GENERAL CHEMISTRY & CHEMISTRY FOR
AGRICULTURALIST
• COURSE CODE : SOAT 101 / 201
• COURSE TITLE : GENERAL CHEMISTRY &
CHEMISTRY FOR AGRICULTURALISTS
• CREDIT
: THREE (3)
• COMPONENT
: THEORY AND PRACTICAL
• INSTRUCTOR
: DR SAMUEL KWESI ASOMANING
1
Objectives :The course is intended to broaden students’
understanding of the principles of physical and
inorganic chemistry.
To enable students to develop basic laboratory skills in
quantitative and qualitative analysis.
2
COURSE OUTLINE
• 1.0 Atomic structure:
*Postulates of Dalton’s atomic theory
*The discovery of the electron, J.J. Thomson,s
experiment.
*Rutheford’s Model and Bohr’s Model of the atom.
*Quantum chemistry
• 2.0 Isotopes:
*Mass Spectrometry and calculation of relative atomic
mass of isotopes
3
*definitions of the atomic mass unit, relative molecular
mass etc
• 3.0 Dimensional Analysis and Units in Chemistry
• 4.0 Amount of substance and the mole concept.
4
COURSE OUTLINE
• 5.0 Periodic classification of elements
*Periodic table (groups and periods). Atomic properties and,
periodic table, properties of alkali and alkaline earth metals,
halogens and transition metals.
• 6.0 Bonding and Intermolecular forces:
*Ionic and covalent bonding & Intermolecular forces such as
Vanderwaals forces and hydrogen bonding, sp3 , sp2
and sp hybridization.
5
COURSE OUTLINE
• 7.0 General concept of Acid and Bases:
*Arrhenius, Bronsted Lowry and Lewis Acids
*Acid strength and simple pH calculations for solutions of
strong and weak acids and bases.
* Salt hydrolysis or Cation and anion hydrolysis.
6
COURSE OUTLINE
*Principle of pH determination.
*Buffer solutions and preparation of buffer solutions
• 7.0 Preparation of solutions in laboratory
• 8.0 Chemical equilibrium
*solubility and solubility products (Ksp)
7
•
+
COURSE OUTLINE
• 9.0 Electrochemical series
*redox systems
*oxidation numbers
*balancing of redox reactions in both acidic and alkaline
solutions or medium.
• 10.0 Colligative properties
*Raoult’s law
8
COURSE OUTLINE
• 11.0 Rates of reactions
*factors affecting rate of reactions
*calculations involving rate of reactions
• 12.0 Structure of organic molecules
*Nomenclature of alkanes, sp3 properties, isomerism and
sources of substitution reactions and halogenation reactions.
*Alkenes; sp2 hybridisation double bond formation geometry
of c=c bond, nomenclature and isomerism: cis and trans
figuration.
9
• 14.0 Brief introduction of Aromaticity
*physical and chemical properties of benzene
*structures and namings : nitrobenzene, toluene, chloro,
bromobenzenes, aniline naphthalene; phenanthrene etc
*some reactions involving the benzene ring; Friedel-Craft
• alkylation, acylation, halogenation, oxidation of toluene, benzene
with halogens (ring splitting)
10
Reading Materials
• Raymond Chang. 2003. General Chemistry, The essential
concepts. 3rd edition. The McGraw-Hill Companies, Inc.,
• Philip Matthews, 2004. Advanced Chemistry. Cambridge
University Press, UK.
• Wong, Y.C., C.T. Wong, S.O.Onyinka and L.E. Akpanisi.
2002. University General Chemistry; Inorganic & Physical.
Manhattan Press (H.K.) Ltd.
• Any Physical Chemistry text book.
11
THE STRUCTURE OF THE ATOM
• The Greek philosopher, Demokritos, in 500 BC
expressed the believe that all matter consists of very
small particles. He named these particles atoms from
the Greek word ‘atomos’ meaning indivisible.
• John Dalton, an English scientist in 1808 also put
forward his atomic theory which also embraced the
idea of matter being made up of atoms.
12
Modified version of postulates of Dalton’s atomic theory :-
1. Atoms are the smallest particles of matter.
2. Atoms can neither be created nor destroy in chemical
reactions.
3. Atoms of the same element do not always have the
same mass and are different from the atoms of other
element.
4. Chemical combination takes place between small whole
number of atoms.
13
Definition of an atom:An atom is the basic unit of an element that can enter into
chemical combination.
14
THE DISCOVERY OF THE ELECTRON
J.J. Thomson’s Experiment, 1897:He discovered a fundamental constituent of all matter- the
first subatomic particle, called the electron using
J.J. Thomson’s Cathode Ray Tube:1. The cathode ray tube is a glass tube from which the air
has been evacuated using a vacuum pump .
2. The two metal plates of the cathode ray tube were
connected to a high-voltage source. The negatively
charged plate, the cathode, emits invisible ray known
as cathode ray.
3. The cathode rays were attracted to the positively
charged plate called the anode.
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CATHODE RAY TUBE
16
4. The ray passed through a thin slit and gave a glow or
bright light as it strike the ZnS fluorescent screen detector.
5. The rays were deflected by both electric and magnetic
fields which were connected to the outside of the cathode
ray tube. The deflection of the rays indicate that they were
charged particles.
6. Since the rays were seen to be repelled by the negative
plate and attracted by the positive plate, confirmed that they
carried negative charges.
