
Atomic
number
Electronic
configuration
Number of
electrons in
outermost
shell
Helium
He
2
2
2
Neon
Ne
10
2,8
8
Argon
Ne
18
2,8,8
8
Krypton
Kr
36
2,8,18,8
8
Xenon
Xe
54
2,8,18,18,8
8
Radon
Rn
86
2,8,18,32,18,8 8
2. Instability of atoms
Atoms whose outermost shells are not filled with either 2 or 8 electrons are instable.
Therefore, atoms of these element, will lose, gain or share electrons in order to become
stable like noble gases.
 Metals lose electrons from their outermost shells to become stable.
 Non-metals gain or share electrons to become stable atoms.
Example
The following diagrams represent electron models of certain elements.
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i) Give the names of elements a, b, c, d, and e.
ii) Comment about the number of electrons in the outermost energy level in each of the
elements.
iii) State whether each element is stable or unstable.
Answer
Name of element
Number of electrons
Stability
a) Hydrogen
1
Unstable
b) Helium
2
Stable
c) Oxygen
6
Unstable
d) Lithium
1
Unstable
e) Neon
8
Stable
3. Formation of ions from atoms
 Atoms are electrically neutral (because the number of protons in the nucleus is equal to
the number of electrons in the shell). This means that atom has no charge.
 When atoms lose or gain electrons, they form ions.
 An ion is a positively or negatively charged atom.
 Atoms of metal elements lose electrons from their outermost shells to form positively
charged ions (Cations).
 Atoms of non-metal elements gain electrons to form negatively charged ions (Anions).
i)
Formation of cations
Atoms of metal elements lose electrons from their outermost shells to form positively
charged ions (Cations).
Examples
 Formation of sodium ion
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 Formation ion calcium ion
 Formation of aluminium ion
ii)
Formation of anions
Atoms of non-metal elements gain electrons to form negatively charged ions (Anions).
 Formation of fluoride ion
 Formation of nitride ion
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4. Chemical bonding
A chemical bond is the force that holds atoms (or ions) together.
Types of chemical bonds
There are three types of chemical bonds:
 Ionic bonds
 Covalent bonds
 Metallic bonds
1. Ionic bonding
An ionic bond is the force that holds the ions together in an ionic compound.
i.
Formation of ionic bond
 Ionic bond is formed between metals and non-metals atoms.
 It is formed by transfer of electrons from metal atoms to non-metal atoms. Metal
atoms lose their outermost electron(s), forming cations and non-metal atoms gain
electron(s) to fill their outermost shell, forming anions.
 The electrostatic force of attraction between the oppositely charged ions that holds the
ions together is called ionic bond.
Examples:
When a hot sodium atom is placed in chlorine gas, a reaction takes place resulting in
formation of sodium chloride.
1. Formation of sodium chloride (using diagram)
The resulting compound is called sodium chloride, NaCl.
A dot ( • ) and a cross (
ionic compound.
× ) diagram is used to represent ionic bonding that results into
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2. Formation of magnesium fluoride
 The magnesium ion and fluoride ions are attracted to one another by ionic bonds.
 The resulting compound is called magnesium fluoride, MgF2.
3. Formation of magnesium oxide (MgO)
The table below shows some of the common ionic compounds, their formulae and ions
present in them.
Names of ionic compounds Formula
Ion present
Aluminium oxide
Al2O3
Al3+ and O2−
Ammonium chloride
NH4Cl
NH4+ and Cl−
Calcium hydroxide
Ca(OH)2
Ca2+ and OH−
Calcium nitrate
Ca(NO3)2
Ca2+ and NO3−
Calcium oxide
CaO
Ca2+ and O2−
Magnesium chloride
MgCl2
Mg2+ and Cl−
Potassium chloride
KCl
K+ and Cl−
Sodium hydroxide
NaOH
Na+ and OH−
Sodium carbonate
Na2CO3
Na+ and CO3−
Cu2+ and SO42−
CuSO4
Copper sulphate
ii.




Physical properties of ionic compounds
Ionic compounds have high melting points and boiling point.
They have high densities.
Most are soluble in water.
Ionic compounds conduct electricity when molten (liquid) or in aqueous solution
(dissolved in water), because their ions are free to move from place to place.
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 Ionic compounds cannot conduct electricity when solid, as their ions are held in
fixed positions and cannot move.
 They are crystalline solids at room temperature.
Note:
 Melting point is the temperature at which a solid turn into liquid.
 Boiling point is the temperature at which a liquid changes into gas.
 The temperature of a solid and a liquid remains the same once melting and boiling has
been started.
Exercises
1) a) What is a chemical bonding?
b) Define an ionic bonding.
c) How ionic bonding is formed?
2) Fill in the missing words in the following sentences.
a) Ionic bonding is a type of attraction between …………….and ........................charged
ions. It is formed when there is a complete transfer of electrons from atoms of a
………………to atoms of a……………………..
b) When atoms lose electrons they form ……………..charged ion called……………..
The lost electrons are gained by other atoms which become...................... charged ions
and are known as ……………………..
3) Choose the best answer.
Ionic bonding usually occurs between what type of atoms?
i) Metal and metal
ii) Non-metal and non-metal
iii) Metal and non-metal.
4) Draw dot and cross diagrams to show formation of ionic bonding in the following
compounds and derive the chemical formulae of the compounds formed.
a) Magnesium oxide
b) Calcium oxide
c) Sodium chloride
d) Sodium sulphide
e) Magnesium chloride
f) Aluminium oxide
g) Sodium oxide
h) Magnesium nitride
5) a) Give five examples of ionic compounds.
b) Explain five properties of ionic compounds.
c) Why do ionic compounds conduct electricity when dissolved in water?
e) Explain the following physical properties of ionic compounds.
i) Ionic compounds conduct electricity in molten and in aqueous form but not in
solid.
ii) Ionic compounds have high melting and boiling points.
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6) Sodium chloride is an ionic compound. It is formed when sodium reacts with chlorine.
The atomic number of sodium and chlorine is 11 and 17 respectively.
a) Draw and label dot and cross diagrams to show the arrangement of the electrons in
the atoms of sodium and chlorine.
b) Draw and label dot and cross diagrams to show the arrangement of the electrons in
the ions formed when sodium reacts with chlorine.
c) Give the symbols of sodium ion and chloride ion formed.
d) Explain why solid sodium chloride does not conduct electricity but when molten it
does conduct electricity.
2. Covalent bonding
A covalent bond is formed by sharing of electrons between atoms.
i.
Formation of covalent bond
 It is formed by sharing of electrons between atoms of non-metal only.
The sharing of electrons between atoms is called a covalent bond, and the electrons that
join atoms in a covalent bond are called bonding pair of electrons. A discrete group of
atoms connected by covalent bond is called a molecule.




Each atom contributes one electrons to the pair that is being shared.
When electrons are shared in this way, molecules are formed, not ions.
The compounds containing covalent bonds are known as covalent compounds.
Covalent bonds are found in:
 Non-metal elements such as Oxygen, Hydrogen, Fluorine, Nitrogen, Chlorine,
Bromine, Carbon, Phosphorus and Sulphur. (Similar elements)
 Compounds made of two or more different non-metal elements such as ammonia
(NH3), Water (H2O), Methane (CH4), Carbon dioxide (CO2), etc.
 A dot (•) and a cross (×) diagram are used to represent covalent bonding that results
into covalent compound.
a) Covalent bonds in elements
One covalent bond, two covalent bonds, three covalent bonds and four covalent bonds.
i)
One covalent bond is formed when one pair of electrons is shared between
atoms.
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Examples:
 Formation of chlorine molecule
ii)
Two covalent bonds are formed when two pair of electrons are shared and Three
covalent bonds are formed when three pair of electrons are shared.
Example:
 Formation of oxygen molecule and nitrogen molecule
b)Covalent bonds in compounds
i.
ii.
Formation of water molecule (H2O)
Formation of ammonia molecule (NH3)
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iii.
Formation of carbon dioxide molecule (CO2)
The table below shows some covalent compounds and their formulae and names.
Names of covalent
Formula of
Elements
compounds
covalent
present
compounds
Methane
Ammonia
Hydrogen sulphide
Carbon dioxide
Carbon tetrachloride
Ethane
Water
Hydrogen chloride
ii.





CH4
NH3
H 2S
CO2
CCl4
C2H6
H 2O
HCl
C and H
N and H
H and S
C and O
C and Cl
C and H
H and O
H and Cl
Properties of covalent compounds
Covalent compounds have low melting point and boiling point.
Covalent compounds do not conduct electricity because they do not contain ions.
They are insoluble in water but soluble in organic solvent like ethanol.
They have low densities.
They are usually liquids, gases or solids. For example, alcohol, water, cooking oil are
liquids. Methane, ethane and chlorine are gases. Iodine, Silicon dioxide, Diamond and
graphite are solid.
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iii.
Differences between ionic compounds and covalent
compounds
Ionic compounds
 Ionic compounds are crystalline
solids
 Ionic compounds have high melting
point and boiling point
 Ionic compound conduct electricity
when dissolved in water or melted
 Ionic compounds are usually soluble
in water
Covalent compounds
 Covalent compounds are usually
solids, liquids and gases
 Covalent compounds have usually
low melting point and boiling point
 Most covalent compounds do not
conduct electricity
 Covalent compounds are usually
insoluble in water (except sugar,
glucose)
Exercises
1) What do you mean by a covalent bond?
2) Choose the correct answer.
a) What does a covalent bond involve?
i) Sharing electrons between atoms
ii) Moving electrons between atoms
iii) Forming free electrons.
b) How many electrons are involved in each covalent bond?
i) One
ii) Two
iii) Three
3) I) Draw dot and cross diagrams to show the bonding in:
a) Fluorine molecule (F2)
b) Hydrogen chloride molecule (HCl)
c) Methane molecule (CH4)
d) Chlorine molecule (Cl2)
e) Ammonia molecule (NH3)
f) Water molecule (H2O)
g) Oxygen molecule (O2)
h) Hydrogen sulphide (H2S)
i) Nitrogen molecule (N2)
j) Carbon dioxide molecule (CO2)
II) How many covalent bonds are there in each molecule above?
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4) Study the following atomic models
a) Name the types of covalent bonds shown in structures (a) and (b).
b) Give the name and the chemical formula of molecules (a) and (b).
5) a) Name five examples of covalent compounds. Also write their chemical formulae.
b) Explain the properties of covalent compounds.
c) Why covalent compounds do not conduct electricity when dissolved in water.
6) a) Distinguish between covalent bond and ionic bond.
b) Explain the differences between ionic compounds and covalent compounds.
5. Giant covalent structures
Structure refer to the arrangement of the bonded atoms, ions or molecules.
A giant covalent structure has a network of covalently bonded atoms.
Example: Graphite and Diamond have giant covalent structure
a) Graphite
i) Bonding in graphite
Each carbon atom is covalently bonded to three other carbon atoms hexagonally arranged
in flat parallel layers.
ii) General properties of graphite
 Graphite is insoluble in water.
 Graphite conducts electricity because there are free electrons available to carry
electric charges.
 It has high melting point (35000C)
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 It has a low density
 It is black in color.
 It is soft and greasy.
iii)
Uses of graphite
 Graphite is used as an electrode in dry cells and during electrolysis. Because it is a
good conductor of electricity.
 It is used as a lubricant. Because it is slippery.
 Graphite is used in the manufacture of “lead” in pencils. Because it is soft.
Note:
“Lead” is the black or grey part of the pencil used to write.
b) Diamond
i)
Bonding in diamond
In diamond, every carbon atom is covalently bonded to four other carbon atoms
tetrahedrally arranged.
ii)
General properties of diamond
 Diamond is insoluble in water.
 Diamond does not conduct electricity because there are no free electrons available to
carry electric charges.
 It has high melting point (37000C)
 It has a high density
 It is colorless, transparent and dazzling (sparkles), hence beautiful.
 It is the hardest naturally occurring substance.
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iii)
Uses of diamond
 Diamond is used in making jewellery. Because it has a sparkling (shining)
surface.
 Diamond is used in the manufacture of glass cutters. Because it is very hard.
 It is used for making drilling bits. Because it is very hard.
 It is used to cut metals. Because it is very hard.
 Making different objects like watches (isaha), ornaments (imitako), etc.
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Exercises
1) a) What structure?
b) Name two substances which have giant covalent structure.
2) a) Draw the structure of graphite and explain the arrangement of atoms.
b) Explain the physical properties of graphite.
c) Write three uses of graphite.
3) a) Draw the structure of diamond and explain the arrangement of atoms.
b) Explain the physical properties of diamond.
c) Write three uses of diamond.
4) The following diagrams show the structure of two forms of carbon (graphite and
diamond). Study them and answer the questions that follow:
a) State what is labelled as I and II in graphite.
b) Give one use of graphite that depends on its slippery nature.
c) Explain how bonding in graphite makes it conduct electricity.
d) State a use of graphite that depends on its electrical conductivity.
e) State two properties and two uses of diamond.
5) The following table gives information about six substances (the letters do not represent
the actual chemical symbols)
Substance Melting point Boiling point Electrical
Electrical
0C
0C
conductivity as conductivity as
a solid
a liquid
A
801
1413
Poor
Good
B
-210
-196
Poor
Poor
C
776
1497
Good
Good
D
-117
78
Poor
Poor
E
1607
2227
Poor
Poor
F
-5
102
Poor
Poor
a)
b)
c)
d)
e)
f)
Which substance has a giant covalent structure?
Which substances are solids at room temperature?
Which substance could be sodium chloride?
Which substance is an ionic compound?
Which substances are liquids at room temperature?
Which substance is most suitable for making of a cooking pot?
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3. Metallic bonding
Metallic bonding is the force of attraction between valence electrons and the metal ions.
The force which binds various metal atoms together is called metallic bond.
In the metal, each metal atom gives up its electron(s) in the outermost energy level to form
a sea of delocalized electrons, which move about freely in the metal structure. This partial
removal of the electron(s) from metal atom creates a positive metal ion or positive core that
is the rest of the atom excluding the outermost energy level electrons.
Consequently, an electrostatic force of attraction develops between the positive metal ion
and the sea of electrons and this is the metallic bond.
Note:
The more electrons that are given up to the sea of electrons, the stronger the metallic bond.
Example:
 The metallic bond in magnesium is stronger than that in sodium metal because
magnesium gives two electrons to the sea of delocalized electrons than sodium which
gives only one electron.
 Aluminium forms stronger metallic bond than magnesium metal.
a. Bonding in metals
Example
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b. Physical properties of metals
 Since electrons are able to move freely, they can easily transfer (conduct) heat and
electricity through the metal. The reason why Metals are good conductors of
electricity and heat.
Note: Conduction of electricity
A substance can conduct electricity if:
 It contains charged particles (electrons or ions) and
 These particles are free to move from place to place.
 In ionic compounds, the particles that carry electric current are called ions while in the
wires, the particles that carry electric current are called electrons.
Examples
Silver metal is the best conductor of electricity; copper is the next best conductor. Gold,
Aluminium and Tungsten are also good conductor of electricity.
 Metals are solids at room temperature except Mercury (Hg) which is liquid at
room temperature.
 Metals have high density and are very heavy.
 Metals are malleable, means that they can be beaten (hammered) into very thin
sheets. This property of metals is called malleability. Gold and Silver are the most
malleable metals.
 Metals are ductile, means that they can be drawn into thin wires. Gold and Silver are
the ductile metals.
 Metals have high melting points and boiling points.
 Metals are sonorous, means that they can make sound when hit.
 Metals are lustrous, means that they have a shining surface.
 Metals are hard except sodium and potassium which are soft and can be cut with a
knife.
c. Uses of metals
 Metals are used for manufacturing of building equipments like doors, windows and
roofing sheets.
 They are used in making saucepans, dishes, spoons, knives, forks, etc.
 Metals like copper and aluminium are used for making electric wires.
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Exercises
1)
2)
3)
4)
5)
Describe the nature of metallic bonding.
Explain the properties of metals.
Explain why metals are good conductor of electricity?
State two general uses of metals.
Explain the terms:
i) Malleable
ii) Ductile
6) State a use of metals based on:
i) Malleability
ii) Ductility
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UNIT 2: TRENDS IN PROPERTIES OF ELEMENTS IN
THE PERIODIC TABLE
1. Classification of elements into metals, metalloids
and non-metals
Elements can be grouped broadly as Metals, Non-metals or Metalloids.
Metals
 Atoms of metal elements have 1, 2 or 3 electrons in their outermost energy levels.
 Metal elements are in groups Ia, IIa, and IIIa of the periodic table.
 Group number: is indicated by number of electrons in the outermost shell of an
atom.
 Period number: is indicated by the number of electrons shells in an atom.
Examples of metal elements are:
Lithium, Sodium, Magnesium and Aluminium, Potassium, etc.
Apart from groups Ia, IIa, and IIIa elements, most metals are transition elements.
Examples of transition metal elements are:
Silver(Ag), Iron (Fe), Copper(Cu) and Chromium(Cr).
Non-metals
 Non-metal elements have 4, 5, 6,7 and 8 electrons in their outermost energy level.
 They are found in groups IVa, Va, VIa, VIIa and VIIIa.
 Non-metals are found in blue color.
Examples of non-metal elements are:
Oxygen (O), Nitrogen (N), Sulphur (S) and Chlorine (Cl), etc.
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Metalloids
Metalloids are elements which have both characteristics of metals and non-metals.
They are found in between metals and non-metals in the periodic table.
Metalloids elements are:
Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te) and
Polonium(Po).
The following periodic table indicates the groups and periods in which each
element belongs.
 Vertical columns of elements are called groups. Elements in the same group have the
same number of electrons in the outermost shell. This number of electrons which are in
the outermost shell is the same as group number.
 Horizontal rows of elements are called periods. Elements in the same period have the
same number of shells (energy level). This number of shells is the same as the number
of period.
a) Elements “A” are called representative elements or main group elements.
From group Ia, IIa, .........................VIIIa.
b) Elements “B” are called Transition metals.
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2. Physical properties of metals and non-metals
Physical properties of metals
Physical properties of non-metals

Metals are good conductors of electricity and
heat because they contain free electrons
which are necessary to conduct electricity

Non-metals do not conduct heat and
electricity because they have no free
electrons which are necessary to conduct
heat and electricity. Except carbon in the
form of graphite which is a good conductor
of electricity.

Metals are malleable, means that they can be
beaten (hammered) into very thin sheets. This
property of metals is called malleability. Gold
and Silver are the most malleable metals.

Non-metals are non-malleable, this means
that they cannot be made into sheets.

Metals are ductile, means that they can be
drawn into thin wires. Gold and Silver are the
ductile metals

Non-metals are not ductile, means that
they cannot be drawn into sheets.

Metals have high melting points and boiling
points.

Non-metals have low melting points and
boiling points. Only one non-metal called
diamond (Allotropic form of carbon) which
has high melting point. The melting point
of diamond is 35000C.

Metals have high density and are very heavy

Non-metals have low density, that is they
are light substance.

Metals are sonorous, means that they can
make sound when hit.

Non-metals are non-sonorous, means that
they do not produce ringing sound when
hit.

Metals are lustrous, means that they have a
shining surface.

Non-metals are not lustrous, means that
they do not have a shining surface. The
only non-metal having a shining surface is
iodine.
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EXERCISES
1)
2)
3)
4)
5)
An element has an electron arrangement 2,8,3. Is the element a metal or non-metal?
What meant by saying that the metals are malleable and ductile?
With the help of example, describe how metals differ from non-metals.
Name one metal and one non-metal which exist in liquid state at room temperature.
a) What are metalloids?
b) Give three examples of elements which are metalloids.
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3. Trends in reactivity for metals and non-metals
Chemical reactivity is the relative tendency of an element to lose or gain electrons in
chemical combinations.
a) Variation of chemical reactivity in a group

While moving from top to bottom in a group of metals (IA, IIA, IIIA) in a periodic
table, the atomic size increases with increment in the number of shells and the force of
attraction between the nucleus and valence shell decreases. This is the reason why
bigger atom/s can lose the valence electron/s more easily than the smaller
atom/s. thus, the tendency of losing the valence electron/s increases and the chemical
reactivity increases on moving from top to bottom in a group of metals.
NOTE: The chemical reactivity in metals depends on the tendency to lose
electrons.
Group 1
Li
Least reactive
Na
K
Reactivity of metals increases on going down in a group
Rb
Cs
Most reactive
Example:
Why is potassium highly reactive than the sodium?
In group 1, potassium is more reactive than sodium which in turn is more reactive than
lithium. This is because a potassium atom loses its valence electron more easily than
sodium atom since the atomic size of potassium is bigger than the atomic size of
sodium. and so on.