7. These negative charges were named electrons.
17
8. J.J.Thomson also determined the electric charge/mass ratio
of the charged particles. (NB: The deflection depends on both
the mass and the charge so only the ratio could be determined.
9. He found out that cathode rays had the same charge/mass
ratio whatever gas was used in the tube and whatever material
was used as the cathode. The number he came out with is 1.76 x 108 C/g
10. He therefore suggested that from his investigations
electrons were present in all atoms (or matter).
18
J.J. Thomson’s model of Atomic structure:1. An atom is a solid uniform, positively charged sphere in
which rings of negatively charged electrons are embedded.
2. The positive and the negative charges balance each other
to make the atom electrically neutral.
The model could not explain the detailed structure of the
atom.
19
Rutherford’s Nuclear Atom:• In 1910, Lord Rutherford with his associate, Hans Geiger and Ernest
Marsden decided to probe the structure of the atom.
• In their experiment, they bombarded various nuclei of thin gold foil with
alpha particles (positive rays) from a radio active source.
20
Diagram of the Rutherford’s experiment
21
Observations made:
1. Nearly or almost all the alpha (α) particles passed through the thin gold
metal sheet unimpeded or without deflection.
2. Few α-particles were deflected through large angles.
3. A small proportion of the α-particles bounced back.
22
Interpretation of Results:He concluded from his experiment that:
1. Most of the volume of the atom is empty space occupied
by electrons which are too light to deflect alpha particles.
2. Deflections through large angles were due to repulsion of
the alpha particles which came very close to a
concentrated region of positive charge.
3. Very few alpha particles bounced back as a result of head
on collision with a massive positively charge centres.
23
Rutherford’s Model of the Atom:1. The atom has a central positive nucleus composed of protons, where all
the mass is concentrated.
2. The electrons in the atom orbit around the nucleus.
3. The greater proportion of the atom is an empty space occupied by
electrons.
24
• James Chadwick in 1932 also discovered the third subatomic particles
called Neutrons. They were found in the nucleus and has no charge.
25
• The difficulty with Rutherford’s theory of orbital electrons
was that, according to classical physics(that is
electromagnetic theory), a negative electron orbiting
around a positive field should continuously radiate
energy.
• Based on classical mechanics, Neil Bohr assumed that
the hydrogen atom could exist in stationary states (or
definite energy levels) where an electron could revolve
around the nucleus without radiating energy.
26
• Niels Bohr’s proposed theory for the atom:
1. Electrons may only move in defined paths called stationary
states (or definite energy levels or orbits) which are at fixed
distances away from the nucleus.
2. Once the electrons are in the stationary states, they do not
lose or gain energy.
3. Energy of radiation is emitted or absorbed only when an
electron moves from one stationary state or orbit to another.
4. The energy difference between the final and the initial
stationary state is equal to one quantum of energy.
27
Present Day Structure of the Atom
1. The atom is made up of three fundamental particles
called protons (P), neutrons (n) and electrons (e)
2. The protons are positively charged (+), elecetrons are
negatively charged (-) and neutrons have no charge (o) i.e
neutral.
3. The nucleus of the atom is made up of positively
charged protons and uncharged neutrons.
4. The negatively charged electrons orbit around the
nucleus; have very small mass compared with protons and
are spread in an area of empty space.
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• In a neutral atom, number of protons = Number of electrons.
Name
Relative Mass Charg Location in the atom
(amu)
e
Representatio
n
Proton
1
+1
in the nucleus
1
1P
Neutron
1
0
in the nucleus
1
0n
Electron
1
-1
around the nucleus
0
−1e
29
Atomic Orbital
• 1. Though we cannot pinpoint an electron in an atom, we can define the
region in which the electron may be at a given time.
• 2. An atomic orbital is the volume of space around the nucleus in which
there is a higher probability of finding an electron.
• 3. Regions of high electron density represent a high probability of locating
the electron.
30
EXERCISE
1. State all the postulates of Dalton’s atomic theory. Which of the
theories had later been disproved.
2. What is the structure of the atom as proposed by Rutherford?
3. Describe briefly Rutherford’s alpha particle scattering
experiment and state the observations made.
4.How are the results of Rutherford’s experiment used to
establish the structure of the atom?
31
5. State two postulates of the Dalton’s atomic theory, which had later been
disproved.
6. Account for the results of Rutherford’s alpha particle scattering
experiment.
32
QUANTUM CHEMISTRY
• Quantum chemistry is a branch of chemistry that
describe how electrons are arranged in atoms and how
the spatial arrangement of electrons are related to their
energies.
• This will allow us to explain how the arrangement of
electrons in an atom enable chemists to predict and
explain the chemistry of an element.
33
QUANTUM NUMBERS
There are four quantum numbers:
Three of the quantum numbers are required to describe
the distribution of electrons in hydrogen and other
atoms.
The three are:
1.Principal quantum number(n).
2. Azimunthal quantum number or orbital shape quantum
number (l)
3.Magnetic quantum number.(ml)
NB: They are used to describe atomic orbitals and to
label electrons that reside in them.
• .
34
• 4. A fourth quantum number – the spin quantum number (ms)
,describes the behaviour of specific electron and completes
the description of electrons in atom.
35
1. The Principal Quantum Number (n):• It has positive integral values 1, 2, 3 etc.
• The energy levels or shells are determined by the n value.
The larger the value of n, the greater is the average energy
of the levels belonging to the shell. As in the Bohr theory,
n= 1 defines the K shell, n=2 defines the L shell, n=3
defines the M shell and so on
• The n also relates to the average distance of an electron in
a particular orbital from the nucleus.
36
• The larger the value of n the greater the average
distance of an electron in an orbital from the nucleus
and therefore the larger and less stable the orbital.
• The principal quantum number, n reflects the effective
volume (or mean radius) of an electron orbital.
37
• Maximum number of electrons in a shell is given by 2(n)2
where n = positive integer
38
2. The Azimuthal Quantum number (l)
• It determines the shape of the orbitals with a particular
n value.
• The values of l depend on the value of the principal
quantum number, n. The value of n determines what
values of l are allowed or possible.
• With a particular n value, l constitute subshells or
sublevels and therefore it measures to some extent the
energy of the electrons in the orbitals.
• For a given n or shell, l may have values of 0,1, 2, 3,…
to a maximum of (n – 1) for that shell. That is l can
ranged from l= 0 to (n-l) inclusive.
39
• This means that for the K shell, with n = 1, the only value of l that
is possible is l = 0.
• When n = 2, two values of l are possible, 0 and 1, resulting in
two subshells for the L shell.
40
• If n = 1 , that is l = n – 1 = 1 – 1 = 0, it means there is only
one possible value of l which is 0;
• If n = 2, l = n-1 = 2 – 1 = 1 , it means there are two values
of l, given by 0 and 1.
• If n = 3, l = n -1 = 3 – 1 = 2, it means there are three
possible values of l, given by 0 , 1 and 2
• The value of l is generally designated by the letters s, p, d…
as follows:
41
l
0
1
2
3
4
5
Name of orbital
s
p
d
f
g
h
42
NB:
The letter s, p, d, and f were from the word’s sharp,
principal, diffuse and fundamental. They were deduced
from the observed spectral lines of some atomic emission
spectra studied.
NB:
• 1. A collection of orbitals with the same principal quantum
number, n is called shell or energy level.
• 2. The energy levels (or main shells or orbitals) K, L, M, N
etc. with principal quantum number, n = 1, n = 2, n = 3, n
= 4 etc ., have subshells called orbitals. It is these orbitals
that actually accommodate the electrons.
43
• 3. If one or more orbitals correspond to the same value n
then the l values constitute subshells or sublevels.
• For example, the shell with n = 2 is composed of 2 subshells,
l = 0 and 1. These subshells are called the 2s and 2p
subshells, where 2 denotes the value of n and s and p
denotes the values of l
44
n
l = (n-1)
subshell
2
0
2s
1
2px 2py 2pz
45
3. The Magnetic Quantum number (ml)
• This restricts the orientation and shape each type of
orbital in space.
• It indicates the number of orbitals in a subshell with a
particular l value and hence the total number of electrons
in a subshell can be determined.
• Within a subshell, the value of ml depends on the value of
l.
• It assumes all values between –l and +l including zero
(0). That is –l,..0,...+l
46
• For a certain value of l there are maximum of (2l + 1) possible
integral values of ml
• The number of ml values indicates the number of orbitals
in a subshell with a particular l value.
• For example, if l = 0, then only one value of m is permitted, ml
= 0 ; this means that the s subshell has only one orbital, the s
orbital.
• If l = 1, then there are [(2x1) + 1)] = 3 or 3 values of ml ,
namely, -1, 0, and 1
• If l = 2, then there are [(2x2) + 1)] = 5 or 5 values of ml
namely -2, -1, 0, +1, +2
• Note: If n = 2 and l = 1, the value of n indicates that we have
2p subshell and in this subshell are three of the 2p orbitals
because there are three values of ml i.e. -1,.0,.1
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n
l (n-1)
(ml)
0
No of
orbitals
1
Atomic orbital
designation
2s
2
0
1
-1, 0, 1
3
2px 2py 2pz
48
4. The Electron Spin Quantum Number (ms)
• A spinning charge generate a magnetic field, and it is this
motion that cause an electron to produce a magnetic effect.
• It denotes the value and direction of spin of the electron.
• The quantum number associated with this spin has only two
possible values, one clockwise (+1 2 ) and the other counter
clockwise (-1 2 ), or an electron can spin only in one of two
possible directions.
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• The relationship between Quantum Numbers and Atomic
Orbitals
Principal
Quantum
Number, n
(Shell)
K
1
l
(n-1)
Subshell
0
1s
L
0
M
2
3
ml
(2l + 1)
No of
Orbitals
Max no of
electrons
in a shell
Atomic Orbital designation
ms
0
1
2
1s
+1 2 , - 1
2
2s
0
1
2
2s
+1 2 , - 1
2
1
2p
-1, 0, +1
3
6
2px 2py 2pz
+1 2 , - 1
2
0
3s
0
1
2
3s
+1 2 , - 1
2
1
3p
-1, 0, +1
3
6
3px 3py 3pz
+1 2 , - 1
2
2
3d
-2, -1, 0, +1, +2
5
10
3d1, 3d2, d3,
+1 2 , - 1
2
3d4, 3d5
50
• Question1.
What is the total number of orbitals associated with the
principal quantum number n = 3?
• Question 2
What is the total number of orbitals associated with the
principal quantum number, n = 4?