While moving from top to bottom in a group of non-metals (VA, VIA, VIIA,
VIIIA), the atomic size increases with the addition number of shells and the force of
attraction between the nucleus and valence shell decreases. The smaller atom/s can
gain the valence electron/s more easily than the bigger atom/s. thus the tendency
of gaining electron/s in the valence shell decreases as well as the chemical reactivity
also decreases on moving from top to bottom in a group of non-metals.
Note: The chemical reactivity in non-metals depends on the tendency to gain
electrons.
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Group VII
F
Most reactive
Cl
Br Reactivity of non-metals decreases on going down in a group.
I
Least reactive
Example:
Why is fluorine highly reactive than the chlorine?
F is more reactive than Cl because fluorine atom can gain one electron more easily
than chlorine since the atomic size of fluorine is smaller than the atomic size of
chlorine. and so on.
b) Variation of chemical reactivity in a period
 While moving from left to right in a period, the chemical reactivity of metal
elements decreases because the number of valence electrons a metal has to lose
increases.
Period 3:

Na
Mg Al
More
reactive
Reactivity decreses
Si
P
S
Cl
Least
More
reactive
reactive
Reactivity increases
Sodium is the most reactive than magnesium because sodium atom can lose one
electron more easily to form cation than magnesium which loses two electrons.
 Magnesium is most reactive than aluminium because a magnesium atom loses two
electrons more easily than aluminium which loses three electrons.
 While moving from left to right in a period, the chemical reactivity of non-metal
elements increases because the number of valence electrons a non-metal has to gain
decreases.
N
Least
reactive
P
Least
Reactive
O
S
F
Most reactive
Cl
Most reactive
 Chlorine is most reactive than Sulphur because chlorine can easily gain one electron
than Sulphur which gains two electrons.
 Phosphorus accept 3 electrons to form anion but it has lower tendency to accept
electrons compared to Sulphur which can accept electrons more easily.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Exercise
1) a) How does the reactivity of metal elements vary down groups and across a period?
b) Give examples to support your ideas in (1) (a).
2) By using examples, explain how reactivity of non-metals vary down groups and vary
across a period?
3) The following figure shows the reactivity of elements in period 3.
From the graph, the following observations are made. Explain each of them.
a) Sodium is more reactive than magnesium.
b) Silicon has a very low reactivity.
c) The reactivity of chlorine is almost the same as that of sodium.
d) The reactivity of Argon is zero.
4) The following is a periodic table showing some elements. Use the table and the elements
shown to answer the questions that follow.
I
II
Groups
III
IV
V
VI
VII
H
VIII
He
C
Na
Mg
K
Ca
TRANSITION METALS
Al
N
O
F
P
S
Cl
Ne
Br
a) How many electrons does an atom of element F contains?
b) Write the electronic configuration of the elements C and K.
c) Give the symbols of two elements the belong to alkali metals? Alkaline earth metals?
Halogens? Noble gases?
d) Give the formula of the compound formed between element Mg and P.
e) i) Which of the elements Na and K is more reactive?
ii) Which of the elements Mg and Al is more reactive?
f) i) Which of the elements Cl and Br is more reactive?
ii) Which of the elements O and N is more reactive?
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
26
4. Chemical properties of metals
a) Reaction of metals with water
Metals react with water to form a metal hydroxide and hydrogen gas.
Examples

Sodium, Potassium and Calcium react with cold water to form metal hydroxide and
hydrogen gas.
Metal + water
Metal hydroxide + hydrogen gas
2Na(s) + 2H2O(l)
2NaOH(aq) + H2(g)
Sodium
2K(s)
Cold water
+
2H2O(l)
Potassium
2Ca(s)
Cold water
+ 2H2O(l)
Calcium
Cold water
Sodium hydroxide
2KOH(aq)
Hydrogen
+
H2(g)
Potassium hydroxide
2Ca(OH)2(aq)
Hydrogen
+
Calcium hydroxide
H2(g)
Hydrogen

Magnesium reacts with both hot water and steam (Very hot gaseous form of water)
but it does not react with cold water.
2Mg(s) + 2H2O(l)
2Mg(OH)2(aq) + H2(g)
Magnesium
2Mg(s)
Hot water
+ H2O(l)
Magnesium
Steam
Magnesium hydroxide
MgO(s)
+
Hydrogen
H2(g)
Magnesium oxide
Hydrogen
When magnesium reacts with hot water, it forms magnesium hydroxide and hydrogen gas
and it reacts with steam to form magnesium oxide and hydrogen gas.
 Metals such as Aluminium, Iron and Zinc do not react with either cold or hot water.
They react with steam to form metal oxide and hydrogen gas.
Al(s) + 3H2O(g)
Al2O3(s) + 3H2(g)
Alumium
Steam
Hydrogen
Zn(s) + H2O(g)
ZnO(s) +
H2(g)
Zinc
Zinc oxide
Hydrogen
Steam
3Fe(s) + 4H2O(g)
Iron

Aluminium oxide
Steam
Fe3O4(s)
Iron oxide
+
4H2(g)
Hydrogen
Some metals do not react with cold water, hot water and even with steam.
Examples: Copper, Gold, Silver and Mercury.
Cu(s) + H2O(g)
No reaction
Copper
water (or steam)
b) Reaction of metals with acids
Metals usually react with dilute acids, to give salts and hydrogen gas.
Metal + Dilute acids
Salt + Hydrogen
2Na(s) + 2HCl(aq)
2NaCl(aq) + H2(g)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
27
2Na(s) + 2H2SO4 (aq)
Mg(s) + 2HCl(aq)
Mg(s) + H2SO4(aq)
Ca(s) + 2HCl(aq)
Zn(s) + 2H2SO4(aq)
2Na2SO4(aq) + H2(g)
2MgCl2(aq) + H2(g)
MgSO4(aq) + H2(g)
2CaCl2(aq) + H2(g)
ZnSO4(aq) + H2(g)
Note:
Copper (Cu), Gold and Silver do not react with dilute acids (Hydrochloric acid,
Sulphuric acid and nitric acid)
Cu(s) + HCl(aq)
No reaction
c) Reaction of metals with halogens
Heated metals react with halogens to form salts (or metal halides)
2Na(s) + Cl2(g)
2NaCl(s)
2Ca(s) + Cl2(g)
CaCl2(s)
2Al(s) + 3Cl2(g)
2AlCl3(s)
2Fe(s) + 3Cl2(g)
2FeCl3(s)
2Cu(s) + Cl2(g)
CuCl2(s)
Zn(s) + Cl2(g)
ZnCl2(s)
Note: All metal chloride are ionic compounds.
d) Reaction of metals with oxygen
Metals react with oxygen to form metal oxides.
4Na(s) + O2(g)
2Na2O(s)
4K(s) + O2(g)
2KO(s)
2Mg(s) + O2(g)
2MgO(s)
2Ca(s) + O2(g)
2CaO(s)
NOTE: Silver(Ag) and gold (Au) do not react with oxygen even at high temperatures.
5. Chemical properties of non-metals
a) Reaction of non-metals with water and dilute acids
Non-metals neither react with water nor dilute acids.
b) Reaction of non-metals with halogens (say chlorine)
Non-metals react with chlorine to form covalent chlorides. Non-metal chlorides are usually
liquids or gases. They do not conduct electricity.
Examples:
C(s) + 2Cl2(g)
CCl4(g)
Carbon tetrachloride
P4(s) + 6Cl2(g)
4PCl3(l)
phosphorus chloride
c) Reaction of non-metals with oxygen
Non-metals react with oxygen to form acidic oxides or neutral oxides.
C(s) + O2(g)
CO2(g)
Carbon dioxide (acidic oxide)
H2(g) + O2(g)
H2O(l)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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6. Differences between chemical properties of metals
and non-metals
Metals
Non-metals


Usually have 1 – 3 valence electrons
Metals form basic oxides




Tend to lose valence electrons
Combine with non-metals to produce
ionic compounds


Usually have 4 – 8 valence electrons
Non-metals form acidic oxides or
neutral oxides
Tend to gain electrons
Combine with non-metals to produce
covalent compounds
7. Uses of some metals and non-metals
 Metals like copper and aluminium are used for making electric wires and cooking
utensils.
 Gold and silver are used for making jewellery.
 Silicon is used in making computer chips.
 Tungsten is used in making filaments of bulb.
 Sulphur is used for making gun powder and sulphuric acid.
 Oxygen is used for the respiration of living things.
 Argon is used in electrical bulbs.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Exercises
1) Complete and balance the following chemical equations.
a) Na(s) + H2O(l)
b) Ca(s) + H2O(l)
c) K(s) + HCl(aq)
d) Mg(s) + H2SO4(aq)
e) Mg(s) + H2O(g)
f) Mg(s) + H2O(l)
heat
g) Na(s) + O2(g)
h) C(s) + O2(g)
i) Mg(s) + O2(g)
2) Magnesium reacts with dilute acids. Name two products of such reactions.
3) Elements M belongs to group IIIa of the periodic table.
a) How many electrons does M have in the outermost shell?
b) Write the formula for:
i) The oxide of M.
ii) The chloride of M.
iii) The nitride of M.
4) a) Name the three sub-atomic particles of an atom.
b) An atom of magnesium is represented as: 24
12𝑀𝑔
i) State the number of each of the three sub-atomic particles that the atom has.
ii) Give its electronic arrangement.
iii) Give the formula and electron arrangement of the most stable ion that the atom
can form.
5) State the number of neutrons, protons and electrons in the aluminium ion shown below:
27𝐴𝑙3+
13
6) Element X and Y (not actual chemical symbols) have atomic number 12 and 16
respectively.
a) To which group of the periodic table do element X and Y belong to?
b) Classify the elements as a metal and a non-metal.
c) Name type of bond expected when element X and Y react.
d) Draw dot and cross diagram to show bonding in the compound formed when element
X and Y react.
e) Deduce the formula of the resulting compound.
7) Study the following table showing data for the atoms A, B, C, D and E.
Atoms
A
B
C
D
E
Electrons
8
13
16
Y
8
Protons
8
13
16
11
Z
Neutrons
8
14
16
11
10
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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a) Work out the values of Y and Z.
b) Work out the approximative relative atomic mass of C.
c) Write the electronic configuration of the following ions and atoms.
i) C
ii) 𝐀𝟐−
iii) 𝐁𝟑+
iv) 𝐃
d) Which atoms:
i) Are isotopes.
ii) Belong to the same group in the periodic table.
iii) Belong to the same period in the periodic table.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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UNIT3: WATER POLLUTION
1) Definition of water pollution
Water pollution is the addition of harmful substances into water sources (e.g. lakes,
rivers, oceans, etc.)
2) Main water pollutants
Pollutants are substances which make water to be harmful.
The main water pollutants include:
a. Sewage: Is waste water from toilets, bathrooms, industries, etc.
b. Fertilizers: fertilizers used by farmers run off by rain water from the fields adding
nutrients to water sources.
c. Faeces and urine: Human and animal wastes such as faeces and urine pollute water.
These wastes contain germs. When passed in water, they make the water unsafe for
drinking and for domestic use.
d. Soaps and detergents: From homes and dry cleaning industries.
e. Chemical wastes (are also called industrial wastes): The water bodies get polluted
with chemicals likes dyes; detergent; compounds of Mercury, Cadmium, Lead, Nickel,
etc.
f. Oil spillage: Oil and oil wastes enter water bodies (from storage tanks, Automobile
waste oil, refineries, and industries) affect aquatic life as well as birds.
g. Plastic: Plastic wastes like polythene bags and other plastic containers when dumped
in a water sources the accumulate in it.
h. Radioactive waste: Is a waste that contains radioactive substance. (Wastes from
nuclear plants, wastes of uranium and thorium during their mining)
i. Hot water: By adding hot water into the water body, it rises the temperature which
results in the death of many aquatic organisms.
3) Dangers of polluted water

When consumed polluted water, people can get water borne diseases like cholera,
diarrhea, dysentery, and typhoid.

Polluted water causes death of aquatic (water) animals like fish, crabs, birds and
dolphins. When acidic fertilizers and other acidic wastes are dumped in water bodies,
they make water acidic. This affects the survival of aquatic plants and animals.

Polluted water also causes eutrophication. Eutrophication is the rapid growth of
water plants due to over fertilization of the water. Rapid growth of these plants
covers the entire surface of water and blocks oxygen from atmosphere and light to reach
under water. The reduced oxygen level can lead to death of aquatic animals.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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4) Prevention of water pollution
 Avoid urinating or defecating in or near the water.
 Encourage people to build toilets and site for wastes.
 Do not throw rubbish/litter in water bodies.
 Do not overuse fertilizers and pesticides in fields.
 Do not throw oils, paints, chemicals and medicine in water bodies.
 Avoid bathing, watering animals and washing clothes in water sources.
 Recycling materials whose production creates pollution.
 Encourage people to use biological manures.
 Practicing farming methods that reduce soil erosion like terrace farming.
Example: In Rwandan regulations, it is prohibited to practice any agriculture activity in
less than 50 m from shores of lakes, rivers and swamps.
EXERCISES
1)
2)
3)
4)
5)
What is water pollution?
Identify four water pollutants
Suggest three ways you can use to avoid pollution of water.
Describe three effects (health hazards) of polluted water to humans
Describe the dangers of polluted water.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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UNIT4: EFFECTIVE WAYS OF WASTE MANAGEMENT
a) Definition of wastes
 Wastes are things that we do not need.
 Waste management is all the processes of handling waste and reduce it.
b) Steps of effective waste management
There are six steps of effective waste management
1. Prevention: Means avoiding waste completely.
2. Minimization: Means reducing the creation of waste material.
3. Reuse: means using a material that has been used before for another purpose.
Example: plastic bottle
4. Recycling: involves reprocessing (or transform) used materials(wastes) into new useful
products.
Example: plastic bottle
5. Energy recovery: means converting waste into usable heat, electricity or fuel.
6. Disposal: refers to dumping of wastes in specific places. These places are called
landfills.
c) Importance and benefits of waste recycling
1. Recycling keeps the environment clean and fresh: When wastes are recycled,
pollution of the air, water and soil is reduced.
2. Recycling conserves natural resources: When materials and products are recycled,
they reduce the exploitation of natural resources.
Examples:
 If paper and other timber products are recycled, there will be reduced need of
harvesting trees.
 If metallic materials are recycled, they will slow down extraction of their ores.
3. Recycling saves energy: It takes much less energy to make products using recycled
materials as compared to making products from raw materials.
4. Recycling creates jobs (employment): People are employed to collect, sort and work
in recycling companies.
d) Effects of wastes and poor disposal
1. Soil pollution: Soil on which sewage and solid wastes are dumped is unsuitable for
cultivation of crops.
2. Air Pollution: When wastes are rotting, bad smell is produced.
3. Water Pollution: When wastes are dumped in water sources, they change its physical
properties and composition. such water is unsuitable for human use.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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4. Spread of diseases: Due to flies, mosquitoes which are carriers of illnesses after
breeding on solid wastes. Also when wastes such as human faeces are not properly
disposed, they can contaminate food and water. This can result into water-borne
diseases such as cholera, dysentery and typhoid.
5. Poor disposal of waste may cause injury.
6. Blockage of waterways: Solid wastes block waterways and this can lead to flooding.
Exercises
1)
2)
3)
4)
5)
Describe the term “waste management”
Describe the steps which can be taken to achieve effective waste management.
Explain the importance and benefits of waste recycling.
Discuss the various effects of waste materials and poor waste disposal.
If well managed, wastes from the kitchen and food leftovers can be beneficial to us and
other organisms. explain
6) State 2 dangers of the materials that do not decay (rot) when they are dumped in
composts
7) Burning is one of the ways of managing wastes. Identify a negative consequence of this
practice on the environment.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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UNIT5: CATEGORIES OF CHEMICAL REACTIONS
1) Types of chemical reaction
Chemical reactions can be classified as follows:
1. Combination reaction: Two or more substances combine to form a single product.
A + B
C
Examples
a) H2 + Cl2
b) 2Na + Cl2
c) CaO(s) + CO2(g)
d) NH3 + HCl
2HCl
2NaCl
CaCO3(s)
NH4Cl
2. Decomposition reactions: A compound breaks into parts. Heat is enough to
cause the reaction of decomposition. AB
A +B
Examples:
a)
b)
c)
d)
CaCO3(s) heat
heat
2KClO3(s)
CuSO4.5H2O(S)
2H2O2(l) sunlight
CaO(S) + CO2(g)
2KCl(s) + 3O2(g)
heat
CuSO4(s) + 5H2O(l)
2H2O(l) + O2(g)
3. Single displacement reactions: A single element replaces an element in a
compound. A + BC
AC + B
Examples:
a) Mg(s) + CuO(s)
b) Zn(s) + 2HCl(aq)
c) Cl2(g) + 2KBr(aq)
d) Cl2(g) + 2NaI(aq)
e) Cu(s) + 2AgNO3(aq)
MgO(s) + Cu(s)
ZnCl2(aq) + H2(g)
2KCl(aq) + Br2(aq)
2NaCl(aq) + I2(s)
Cu(NO3)2(aq) +2Ag(s)
Note: In single replacement reactions, more active metals displace less active metal (or
hydrogen) from their compounds. Below is the arrangement of elements in decreasing order
of their ability to replace elements (metal ion) in aqueous solution. This series is known as
reactivity series.
K > Na > Ca > Mg > Al > Zn > Cr > Fe > Ni > Sn > Pb > H > Cu > Ag > Au
The above series show that potassium (K) is the most reactive metal and gold (Au) is the
least reactive metal.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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4. Double displacement reactions: Two compounds (reactants) exchange their
cations and anions.
AB + CD
AD + CB
Double displacement reactions include:
 Precipitation reactions and
 Neutralization reactions
i. Precipitation reactions: Two soluble salts (or ionic compounds) are mixed and
form an insoluble salt called precipitate.
Examples:
a) BaCl2(aq) + Na2SO4(aq)
BaSO4(s) + 2NaCl(aq)
b) AgNO3(aq) + NaCl(aq)
AgCl(s) + NaNO3(aq)
c) Pb(NO3)2(aq) + 2KI(aq)
PbI2(s) + 2KNO3(aq)
Note: Precipitate is a solid that is thrown down when two aqueous solutions are mixed
together.
ii.
Neutralization reactions: Neutralizations reaction is a reaction in which an
acid and a base react to form salt and water only.
Acid + Base
Examples:
a) NaOH(aq) + HCl(aq)
b) 2KOH(aq) + H2SO4(aq)
c) KOH(aq) + HNO3(aq)
Salt + water
NaCl(aq) + H2O(l)
K2SO4(aq) + H2O(l)
KNO3(aq) + H20(l)
5. Combustion reactions: are the reactions of an element or compound with oxygen.
Heat energy and light are given out during combustion.
A + B
AB Where B is oxygen
Examples:
a)
b)
c)
d)
C(s) + O2(g)
2H2(g) + O2(g)
CH4(g) + O2(g)
C6H12O6 + 6O2
CO2(g)
H2O(l)
CO2(g) + 2H20(l)
6CO2 + 6H20
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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2) Classification of chemical reactions as endothermic and
exothermic reactions
a) Endothermic reaction: Is a chemical reaction in which heat energy is
absorbed.
Examples:
a)
b)
c)
d)
CaCO3(s) heat
CaO(S) + CO2(g)
heat
2KClO3(s)
2KCl(s) + 3O2(g)
heat
CuSO4.5H2O(s)
CuSO4(S) + 5H2O(l)
sunlight
6CO2 + H2O
C6H12O6 + 6O2
b) Exothermic reaction: Is a chemical reaction in which heat energy is released.
Examples:
a) H2O(l) cool
b) N2(g) + 3H2(g)
c) H2 + Cl2
H2O(s)
2NH3(g)
2HCl
∆H= -46.2 KJ/mol
∆H= -860 KJ
3) Ionic equations and rules of writing ionic equations
There are three types of chemical equations:
 Word equations
 Formula equations
 Ionic equations
In this unit we are going to look at the ionic equations. Remember that word equations
and formula equations have been studied in S1.
Ionic equation is the equation that shows the ions actually participating in the
reaction. During the reaction, there are some ions which simply watch the reaction and
we can refer to them as spectator ions. Spectator ions are the ions that appear exactly the
same on each side of the ionic equation.
Rules of writing an ionic equation
1)
2)
3)
4)
Write formula equation and balance it if necessary.
Break the ions for aqueous compounds (write the ions separately)
All solids, liquids and gases do not form ions. (write the full formula)
Cancel out spectator ions (ions that appear the same on each side of the ionic equation)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Example1:
Write ionic quation for the reaction of magnesium and dilute hydrochloric acid to form
magnesium chloride and hydrogen.
Solution:
Rule1: Mg(s) + 2HCl(aq)
MgCl2(aq) + H2(g)
Rule2 and 3: Split dissolved ionic substances into separate ions.
Mg(s) + 2H+(aq) + 2Cl−(aq)
Mg2+(aq) + 2Cl−(aq) + H2(s)
Rule4:Concel out spectator ions (ions that appear the same on both side)
Mg(s) + 2H+(aq) + 2Cl−(aq)
Mg2+(aq) + 2Cl−(aq) + H2(s)
Then balanced ionic equation is
Mg(s) + 2H+(aq)
Mg2+(aq) + H2(g)
Example2:
Write ionic equation for the reaction between sodium hydroxide and dilute hydrochloric
acid to form sodium chloride and water.
Solution:
1. Formula equation:
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
2. Split aqueous compounds into ions (ionic compounds)
Na+(aq)+ OH−(aq) + H+(aq) + Cl−(aq)
Na+(aq) + Cl(aq) + H2O(l)
3. Concel out spectator ions:
Na+(aq)+ OH−(aq) + H+(aq) + Cl−(aq)
Na+(aq) + Cl−(aq) + H2O(l)
Then balanced ionic equation is
OH−(aq) + H+(aq)
H2O(l)
Example3:
Write the ionic equation for the following reaction formula equation:
Pb(NO3)2(aq) + 2KI(aq)
PbI2(s) + 2KNO3(aq)
Solution:
Split aqueous compounds(ionic compounds) into ions
Pb2+(aq) + 2NO3−(aq) + 2K+(aq) + 2I−(aq)
PbI2(s) + 2K+(aq) + 2NO3−(aq)
Therefore, balanced ionic equation is:
Pb2+(aq) + 2I−(aq)
PbI2(s)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Exercises
1) a) What is a chemical reaction?
b) Name six types of chemical reaction.
2) a) What is combination reaction?
b) Write three examples of combination reactions.
3) a) What is decompositin reaction?
b) State three examples of decomposition reactions.
4) a) Explain what a single displacement reaction is?
b) Give three examples of single displacement reactions.
5) a) What is a neutralisation reaction?
b) Give three examples of neutralisation reactions.
6) a) What is a preciptation reaction?
b) Write three examples of precipitation reactions.
7) a) What is a combustion reaction?
b) Give three examples of combustion reactions.
8) Classify the following reactions into decomposition, combustion, single
displacement, precipitation, neutralisation and combination.
a) N2(g) + 3H2(g)
2NH3(g)
b) 2NH3(g)
N2(g) + 3H2(g)
2KNO2(s) + O2(g)
c) 2KNO3(s)
Fe
+
CuCl
FeCl2(aq) + Cu(s)
d)
(s)
2(aq)
K2SO4(aq) + 2H2O(l)
e) 2KOH(aq) + H2SO4(aq)
f) CH4(g) + O2(g)
CO2(g) + 2H2O(l)
g) Pb(NO3)2(aq) + Na2SO4(aq)
PbSO4(s) + 2NaNO3(aq)
9) a) What is an endothermic reaction?
b) State two reactions that are endothermic.
10) a) What is an exothermic reaction?
b) State two reactions that are exothermic.
11) Use the words given below to fill in the following sentences correctly.(release,
exothermic, absorb, endothermic).
Exothermic reactions ..........................heat to the surrounding whereas endothermic
reactions......................... heat from the surrounding. If the temperature rises during a
reaction, it is ........................ reaction. If the temperature drops during a raction then it
is …………………….
12) a) what are ionic equations?
b) what is the name of the ions that do not undergo any change during chemical
reaction?
c) work out the ionic equations for the reactions below:
i) NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
ii) Cu(s) + 2AgNO3(aq)
Cu(NO3)2(aq)
+ 2Ag(s)
iii) BaCl2(aq) + (NH4)2SO4(aq)
BaSO4(s) + NH4Cl(aq)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
40
iv) 2FeCl2(aq) + Cl2(g)
2FeCl3(aq)
13) Dilute sulphuric acid is added to aqueous barium nitrate. An insoluble precipitate of
barium sulphate is formed.
a) Which type of reaction takes place.
b) Write a full chemical equation for the reaction.
c) Write an ionic equation for the reaction.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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UNIT6: PREPARATION OF SALTS AND
IDENTIFICATIONS OF IONS
1) Solubility of salts
Sodium chloride (NaCl) mixes with water uniformly to form a homogeneous mixture. In this
sense, sodium chloride salt is soluble in water and dissolves in it. The sodium chloride
(NaCl) is a solute, the water is a solvent and the mixture is a solution.
Defition of some terms:
 A solute is a substance (Salt) that dissolves in a solvent.
 A solvent is a liquid that dissolves the solute (say water)
 A solution is a uniform mixture of solute and solvent.
 Dissolve refers to crystals or solid breaking up and disappering in a solvent (water)
 Solubility is the amount of solute (in grams) required to form a saturated solution in
100 grams of solvent(water) at a given temperature.
 Soluble: when a solute dissolves in a solvent.
 Insoluble: when a solute does not dissolve in a solvent.
2) Concept of saturated, unsaturated and
supersuturated.
Activity
Material required:
 Three hard glass beakers (500 ml)
 Sugar
 Water
 Spoon
Procedure
 Fill all three beakers with equal amount of water.
 Label them A, B and C.
 Add one teaspoons sugar in beaker A, Two teaspoons sugar in beaker B, and Three
teaspoons in beaker C.
 Stir with spoon.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Observe the beaker
Observation: In this activity,
 Solution in beaker A is unsaturated. (contains less solute than it could hold)
 Solution in beaker B is unsaturated. (contains less solute than it could hold)
 Solution in beaker C is saturated. (contains what it should hold)