51
• Solution
n
4
L
(n-1)
Sub
Shell
ml
(2l + 1)
No of
Orbitals
0
1
2
4s
4p
4d
0
-1, 0, +1
-2, -1, 0, +1, +2
1
3
5
3
4f
-3, -2, -1, 0, +1,+ 2,+
3
7
Total number of orbitals = 1 + 3 + 5 +7 = 16
52
Give the values of the possible quantum numbers
associated with the orbitals in the 3p subshell.
Solution;
n
3
l
(n-1)
ml
(2l + 1)
Sub
shell
ms
0
0
3s
+ 1 2, -1
2
1
-1, 0, +1
3p
+ 1 2, -1
2
2
-2, -1, 0, +1, +2
3d
+ 1 2, -1
2
∴ for the orbitals in the 3p subshell, n = 3, l = 1, ml = -1, 0,
+1 , ms = +1 2, -1 2
53
• Question 4
List the values of n, l and ml and ms for orbitals in the 4d subshell.
• Question 5
Explain the significance of the quantum numbers n, l, ml and ms in describing
individual electrons in atomic orbitals and illustrate your answer with at least
2 examples of electron configurations
Solution:
• Refer to the notes for the significance of the quantum numbers.
54
Solution:
Refer to the notes for the significance of the quantum numbers.
Example a 3Li
1s2
2s1
n
1
2
l
0
0
ml
0
0
ms
1
3Li
2,
-1
2
-1
2
55
Example b;
8O
1s2
2s2
n
1
l
ml
8O
ms
2p2x
2p1y
2
2
2
2
0
0
1
1
1
0
0
-1
0
+1
1
2,
-1
2
1
2,
-1
2
1
2,
-1
2
1
2,
2p1z
1
2,
56
Question 6
• Give the maximum number of electrons occupying each of
the subshells (the orbitals) for which n, the principal
quantum number is 3
• Solution:
Refer to the table under the relationship between Quantum
numbers and atomic orbitals.
57
Question 7
What are the n and l quantum number designations for the
following subshells: 3s, 4p and 6d ?
Solution:
Subshell
n
l
3s
3
0
4p
4
1
6d
6
2
58
Orbitals in Different Energy Levels
1. Orbitals with the same principal quantum number (n) all
have the same energy.
2. Energy levels in atoms are arranged in series; the levels
get closer together as the series is ascended.
3. With the exception of the first, each energy level is
associated with more than one sub- level or orbitals .
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Energy levels (n)
5
5s
5p
5d
5f
4
4s
4p
4d
4f
3
3s
3p
3d
2
2s
2p
1
1s
s
N shell
M shell
L shell
K shell
p
d
f
Types of orbitals
60
All S sub –levels have only one (1) orbital
All p sub-levels have three (3) orbitals
All d sub-levels have five (5) orbitals
All f sub-levels have seven (7) orbitals.
Energy of the subshell increases with l ( s < p < d < f ) . Orbitals
closer to the nucleus have lower energy than those far away from
the nucleus and therefore more stable.
61
Question 8
How many orbitals are there in the third energy level?
62
2.4 Shpes of Orbitals
A: S – orbital
S orbital is represented by a spherical shape.
1s, 2s, 3s etc. orbitals only differ in radii (or size) which
increases as the principal quantum number increases
63
The S orbital is non – directional or spherically symmetrical
about the nucleus.
The probability of finding an electron at a distance from the
nucleus is the same in all directions.
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B: P – Orbital
All p – orbitals have the shape like dumb-bells or two pears
joined together at the nodal plane.
The three P-orbitals which are of the same energy (i.e.
degenerate orbitals) are represented as px py pz indicating
they are directed in space towards x, y, and z axes.
The three P-orbitals are perpendicular to each other.
65
NB: The 3p orbitals, Px , Py and Pz differ in orientation only.
66
Question
Explain, with the aid of a diagrams, the shapes and directions
of the S and P orbitals.
67
Guiding Principles Used To Explain The Arrangement of
Electrons In Atomic Orbitals
• Three rules or principles guide us in writing the electronic
configuration of atoms or elements.
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1. AUFBAU PRINCIPLE:
“Aufbau” means “building up”.
It states that orbitals of lower energies must be filled with
electrons first before orbitals of higher energies
or electrons always occupy the lowest energy first.
2. PAULI’S EXCLUSION PRINCIPLE:
• It states that the total number of electrons that can fill any
one orbital must be two and they must have opposite spins.
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OR
No two electrons in an atom must have the same set of the
four quantum numbers.
3. HUND’S RULE OF MAXIMUM MULTIPLICITY:
Degenerate orbitals (orbitals of the same energy) are first
filled singly and with parallel spins before they begin to pair
up with opposite spins
70
Electron Spin
1. Each orbital can hold a maximum of two electrons, so long
as their spins are opposite.
2. Paired electrons with opposite spins tend to be more stable
than the unpaired electrons.
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Order of Filling Electrons in Energy Levels
1. When writing electronic configuration, the 3d orbital is
written before the 4s.
1s 2s 2p 3s 3p 3d 4s 4p 4d
2.The order of filling is 1s 2s 2p 3s 3p 4s 3d 4p 4d etc
3. Note: The 4s orbital is lower in energy than the 3d orbital.
This is because the 4s orbital penetrates deep into the
nucleus than the 3d orbital. The nearer an orbital is to the
nucleus, the lower its energy.
72
73
The detailed electronic configuration of the
elements H to Zn are illustrated below:
It is observed that, 10Ne fill the L shell (with principal
quantum number no ,. n = 2) completely and 11Na
begin to fill the M-shell (n=3).
Therefore, conveniently one can represent the filling
of the L – shell by the neon core, 10Ne as :
2 2s2 2p2
2
2
10Ne = 1s
x 2p y 2pz
Hence:
74
75
76
• Note that, for Cr and Cu the configuration B are assigned to Cr and Cu
are correct and not configuration A. This is because there is extra
stability associated with atoms with fully filled orbitals as well as half
filled orbitals.
77
Question
Draw the electronic configuration of the following atoms using
boxes
1. Lithium
3Li
2. Carbon
6C
3. Oxygen
8O
4. Scandium
21Sc
5. Manganese 25Mn
Question 2
Give the detailed electron configuration of the elements with the
following atomic numbers
i). 7 and ii). 24
78
Question
Give the electron configuration of each of the ions, F- and Cr3+
[ Atomic numbers: F = 9, Cr = 24]
Question
a. What principles or values are violated in the following
electronic configurations?
i) Boron:
1s2 2s3
ii) Nitrogen: 1s2 2s2 2p2x 2p1y
iii) Berylliun: 1s2 2p2x
b. Write the correct configuration for each of the elements.
79
MASS SPECTROMETRY
Definitions:1. Isotopes:
They are atoms of the same element with the same atomic
number but with different mass number.
Or
Atoms of the same element with the same number of protons
but different number of neutrons.
Isotopes of an element have the same number of protons
and hence have the same number of electrons since the
electrons are the ones responsible for chemical properties,
isotopes of an element, show similar chemical properties.
80
2. Isotopy:
An element which possess atoms of similar chemical
properties but different masses is said to show isotopy.
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1. In mass spectrometry, masses and relative abundance are
measured when charged particles of isotopes of different
masses are separated.
2. The instrument used is called mass spectrometer.
Uses of Mass Spectrometer
1. Determination of Isotopes and relative atomic masses:
Mass spectrometer gives information about relative abundance
and the
masses of isotopes of elements. Thus for an
element, the number of lines drawn or peaks gives the number
of isotopes.
82
For example, Cl exit as two isotopes with mass 35 and 37 and
relative abundance of 75.53 and 24.47% respectively. A mass
spectrum of Cl will therefore reveal peaks exactly at masses 35
and 37 and peak height (intensity ) at 75.53 and 24.47%
respectively. The relative atomic mass of the element can
therefore be calculated.
2. It is used to determine the relative atomic masses and
relative molecular masses molecular masses of compounds:
A compound whose elements exist as isotopes does not have
the one molecular.
83
Functions of the main parts of the mass Spectrometer:
84
The mass spectrometer consists of :1. Vaporisation chamber
2. Ionization chamber
3. Accelerating electric field
4. Magnetic field at right angles to the electric field.
5. Detector chamber
85
1. Vaporisation chamber:
The sample of the atom is heated in this chamber to produce it
gaseous vapour.
2. Ionization chamber:
The gaseous atom is then passed through this chamber. High
energy electrons from the cathode are attracted to the
positively charged anode and in the process collide with atoms of
the vapourised sample and knock off electron(s) from them to
become positive ions. The main part of the mass spectrometer in
which the vapourised sample enters is always kept at very low
pressure.
86
3. Electric field:
The specific ions are attracted and accelerated by the negative
electric plate and move through a narrow slit into the magnetic
field arranged perpendicularly to the electric field
4. Magnetic field:
It deflect ions according to their charge to mass ratio into a
semi-circular paths The greater the charge carried.by the
particle, the greater the deflection and vice versa. Particles of
the same mass to charge ratio are deflected to follow the same
path. those with different masses deflect to follow different
paths.
87
5. Detector chamber
The detector collects the ions of different masses separately
and the collected charges which constitute an electric current
is proportional to the relative abundance of each species are
shown as peaks in a recorder (or read out system). The
instrument records the information as a mass spectrum,
Relative intensities or peak heights are then measured.
The mass spectrum is a plot of the relative abundances
(intensities or peak heights) versus the mass of the charged
particles
88
Relative atomic mass of the element is then computed as
follows:
∑(𝑀𝑎ℎ𝑎 + 𝑀𝑏 ℎ𝑏 + ⋯ 𝑀𝑛ℎ𝑛)
ℎ𝑎 + ℎ𝑏 + ⋯ + ℎ𝑛𝑏𝑛.
Where Ma and Mb are the isotopic masses of the atoms and ha
and hb are their respective relative abundances
89
Calculating Relative Atomic Masses of Isotopes
Question 1
69
Naturally occurring Gallium consists of two isotopes 31
Ga and
74
31Ga with relative abundance of 60% and 40% respectively. If
their masses are 68.93 amu (atomic mass unit) and 70.92 amu,
what is the average mass of the Ga atom.
Solution
The average mass of Ga is the Relative atomic mass of Ga.