Heat the beaker C and observe the beaker.
Let beaker C sit uncovered for some days to evaporate ½ of water.
Conclusion:
 After heating, sugar left in beaker C and also dissolves completely, so the solution in
beaker C is unsaturated.
 After evaporating ½ of water, beaker C contains 3 teaspoons of dissolved sugar. Now
the solution in beaker C is supersaturated (contains more solute than saturated
solution).
3) Definitions of some terms
 Saturated solution is a solution which cannot dissolve anymore solute at a given
temperature.
 Contains maximum amount of solute at given temperature.
 Contains what it should hold.
 If additional solute is added it will not dissolve.
 Unsaturated solution is a solution which can dissolve anymore solute at a given
temperature.
 Contains less solute than saturated solution.
 Contains less solute than it could hold.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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
Additional solute can dissolve in an unsaturated solution.
 Supersaturated solution is a solution in which more solute is dissolved by
increasing the temperature of a saturated solution.
 Contains more solute than a saturated solution.
 Contain more than it should hold.
 Additional of more solute causes the excess solute to precipitate.
Note
 When a saturated solution is heated, the solution becomes unsaturated.
 When more water is added to a saturated solution, the solution also becomes
unsaturated.
 When a saturated solution cools down, a supersaturated solution is formed.
4) Factors influencing solubility of different salts
a) Nature of solute: A solution can easily dissolve in a solvent and dissolve with
difficulty into another.
Example: Some solutes such as NaCl, KCl, KNO3, etc are soluble in water. On the
other hand they are not soluble in ethanol, CCl4.
b) Temparature: The solubility of most of the ionic compounds increases with increase in
temperature. On the other hand, some compounds dissolve better by decreasing the
temperature.
Note:
 The solubility od solids and liquids increases with increase in temperature.
 The solubility of gases always decreases with increase in temperature.
 Temperature increases the amount of solute that can be dissolved in a solvent.
5) Solubility curve
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Activity 1
Study the solubility curve of a salt shown below and answer the questions that follow:
Solubility of salt X is 60 g/100g of water at 400C. 60 g of this salt was dissolved in 100g of
water and allowed to cool.
a) At what temperature from the graph is the solution unsaturated? Explain.
b) At what temperature from the graph is the solution saturated? Explain.
c) At what temperature from the graph is the solution supersaturated?
d) In which beaker can crystals be formed when allowed to stay overnight? Explain.
Answer
a) At 70 0C (Point below the line)
b) At 40 0C (Point on the line)
c) At 10 0C (Point above the line)
d) Crystals can be formed in beaker a, where the solution is supersaturated and in
beaker b, where the solution is saturated.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Activity2:
Study the solubility curves for potassium nitrate and potassium chloride below.
a)
b)
c)
d)
What is the solubility of potassium nitrate at 200C?
What is the effect of temperature on the solubility of potassium chloride in water?
Were the solubility of two salts determined at the same temperatures?
What is the importance of solubility curves?
Activity3:
Study the solubility curves for substance X and Y.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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a) What mass of substance X will dissolve in 100grams of water at 300C.
b) What mass of substance Y will dissolve in 100grams of water at 600C.
c) At what temperature will 40 grams of substance X form a saturated solution when
added to 100 grams of water?
d) What is the total mass of compound X that can be dissolved in 20 grams of water at
260C.
e) What is the total mass of compound Y that can be dissolved in 20 grams of water at
600C.
Answers
a) 46 grams/ 100 grams of water.
b) 18 grams/ 100 grams of water.
c) At 260C.
d) At 260C, 100g of water can only hold 40g of compound X.
𝑋 𝑔𝑟𝑎𝑚𝑠
40𝑔𝑟𝑎𝑚𝑠
So at 260C ,
=
, Then X= 10 grams
100𝑔𝑟𝑎𝑚𝑠 𝑜𝑓
𝑤𝑎𝑡𝑒𝑟
25 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓
𝑤𝑎𝑡𝑒𝑟
e) At 600C, 100g of water hold 18g of compound Y.
1g of water hold 18𝑔 of compond Y
100𝑔
20g of water hold
18𝑔 ×20𝑔
100𝑔
of cpompound Y
= 3.6 grams of compound Y
6) Calculation of solubility
Solubility =
𝐖𝐞𝐢𝐠𝐭 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 ( 𝐢𝐧 𝐠𝐫𝐚𝐦)
𝐖𝐞𝐢𝐠𝐭 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 ( 𝐢𝐧 𝐠𝐫𝐚𝐦𝐬)
× 100
(At a particular temperature)
Note: Solubilty has no unit.
Examples:
1) The solubility of a solute at 300C is 40. What amount of water is rquired to make
saturated solution of 80 grams of a solute?
Answer
Weigt of solute = 80grams
Solubility at 300C = 40
Weigt of solvent (say water) = ?
Solubility =
𝐖𝐞𝐢𝐠𝐭 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 ( 𝐢𝐧 𝐠𝐫𝐚𝐦)
𝐖𝐞𝐢𝐠𝐭 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 ( 𝐢𝐧 𝐠𝐫𝐚𝐦𝐬)
× 100
Weigt of solvent = Weigt of solute × 100 =
Solubility
𝟖𝟎×100
𝟒𝟎
= 200grams
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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2) At 300C, 7grams of sugar dissolves in 5 grams of water to form a saturated solution.
Find the solubility of sugar.
Answer
Mass of solute = 7g
Mass of solvent (water) = 5g
Solubility =
=
𝐖𝐞𝐢𝐠𝐭 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 ( 𝐢𝐧 𝐠𝐫𝐚𝐦)
𝐖𝐞𝐢𝐠𝐭 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 ( 𝐢𝐧 𝐠𝐫𝐚𝐦𝐬)
7g
5g
× 100
× 100 = 140
The solubility of sugar at 300C is 140
3) 7 grams of saturated solution of salt saturated at 600C is evaporated to dryness; 2grams
of white residue is left behind. What is the solubility of salt at that temperature?
Answer
Weigt of saturated solution = 7 grams
Weigt of solute (salt) = 2 grams
Solubility of salt at 600C =?
Weigt of saturated solution = Weigt of solute + Weigt of solvent
Weigt of solvent = Weigt of solution − Weigt of solute
= 7 g −2g = 5g
Solubility =
=
𝐖𝐞𝐢𝐠𝐭 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 ( 𝐢𝐧 𝐠𝐫𝐚𝐦)
𝐖𝐞𝐢𝐠𝐭 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 ( 𝐢𝐧 𝐠𝐫𝐚𝐦𝐬)
𝟐
× 𝟏𝟎𝟎 = 𝟒𝟎
𝟓
× 100
The solubility of salt at 600C is 40.
7) Different ways of preparing normal salts
i) Preparing salts from reaction of acid with active metals
Metal + Dilute acids
Salt + Hydrogen gas
2Na(s) + 2HCl(aq)
2Na(s) + 2H2SO4 (aq)
Mg(s) + 2HCl(aq)
2NaCl(aq) + H2(g)
2Na2SO4(aq) + H2(g)
2MgCl2(aq) + H2(g)
ii) Reacting a dilute acid with metal oxide
Metal oxide + Acid
Example:
CuO(s) + H2SO4 (aq)
Salt + Water
CuSO4(aq) + H2O(l)
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iii) Reaction with bases
Dilute hydrochloric acid reacts with basic oxides and alkalis to give salt and water.
MgO(s) + 2HCl(aq)
MgCl2(aq) + H2O(g)
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
Note: The reaction between an acid and a base to form a salt and water only is known as
neutralization reaction.
iv) Reaction with carbonates and hydrogen carbonates
Dilute hydrochloric acid reacts with carbonates and hydrogen carbonates to give metal
carbonate, water and carbon dioxide.
Na2CO3(s) + 2HCl(aq)
2NaCl(aq) + H2O(l) + CO2(g)
NaHCO3(s) + HCl(aq)
NaCl(aq) + H2O(l) + CO2(g)
v) Precipitation method
Two soluble salts (or ionic compounds) are mixed and form an insoluble salt called
precipitate.
Examples:
BaCl2(aq) + Na2SO4(aq)
AgNO3(aq) + NaCl(aq)
Pb(NO3)2(aq) + 2KI(aq)
BaSO4(s) + 2NaCl(aq)
AgCl(s) + NaNO3(aq)
PbI2(s) + 2KNO3(aq)
vi) Reacting metal oxide with non-metal oxide
Na2O(s) + CO2(g)
CaO(s) + SO3(g)
MgO(s) + CO2(g)
vii)
Na2CO3(s)
CaCO4(s)
MgCO3(s)
Reacting base with non-metal oxide
Pass carbon dioxide gas through limewater (calcium hydroxide) and sodium hydroxide.
Ca(OH)2(aq) + CO2(g)
CaCO3(s) + H2O (l)
Lime water
CO2(g) + 2 NaOH(aq)
white ppt
Na2CO3(aq) + H2O(l)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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8) Uses of salts
Salt
Uses
Sodium chloride (NaCl)
 Flavouring food
 Food preservative
Sodium carbonate
decahydrate
 Making glass
 Softening hard water
 Manufacture of soap
(Na2CO3 .10H2O) also called
washing soda.
Sodium hydrogencarbonate
(NaHCO3) also called
baking soda
Potassium carbonate
(K2CO3)
 Used in medicines
 It is added to dough
to raise it
 Making soft soaps
 Making hard glass
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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9) Identification of cations and anions
i) Identification of cations
Cations are positively charged ions. The cations you are expected to identify at this level
are: 𝐂𝐚𝟐+, 𝐌𝐠𝟐+, 𝐙𝐧𝟐+, 𝐀𝐥𝟑+,𝐏𝐛𝟐+,𝐂𝐮𝟐+, 𝐅𝐞𝟐+, 𝐅𝐞𝟑+, NH4+.
These cations are identified by using Sodium hydroxide solution, NaOH(aq), as
reagent.
Dilute sodium hydroxide (NaOH) solution is added to a solution containing the suspected
ion. The bable below shows the observations made.
Cation Add few drops of NaOH(aq) Add excess NaOH(aq)
𝑪𝒂𝟐+
𝑴𝒈𝟐+
𝒁𝒏𝟐+
white precipitate is formed
white precipitate is formed
white precipitate is formed
White precipitate is insoluble
White precipitate is insoluble
𝑨𝒍𝟑+
white precipitate is formed
White precipitate dissolves
𝑷𝒃𝟐+
white precipitate is formed
White precipitate dissolves
𝑪𝒖𝟐+
blue precipitate is formed
blue precipitate is insoluble
𝑭𝒆𝟐+
green precipitate is formed
green precipitate is insoluble
𝑭𝒆𝟑+
brown precipitate is formed
brown precipitate is insoluble
Ammonia gas is released when the mixture is heated
𝑁𝐻4+

White precipitate dissolves
To distinguish between 𝐀𝐥𝟑+, 𝐙𝐧𝟐+ and 𝐏𝐛𝟐+
Reagent: ammonia solution, NH3(aq) is used as reagent
Cation
𝑨𝒍𝟑+
𝑷𝒃𝟐+
𝒁𝒏𝟐+
Add few drops of 𝐍𝐇𝟑(𝐚𝐪)
white precipitate
Add excess 𝐍𝐇𝟑(𝐚𝐪)
White precipitate is insoluble
white precipitate
White precipitate is insoluble
white precipitate
White precipitate is soluble
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
To distinguish between 𝐏𝐛𝟐+ and 𝐀𝐥𝟑+
Cation
𝐏𝐛𝟐+ ions
𝐀𝐥𝟑+ ions