69
Ar(Ga) = atomic mass unit of 31
Ga 𝑥 𝑖𝑡𝑠 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒
abundance/100+ atomic mass unit of 74
31Ga x its relative
abundance /100
90
Ar(Ga) =
68.93 𝑥 60
100
70.92 𝑥 40
+
100
=
68.93 𝑥 60+ 70.92 𝑥 40
100
=
4135.8+2836.8
100
=
69.73
91
Note that: atomic mass = isotopic mass ≈ mass number, Hence
if the atomic masses (isotopic masses) of the various isotopes
are not given, their mass numbers may be used. However, if
both are given, then the atomic (isotopic) masses should be
used and not the mass numbers.
Question 2
35
The
element
Q
has
two
naturally
occurring
isotopes,
17Q and
37
17Q with relative abundance of 75.5% and 24.5 % respectively.
Calculate the relative atomic mass of the element.
Relative Atomic mass of (Ar) of Q
∑( Mass number x Relative abundance)
=
Total abundance
92
35 𝑥 75.5+ 37 𝑥 24.5
=
100
= 35.5
93
Question 2
The element Q has two naturally occurring isotopes, 35
17Q and
37
17Q with relative abundance of 75.5% and 24.5 % respectively.
Calculate the relative atomic mass of the element.
Solution
Relative Atomic mass of (Ar) of Q
=
Ar(Q)
=
∑( Mass number x Relative abundance)
Total abundance
35 𝑥 75.5
100
37 𝑥 24.5
+
100
35 𝑥 75.5+ 37 𝑥 24.5
=
100
= 35.5
94
Question 3
There are two isotopes of Bromine with mass numbers 79 and
81 respectively. If the isotopes exist in the ratio of 2: 1, the
lighter isotope being the most abundant, what is the relative
atomic mass of bromine? Account for why the value is not a
whole number.
95
Solution:
79 x 2 +( 81 x 1)
Ar (Br) =
3
= 239/3
= 79.7
The value of the atomic mass is not a whole number since Br is
isotopic and the weighted average of it isotopic masses gives a
fractional value.
96
Question 4
• Chlorine has two isotopes with mass numbers 35 and 17
respectively. If the relative atomic mass of Cl is 35.5.
Calculate the relative abundance of each of the isotopes.
97
Question 5
The element Q has two naturally occurring isotopes, 35
17Q and
37
17Q with relative abundance of 75.5% and 24.5 % respectively.
Calculate the relative atomic mass of the element.
Question 6
35
Chlorine has two naturally occurring isotopes, 37
Cl
and
17
17Cl .
The accurate mass and relative abundance of the first isotope
37
17Cl are 36.97 and 24.47% respectively. Calculate the relative
abundance and the accurate mass of the second isotope.[
Relative atomic mass of chlorine is 35.45].
98
Solution:
The relative abundance of the second isotope
= 100 – 24.7 = 75.53
Let y represent the relative atomic mass of
∴ Ar (Cl)
36.97 𝑥 24.47
=
100
35.45 =
35.45 x 100
=
35
17Cl
35
17Cl
75.53 𝑥 𝑦
+
100
904.66 +( 75.53 𝑦)
100
904.66 + 75.53y
99
3545 =
y =
∴y =
904.66 + 75.53y
3545−904.66
75.53
34.96
100
Question 5
• Two of the three stable isotopes of Mg has atomic masses of 24 and 25
and percent abundances of 78.6% and 10.1% respectively. If the
relative atomic mass of Mg is 24.3. Calculate the relative abundance of
the third isotope and hence its atomic mass
Question 6
• An
element of X with relative atomic 210.2 has two isotopes
212
X. What is the relative abundance of each isotope.
210
X and
101
Representation of an atom:
For an atom X, Mass number is placed as a left
superscript. Atomic number Z, is placed as a left subscript.
i.e 𝑨𝒁𝑿
eg.
35
17Cl
35 = mass number and 17 = Atomic number
102
Important Terms to Note and Calculation involving the
particulate entities of the atom:
Isotope
Atomic
Mass Relative
Atomic
Mass Mass
(Ma)
(Ar)
(A)
35
17Q
34.97u
34.97
35
37
17Q
36.96u
36.96
37
number
103
1. Relative Atomic Mass (Ar)
• Relative atomic mass of an element Ar, is defined
as the weighted average of the isotopic masses on
the scale on which the mass of one atom of carbon
is taken as exactly 12. It is a pure number and has
no units.
• The relative atomic mass of an element represents
the average mass of one atom taking into account
the different isotopes and their relative proportions.
104
NB: The nearest whole number to relative isotopic (atomic )
mass of an isotope is the mass number.
• Eg Relative isotopic mass of a single chlorine isotope 35
17Cl is
34.97. Its nearest whole number is 35 which is the mass
number.
105
2.
It can also be defined 1as the average mass of one atom of an
element compared with of the mass of one atom of 12C, or
12
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑎𝑡𝑜𝑚 𝑜𝑓 𝑎𝑛 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
1
12 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑎𝑡𝑜𝑚 𝑜𝑓 𝐶−12
=
Question
The atomic mass unit of Carbon – 12 is 1.6603 x 10-24 g. If the
mass of an atom is 5.313 x 10-23 g determine its relative mass.
−
Ar
=
5.313 x 10 23
−
1.6603 x 10 24
= 32
106
Question
The ratio of the mass of an aluminium atom to Carbon – 12 atom is
2.248:1. Calculate the relative atomic mass of aluminium.
Solution
Ar Al =
𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑎𝑡𝑜𝑚 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
1
12 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑎𝑡𝑜𝑚 𝑜𝑓 𝐶−12
=
2.248
1 (112 )
= 26.976
3. Ar (Mg) = 24. This means, the relative atomic mass of naturally
occurring magnesium is 24. which also implies one atom of
1
magnesium is 24 times as heavy as of the mass of one atom of
12
C-12
107
4. The atomic mass unit (amu),
1
is defined as
of the mass of an atom of Carbon – 12. It is
12
represented by u (amu). It is measured on the carbon – 12
scale and has units of U
5. Atomic Number or Proton Number.
It is the number of protons in the nucleus of an atom. Symbol Z
For a neutral atom the atomic number which is the number of
protons in the nucleus = number of electrons in the atom.
108
6. Nucleons
• They are the protons and neutrons in the nucleons of an
atom ( i.e the particles making up the nucleus
7. Nuclide
• It is a nuclear species (or atom) with a specific atomic
number and a specific mass number.
• Eg
16
8O
17
8O
14
7O
109
8. Neutron Number
• It is the number of neutrons in the nucleus of the atom
N
The symbol is
9. Mass Number or Nucleon Number
• It is the sum of the number of protons and neutrons in an atom. Symbol is
A
• Mass number = atomic number + number of neutrons
i.e A = Z + N
110
Worked Examples
Question 1
• Atomic number of Na is 11 and its mass number is 23. Find
the number of
i) Protons
ii). electrons and
iii) neutrons in one of its atoms
• Solutions
i) Atomic number = Number of Protons
∴ No of protons = 11
ii) No of protons
= No of electrons = 11
111
iii) Mass number = Atomic No + No of neutrons
23
= 11 + N
∴ No of neutrons = 23 – 11
= 12
Question 2
An atom of an element has 69 neutrons and 50 electrons.
What is its mass number ?.
Solution
Atomic number = Number of protons = number of electrons
= 50
∴ Mass number = 50 + 69
= 119
112
Question 3
How many protons, neutrons and electrons are present in each
of the following species
) 35
39
i)
K
ii
19
17Cl
Solution
39
19K
i)
Number of protons = atomic number = 19
Number of neutrons = mass number – number of protons
= 39 – 19 = 20
113
number of electrons = number of protons = 19
–
ii 35
17Cl
Number of protons = atomic number = 17
Number of neutrons = mass number – number of protons
= 35 – 17 = 18
Number of electrons = number of protons + 1 electron gained
= 17 + 1 = 18
114
Question 4
The total number of electrons and protons in a neutral atom is 28. The mass
number is 32. What is the number of neutrons ?
Mass number = Number of protons
32 = Number of protons
+
+
number of neutrons
N
Let the number of electrons in the atom be x
∴ the number of protons in the atom is also = x , since the number of protons
= number of electrons
∴ x + x = 28
i.e 2x
= 28
x
= 28/2 = 14
the number of protons = number of electrons = 14
32 = Number of protons
∴N = 32 – 14
= 18
+
N
115
DIMENSIONAL ANALYSIS AND UNITS IN
CHEMISTRY
OBJECTIVES:
By the end of the lecture, it is expected that students will be
able to explain what a unit is.
ii. Know the basic units for mass, length, volume and amount of
substance.
iii. Understand the prefixes for fractions or multiples of units.
iv. Be able to explain what % and ppm mean
v. Be able to convert % to ppm and vice versa
116
• Most properties in science are given in units to make
comparisons, interpretations and therefore, draw conclusions
or make recommendations.
• A unit is a quantity by reference to which others are measured.
• Properties in the various fields of Agriculture and consumer
sciences are measured in units related to
• le Système International d’ Unitès, abbreviated SI.
117
Table 1. Basic units of some measurements
The basic units of the quantities mass, length, volume and
amount of substance are given in Table 1.
Property
Basic unit
Symbol
Mass
gramme
g
Length
metre
m
Volume
litre
L
Amount of
substance
mole
mol
118
In measurements, there are instances when there are fractions
or multiples of the units. In instances like that, prefixes in Table
2, accepted worldwide, are assigned.
119
Fraction
Prefix
Symbol
Multiples Prefix
Symbol
10-1
deci
d
10
deca
da
10-2
centi
c
102
hecto
h
10-3
milli
m
103
kilo
k
10-6
micro
m
106
mega
M
10-9
nano
n
109
giga
G
10-12
pico
p
1012
tera
T
120
From Table 2,
• micro is 10-6
therefore 1micrometer is 10-6 of a meter
The symbol of micrometer (μm).
Please note the symbol for micro. It is ‘μ’ and not ‘u’.
•
milli is 10-3
therefore 1milligrame is 10-3 of a gramme
Note the difference in symbols between Mega (M) and milli (m).
121
• Megagramme is 106
Therefore a Megagramme (Mg) is One million (106 ) gramme
• Note the difference in symbols between Mega (M) and milli
(m).
• Mg, g and kg are units which refer to measurements of
weight. For ease of comparison in weights, the same unit
could be used and conclusions drawn as to which weight is
heavier or lighter.
122
• Write and explain the following units;
1. Centimole
2. Megabyte
3. Millilitre
4. Decijoule
5. Microvolt
6. Nanometre
7. Hectomole
123
A milk tin full of soil sampled from a field on a rainy day may
weigh 60 g. Three days later, the soil sample could weigh
45,000 mg.
The units ‘g’ and ‘mg’ are all quantities by reference to which
the soil is measured. However, for ease of comparison, the
units will have to be the same to ascertain which soil is heavier.