Reagent
Observation
Add potassium iodide solution , KI(aq)
yellow ppt is formed
Add potassium iodide solution , KI(aq)
no precipitate is formed
To distinguish between 𝐌𝐠𝟐+ and 𝐂𝐚𝟐+
Cation
𝐌𝐠𝟐+ ions
𝐂𝐚𝟐+
ions
Reagent
Observation
Add ammonia solution , NH3(aq) also
written as NH4OH(aq)
Add ammonia solution , NH3(aq) also
written as NH4OH(aq)
white ppt is formed
no precipitate is formed
(no change)
ii) Identification of anions
a) Tests for sulphate 𝐒𝐎𝟐− ions and 𝐒𝐎𝟐− ions in solution
𝟑
Anion
𝟒
Reagent
𝐒𝐎𝟐−
ions
Add barium chloride (BaCl2) solution
followed by addition of dilute
hydrochloric acid (HCl)
𝐒𝐎𝟐−
ions
Add barium chloride (BaCl2) solution
followed by addition of dilute
hydrochloric acid (HCl)
𝟑
𝟒
Observation
white precipitate which
dissolves in dilute
hydrochloric acid.
white precipitate which
does not dissolve in dilute
hydrochloric acid.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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b) Tests for chloride 𝐂𝐥− ions, iodide 𝐈− ions and nitrate 𝐍𝐎−
𝟑 ions in
solution
Anion
𝐂𝐥− ions
Reagent
Add Silver nitrate (AgNO3) solution
followed by excess dilute nitric acid
(HNO3) to salt solution.
ions
Add Silver nitrate (AgNO3) solution
followed by excess dilute nitric acid
(HNO3) to salt solution.
𝐍𝐎− ions
Add Silver nitrate (AgNO3) solution
followed by excess dilute nitric acid
(HNO3) to salt solution.
𝐈−
𝟑
Observation
𝐂𝐥− ions form white
precipitate with silver
nitrate. The precipitate does
not dissolve in excess dilute
nitric acid
𝐈− ions form yellow
precipitate with silver
nitrate.
No precipitate is formed
with NO−
3
c) Tests for carbonate 𝐂𝐎𝟑𝟐− ions in solution
 Reagent: Use Dilute hydrochloric acid (HCl)
 Observation: Effervescence (bubbles) of colorless gas of carbon dioxide forms when
carbonates react with dilute hydrochloric acid.
d) Tests for nitate ions 𝐍𝐎−
𝟑 ions in a solution
 Reagent: Add iron (II) sulphate solution (FeSO4(aq)) followed by concentrated
sulphuric acid (conc. H2SO4)
 Observation: A brown ring forms between sulphuric acid and the rest of the
solution (i.e mixture of FeSO4 solution and nitrate solution). Then, the ring
disappears when the solution is shaken.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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iii) Identification of gases
a) Test for oxygen (O2) gas
 Reagent: Use a glowing splint of wood.
 Observation: A glowing wooden splint relights in a test tube of oxygen.
b) Test for hydrogen (H2) gas
 Reagent: Use a lit splint of wood.
 Observation: “Pop” sound with a lighted splint in a test tube of hydrogen is
observed.
c) Test for chlorine (Cl2) gas
 Reagent: Use moist blue litmus paper.
 Observation: Blue litmus paper turns red and then is bleached white.
d) Test for ammonia (NH3) gas
 Reagent: Use moist red litmus paper.
 Observation: Red litmus paper turns blue.
Or
 Reagent: Use concentreted hydrochloric acid.
 Observation: Ammonia forms dense white fumes of NH4Cl with hydrogen
chloride gas.
e) Test for carbon dioxide (CO2) gas
 Reagent: Use limewater, Ca(OH)2(aq) (It is colorless)
 Observation:
 CO2 turns limewater milky due to formation of white precipitate of calcium
cabonate.
 White precipitate disappear if CO2 is passed through the solution in excess.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Exercises
1) a) Define the term “ solubility”
b) Brine is a concentrated solution of sodium chloride. Identify the solute and solvent in
brine
c) What happens to the salt dissolved when the amount of water in which it is dissolved
is reduced by evaporation?
d) Classify the following salts as soluble or insoluble.
NaCl, NH4NO3, ZnCO3, PbCl2, Na2CO3.
2) Define the following terms:
a) Saturated solution
b) Unsaturated solution
c) Supersaturated
3) Choose the best answer:
a) The solubility of many salts ........................ when temperature decreases.
i) Increases
ii) Decreases
iii) Double
b) Solubility of salt increases when…………………
i) More salt is added to the solution
ii) Salt solution is heated.
iii) The solution is stirred.
c) A solution which can not dissolve any more solute at a given temperature is
called……………..
i) Supersaturated
ii) Unsaturated solution
iii) Saturated solution.
4) What are the factors that influence the solubility of solts?
5) a) Answer the following questions using the solubility graph below:
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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i)
ii)
iii)
iv)
How much sodium nitrate will dissolve at 30 0C?
Which salt is most soluble at 60 0C?
Which salt is least soluble at 40 0C?
At what temperature will 60 g of sodium sulphate dissolve in 100 g of water.
b) The solubility of a salt at 30 0C is 40. What amount of water is required to make
saturated solution of 80 g grams of a solute?
c) At 30 0C, 14 grams of sugar dissolves in 10 grams of water to form a saturated solution.
Find the solubility of sugar
6) Complete the following table.
Experiment
Add NaOH solution to
solution X
Observation
A blue precipitate
Solution Y contains 𝐅𝐞𝟐+
Add NaOH solution to
solution Y
Add NaOH solution to
solution Z
A white precipitate
that dissolves in
excess NaOH
Solution W contains 𝐂𝐎𝟐−
𝟑
Add HCl solution to
solution W
Add barium chloride
solution followed by
dilute HCl to solution D
Conclusion
A white precipitate
tha dissolves in
dilute HCl
7) State a reagent that can be used to distinguish between the following pairs of ions and
state the observable change in each case:
a) CO32− and Cl−
b) Cu2+ and Ca2+
c) Fe3+ and Fe2+
d) Zn2+ and Fe2+
8) With the help of equations wher possible, state the chemical test that would be used to
distinguish each pair of the following substances and the observation in each case:
a) Zn(NO3)2(aq) and Fe(NO3)2(aq)
b) NaCl(aq) and Na2NO3(aq)
c) Pb(NO3)2(aq) and Cu(NO3)2(aq)
d) CuSO4(aq) and FeSO4(aq)
e) FeSO4(aq) and Fe2(SO4)3(aq)
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9) Give a reagent that can be used to test for the presence of 𝒁𝒏𝟐+ ions in a solution and
the observation made.
10) Drinking water was suspected to be contaminated with the following ions.
Cu2+, Fe3+, SO2− and CO2−. A sample of the water was divided into several portions and
4
3
tested for the presence of the above ions.
a) The first portion was mixed with nitric acid and there was no observable change.
Wthat conclusion can be made from this observation and explain your answer.
b) A second portion was tested using aqueous ammonia solution. A few drops of
ammonia solution were added, followed by excess ammonia. Describe what would be
observed if Cu2+ ions were present.
c) How would you test for the presence of SO42− ? State the reagent and the expected
observation for a positive result.
d) Another portion was mixed with a reagent which removed Cu2+. If the remaining
solution contained Fe3+, what test would confirm the presence of Fe3+ ? State the
reagent and observation.
e) Rust contains a compound of iron (III).
i) State the conditions necessary for rusting to take place.
ii) Give two methods of preventing rusting.
iii) Give one similarity and one difference between rusting and combustion.
( N. E: 2013)
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Practice question
11 You are provided with solid Q, which contains a single cation.perform the following
tests on the solid and hence identify the cation present in Q.
Test
a) Dissolve Q in about
5 cm3 of water. Divide
the resulting solution
into three equal
portions of about
1 cm3 each in separate
test tubes.
i) To the first portion
add aqueous NaOH
until in excess.
ii) To the second portion
add aqueous NH3
until in excess.
iii) To the third portion
add aqueous KI.
Observation
Solid dissolved easily.
Deduction / Conclusion
Q is a soluble salt. Possibly
a nitrate.
White precipitate formed,
dissolved in excess to
give a colourless solution.
White precipitate formed,
insoluble in excess.
Yellow precipitate formed
Identify the cation present in solid Q.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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UNIT7: THE MOLE CONCEPT AND GAS LAWS
1. Avogadro number and the mole concept
In science, the mole is a term referring to a definite quantity. In everyday life we use
various terms to refer to definite quantities of things. The table below shows some of
these terms we use in everyday life.
Term
Pair
Dozen
Decade
Tray
Number of items
2
12
10
30
Counted items
Shoes, Socks, Hand gloves, earrings
Books, Pens, Pencils, Cups, Plates
Years
eggs
If you request to be sold a dozen of exercises books; a dozen of pencils; a dozen of pen; or a
dozen of cup; the actual number of each item will be 12. However, their masses will be
different.
 Mole is a unit just like dozen that is used to describe a certain number of elementary
particles such as atoms, molecules, ions and electrons that are involved in chemical
reactions.
 These particles are very small and cannot be counted industrially because they cannot
be seen with naked eye.
 The term “mole” refers to a particular number of particles, known as the Avogadro’s
number or Avogadro’s constant.
 Avogadro’s number, NA is equal to 6.022×2023 particles.
 The particles in consideration may be electrons, ions, molecules, protons or atoms.
Example
6.022×2023 atoms of the same kind make one mole of that particular substance.
2. Definition of mole
The mole is the amount of substance that contains 6.022×2023 particles.
The amount of substance can be mass or volume of a gas or volume of solution.
3. Calculation of the number of moles
The number of moles =
Then n
=
𝑵
𝑵𝑨
where
𝐆𝐢𝐯𝐞𝐧 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬
𝐀𝐯𝐨𝐠𝐚𝐝𝐫𝐨 𝐧𝐮𝐦𝐛𝐞𝐫
=
𝑵
𝑵𝑨
n: Number of moles
N: Given number of particles
NA: Avogadro’s number, equal to 6.022 × 2023
The SI unit of mole is “ mol ”
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Example1
Calculate the number of moles of 12.044 × 1023 helium atoms.
Answer
number of moles =
𝐆𝐢𝐯𝐞𝐧 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬
𝐀𝐯𝐨𝐠𝐚𝐝𝐫𝐨 𝐧𝐮𝐦𝐛𝐞𝐫
=
𝟏𝟐.𝟎𝟒𝟒×𝟏𝟎𝟐𝟑
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑
= 2 mol
Example2
How many moles are in 3.011 × 1023 hydrogen atoms?
Solution:
number of moles =
𝐆𝐢𝐯𝐞𝐧 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬
𝐀𝐯𝐨𝐠𝐚𝐝𝐫𝐨 𝐧𝐮𝐦𝐛𝐞𝐫
=
𝟑.𝟎𝟏𝟏×𝟏𝟎𝟐𝟑
=
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑
0.5 mol
Example
How many atoms are there in 0.1 moles?
Answer
Number of atoms = number of moles × Avogadro number
= 0.1 × 6.022 × 1023
= 0.6022 × 1023 atoms
Exercises
1) Calculate the number of moles are there in:
a) 3.011 × 1023 𝑎𝑡𝑜𝑚𝑠
b) 12.04 × 1023 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
c) 1.8066 × 1023 𝑖𝑜𝑛𝑠
d) 1.5055 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
2) Calculate the number of moles of 12.044 × 1023 helium atoms.
3) Calculate the number of particles in:
a) 0.1 moles of carbon atoms
b) 0.4 moles
c) 5 moles
4. Definition of relative atomic mass
The relative atomic mass(RAM) is the average mass of all atoms in an element compared to
𝟏
one twelfth ( ) the mass of carbon-12 isotope.
𝟏𝟐
It is expressed mathematical as follows:
RAM of an element =
𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐚𝐧 𝐚𝐭𝐨𝐦 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
𝟏 (𝐭𝐡 𝐭𝐡𝐞 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐜𝐚𝐫𝐛𝐨𝐧−𝟏𝟐 𝐢𝐬𝐨𝐭𝐨𝐩𝐞)
𝟏𝟐
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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 The relative atomic mass is a pure number and has no unit
Table below shows the relative atomic masses of some elements.
Name of element
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Sodium
Magnesium
Aluminium
Silicon
Phosphorus
Sulphur
Chlorine
Argon
Potassium
Calcium
Symbol
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Relative atomic mass
1.008
4.003
6.941
9.012
10.81
12.01
14.01
16.00
19.01
22.18
22.99
24.31
26.98
28.09
30.97
32.06
35.45
39.95
39.10
40.08
Note:
 Rounded relative atomic masses are often used in calculations.
Example: K=39 , Na=23 , Mg=24 , Cl=35.5 , H=1
 RAM has no units
5. Definition and calculation of relative molecular
mass(RMM)
The relative molecular mass is defined as the sum of all the individual atomic masses of all
atoms in the formula.
Example:
a) RMM of hydrogen gas (H2) is 2×RAM of H =2×1=2
b) RMM of water (H2O) is (2×RAM of H) + (1×RAM of O) = (2×1) + (1×16)
= 2+16 =18
c) RMM of glucose(C6H12O6) is (6×RAM of C) + (12×RAM of H) + (6×RAM of O)
= (6×12) + (12×1) + (6×16) =72 + 12 +96 =180
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Note:
 RMM has no units
 RMM is used for molecular or covalent compound or elements.
6. Definition and calculation of relative formula
mass(RFM)
Relative formula mass is the sum of relative atomic masses of atoms in a formula unit.
It applies to both ionic and molecular compounds.
Example
a) Calculate the RFM of NaCl. (Na=23, Cl=35.5 )
Answer:
RFM of NaCl= (1×RAM of Na) + ( 1×RAM of Cl)
=(1×23) + (1×35.5) =58.5
b) Calculate the RFM of K2CO3 ( K=39, C=12, O=16)
RFM of K2CO3 = (2×RAM of K) + (1×RAM of C) + (3×RAM of O)
=(2×39) + (1×12) + (3×16) = 78+12+ 48 = 138
Note:
 RFM has no units
Exercise
Calculate the relative formula mass, RFM of the following:
a)
b)
c)
d)
e)
Lead(II) nitrate, (RAM of Pb = 207, N = 14, O =16)
Sulphuric acid, (RAM of H = 1, S = 32, O =16)
Calcium hydrogen carbonate, (RAM of Ca = 40, H= 1, C =12, O =16)
Zinc oxide, (RAM of Zn = 65, O =16)
Sodium carbonate, (RAM of Na = 23, C= 12, O =16)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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7. Calculation of molar mass
Definition
Molar mass is the mass of one mole of an element or a compound. It is expressed in units of
grams per mole.
i) Molar mass for an element = Relative atomic mass of that element in grams per mol if
element is made of atoms.
Examples
 Molar mass of carbon is 12g/mol
 Molar mass of magnesium is 24g/mol
ii) Molar mass for an element= Relative molecular mass in grams per mol if the element
is made of molecule.
Examples
 Molar mass of water (H2O) =(2×RAM of H) + (1×RAM of O)
= (2×1) + (1×16)=18g/mol
 Molar mass of ammonia (NH3) = (1×14) + (3×1) = 14+ 3=17g/mol
 Molar mass of oxygen (O2) = (2×16) =32g/mol
iii) Molar mass for a compound = Relative formula mass of that compound in grams per
mol. Means that for a compound, molar mass is the sum of relative atomic masses of all
the atoms in the formula unit in grams per mol.
Examples
 Molar mass of sodium chloride (NaCl) is
= (1×RAM of Na) + (1×RAM of Cl) = (1×23) + (1×35.5)= 23 + 35.5 = 58.5g/mol
 Molar mass of potassium carbonate (K2CO3) is
= (2×RAM of K) + (1×RAM of C) + (3×RAM of O)
= (2×39) + (1×12) +( 3×16) = 72 +12+ 48 =138g/mol
Molar mass of sulphuric acid (H2SO4) is

= (2×1) + (1×32) + (4×16) = 98g/mol
Exercise
Calculate the molar masses of the following substances:
a)
b)
c)
d)
e)
Hydrochloric acid, HCl
Sodium carbonate, Na2CO3
Nitric acid, HNO3
Copper sulphate pentahydrate, CuSO4•H2O
Sodium hydroxide, NaOH.
(Atomic masses of H =1, Cl = 35.5, Na =23, C = 12, O = 16, N = 14, Cu =63.5, s =32)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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8. Relationship between number of moles, mass and
molar mass
Number of moles of substance X =
n=
𝐦
𝐌𝐦
𝐌𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐠𝐫𝐚𝐦𝐬
𝐌𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐠𝐫𝐚𝐦𝐬 𝐩𝐞𝐫 𝐦𝐨𝐥𝐞
n= number of moles in mol
where
m= mass in g
Mm= molar mass in g/mol
This means that mass= moles× molar mass, m= n×Mm
and
molar mass =
𝐦𝐚𝐬𝐬
𝐦𝐨𝐥𝐞𝐬
;
Mm = 𝐦
𝐧
Examples
1. Calculate the number of moles of 16.2 g of water (H=1, O=16).
Answer
m = 16.2g
Mm of H2O = (2×1) + (1×16) = 18 g/mol
Number of moles =
n=
𝟏𝟔.𝟐𝐠
𝟏𝟖𝐠/𝐦𝐨𝐥
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐇𝟐𝐎 𝐢𝐧 𝐠𝐫𝐚𝐦𝐬
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬
= 0.9 mol of water
2. Calculate the number of moles of magnesium atoms in 2.4 g. (Mg=24)
Answer
n= 𝒎 = 𝟐.𝟒𝒈 = 0.1 mol of Mg atoms.
𝑴𝒎
𝟐𝟒𝒈/𝒎𝒐𝒍
3. What is the mass of 0.04 moles of potassium? (K=39)
Answer
n= 𝒎
m = n×Mm
𝑴𝒎
m = 0.04 mol × 39 g/mol
m = 1.56 g of K
Note:
The following relationship is used to calculate the number of molecules or
number of atoms in a certain mass of a substance.
𝐌𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞
Number of molecules of a substance = 𝐌𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 × Avogadro number
𝑵
𝑵𝑨
=
𝒎
𝑴𝒎
then N =
𝒎
𝑴𝒎
× NA Or
N = n × NA where N = Number of molecules, atoms.
m= mass of substance
Mm = molar mass
NA = Avogadro number.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
64
Number of molecules or atoms = Number of moles × Avogadro number
Example
The antibiotic penicillin has the molecular formula C 16H18N2SO4. One injection of penicillin
contains 500 mg of penicillin. When one intra-muscular injection of penicillin is
administered.
a) How many moles of penicillin are administered?
b) How many molecules of penicillin are administered?
c) How many atoms of nitrogen are injected?
Answer
Molar mass of penicillin (C16H18N2SO4) = (16×12) + (18×1) + (2×14) + (1×32) + (4×16)
= 334g/mol
Mass of penicillin = 500mg = 0.500g
𝟎.𝟓𝟎𝟎𝒈
a) Number of moles of penicillin =
= 0.001497006 mol
𝟑𝟑𝟒𝒈/𝒎𝒐𝒍
b) Number of molecules = Number of moles × Avogadro number
= 0.001497×6.022×1023 = 0.00901497 ×1023 molecules of penicillin
c) 1molecule of penicillin contains 2 atoms of nitrogen
0.00901497×1023 molecules of penicillin contain 2×0.00901497×1023
= 0.01802994×1023 molecules of nitrogen.
Exercises
1) Calculate the number of moles of the following:
a) 84 g of nitrogen gas, N2 (N =14)
b) 96 g of oxygen gas, O2
(O =16)
c) 74.2 g of sodium carbonate, Na2CO3, (Na =23, C =12, O =16)
d) 50 g of copper (II) sulphate pentahydrate, CuSO 4•5H2O
(Cu =63.5, S =32, O =16, H =1)
e) 114.4 g of sodium carbonate decahydrate Na2CO3•10H2O
(Na =23, C =12, O =16, H = 1)
f) 4.83 g of sodium nitrite, (Na =23, N =14, O =16)
2) Calculate the mass of the following:
a) 6.9 moles of carbon dioxide, CO2 (C =12, O =16)
b) 500 moles of ammonium nitrate, NH4NO3 (N =14, H =1, O =16)
c) 1.5 moles of ethanol, C2H5OH (C =12, H =1, O =16)
d) 8.5 moles of copper (II) sulphate pentahydrate, CuSO4•4H2O
(Cu =63.5, S =32, O =16, H =1)
3) How many atoms and S8 molecules are present in 50 g of Sulphur? The relative atomic
mass of Sulphur is 32.
4) Calculate the number of molecules of chloroform (CHCl3) weighing 0.0239 g
(H =1, C =12, Cl = 35.5)
5) Find the number of atoms in the following:
a) 52 mol of Ne
b) 52 g of Ne (Ne = 20, Avogadro’s number = 6.022 × 1022)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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6) How many molecules are present in:
a) 9 g of water
b) 17 g of ammonia
(H= 1, O= 16, N= 14, Avogadro’s number = 6.022 × 1022)
9. Molar gas volume
Definition
The molar gas volume is the volume occupied by one mole of a gas under standard
conditions of temperature and pressure.
These conditions are:
 At standard condition of temperature and pressure (STP), (temperature is 00C or
273K and pressure is 1atm or 760mmHg), 1mole of any gas occupies 22.4 dm3 or
22400cm3 of volume.
 At room temperature and pressure, (rtp), (temperature is 250C or 298K and
pressure is 1atm or 760mmHg), 1mole of any gas occupies a volume of 24dm3 or
2400cm3.
Example
 At



standard temperature and pressure(stp):
1mole of oxygen gas (O2) weighs 32g and will occupy a volume of 22.4dm3.
1mole of ammonia (NH3) gas weighs 17g and will occupy a volume of 22.4 dm3
1mole of carbon dioxide (CO2) gas weighs 44g and will occupy a volume of 22.4dm3.
Converting known volumes of gases to moles
Number of moles (n) of a given gas =𝐆𝐢𝐯𝐞𝐧 𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐚 𝐠𝐚𝐬 =
𝐌𝐨𝐥𝐚𝐫 𝐠𝐚𝐬 𝐯𝐨𝐥𝐮𝐦𝐞
𝐕
𝐕𝐦
NOTE:
 At stp, Vm =22.4 dm3/mol
 At rtp, Vm = 24 dm3/mol
Then n =
or
n=
𝐕 ( 𝐢𝐧 𝐝𝐦𝟑 𝐨𝐫 𝐥
𝟐𝟐.𝟒𝐝𝐦𝟑/𝐦𝐨𝐥
𝐕 (𝐢𝐧 𝐝𝐦𝟑 𝐨𝐫 𝐥)
𝟐𝟒 𝐝𝐦𝟑/𝐦𝐨𝐥
and
V = n × 22.4 dm3/mol
and
V= n × 24 dm3/mol
at STP
at rtp
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Examples
1. Calculate the number of moles of 5.6 dm3 of NH3 gas at stp.
Answer
Given: V = 5.6 dm3
Vm = 22.4 dm3/mol ( at stp)
Then n = 𝑽 = 𝟓.𝟔𝒅𝒎𝟑
= 0.25 mol
𝟐𝟐.𝟒𝒅𝒎𝟑/𝒎𝒐𝒍
𝑽𝒎
2. Calculate the moles of 1.12 dm3 of hydrogen gas at stp.
Answer
n=
𝑽
𝑽𝒎
=
𝟏.𝟏𝟐𝒅𝒎𝟑
𝟐𝟐.𝟒𝒅𝒎𝟑/𝒎𝒐𝒍
=0.05mol
3. How many molecules are there in 5.6 dm3 of ammonia gas at stp?
Answer
Number of moles in 5.6 dm3 of NH3
n=
𝑽
𝑽𝒎
=
𝟓.𝟔𝒅𝒎𝟑
𝟐𝟐.𝟒𝒅𝒎𝟑/𝒎𝒐𝒍
= 0.25 mol
 Then, number of molecules = moles × Avogadro number
= 0.25 mol × 6.022× 1023molecules /mol
= 3.01× 1023molecules
4. Find out the volume of 14 g of nitrogen gas at stp. (N=14)
Answer
 Number of moles =

Then, n =
𝒎𝒂𝒔𝒔
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
𝑽 ( 𝒊𝒏 𝒅𝒎𝟑 𝒐𝒓 𝒍)
=
𝟏𝟒𝒈
𝟐𝟖𝒈/𝒎𝒐𝒍
= 0.5 mol
and V= n × 22.4 dm3/mol
𝟐𝟐.𝟒𝒅𝒎𝟑/𝒎𝒐𝒍
V= 0.5 mol× 22.4 dm3/mol
V= 11.2 dm3
5. Find out the volume of 6.022 × 1022 molecules of ammonia (NH3) gas at stp.
Answer
 Number of moles =
𝑮𝒊𝒗𝒆𝒏 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔
𝑨𝒗𝒐𝒈𝒂𝒅𝒓𝒐 𝒏𝒖𝒎𝒃𝒆𝒓

Then n =
𝑽 ( 𝒊𝒏 𝒅𝒎𝟑 𝒐𝒓 𝒍)
𝟐𝟐.𝟒𝒅𝒎𝟑/𝒎𝒐𝒍
=
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟐
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑
= 0.1 mol
V = n × 22.4
V = 0.1 mol × 22.4 dm3/mol
V = 2.24 dm3
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
67
Exercises
1) Calculate the number of moles in each of the following:
a) 11 g of CO2
(RAM of C =12, O =16)
b) 3.01 × 1022 molecules of CO2
(Avogadro’s number = 6.022 × 1023)
c) 1.12 litres of CO2 at STP
(molar gas volume at stp = 22.4 l)
2) Find out volume of the following at STP.
a) 14 g of nitrogen gas.
b) 6.022 × 1022 molecules of ammonia (NH3)
c) 0.1 moles of Sulphur dioxide (SO2)
(N =14, Avogadro’s number 6.022 × 1023, molar gas volume at STP = 22.4 l)
10. Calculation of mass percentage composition of
an element in a compound
Percentage by mass of an element
y=
Examples
1.
a)
b)
c)
(𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐚𝐭𝐨𝐦𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐲 × 𝐑𝐀𝐌)
𝐑𝐞𝐥𝐚𝐭𝐢𝐯𝐞 𝐟𝐨𝐫𝐦𝐮𝐥𝐚 𝐦𝐚𝐬𝐬
× 100
What is the percentage by mass of nitrogen in the following compounds?
NH3
(NH4)2SO4
NaNO3
(N= 14, H=1, S=32, O=16, Na=23)
Answers
a) RFM of NH3 = (1×14) + (3×1) = 17
RAM of N=14
Number of atoms of N × RAM of N = 1×14 =14
% of N = 𝟏𝟒 ×100 = 83.35 %
𝟏𝟕
b) RFM of (NH4)2SO4 = (2×14) + (8×1) + (1×32) + (4×16) = 132
Number of atoms of N × RAM of N = 2×14 = 28
% of N = 𝟐𝟖 ×100 = 21.21 %
𝟏𝟑𝟐
c) RFM of NaNO3 = (1×23) + (1×14) + ( 3×16) = 85
Number of atoms of N × RAM of N = 1× 14 = 14
% of N = 𝟏𝟒 × 100 = 16.47 %
𝟖𝟓
2. Calculate the percentage of oxygen in aluminium sulphate, Al2(SO4)3
(Al=27, S=32 and O=16)
Answer
RFM of Al2(SO4)3 = (2×27) + (3×32) + (12×16) = 342
Number of atoms of O × RAM of O = 12 × 16= 192
% of O = 𝟏𝟗𝟐 ×100 = 56.1 %
𝟑𝟒𝟐
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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3. Calculate the percentage of water of crystallization in washing soda whose formula is
Na2CO3.10H2O. (Na=23, C=12, O=16, H=1)
Answer
RFM of Na2CO3.10H2O = (2×23) + (1×12) + (13×16) + (20×1) = 286
Mass of H2O = 10 × {(2×1) + 16} = 10 + 18= 180
% of H2O = 𝟏𝟖𝟎 × 100 = 62.94 % by mass
𝟐𝟖𝟔
4. What is the percentage composition of oxygen in hydrated sodium sulphate,
Na2SO4.10H2O? ( Na=23, S=32, O=16, H=1)
Answer
RFM of Na2SO4.10H2O = (2×23) + (1×32) + (14×16) + (20×1) = 322
Number of atoms of O × RAM of O = 14 × 16 = 224
% of O = 𝟐𝟐𝟒 × 100 = 69.56 %
𝟑𝟐𝟐
Exercises
1) Calculate the percentage of carbon in the following compounds.
a) Carbon dioxide, CO2 (RAM of C = 12, O =16)
b) Methane, CH4 (RAM of C = 12, H =1)
c) Calcium carbonate, CaCO3 (RAM of Ca = 40, C = 12, O =16)
d) Ethanol, C2H5OH (RAM of C = 12, O =16, H = 1)
2) Calculate the percentage composition of oxygen in sodium sulphate decahydrate,
Na2SO4•10H2O. (Na =23, S =32, O =16, H = 1)
3) Calculate the percentage composition of phosphorus in calcium phosphate, Ca3(PO4)2.
(Ca = 40, P =31, O = 16)
4) Hydrated magnesium sulphate has the formula, MgSO4•7H2O.
a) Determine the percentage composition of each element present by mass.
b) Calculate the percentage composition of water of crystallization by mass.
(Mg =24, S = 32, O =16, H = 1)
11. Empirical formula and molecular formula
a. Empirical formula shows the simplest whole number ratio of the atoms present in a
compound.
b. Molecular formula shows the exact number of atoms of the various elements present
in a molecule of a compound.
Empirical formula can be determined from the composition of the compound given in
terms of % composition by mass of the elements.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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 Steps for writing the empirical formula
1) Write the symbols of the elements of the compound.
2) Write the percentages composition of each element.
3) Divide the percentage or mass of each element by its atomic mass (to find mole of each
element).
4) Divide each result by the smallest result (to find smallest atomic ratio).
NOTE: For step 4
 If the atomic ratios obtained in 4 are not the whole number, they should be multiplied
by a suitable common factor to convert each of them to the whole numbers (or
approximatively equal to the whole numbers). Minor fractions are ignored by rounding
up or down (ex: 7.95 = 8 or 2.01=2).
 If your result ends in one of the following, multiply all results by the same factor (same
whole number).
0.5
1.5
?.20
2.5
?.5 × 2
?.33
?.25
?.40
3.5
?.66
×3
?.75 ×4
?.60
×5
4.5
?.80
….

Then, if the ratio is very near to a whole number, rounding it up or down to the
whole number.
5) Write down the symbols of the various elements side by side and put the above numbers
as the subscripts to the lower right hand corner of each symbol. This will represent the
empirical formula of the compound.
 Writing molecular formula
After calculating the empirical formula as described above;
6) Find out the empirical formula mass by adding the atomic masses of all the atoms
present in the empirical formula of the compound.
7) Divide the molecular mass by the empirical formula mass and find out the value of
n
8) Multiply the empirical formula of the compound with value of
molecular formula of the compound.
n so as to find out the
Example1
A substance, on analysis, give the following percentage composition:
Na= 43.4%, C= 11.3%, O=45.3%. Calculate its empirical formula. (Na=23, C=12, O=16)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Answer
Elements
Na
Mass (%)
43.4
Moles
43.4
Simplest mole ratio
1.887
0.942
Simplest atomic ratio
23
C
O
11.3
=1.887
11.3
12
= 0.942
0.942
=
0.942
=2
2
45.3
45.3=2.831
16
2.831
=
0.942
1
1
3
3
Empirical formula Na2CO3
Example2:
An organic compound on analysis gave the following data: C =57.82%, H = 3.6% and the
rest is oxygen. Its vapour density is 83. Find its empirical and molecular formula.
C =12, H =1, O=16
Answer
Elements
C
Mass (%)
57.82
3.6
Moles
57.82
3.6
12
Simplest mole ratio
(Divide by the smallest
result)
Simplest atomic ratio
(Multiply all the above
results by 2 to get whole
number)
4.8
2.4
H
= 4.8
1
=2
2 ×2 = 4
100-(57.82+3.6) =38.58
=3.6
3.6
2.4
O
= 1.5
1.5 × 2 =3
38.58
=2.4
16
2.4
=1
2.4
1× 2 =2
Empirical formula C4H3O2
Calculation of molecular formula:
 Finding the empirical formula mass= (4×12) + (3×1) + (2×16) =83
 Molecular mass = 2 × vapour density = 2×83 =166
 Then, empirical formula mass × n = molecular formula mass
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
71
83n =166
n=
n=2
𝟏𝟔𝟔
𝟖𝟑
Therefore, molecular formula = empirical formula × n
= (C4H3O2)2
or C8H6O4
Example 3
An analysis of organic compound showed that it has 39.13% carbon, 52.23% oxygen and the
remaining is hydrogen. Determine the empirical formula of the compound.
Answer
Elements
C
Mass (%)
H
39.13
39.13
12
Moles
Simplest mole ratio
(Divide by the smallest
result)
Simplest atomic ratio
(Multiply all the above
results by 2 to get whole
number)
3.26
3.26
O
100-(39.13+52.23)= 8.64
8.64
= 3.26
1
8.64
=1
3.26
1×3=3
= 8.64
52.23
52.23
16
3.26
3.26
= 2.65
2.65×3=7.95 ~8
= 3.26
=1
1×3=3
Empirical formula C3H8O3
Example4:
A compound has the following composition: Mg =9.76% , S =13.01% , O = 26.01%
H20= 51.22%. What is its empirical formula? (Mg =24, S=32, O=16 , H=1)
Answer
Mg
9.76 = 0.406
24
0.406
0.406
=1
S
13.01
O
=0.406
32
0.406
0.406
26.01
16
=1
= 1.625
1.625
0.406
=4
H 2O
=2.846
51.22
18
2.846
0.406
=7
Therefore, empirical formula is MgSO4 .7H20
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
72
Example5
A compound has the following composition, 69.42% carbon, 4.13% hydrogen and the rest
oxygen.
a) Determine the empirical formula of the compound.
b) If the relative molecular mass is 242, determine its molecular formula.
(C =12, H= 1, O=16)
Answer
a)
Elements
C
Mass (%)
Moles
69.42.82
69.42
12
Simplest mole ratio
(Divide by the smallest
value)
Simplest atomic ratio
(Multiply all the above
results by 2 to get whole
number)
H
= 5.785
5.785
1.65
4.13
26.45
4.13=4.13
26.45
1
16
4.13
= 3.5
3.5 ×2 = 7
O
1.65
1.65
= 2.5
2.5 × 2 = 5
1.65
= 1.65
=1
1× 2 =2
Empirical formula C7H5O2
b) (Empirical formula mass) × n = Molecule formula mass
{(7×12) + (5×1) + (2×16)} × n = 242
(84 + 5 +32)n =242
121n = 242
𝟐𝟒𝟐
n = 𝟏𝟐𝟏
n =2
Then, molecular formula is (C7H502 )2
or C14H10O4
Example 6
Octane is a hydrocarbon, it contains only carbon and hydrogen. It is 84.2% carbon and
15.8% hydrogen by mass. Its molecular mass is 114. What is its molecular formula?
Answer
First find the empirical formula for the compound
From the %, we can say that in 100 g of octane, 84.2 g is carbon and 15.8 g is hydrogen.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
73
So 84.2 g of carbon combines with 15.8 g of hydrogen.
Elements
Mass (%)
Moles
Simplest mole ratio
(Divide by the smallest
value)
Simplest atomic ratio
(Multiply all the above
results by 4 to get whole
number)
C
H
84.2
84.2
12
19.8
= 7.02
7.02
7.02
=1
1 ×4 = 4
15.8
1
= 15.8
15.8
7.02
= 2.25
2.25 × 4 = 9
Empirical formula C4H9
Then use molecular mass to find the molecular formula:
Finding the empirical formula mass= (4×12) + (9×1)=57
Then, empirical formula mass × n = molecular formula mass
57n = 114
n = 114 = 2
57
Therefore, molecular formula = empirical formula × n
= (C4H9)2
or C8H18
Exercises
1) Write the empirical formula of the compounds having molecular formulae:
a) C6H6
b) C6H12
c) H2O2
d) H2O
e) N2O4
f) Fe2O3
2) The empirical formula of a hydrocarbon is C2H3. The hydrocarbon has a relative
molecular mass of 54. Determine its molecular formula. (C = 12, H = 1)
3) A compound has an empirical formula of C3H6O. its relative molecular mass is 116.
a) Determine its molecular formula.
b) Calculate the percentage composition of carbon by mass in the compound.
(C =12, H =1, O =16)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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4) What is the simplest formula of the compound which has the following percentage
composition: carbon 80 %, hydrogen 20 %? If the molecular mass is 30, calculate its
molecular formula. (C =12, H =1)
5) An organic compound on analysis gave the following data: C = 57.82 %, H = 3.6 % and
the rest is oxygen. Its vapour density is 83. Find its empirical formula and molecular
formula.
(C =12, H =1, O =16)
6) 2.746 g of a compound gave on analysis 1.94 g of silver, 0.268 g of Sulphur and 0.538 g
of oxygen. Find the empirical formula of the compound.
(atomic masses: Ag = 108, S = 32, O =16)
7) The composition of a compound is 24.24 % carbon, 4.04 % hydrogen and 71.72 %
chlorine.
a) Determine the empirical formula of the compound.
b) If the relative molecular mass is 99, determine its molecular formula.
(C =12, H = 1, Cl =35.5)
8) A compound contains 40 % carbon, 6.67 % hydrogen and the rest oxygen. Determine its
empirical formula and hence, its molecular formula given that its relative molecular
mass is 180.
(C =12, H =1, O =16)
9) An organic compound X was analyzed and found to be constituted of the following
elements with their percent composition by mass:
Mg=28.03%, Si=21.60%, H=1.16%, O=49.21%
The molecular mass of compound X is 521 g/mole.
(Atomic mass: Mg=24, Si=28, H=1, O=16)
a) Determine the empirical formula of compound X.
b) Determine the molecular formula of compound X.
12. Stoichiometric calculations
Stoichiometric calculation is the calculation that shows the relationship between the
amounts of reactants and of the products in an equation.
The numbers appearing before the formula units in the equation show mole ratio of the
reactants and of the products.
Example
The equation for the thermal decomposition of sodium nitrate is;
heat
2NaNO3(s)
2NaNO2(s) + O2(g)
Mole ratios are 2: 2 : 1
The mole ratio of the reactants and of the products in an equation can be used to calculate
reacting masses or volumes.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Steps that must be followed in stoichiometric calculations are:
1) Write a balanced equation.
2) Convert the known mass or volume from mass (in g) or volume (in l) to moles.

Moles of given substance =
Given mass
Molar mass
or Moles of given substance =
Given
volume
Molar gas volume
3) Determine the mole ratio from the coefficient in the balanced equation to convert from
moles of known to moles of unknown.
Mole ratio
=
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐝 (𝐮𝐧𝐤𝐧𝐨𝐰𝐧) 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐠𝐢𝐯𝐞𝐧( 𝐤𝐧𝐨𝐰𝐧) 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞
or
Mole ratio =
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
4) Convert from moles of the given substance to moles of the unknown
substance, by multiplying the amount of moles of the given substance by the mole
ratio. (mole to mole).
Moles of desired (unknown) substance = 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧 ×
𝐦𝐨𝐥𝐞 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧
5) Convert the calculated moles (moles of the unknown substance) from moles to
the required mass or volume.

Mass = Moles of desired substance × molar mass of desired
substance

or
Volume = Moles of desired substance × Molar gas volume
i) Mass-mass relationship
In mass-mass problems, you are given the mass of a compound in the problem and asked to
find the mass of another compound.
Example
Reduction of copper(II) oxide by carbon is as follows:
heat
2CuO(s) + C(s)
2Cu(s) + CO2(g)
a) Calculate the mass of carbon that reduces 31.8 g of copper (II) oxide.
b) Determine the mass of carbon dioxide formed. (Cu = 63.5, O = 16, C =12)
Answer
a) Calculation of mass of C
1. Balanced equation
heat
2CuO(s) + C(s)
2Cu(s) + CO2(g)
31.8𝑔
𝑀𝑎𝑠𝑠
2. Moles of CuO =
=
= 0.4mol
79.5𝑔/𝑚𝑜𝑙
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
3. Mole ratio
=
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
=
𝟏
𝟐
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4. Moles of the desired reactant which is C are 0.4mol × 𝟏 = 0.2 mol
𝟐
5. Mass of C = Moles × molar mass
= 0.2 mol ×12g/mol = 2.4 g of C
Or you can also use the following expression
𝐦 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 = 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐤𝐧𝐨𝐰𝐧 × 𝐦𝐨𝐥𝐞 𝐫𝐚𝐭𝐢𝐨 × 𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐮𝐧𝐤𝐧𝐨𝐰𝐧
This means that:
𝐦 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 =
Then,
𝒎 𝒐𝒇 ( 𝑪 ) =
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐤𝐧𝐨𝐰𝐧
𝐌𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐤𝐧𝐨𝐰𝐧
𝟑𝟏.𝟖 𝒈
𝟕𝟗.𝟓 𝒈/𝒎𝒐𝒍
×
𝟏 𝒎𝒐𝒍
𝟐 𝒎𝒐𝒍
×
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧
× 𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐮𝐧𝐤𝐧𝐨𝐰𝐧
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧
from balanced
chemical equation
× 𝟏𝟐 𝒈/𝒎𝒐𝒍 = 𝟐. 𝟒 𝒈 𝒐𝒇
b) Calculation of mass of CO2
 Mole ratio
=
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
 Moles of CO2 = 0.4 mol ×
1 𝑚𝑜𝑙
2𝑚𝑜𝑙
=
𝟏
𝟐
= 0.2 mol
 Mass of CO2 = Moles of CO2 × Molar mass of CO2
= 0.2mol × 44g/mol
= 8.8 g of CO2
Or you can also use the following expression
𝒎 𝒐𝒇 (𝑪𝑶𝟐 ) =
𝟑𝟏.𝟖 𝒈
𝟕𝟗.𝟓
𝒈/𝒎𝒐𝒍
×
𝟏 𝒎𝒐𝒍
𝟐 𝒎𝒐𝒍
× 𝟒𝟒 𝒈/𝒎𝒐𝒍 = 𝟖. 𝟖 𝒈 𝒐𝒇 𝑪𝑶𝟐
ii) Mass - Volume relationships
In this case, either the mass of a compound will be given and the volume of another is
asked, or the volume of a gas will be given and the mass of another compound will be asked.
Example
Consider the following equation.
Fe2O3(s) + 3 CO(g)
2Fe(s) + 3CO2(g)
Determine the volume of carbon dioxide gas that will be produced from 112.5 g of iron at
stp.
(Fe = 56, O = 16, C = 12 , Molar gas volume at stp is 22.4 dm3/mol)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
77
Solution
1. Fe2O3(s) + 3 CO(g)
2. Moles of Fe = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐹𝑒
2Fe(s)
=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐹𝑒
3. Mole ratio
=
112.5𝑔
56𝑔/𝑚𝑜𝑙
+
3CO2(g)
= 2.008 mol
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
4. Moles of CO2 = 2.008 mol ×
3𝑚𝑜𝑙
=
𝟑
𝟐
= 3.012 mol
2𝑚𝑜𝑙
5. Volume of CO2 = Moles of CO2 × Molar gas volume
= 3.012 mol ×22.4 dm3/mol = 67.5 dm3 of CO2
Or you can use the following expression
𝑉 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑘𝑛𝑜𝑤𝑛 ×
𝑚𝑜𝑙𝑒 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
× 𝒎𝒐𝒍𝒂𝒓 𝒈𝒂𝒔 𝒗𝒐𝒍𝒖𝒎𝒆
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑘𝑛𝑜𝑤𝑛
Then,
𝑽(𝑪𝑶𝟐 ) =
𝟏𝟏𝟐.𝟓 𝒈
𝟓𝟔 𝒈/𝒎𝒐𝒍
×
𝟑 𝒎𝒐𝒍
𝟐 𝒎𝒐𝒍
× 𝟐𝟐. 𝟒 𝒅𝒎𝟑/𝒎𝒐𝒍 = 𝟔𝟕. 𝟓 𝒅𝒎𝟑
iii) Volume- Volume relationships
A volume-volume problem concerns only the gaseous compounds of a reaction. In this case,
the volume of one compound will be given and the volume of another is asked.
Example
Consider the reaction below:
4 NH3(g) + 3 O2(g)
2 N2(g) + 6 H2O(g)
What is the volume of ammonia gas will react with 22.4 L of oxygen gas at stp?
Solution
Moles of O2 =
Mole ratio
22.4 𝑙
22.4 𝑙/𝑚𝑜𝑙
=
= 1mol
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
=
𝟒
𝟑
Moles of NH3 = 1mol × 4𝑚𝑜𝑙 = 1.33 mol
3𝑚𝑜𝑙
Volume of NH3 = Moles of NH3 × Molar gas volume
= 1.33 mol × 22.4 l/mol = 29.7 l of NH3
Or you can use the following expression
𝑉 𝑜𝑓 𝑡ℎ𝑒 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑘𝑛𝑜𝑤𝑛 ×
𝑚𝑜𝑙𝑒 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑘𝑛𝑜𝑤𝑛
× 𝒎𝒐𝒍𝒂𝒓 𝒈𝒂𝒔 𝒗𝒐𝒍𝒖𝒎𝒆
Then,
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
78
𝑽(𝑵𝑯𝟑 ) =
𝟐𝟐.𝟒 𝒍
𝟐𝟐.𝟒
𝒍/𝒎𝒐𝒍
×
𝟒 𝒎𝒐𝒍
𝟑 𝒎𝒐𝒍
× 𝟐𝟐. 𝟒 𝒍/𝒎𝒐𝒍 = 𝟐𝟗. 𝟕 𝒍
Exercises
1) Potassium chlorate (V) decomposes when heated as below
2 KClO3(s)
2 KCl(s) + 3 O2(g)
a) Calculate the mass of potassium chloride formed when 49 g of potassium chlorate
(V) completely decomposes.
b) Determine the mass of oxygen gas liberated. (K =39, Cl = 35.5, O = 16)
2) Lead (II) nitrate when heated decomposes to lead (II) oxide, nitrogen dioxide and oxygen
gas.
a) Write a balanced equation for the reaction.
b) Calculate the mass of lead (II) oxide formed when 463.4 g of lead (II) nitrate
decomposes.
c) Calculate the mass of nitrogen (II) oxide (nitrogen dioxide) formed.
(Pb = 207, N = 14, O = 16)
3) Copper (II) nitrate decomposes when heated to copper (II) oxide, nitrogen dioxide and
oxygen gas.
a) Write a balanced chemical equation to represent the reaction.
b) Calculate the mass of copper (II) oxide formed when 15.04 g of copper (II) nitrate
decomposes completely.
c) Determine the mass of nitrogen dioxide formed during the reaction.
(Cu = 63.5, N = 14 O = 16)
4) Sodium reacts with water to form sodium hydroxide and hydrogen gas.
a) Write a balanced chemical equation to represent the reaction
b) Calculate the mass of sodium that would react with water to form 300 g of sodium
hydroxide.
c) Calculate the mass of hydrogen gas that would be liberated during the reaction.
(Na = 23, H = 1, O =16)
5) When calcium carbonate is heated strongly, it decomposes as shown below:
Heat
CaCO3(s)
CaO(s) + CO2(g)
a) Calculate the relative formula mass of:
i) CaCO3
ii) CaO
b) Calculate the mass of calcium oxide formed when 150 g of calcium carbonate is
completely decomposed.
c) Determine the volume of carbon dioxide evolved at standard conditions.
(Ca = 40, C = 12, O = 16 , Molar gas volume at STP = 22.4 Litres)
6) Sodium hydrogen carbonate decomposes when heated as shown in the equation below:
NaHCO3((s) Heat
Na2CO3(s) + H2O(l) + CO2(g)
a) 67.2 g of sodium hydrogen carbonate is completely decomposed by heat. Calculate
the volume of carbon dioxide gas formed at standard temperature and pressure.
b) Determine the mass of the residue (Na2CO3)
(Na = 23, C = 12, O =16, Molar gas volume at STP =22.4 litres)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
79
7) Magnesium reacts with dilute sulphuric acid to form magnesium sulphate and hydrogen
gas.
a) Write a balanced equation to represent the reaction.
b) Calculate the volume of hydrogen gas liberated at STP when 4.8 g of magnesium
reacts completely with dilute sulphuric acid.
c) Determine the mass of magnesium sulphate formed in the reaction.
(Mg = 24, H = 1, S =32, O = 16)
8) What volume of nitrogen would react with excess hydrogen to produce 10 cm3 of
ammonia at STP? (Molar gas volume at STP = 22.4 dm3)
13. Limiting reactants
In a chemical reaction,
 The limiting reactant is the reactant that is used up completely (runs out
first). This stops the reaction and no further products are made. The limiting reactant
determines how much product you can make.
 Excess reactant is the reactant that is not completely used up during the chemical
reaction. There is some of this reactant leftover.
Deciding which reactant is the limiting reactant and the reactant in
excess
Below are the steps that must be followed:
1. Write balanced chemical equation.
2. Identify moles of each reactant present.
3. Divide moles of each reactant by its stoichiometric coefficient.
4. Smallest number indicates limiting reactant.
5. Use the amount of limiting reactant to calculate the amount of product produced
(if asked)
6. If necessary, calculate how much is left in excess of non-limiting reactant.
Example1
Calculate the limiting reactant when 2.4 g of magnesium is burnt in 10.0 g of oxygen.
Answer
1. Balanced equation: 2Mg(s) + O2(g)
2.4𝑔
2. Moles of Mg = 𝑚𝑎𝑠𝑠
=
= 0.1mol
Moles of O2 =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
=
24𝑔/𝑚𝑜𝑙
10𝑔
32𝑔/𝑚𝑜𝑙
2MgO
= 0.3125 mol
3. Divide moles of each reactant by its stoichiometric coefficient.
 For Mg : 0.1 𝑚𝑜𝑙 = 0.05
2 𝑚𝑜𝑙
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
80
 For O2 : 0.3125 𝑚𝑜𝑙 = 0.3125
1 𝑚𝑜𝑙
Smallest number indicates limiting reactant. then,
Limiting reactant is Mg
Excess reactant is O2
Example2
Consider the reaction:
Mg + S
MgS
If 2.00 g of magnesium is reacted with 2.00 g of Sulphur
a) Which is the limiting reagent?
b) How much magnesium sulphide (MgS) can be produced?
c) Calculate the amount of one the reactants which remains unreacted?
(Mg =24, S= 32)
Answer
a) Finding of limiting reagent
1. Balanced equation is, Mg + S
MgS
2𝑔
2. Moles of Mg : 𝑚𝑎𝑠𝑠
=
=0.08mol
24𝑔/𝑚𝑜𝑙
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
Moles of S :
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
=
2𝑔
32𝑔/𝑚𝑜𝑙
= 0.0625 mol
3. Divide moles of each reactant by stoichiometric coefficient
 For Mg : 0.08 𝑚𝑜𝑙 = 0.08

1𝑚𝑜𝑙
for S : 0.0625 𝑚𝑜𝑙 = 0.0625
1 𝑚𝑜𝑙
Limiting reagent is S
b) Mole ratio
=
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
Moles of MgS : O.0625mol ×
1 𝑚𝑜𝑙
=
𝟏
𝟏
= 0.0625 mol of MgS
1 𝑚𝑜𝑙
Mass of MgS formed = Moles of MgS × molar mass of MgS
= 0.0625 mol × 56g/mol = 3.5g of MgS
To calculate moles of excess reactant = Initial quantity (mole or mass) of the
excess reactant – amount consumed by the complete consumption of the limiting
reactant.
c) Moles of Mg left unreacted = 0.08 − 0.0625 = 0.0175 mol
Mass of Mg left unreacted = Mole of Mg left unreacted × Molar mass of Mg
= 0.0175 mol × 24g/mol = 0.42 g of Mg left unreacted
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
81
Example 3
Consider the reaction
2 H2(g) + O2(g)
2 H2O(g)
If 20 g of H2 gas is reacted with 96 g of O2 gas.
a) Which reactant is the limiting reactant?
b) How much H20 is produced?
c) How much of the excess reactant remains?
Answer
a) Finding of limiting reagent
 Balanced equation is, 2 H2(g) + O2(g)

Moles of H2 :

Moles of O2 :
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
=
=
20𝑔
2𝑔/𝑚𝑜𝑙
96𝑔
2 H2O(g)
= 10 mol
32𝑔/𝑚𝑜𝑙
= 3 mol
 Divide moles of each reactant by stoichiometric coefficient
 For H2 : 10 𝑚𝑜𝑙 = 5

2𝑚𝑜𝑙
for O2 : 3 𝑚𝑜𝑙 = 3
1 𝑚𝑜𝑙
Limiting reagent is O2
b) Use the equation to get the mole ratio of the required (unknown) substance to the
known (limiting reactant)
Mole ratio
=
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐮𝐧𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐤𝐧𝐨𝐰𝐧 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞 𝐢𝐧 𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
Moles of H2O : 3 mol ×
2𝑚𝑜𝑙
=
𝟐
𝟏
= 6 mol of H2O
1 𝑚𝑜𝑙
Mass of H2O formed = Moles of H2O × molar mass of H2O
= 6 mol × 18 g/mol = 108 g of H2O
Or you can also use the following expression
𝒎 𝒐𝒇 (𝑯𝟐𝑶) 𝒇𝒐𝒓𝒎𝒆𝒅 =
96𝑔
32𝑔/𝑚𝑜𝑙
×
𝟐 𝒎𝒐𝒍
𝟏 𝒎𝒐𝒍
× 𝟏𝟖 𝒈/𝒎𝒐𝒍 = 𝟏𝟎𝟖 𝒈
c) Determine grams of hydrogen that react.
Use the equation to get the mole ratio of the required (unknown) substance to the
known (limiting reactant).

The mole ratio between H2 (required) and O2 (limiting reactant) is

Moles of hydrogen used: 3 mol ×

Mass of hydrogen used = Moles of H2 × molar mass of H2
= 6 mol × 2g/mol = 12g of H2
2𝑚𝑜𝑙
= 6 mol
𝟐
𝟏
1 𝑚𝑜𝑙
Excess mass of hydrogen: 20 g ― 12 g = 8 g of H2 remaining
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
82
Or you can also use the following expression
𝒎 𝒐𝒇 (𝑯𝟐 ) 𝒖𝒔𝒆𝒅 =
96𝑔
32𝑔/𝑚𝑜𝑙
×
𝟐 𝒎𝒐𝒍
𝟏 𝒎𝒐𝒍
× 𝟐 𝒈/𝒎𝒐𝒍 = 𝟏𝟐 𝒈
Excess mass of hydrogen: 20 g ― 12 g = 8 g of H2 remaining
Exercises
1) Take the reaction:
NH3 + O2
NO + H2O
In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2.
a) Which reactant is the limiting reactant? Answer: O2
b) How many grams of NO are formed?
Answer: 2.63 g NO
c) How much of the excess reactant remains after the reaction?
Answer: 1.76 g NH3 left
2) Copper reacts with silver nitrate solution according to the equation:
Cu(s) + 2AgNO3(aq)
Cu(NO3)2(aq)
+ 2Ag(s)
If 0.50 mole of copper is added to 1.5 mole of silver nitrate, which is the limiting reagent
and how many moles of silver are formed?
3)
Iron reacts with chlorine gas to form iron (III) chloride.
a) Write a balanced chemical equation for the reaction.
b) 44.8 g of iron is reacted with 30 litres of chlorine at standard conditions.
i) Determine the limiting reactant.
ii) Determine the mass of iron (III) chloride formed.
(Fe = 56, Cl = 35.5 , Molar gas volume at STP = 22.4 litres)
4) Methane, CH4 is a hydrocarbon that burns in oxygen completely to form carbon dioxide
and steam (gaseous water)
a) Write a balanced equation to represent the reaction.
b) 4.48 litres of methane is burnt in 10 litres of oxygen at STP.
i) Calculate the limiting reactant.
ii) Determine the volume of carbon dioxide formed.
5) Propane gas is used as a gaseous fuel. A mixture of 13.44 litres of propane and 72 litres
of oxygen was ignited.
The equation for the reaction is:
C3H8(g) + 5O2(g)
3CO2(g)
+ 4H2O(g)
a) Determine the limiting reactant.
b) Calculate the volume of the excess reactant that remained unreacted.
c) Calculate the volume of carbon dioxide formed at standard conditions.
(Molar gas volume at STP = 22.4 l)
6) A mixture of 5.6 litres of ethane, C2H6 gas and 16.8 litres of oxygen was ignited. Carbon
dioxide and water were formed.
The equation for the reaction is:
2C2H6(g) + 7O2(g)
4CO2(g)
+ 6H2O(g)
a) Calculate the limiting reactant.
b) Determine the volume of carbon dioxide formed at standard conditions.
(Molar gas volume at STP = 22.4 l)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
83
14. GAS LAWS
i) Boyle’s law
Boyle’s law describes the relationship between volume and pressure of gas at constant
temperature.
The law states that:
The volume of a fixed mass of a gas is inversely proportional to pressure at
constant temperature.
This means that, if volume of the gas is increased from V i to Vf, its pressure will decrease
from Pi to Pf.
Thus
Pi × Vi = Pf × Vf
(at constant temperature)
Units of pressure are:
 Pascal (Pa)
 Atmosphere (atm)
 Millimeter of mercury (mmHg)
Note: 1 atm = 760 mmHg = 101326 Pa
Example
A certain gas occupies a volume of 80 cm3 at a pressure of 420 mmHg. Calculate the volume
it will occupy when the pressure is increased to 800 mmHg at a constant temperature.
Answer
Pi = 420 mmHg
Pf = 800mmHg
Vi = 80cm3
Vf = ?
It is known that Pi×Vi=Pf×Vf means that Vf = 𝑷𝒊 ×𝑽𝒊 = 𝟒𝟐𝟎 ×𝟖𝟎 = 42cm3
𝑷𝒇
𝟖𝟎𝟎
Exercises
1) State Boyle’s law.
2) 20 cm3 of oxygen gas was compressed from a pressure of 840 mmHg to 1600 mmHg at a
constant temperature. Determine the new volume of the gas.
3) The volume of a gas at 180 Pa is reduced from 100 cm3 to 60 cm3. What is the pressure.
Temperature remains constant.
4) In an experiment, 60 cm3 of gas X at a pressure of 100 Pa had its volume increased to
150 cm3 at a constant temperature. Determine the new pressure of the gas.
5) A certain gas occupies 30 dm3 at 760 mmHg pressure. Find the volume occupied by the
same gas at 800 mmHg if the temperature is kept constant.
6) A certain mass of a gas occupies 48 ml, at a pressure of 720 mmHg. What is the volume
when pressure in increased to 960 mmHg? Temperature remains constant.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
84
ii) Charles’s law
Charles’s law describes the relationship between volume and temperature of a gas at
constant pressure.
Charles’s law states that:
The volume of a fixed mass of a gas is directly proportional to its absolute
temperature at a constant pressure.
This means that, if temperature is increased from Ti to Tf, then its volume will increase
from Vi to Vf.
Thus
𝑽𝒊
𝑻𝒊
=
𝑽𝒇
𝑻𝒇
(at constant pressure)
Example
A fixed mass of a gas has a volume of 22.4 cm3 at 0 0C. The gas is warmed to room
temperature, 25 0C. Calculate the new volume occupied by the gas if pressure remains
constant.
Answer
Vi = 22.4cm2
Ti = (0 + 273)=273K
𝑽𝒊
𝑻𝒊
=
𝑽𝒇
𝑻𝒇
Vf =
Vf = ?
Tf = (25 + 273)K = 298K
𝑽𝒊 ×𝑻𝒇
𝑻𝒊
=
𝟐𝟐.𝟒 ×𝟐𝟗𝟖
𝟐𝟕𝟑
= 24.45cm3
Exercises
1) State Charle’s law.
2) A certain gas occupies a volume of 25 cm3 at 50 0C. What will be the volume occupied by
the same gas at 75 0C at a constant presuure?
3) A sample of helium has volume of 520 ml at 100 0C. Calculate the temperature at which
the volume will become 260 ml. Assume the pressure is constant.
4) A fixed mass of a gas has volume of 250 cm3 at a temperature of 27 0C and 750 mmHg
pressure. Determine the temperature at which the gas would occupy a volume of 262.5
cm3. Pressure remains constant at 750 mmHg.
5) A gas occupies 2.32 litres at 40 0C. What will be the new volume if the temperature is
raised to 75 0C while pressure remains constant?
6) What will be the volume of a gas at 0 0C which occupies 200 ml at 27 0C?
Assume no change in pressure.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
85
iii) Gay-Lussac’s law
Gay-Lussac’s law relates temperature and pressure of a gas at constant volume.
The law states that:
For a fixed mass of a gas, pressure is directly proportional to temperature at
a constant volume.
This means that, if pressure is increased from Pi to Pf, then its temperature will increase
from Ti to Tf.
𝑷𝒊
Thus,
=
𝑻𝒊
𝑷𝒇
𝑻𝒇
(at constant volume)
Example
A flask containing air is corked when the pressure is 760 mmHg pressure at a temperature
of 17 0C. The temperature of the flask is raised gradually. The cork blows out when
pressure is 900 mmHg. Work the temperature.
Answer
Pi = 760mmHg
Ti = 170C = (17 + 273) = 290K
𝑷𝒊
𝑷𝒇
𝑷𝒇 ×𝑻𝒊
=
Tf =
=
𝑻𝒊
𝑻𝒇
𝑷𝒊
𝟗𝟎𝟎 ×𝟐𝟗𝟎
𝟕𝟔𝟎
Pf = 900mmHg
Tf = ?
= 343.4K
Exercises
1) State Gay-Lussac’s law
2) A car tyre contains 200 cm3 of air at a pressure of 300 kPa and a temperature of 15 0C.
after driving some distance the temperature of the tyre is found to be 41 0C. calculate
the pressure of the tyre if the volume remains constant.
3) A gas is confined in a rigid container exerting a pressure of 250mmHg at a temperature
of 17 0C. to what temperature must the gas be cooled in order for its pressure to become
216 mmHg, volume remaining constant.
4) The pressure of oxygen gas in a steel cylinder is241 kPa at 15 0C. calculate the pressure
of the gas in the cylinder when the temperature rises to 28 0C.
5) A flask containing air is corked when the pressure is 760 mmHg pressure at a
temperature of 17 0C. The temperature of the flask is raised gradually. The cork blows
out when pressure is 900 mmHg pressure. Work the temperature.
iv) Avogadro’s law
This law states that;
The volume(v) is directly proportional to the amount of gas (n) at a constant
temperature and pressure.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
86
This means that, if the amount of gas in a container is increased, the volume increases and
if the amount of gas in a container is decreased, the volume decreases.
𝑽𝒊
Then
𝒏𝒊
=
𝑽𝒇
𝒏𝒇
Example
A 3.80g of oxygen gas in a pump has volume of 150ml at constant temperature and
pressure. If 1.20g of oxygen gas is added into the pump, what will be the new volume of
oxygen gas in the pump if temperature and pressure held constant?
Answer
m1= 3.80g
Moles (n1) =
m2 = (3.80 + 1.20) = 5g
Moles(n2) =
𝑚𝑎𝑠𝑠 1
𝑀𝑚 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛
𝑔𝑎𝑠
𝑚𝑎𝑠𝑠 2
=
=
3.80 𝑔
32𝑔/𝑚𝑜𝑙
5𝑔
32𝑔/𝑚𝑜𝑙
= 0.11875 mol
= 0.15625 mol
𝑀𝑚 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛 𝑔𝑎𝑠
V1 = 150 ml
V2 = ?
𝑽𝟏
𝒏𝟏
𝑽𝟐
𝒏𝟐
V2 =
=
𝑽𝟏 ×𝒏𝟐
𝒏𝟏
=
𝟏𝟓𝟎 𝒎𝒍 ×𝟎.𝟏𝟓𝟔𝟐𝟓 𝒎𝒐𝒍
𝟎.𝟏𝟏𝟖𝟕𝟓 𝒎𝒐𝒍
= 197.3 ml
v) The combined gas law
Combined gas law shows the relationship between volume, pressure and absolute
temperature of a gas.
Its equation is
𝑷𝒊 ×𝑽𝒊
=
𝑷𝒇 ×𝑽𝒇
𝑻𝒊
𝑻𝒇
Example
200 cm3 of nitrogen dioxide(NO2) gas at 30 0C exerts a pressure of 500 mmHg. If the gas is
cooled to 18 0C at 200 mmHg, what volume will the gas occupy?
Answer
Pi =500mmHg
Vi =200cm3
Pf = 200mmHg
Vf = ?
Ti = (30 + 273) K = 303K
𝑷𝒊 ×𝑽𝒊
𝑻𝒊
=
𝑷𝒇 ×𝑽𝒇
𝑻𝒇
Vf =
Tf = (18 + 273)K = 291K
𝑷𝒊 ×𝑽𝒊 ×𝑻𝒇
𝑷𝒇 ×𝑻𝒊
𝟓𝟎𝟎𝒎𝒎𝑯𝒈 ×𝟐𝟎𝟎𝒄𝒎𝟑×𝟐𝟗𝟏𝑲
=
𝟐𝟎𝟎𝒄𝒎𝟑 × 𝟑𝟎𝟑𝑲
= 480.2K
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
87
Exercises
1) A gas occupies a volume of 400 cm3 at 500 K and 760 mmHg pressure. What will be the
temperature of the gas when the volume is 100 cm3 and the pressure is 380 mmHg.
2) A sample of a gas has a volume of 850 cm3 at a temperature of 20 0C and a pressure of
760 mmHg. At what pressure would the same mass of the gas occupy a volume of 500
cm3 if cooled.
3) A certain mass of a gas occupies 0.15 dm3 at 293 K and 98600 Pa. Calculate its volume
at 101000 Pa and 273 K.
4) A balloon contains 80 cm3 of gas at 30 0C and 4 atmospheres. Calculate the volume of
the balloon at 50 0C and 2 atmospheres.
5) A certain gas occupies 600 cm3 at room temperature and 1 atmosphere pressure. At
what temperature will the same gas occupy a volume of 400 cm3 and exert a pressure of
2 atmospheres.
vi) Ideal gas law
This law states that:
The product of the volume and pressure is directly proportional to the
absolute temperature and the amount of substance.
The ideal gas equation combines Avogadro law with the combined gas law.
PV = nRT
Where P= Pressure of ideal gas
V= Volume of ideal gas
n = The amount of gas
T = The absolute temperature
R = The gas constant
(R= 0.082057 ℓ atm /mol K or R= 8.3145 J/ mol K or R= 8.3145 m3 Pa/ mol K )
Example1
5 g of Neon is at 256 mmHg and at a temperature of 35 0C. What is the volume?
R = 0.082057 ℓ atm/ mol K
Answer
P = 256 mmHg
T = 35 0C
m=5g
R = 0.082057 ℓ atm/ mol K
V= ?
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
88
Solution:
We know that
760mmHg
1mmHg
1atm
𝟏 𝒂𝒕𝒎
𝟕𝟔𝟎 𝒎𝒎𝑯𝒈
𝟏 𝒂𝒕𝒎 ×𝟐𝟓𝟔 𝒎𝒎𝑯𝒈
𝟕𝟔𝟎 𝒎𝒎𝑯𝒈
256mmHg
Means P = 0.3368 atm
𝒎𝒂𝒔𝒔
Moles of Neon =
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
=
𝟓𝒈
𝟐𝟎𝒈/𝒎𝒐𝒍
= 0.3368 atm
= 0.25 mol of Neon.
Temperature = (35 + 273) = 308 K
V=
𝒏𝑹𝑻
𝑷
=
= 19 ℓ
𝟎.𝟐𝟓 ×𝟎.𝟎𝟖𝟐𝟎𝟓𝟕 ×𝟑𝟎𝟖
𝟎.𝟑𝟑𝟔𝟖
Example2
A 655 mmHg and 25 0C, a sample of chlorine gas has volume of 750 ml. How many moles of
chlorine gas at this condition? R = 0.082057 l atm/ K mol
Solution
P= 655 mmHg
T = (25 + 273)K = 298 K
V = 750 ml = 0.75 ℓ
n=?
n =
𝑷𝑽
𝑹𝑻
=
(
𝟔𝟓𝟓
)×𝟎.𝟕𝟓
𝟕𝟔𝟎
𝟎.𝟎𝟖𝟐𝟎𝟓𝟕 ×𝟐𝟗𝟖
= 0.026 mol
vii) Graham’s law of diffusion
Diffusion of a gas is the spreading out of gas until it is evenly distributed.
Graham’s law of diffusion states that:
 The rates of diffusion of gases are inversely proportional to the square
roots of their densities at constant temperature and pressure.
In mathematical terms:
R
1
√𝜌
where 𝜌 = Density of the gas
R= rate of diffusion
Now, if there are two gases A and B having RA and RB as their rates of diffusion and 𝜌𝐴 and
𝜌𝐵 as their densities respectively. Then
1
1
R1
and R2
or
√𝜌𝐴
𝑹𝑨
𝑹𝑩
√𝜌𝐵
𝝆𝑩
=√
𝝆𝑨
(1)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
89
Where RA = Rate of diffusion of gas A
RB = Rate of diffusion of gas B
𝜌𝐴 = Density of gas A
𝜌𝐵 = Density of gas B
Example
The density of Sulphur dioxide is 2.9 g/dm3 while that of carbon dioxide is 1.98 g/dm3.
Compare their rates of diffusion.
Answer
𝑅𝐶𝑂2
𝑅𝑆𝑂2
=√
𝜌𝑆𝑂2
𝜌𝐶𝑂2
= √
2.9
1.98
= 1.21
𝑅𝐶𝑂2 = 1.21 𝑅𝑆𝑂2
This means that, the rate of diffusion of CO2 is 1.21 times faster than SO2
We know that: Molecule mass = 2 × vapour density
Therefore, the above expression may be written as:
𝑅𝐴
𝑅𝐵
𝜌𝐵
𝑅𝑀𝑀𝐵
= √𝜌𝐴 = √
𝑅𝑀𝑀
𝑹
𝑨
𝑹𝑩
=√
𝝆𝑩
𝝆𝑨
𝑹𝑨
Then
𝑹𝑩
Thus,
=√
𝐴
𝑹𝑴𝑴𝑩
𝟐
𝑹𝑴𝑴𝑨
𝟐
=√
𝑹𝑴𝑴
𝑩
𝑹𝑴𝑴𝑨
𝑹𝑴𝑴𝑩
=√
(2)
𝑹𝑴𝑴𝑨
Graham’s law may also be states as:
 The rates of diffusion of gases are inversely proportional to the square
root of their molecular masses at constant temperature and pressure.
Example
Compare the rate of diffusion of hydrogen, H2 and carbon dioxide, CO2. (H=1, C=12 , O=16)
Answer
𝑅 𝐻2
𝑅𝐶𝑂2
=√
𝑅𝑀𝑀𝐶𝑂2
𝑅𝑀𝑀𝐻 2
= √
44
2
= 4.69
𝑹𝑯𝟐 = 4.69 × CO2
Hydrogen gas diffuse 4.69 times faster than carbon dioxide.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
90
Again, Graham’s law may also be states as:
 The rate of diffusion of a gas is inversely proportional to time.
The shorter the time the faster the rate of diffusion and vice versa.
Rate =
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐠𝐚𝐬
𝐓𝐢𝐦𝐞 𝐭𝐚𝐤𝐞𝐧
=
𝐕
𝐭
For two gases A and B, the rate of diffusion is RA =𝑉𝐴 and RB = 𝑉𝐵
For equal volumes of gases, VA = VB = V
We get 𝑅𝐴 = 𝑉 × 𝑡𝐵
𝑅𝐵
Then
𝐑𝐀
𝐑𝐁
𝑡𝐴
=
𝑡𝐴
𝑡𝐵
𝑉
𝐭𝐁
(3)
𝐭𝐀
Based on the above formulae (1), (2) and (3) we can conclude that:
𝐑𝐀
𝐑𝐁
𝛒
𝐭
= 𝐁 = √𝐁
𝐭𝐀
𝐑𝐌𝐌𝐁
𝛒𝐀
=√
𝐑𝐌𝐌𝐀
The above formulae also show that the time taken for the diffusion of two gases is
directly proportional to the square root of their densities or molecular
masses at constant temperature and pressure.
Example
It takes 50 seconds for oxygen gas to diffuse through a porous pot. Calculate how long it
takes an equal volume of Sulphur dioxide to diffuse through the same porous pot.
Answer
For equal volume of gases
𝒕𝑶𝟐
𝒕𝑺𝑶𝟐
𝑹𝑴𝑴𝑶𝟐
= √
𝑹𝑴𝑴
𝑺𝑶𝟐
𝒕𝑺𝑶𝟐 = 𝟓𝟎 = 70.71
√𝟏
𝟓𝟎
𝒕𝑺𝑶𝟐
𝟑𝟐
𝟔𝟒
=√
=√
𝟏
𝟐
𝟏
50 = √
𝟐
𝒕𝑺𝑶𝟐
seconds
𝟐
It takes 70.71 seconds for SO2 gas to diffuse through the porous pot.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
91
Exercises
1) Compare the rates of diffusion of oxygen gas and nitrogen gas using their relative
molecular masses. (N =14, O =16)
2) Two gases A and B have densities of 0.18 g/dm3 and 2.90 g/dm3 respectively. If they
diffuse under the same conditions, what are their relative rates of diffusion?
3) Hydrogen gas takes 10 seconds to diffuse through a room. How long would an equal
volume of methane (CH4) take to diffuse. (C =12, H =1)
4) The rate of diffusion of methane through a porous pot is 12 cm3/s. Calculate the rate of
diffusion of carbon dioxide through the same porous pot. (C =12, H =1, O =16)
5) The rate of methane (CH4) and gas X is in the ratio of 2:1. Calculate the relative formula
mass of gas X.
6) The rate of diffusion of two gases A and B are in the ratio of 2:1. If the RMM of gas A is
16, calculate the RMM of gas B.
7) It took 120 second for 100 cm3 of oxygen to diffuse through a small hole. How long will
100 cm3 of another gas with a RMM of 64 take to diffuse through the same hole. (O =16)
8) It takes 110 seconds for a sample of carbon dioxide to diffuse through a porous plug and
275 seconds for the same volume of an unknown gas to diffuse under the same
conditions. What is the molar mass of the unknown gas (in g/mol)?
Answer: 𝒙 = 𝟐𝟕𝟓 𝒈/𝒎𝒐𝒍
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
92
UNIT 8: PREPARATION AND CLASSIFICATION OF
OXIDES
1. Definition of oxides
Oxides are compounds of oxygen with another element.
Examples:





MgO : Magnesium oxide
Na2O : Sodium oxide
NO2 : Nitrogen dioxide
CO2 : Carbon dioxide
SO2 : Sulphur dioxide
2. Preparation of oxides
Oxides can be prepared by the following methods:
a) Direct combination of an element with oxygen.
b) Thermal decomposition of hydroxides, Carbonates and nitrates.
a) Preparation of oxides by direct combination with oxygen
i) Combination of metals with oxygen
Metals react with oxygen to produce metal oxide.
Metal + Oxygen
Metal oxide
Examples:
2Mg(s)
+ O2(g)
2MgO(s)
Magnesium oxide
4Na(s)
+ O2(g)
2Na2O(s)
sodium oxide
4K(s)
+ O2(g)
2K2O(s)
potassium oxide
4Al(s)
+ 3O2(g)
2Al2O3(s)
Aluminium oxide
ii) Combination of non-metal with oxygen
Non-metals react with oxygen to produce non-metal oxide.
Non-metal + Oxygen
Non-metal oxide
Examples:
S(s) +
C(s) +
O2(g)
O2(g)
SO2(g)
CO2(g)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
93
2H2(s) + O2(g)
4P(s) + 5O2(g)
H2O(g)
2P2O5(g)
b) Preparation of oxides by thermal decomposition of
Hydroxides, Carbonates and Nitrates
i) Thermal decomposition of hydroxides
Metal hydroxide decomposes by heat to form metal oxide and water.
heat
Metal hydroxide
Metal oxide + water
Examples:
Mg(OH)2(s)
Ca(OH)2(s)
2Al(OH)3(s)
MgO(s) + H2O(l)
CaO(s) + H2O(l)
Al2O3(s) + 3H2O(l)
heat
heat
heat
Note: Sodium hydroxide (NaOH) and Potassium hydroxide (KOH) are not
decomposed by heat.
ii) Thermal decomposition of carbonates
All metal carbonates undergo thermal decomposition to give metal oxide and carbon dioxide
except sodium carbonate and potassium carbonate which do not decompose on heating.
Example:
Metal carbonate
heat
CaCO3(s)
heat
MgCO3(s)
heat
ZnCO3(s)
heat
Li2CO3(s)
heat
CuCO3(s)
heat
CaO(s)
MgO(s)
ZnO(s)
Li2O(s)
CuO(s)
Metal oxide
+ CO2(g)
+ CO2(g)
+ CO2(g)
+ CO2(g)
+ CO2(g)
+ carbon dioxide
iii) Thermal decomposition of nitrates
 Most metal nitrates decompose on heating to give metal oxide, brown fumes of nitrogen
dioxide and oxygen gas.
Examples
heat
2Cu(NO3)2(s)
2CuO(s) + 4NO2(g) + O2(g)
heat
2Zn(NO3)2(s)
2ZnO(s) + 4NO2(g) + O2(g)
heat
2Pb(NO3)2(s)
2PbO(s) + 4NO2(g) + O2(g)
 Sodium nitrate and potassium nitrate decompose on heating to form metal nitrite and
oygen gas.
heat
2KNO3(s)
2KNO2(s) + O2(g)
heat
2NaNO3(s)
2NaNO2(s) + O2(g)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
94
3. Classification of oxides
On the basis of acid-base nature, oxides can be classified into four groups.
a. Basic oxides
b. Acidic oxides
c. Amphoteric oxides
d. Neutral oxides.
a) Basic oxides
Basic oxides are oxides of metals. Generally, Group 1 and Group 2 elements form
bases called basic oxide
i) Example of basic oxides:





Sodium oxide (Na2O)
Potassium oxide(K2O)
Magnesium oxide(MgO)
Calcium oxide(CaO)
Barium oxide(BaO)
ii) Properties of basic oxides
1) Some basic oxides dissolve in water to form a base (alkaline solution) (metal
hydroxide)

Reaction with water
Na2O(s) + H2O(l)
2NaOH(aq)
base
K2O(s) + H2O(l)
MgO(s) + H2O(l)
CaO(s) + H2O(l)
2KOH(aq)
Mg(OH)2(aq)
2Ca(OH)2(aq)
2) Basic oxides also react with an acid to form a salt and water.

Reaction with acids
MgO(s) + 2HCl(aq)
MgCl2(aq)
+ H2O(l)
Note: Basic oxides do not react with bases.
b) Acidic oxides
Acidic oxides are the oxides of non-metals. (Groups 14 – 17)
i) Examples of acidic oxides



Carbon dioxide (CO2)
Sulphur dioxide (SO2)
Sulphur trioxide (SO3)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
95


Nitrogen dioxide (NO2)
Phosphorus (III) oxide (P2O3)
ii) Properties of acidic oxides
1) Acidic oxides react with water to form acidic solutions.

Reaction with water
CO2(g) + H2O(l)
H2CO3(aq)
Carbonic acid
SO2(g) + H2O(l)
H2SO3(aq)
Sulphurous acid
SO3(g) + H2O(l)
H2SO4(aq)
Sulphuri acid
3NO2(g) + H2O(l)
2HNO3(aq)
+
NO(g)
Nitric acid
2) Acidic oxides also react with bases (alkali) to form a salt and water.

Reaction with bases
CO2(g) + 2NaOH(aq)
Na2CO3(aq) + H2O(l)
Salt
SO2(g) + 2NaOH(aq)
Na2SO3(aq) + H2O(l)
 They are usually gases at room temperature.
Note: Acidic oxides do not react with acids.
c) Amphoteric oxides
Amphoteric oxides are oxides of certain metals. Amphoteric oxides are oxides that
react with both acids and bases to form salt and water. This means that they
react as either acid or base.
i) Examples:




Aluminium oxide (Al2O3)
Zinc oxide (ZnO)
Lead oxide (PbO
Beryllium oxide (BeO)
ii) Properties of amphoteric oxides
Amphoteric oxides are oxides that react with both acids and bases to form salt and water.
Equations:
𝐴𝑙2𝑂3(𝑠) + 6𝐻𝐶𝑙(𝑎𝑞) → 2𝐴𝑙𝐶𝑙3(𝑎𝑞) + 3𝐻2𝑂(𝑙)
{𝐴𝑙2𝑂3(𝑠) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 2𝑁𝑎𝐴𝑙𝑂2(𝑎𝑞) + 𝐻2𝑂(𝑙)
𝑏𝑎𝑠𝑒
𝑠𝑜𝑑𝑖𝑢𝑚 𝑎𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑒
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
96
{
𝑍𝑛𝑂(𝑠) + 2𝐻𝐶𝑙(𝑎𝑞) →
𝑍𝑛𝐶𝑙2(𝑎𝑞) + 𝐻2𝑂(𝑙)
𝑍𝑛𝑂(𝑠) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎2𝑍𝑛𝑂2 (𝑎𝑞) + 𝐻2𝑂(𝑙)
𝑏𝑎𝑠𝑒
(𝑍𝑛𝑂 𝑒𝑥ℎ𝑖𝑏𝑖𝑡𝑠 𝑏𝑎𝑠𝑖𝑐 𝑏𝑒ℎ𝑎𝑣𝑖𝑜𝑢𝑟 𝑤𝑖𝑡ℎ 𝐻𝐶𝑙)
(𝑍𝑛𝑂 𝑒𝑥ℎ𝑖𝑏𝑖𝑡𝑠 𝑎𝑐𝑖𝑑𝑖𝑐 𝑏𝑒ℎ𝑎𝑣𝑖𝑜𝑢𝑟 𝑤𝑖𝑡ℎ 𝑁𝑎𝑂𝐻)
𝑠𝑜𝑑𝑖𝑢𝑚 𝑧𝑖𝑛𝑐𝑎𝑡𝑒
d) Neutral oxides
Neutral oxides are oxides of some non-metals. Neutral oxides show neither basic nor
acidic properties and hence do not form salts when reacted with acids or bases.
i)




Examples
Carbon monoxide (CO)
Water (H2O)
Nitrogen monoxide (NO)
Dinitrogen oxide (N2O)
ii) Properties of neutral oxides
 Neutral oxides do not react with both acids and bases. This means that they do not have
basic or acidic properties.
 They have no effect on litmus solution.
4. Uses and production of slaked lime (Ishwagara)
Raw material: Calcium carbonate (Also known as limestone)
Limestone decomposes when heated to form calcium oxide (quicklime) and carbon dioxide
gas.
CaCO3(s) heat
CaO(s) + CO2(g)
Limestone
Quicklime
Then, solid calcium hydroxide (slaked lime) is obtained by dissolving calcium oxide in a few
drops of water.
CaO(s) + H20(l)
Ca(OH)2(s)
Quicklime
Slakedlime
5. Uses of slaked lime (Ishwagara)
a. A solution of slaked lime is used for white washing walls.
b. Slaked lime is used to reduce soil acidity.
c. Slaked lime is used in the formation of calcium carbonate.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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Exercises
1) On the basis of acid-base nature, oxides can be classified into four groups.
a) List the four groups.
b) Classify the following oxides into the named groups:
i) Sodium oxide
ii) Sulphur dioxide
iii) Copper(II) oxide
iv) Aluminium oxide
v) Water
vi) Carbon monoxide
vii) Nitrogen dioxide
viii) Magnesium oxide.
2) a) One method of preparation of oxides is by direct combination. Name the products
formed when the following elements combine with oxygen:
i) Magnesium
ii) Carbon
iii) Sulphur
iv) Sodium
b) Classify the element in (a) above as metals and non-metals.
c) Which of the elements will form
i) Gaseous oxides
ii) Solid oxides
3) Choose from the list of oxides to answer the questions below. You can use each oxide
once, more than once or not at all.
 Carbon dioxide
 Nitrogen dioxide
 Water
 Carbon monoxide
 Sulphur dioxide
 Magnesium oxide
 Calcium oxide
a) Which of these oxides are basic oxides?
b) Which two oxides cause acid rain?
c) Which two oxides are formed when a hydrocarbon undergoes complete combustion?
d) Which one of these oxide turns anhydrous white copper (II) sulphate blue?
e) Which oxide is formed when calcium carbonate undergoes thermal decomposition.
4) An oxide of element X dissolves in water to form a solution of PH = 5. Is element X a
metal or a non-metal?
5) When copper (II) nitrate is heated, it decomposes to copper (II) oxide, nitrogen dioxide
and oxygen.
a) What is the colour of :
i) Copper (II) nitrate
ii) Copper (II) oxide
iii) Nitrogen dioxide
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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b) Write a balanced equation for the decomposition of copper (II) nitrate.
6) Oxides can be prepared by direct combination between an element and oxygen. The can
also be prepared through thermal decomposition of metallic hydroxide, nitrates and
carbonates.
a) What do you understand by the term decomposition?
b) Name all the products formed when the following compounds are decomposed by
heat.
i) Aluminium hydroxide
ii) Copper (II) carbonate
iii) Calcium nitrate.
7) Consider the following oxides:
Sodium oxide, Sulphur dioxide and aluminium oxid.
a) Select the oxide that reacts with water to give an acidic solution.
b) Select the oxide which when shaken with water, gives a solution with pH greater
than 7.
c) Select the oxide which reacts with both dilute hydrochloric acid and sodium
hydroxide solution.
8) a) Calcium oxide is obtained in large scale through thermal decomposition of calcium
carbonate (limestone). State two large scale uses of calcium oxide.
b) State one use of the oxides given below:
i) Magnesium oxide
ii) Carbon dioxide
iii) Sulphur dioxide.
9) Some oxides are listed below.
Calcium oxide, phosphorus (III) oxide, water, carbon dioxide, sodium oxide,
carbon monoxide, Sulphur dioxide
a) Which one of these oxides will most likely cause acid rain?
b) Which one of these oxides is a product of the reaction between an acid and a
carbonate?
c) Which one of these oxides is formed by the incomplete combustion of carbon?
d) Which one of these oxides is a good solvent?
e) Which one of these oxides is used to neutralize acidic industrial waste products?
f) Which two of these oxides react with water to form an alkaline solution?
10) Calcium nitrate decomposes when heated.
2Ca(NO3)2(s)
2CaO(s) + 4NO2(g) + O2(g)
a) The solid product, CaO, is slightly soluble in water and reacts to form a solution of
Ca(OH)2.
Explain what happens when blue and red litmus paper are separately dipped in the
resulting solution of Ca(OH)2
b) Nitrogen (IV) oxide (NO2) formed is also soluble in water. Explain what happens
when blue and red litmus papers are separately dipped in the resulting solution.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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UNIT 9: ELECTROLYTES AND NON-ELECTROLYTES
1. Electrolyte
Definition:
Electrolyte is a compound that conducts electricity when molten or in aqueous solution.
Examples: H2SO4, HCl, NaCl, KNO3, KOH etc.
Electrolyte conducts an electric current when dissolved in water or molten form.
Types of electrolytes
Electrolytes can be classified into strong electrolytes and weak electrolytes.
a) Strong electrolytes:
Are electrolytes that dissociate (or separate) completely into
ions when dissolved in water or when in molten state.
Strong electrolytes have high electrical conductivity because of high concentration of ions in
their solution.
Examples of strong electrolytes
 Strong acids: H2SO4, HCl, HNO3
 Strong bases: KOH, NaOH, Ca(OH)2
 Salts: NaCl, KNO3, MgCl2 ,etc
Dissociation equation when an electrolyte is dissolved in water is:
H2O(l)
HCl(l)
H+(aq) + Cl−(aq)
H2O(l)
H2SO4(l)
2H+(aq) + SO42−(aq)
H2O(l)
NaCl(s)
Na+(aq) + Cl−(aq)
H2O(l)
NaOH(s)
Na+(aq) + OH−(aq)
b) Weak electrolytes: Are electrolytes that dissociate (or separate) partially into ions
when dissolved in water or when in molten state.
Examples of weak electrolytes
𝐸𝑡ℎ𝑎𝑛𝑜𝑖𝑐 𝑎𝑐𝑖𝑑(𝐶𝐻3𝐶𝑂𝑂𝐻)
 Organic acids:{ 𝑀𝑒𝑡ℎ𝑎𝑛𝑜𝑖𝑐 𝑎𝑐𝑖𝑑(𝐻𝐶𝑂𝑂𝐻)
 Ammonium hydroxide (ammonia solution): NH4OH
Weak electrolytes have low electrical conductivity because of low concentration of ions in
their solution.
Dissociation equation when a non-electrolyte is dissolved in water is:
H2O(l)
CH3COOH(l)
CH3COO−(aq) + H+(aq)
H2O(l)
HCOOH(l)
HCOO−(aq) + H+(aq)
H2O(l)
NH4OH(l)
NH4+(aq) + OH−(aq)
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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2. Non-electrolytes
Non-electrolytes are covalent compounds that do not dissociate (or separate) into ions when
dissolved in water. This means that they do not conduct electricity when dissolved in water
or when in molten state. They remain molecular when added in water.
Examples
 Sugar
 Ethanol
 Urea
3. Definition of electrolysis
Electrolysis is the decomposition of an electrolyte using electricity.
The diagram below shows how electrolysis is carried out:
Definitions of some terms used in electrolysis
Electrode is a piece of metal used to carry an electric current into or out of electrolyte.
They are two types of electrodes namely:
 Anode is an electrode connected to the positive terminal of battery.
 Cathode is an electrode connected to the negative terminal of battery.
During electrolysis:
An electrolyte decomposes into ions (positive ions and negative ions)
Then, positive ions (cations) move towards the cathode while Negative ions (anions) move
towards the anode.
NOTE:
 In an electrolyte, the particles that carry electric current are called ions while
 In the wire, the particles that carry electric current are called electrons.
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4. Difference between electrolytes and
non-electrolytes
a) Electrolytes conduct electricity in aqueous or molten state because the ions are free to
move while non-electrolytes do not conduct electricity in aqueous or molten state
because they have no ions.
b) Electrolytes are ionic compounds while non-electrolytes are covalent compounds.
5. Difference between strong electrolytes and weak
electrolytes
a. Strong electrolytes dissociate (or separate) completely into ions while weak electrolytes
dissociate (or separate) partially into ions.
b. Strong electrolytes have high electrical conductivity while weak electrolytes have low
electrical conductivity.
6. Application of some common electrolytes
a.
b.
Sulphuric acid (H2SO4) is used in car batteries
Ammonium chloride (NH4Cl) is used in Leclanché cell (Dry cell)
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Exercises
1) a) What is an electrolyte?
b) Give two examples of electrolytes.
2) a) What is a non-electrolyte?
b) Give two examples of non-electrolytes.
3) What do you understand by the term electrolysis?
4) An electrolyte can be described as a “strong electrolyte”, or a “weak
electrolyte”
a) Distinguish between a strong and a weak electrolyte.
b) State two examples of each.
5) Differentiate the following terms:
a) Electrolyte and non-electrolyte.
b) Cations and anions
c) Cathode and anode.
6) Describe the movement of cations and the anions in the electrolyte during
electrolysis.
7) Classify the substance given below under the three headings:
Strong electrolytes, weak electrolytes and non-electrolyte
 Ethanoic acid
 Common salt solution
 Ammonium hydroxide
 Ethanol
 Dilute hydrochloric acid
 Sugar solution
 Dilute sulphuric acid
 Sodium hydroxide
 Distilled water
8) Name the particles that conduct electricity during electrolysis in the following:
a) In the conducting wire in the external circuit.
b) In the electrolyte.
9) Metals (wires) conduct an electric current through delocalized electrons
while electrolytes conduct an electric current through ...................... Electrolytes
undergo… ..................... around the electrodes whereas metals are not affected by
electric current.
10) The apparatus shown in the diagram below may be used to distinguish between
an electrolyte and non-electrolyte.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
103
a) State the observations made when the substance (liquid) is
i) An electrolyte
ii) A non-electrolyte
b) Describe how the same set up could be used to distinguish between a strong
electrolyte and a weak electrolyte.
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UNIT 10: PROPERTIES OF ORGANIC COMPOUNDS
AND USES OF ALKANES
1. Definition of organic chemistry
Organic chemistry is a branch of chemistry that deals with the study of compounds
containing carbon.
However, it does not deal with the study of oxides of carbon, carbonates and
hydrogen carbonates or carbonic acid.
2. Difference between organic and inorganic
compounds
The following table gives a summary of differences between organic and inorganic
substances.
Properties
Organic substances
Inorganic substances
Solubility
Flammability
Insoluble in water but soluble in organic
solvents such as ethanol
Are flammable (They catch fire easily)
Volatility
Most are volatile liquids
Ions
Always contain carbon and hydrogen
Occurrence
Always found in living things
Boiling point
Have low boiling points
Soluble in water but
insoluble in organic solvents
Are usually nonflammable(They do not burn
easily)
Many are non-volatile except
for concentrated acids
Usually contain cations and
anions
Always found in non-living
things
Many have high boiling
points
3. Occurrence of organic compounds
 Organic compounds are found in many living things (Plant and animal bodies)
They make up many important biological molecules for example proteins, lipids, cellulose
and carbohydrates. (Maize, Meat, Wood, Dress are materials containing organic
compounds)
 They also occur in crude oil and in products such as diesel, kerosene, petrol, etc.
 Organic compounds also occur in natural gases and biogas.
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4. Homologous series
Homologous series: This is a series of organic compounds in which adjacent members
differ by the – CH2 – group.
Or
A homologous series is a series of compounds that have the same function group, and each
member differ from the next member by a –CH2– unit in their formulae.
 Members of a homologous series have the following in common:
1)
2)
3)
4)
5)
Show similar chemical properties because they have the same functional group.
Have physical properties which vary gradually from one member to another.
Can be prepared in the same general way.
Differ from the next by a – CH2 – group.
Have same general formula
Examples:
CnH2n+2 for alkanes
CnH2n for alkenes
CnH2n+1OH, for alcohols etc.
Example1:
Study the following molecules
CH4, CH3 −CH3 , CH3 –CH2 −CH3
i) Identify common characteristics.
ii) What is the difference between?
Answer
CH4 , CH3−CH3 , CH3−CH2−CH3 are homologous series (because they have same general
formula CnH2n+2 ).
 Differ from the next by a – CH2 – group
 They have same function group: Saturated hydrocarbons (C- C single bonds)
Example2
CH3OH , CH3CH2OH , CH3CH2CH2OH are homologous series (because they have
same general formula CnH2n+1OH ).
 They have same function group: hydroxyl (OH) for each
 Member differs by CH2
5. Hydrocarbons
Hydrocarbons are organic compounds containing hydrogen and carbon only.
Example: Alkanes and alkenes
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6. ALKANES
Alkanes are hydrocarbons contain only single carbon -- carbon, -C – C- bonds, and thus
they are referred to as saturated hydrocarbons.
They are represented by general formula CnH2n+2 where n is the number of carbon
atoms, and n can be 1, 2, 3, 4, ……….
Each successive member varies from the previous one by a constant group of atoms – CH2 –
a) Nomenclature of alkanes
Nomenclature refers to naming. IUPAC (international union of pure and applied chemistry)
system of naming is recommended. In the IUPAC system, the alkane’s name consists of a
“prefix” and a “suffix”.
 The prefix indicates the number of carbon atoms in a molecule
 The suffix for alkanes is –ane.
Examples:
1. Methane: Prefix is meth; Suffix is ane
2. Ethane: Prefix is eth; Suffix is ane
3. Nonane: Prefix is non; Suffix is ane
The table below gives
Number Alkane
of
name
carbon
atoms
1
Methane
the names
Molecul
ar
formula
of the first ten alkanes and their formulae
Displayed
Condensed structure formula
structural
formula
CH4
H
H
C
CH4
H
H
2
Ethane
C2 H 6
HH
CH3CH3
H−C−C−H
3
Propane
C3 H 8
HH
HHH
CH3CH2CH3
H−C−C−C−H
HH H
4
5
6
7
Butane
Pentane
Hexane
Heptane
C4H10
C5H12
C6H14
C7H16
CH3CH2CH2CH3
CH3CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH3
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8
9
10
Octane
Nonane
Decane
C8H18
C9H20
C10H22
CH3CH2CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
Structural formulas
Definition:
Structural formula shows how the atoms are bonded in a molecule (i.e. linked or
connected)
 There are three types of structural formulas:
 Displayed formulas: show each bond.

Condensed formulas: show each carbon atom and its attached
hydrogen atoms and
 Skeletal (stick) formulas.
In this unit we learn displayed structure formula and condensed structure formula
only. Examples are shown in the above table.
b) Structural isomerism
Isomerism: Refers to the existence of compound with the same molecular formula but with
different structural formula.
 Isomers are compounds with the same molecular formula but different structural
formula. isomers have different physical properties.
In alkanes, methane, CH4, ethane, C2H6 and propane, C2H8 do not have isomers.
Butane has the following isomers
 CH3−CH2−CH2−CH3 : Butane
CH3
 CH3
C
CH3 : 2-methylpropane
H
c) Physical and chemical properties of alkanes
1) Physical properties of alkanes
 The lower members of alkanes (C1 to C4) are gases at room temperature.
 Alkanes from C5 to C16 are liquids. Those with more than 16 carbon atoms are waxy
solids.
 The melting point and boiling points of alkanes increase with the increasing number of
carbon atoms in the molecule.
 All alkanes are insoluble in water but soluble in organic solvents such as ethanol and
benzene.
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 Their densities increase gradually as the number of carbon atoms increase.
The table below gives a summary of the physical properties of the first ten
alkanes
Name of
alkane
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane
Octane
Nonane
Decane
Molecular
formula
CH4
C 2H 6
C 3H 8
C4H10
C5H12
C6H14
C7H16
C8H18
C9H20
C10H22
Physical
state at r.t.p
Gas
Gas
Gas
Gas
Liquid
Liquid
Liquid
Liquid
Liquid
Liquid
Density
g/cm3
0.42
0.57
0.59
0.60
0.63
0.66
0.68
0.70
0.72
0.73
Melting
point (0C)
−182
−184
−190
−138
−130
−95
−91
−57
−54
−30
Boiling
point (0C)
−162
−89
−42
−0.5
−36
−69
−98
−126
−151
−174
2) Chemical properties of alkanes
In this sub-unit, we will study some chemical properties of alkanes such as:
i) Combustion
ii) Reaction of alkanes with halogens
iii) Thermal cracking
i) Combustion reactions
 Alkanes burn in excess air with a pale blue flame forming carbon dioxide and water.
Heat energy is released in the process making them suitable for use as fuels.
The general chemical equation for complete combustion of alkanes may be written as
CnH2n+2 +
3𝑛+1
2
O2
n CO2 + (n+1) H2O
Example:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g) + Heat
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(g) + Heat
 In limited supply of air, alkanes burn with a luminous flame forming a mixture of
carbon monoxide gas and steam
2CH4(g) + 3O2(g)
2CO(g) + 4H2O(g)
ii) Reaction of alkanes with halogens
Reactions between halogens and alkanes are generally termed as substitution reactions.
In the presence of sunlight, alkanes react with halogens forming a series of products
depending on the amount of halogen present.
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Example:
CH4(g) + Cl2(g)
U.V light
CH3Cl(g) + HCl(g)
Chloromethane
CH3Cl(g) +
Cl2(g)
U.V light
CH2Cl2(g) + HCl(g)
Dichloromethane
CH2Cl2(g) +
Cl2(g)
U.V light
CHCl3(g) + HCl(g)
Trichloromethane
CHCl3(g) +
Cl2(g)
U.V light
CCl4(g) + HCl(g)
Tetrachloromethane
Note: If the amount of chlorine present is the same as that of methane (1: 1 ratio), then
only the first reaction occurs.
iii) Thermal cracking of alkanes
Cracking refers to the process of breaking down larger alkanes into smaller more useful
alkanes and alkenes. Thermal cracking occurs at a high temperature (4500C−7500C) and
pressure (200 atmospheres).
Example:
C9H18
C4H6 + C5H12
Large alkane
Alkene
CH3−CH2−CH2−CH2−CH3
small alkane
CH2 =CH2
+
CH3−CH2−CH3
d) Laboratory preparation of methane
Generally, alkanes are prepared in the laboratory by the action of heat on a mixture of an
alkanoate and soda lime.
Note: Soda lime is a solid mixture of NaOH and CaO. It is prepared by soaking quicklime,
CaO in caustic soda, NaOH solution. CaO dries the resultant product.
Example:
To prepare methane, CH4, sodium ethanoate is used.
CH3COONa(s) + NaOH(s)
CH4(g) + Na2CO3(s)
Sodium ethanoate
Methane
The figure below shows how methane is prepared in the laboratory.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO
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e) Uses of alkanes (methane)
 Alkanes are used as fuel for lighting, cooking, running vehicles and machines.
 Alkanes are used as solvents.
Examples: Dichloromethane, Trichloromethane and tetrachloromethane are used as
solvents for non-polar compounds
 Alkanes are also used as lubricants.
 Biogas is used as a fuel. It contains methane gas.
 Methane is used to Produce carbon black that is used in paints, printing inks and
automobile tyres.
 Trichloromethane, CHCl3 (Chloroform) was used as an anaesthetic.
Exercises
1) Methane gas is collected over water. What does this indicate about the solubility of
alkanes in water?
2) Distinguish between organic and inorganic chemistry.
3) Write a balanced chemical equation to show laboratory preparation of methane.
4) a) What is meant by the term “hydrocarbon”?
b) C4H10 is a hydrocarbon belonging to the family of alkanes.
i) Give the general name of the above alkane.
ii) Write down the structure formulae of two isomers of C4H10 and name the
branched isomer.
5) a) What are alkanes?
b) State two physical properties of alkanes.
c) Alkanes are very useful. Describe briefly two uses of alkanes in daily life.
6) a) Define thermal cracking of alkanes
b) Complete the following equation
C10H22
? +?
0
600 C
7) Write the structural formula of:
a) Hexane
b) Octane
c) Nonane
8) Pentane is an alkane with five carbon atoms.
a) Write down the molecular formula of pentane.
b) What physical state (solid, liquid or gas) would you expect pentane to be in at room
temperature.
c) Write an equation for the complete combustion of pentane.
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9) Draw the displayed formula for butane:
a) C4H10
b) C7H16
10) Study the table below and use it to answer the questions that follow.
Structure formulae
Melting
Boiling
Molecular
point
point
formulae
(0C)
(0C)
−138
−0.5
−160
−11.5
i)
ii)
a)
b)
c)
d)
Write the name of the compound represented by structure labelled (i) and (ii).
Write the molecular formula of the two compounds in the table.
What is the relationship between the two compounds?
Comment on the molting point and boiling point of the two compounds.
PREPARED BY NIYONKURU Jean, TEACHER OF CHEMISTRY AT GS CURAZO