124
• milli = 10-3
103 mg = 1g
Therefore, 45,000 mg =
(45,000 mg / 1000mg) x 1g
= 45 g.
The soil sampled on the rainy day is therefore,
heavier by 15 g than that three days later.
125
ASSIGNMENT
1. If 1000 kg = 1 ton, show that 1 ton is the same as 1 Mg.
2. Express 2 microns in millimetres.
3. Change 0.2 m into cm.
4. How many milligrammes will equal 2 kg.
5. Change 2 mg into kg
6. Change 2 moles into centimole.
126
Concentration is the measure of the amount or quantity of a
given substance/ element/ ion in another.
* Note that concentration is different from quantity.
Quantity is amount only.
Concentration is quantity in another quantity.
As a guide, concentration is quantity per quantity.
127
 Percent
This is defined as part(s) of a given material in hundred parts of
a whole. For example, parts of yeast in 100 parts of bread or
parts of sugar in 100 mL of tea.
When 10 g of NaCl is dissolved and topped up to 100 mL with
water, the concentration of the solution is 10% NaCl.
Note that there is no space between the unit ‘%’ and the
figure or number unlike the units discussed earlier.
128
• parts per million (ppm)
This is the part of a material in one million parts of a whole.
A loaf of bread with 2 ppm NaCl means 106 parts of the loaf contains
2 parts NaCl. It could also mean that 106 g of the bread contains 2 g
NaCl.
Percent and ppm are units of concentrations. These are fractions with
the denominator referring to the material as a whole in 100 (%) or in
1,000,000 (ppm).
The numerator, is the amount of the element or chemical of interest.
Thus, if the concentration of salt in kenkey is 4%, it implies that the
ball of kenkey is the whole i.e.100 with the salt component being 4.
This could, therefore , be interpreted as;
100 g ball of kenkey contains 4 g salt or,
100 kg ball of kenkey contains 4 kg salt
129
• Similarly, 4 ppm salt in kenkey implies 1,000,000 g kenkey contains
4 g salt
or 1,000,000 hg kenkey = 4 hg salt
Note that for ppm and % both denominator and numerator must
have the same units to cancel out.
130
Conversion of % to ppm
Example
Express 0.001% Mn in soil as ppm.
0.001% Mn implies 100 parts of soil contains 0.001 part Mn
But ppm means parts in 106
Thus if 102 whole soil contains 0.001 part Mn
Then 106 whole soil will contain more:
= 106 / 102 x 0.001
= 0.001 x 104
= 10 ppm
Thus ppm = % x 104
131
Conversion of ppm to %
Example
The P concentration in a plant is 20 ppm.
What is the concentration in percent?
Solution:
20 ppm P in plant means
106 whole plant contains 20 parts P
But % means whole plant should be 100
Thus, if 106 whole plant = 20 parts P
132
Then 102 whole plant = ?
= (102 / 106) x 20 P
= 10-4 x 20 P
= 0.002% P
% = ppm x 10-4
For solids and solid-solid solutions, % and ppm could be vague
as the two units can be interpreted in so many ways. For
example,
2 ppm Na in chocolate could be interpreted in
• as;
133
i. 106 kg chocolate contains 2 kg Na
ii. 10 6 g chocolate contains 2 g Na
iii. 106 mg chocolate contains 2 mg Na, or
iv. 106 lbs (pounds) chocolate contains 2 lb. Na.
To avoid this ambiguity, in agriculture and Consumer Sciences,
the unit mg/kg (preferred) or μg /g are used INSTEAD of ppm.
A soil of 2ppm P is the same as 2 mg P/ kg soil or 2 μg P / g
soil
134
Example
Show that ppm = mg/kg = μg /g
Solution
103 mg = 1 g
1mg 𝑥 1 g
∴ 1 mg =
1000 mg
= 10-3g
Also 1 kg = 103 g
∴ mg / kg (or mg kg-1 ) = 10-3g in 103 g
converting 10-3g to µg
1 g = 106 µg
∴ 10-3g = 10-3g / 1g x 106 µg
= 103 µg
135
⇒ mg/kg = 103 µg /103 g
= ug/g
TRIAL
1. Show that μg/g = ppm.
In solids and solid-solid solutions too, % is vague.
2. Give four possible interpretations that could be given to a piece of
meat with 5% Ca
136
the units ‘g/kg’ (preferred) and ‘mg/g’ are used in place of %.
* Note unlike ppm which is equal to mg/kg or μg/g,
g/kg ≠ %.
There is a conversion factor
Changing from % to g/kg
Here again, the denominator refers to the whole material and the
numerator the element or chemical of interest.
137
Example
The carbon concentration of soil is 1%. What is the concentration in g (C)
/ kg soil or g/kg?
Because the unit to which % is being converted to, is g/kg and the
numerator which is the
weight of the carbon is in ‘g’, it is prudent to Interpret 1% C in soil to mean
100 g soil contains 1 g carbon
i.e. 1 g C / 100 g soil = 1/ 100 = 1%
If 100 g soil = 1 g C
Then 1000 g (1 kg) = (1000 g / 100 g) x 1 g C
= 10 g C
= 10 g C / 1kg soil
138
Alternatively, by what factor must the 100g be multiplied to get 1000 g
which is 1 kg of soil.
If the denominator is multiplied by 10, then the numerator also has to be
multiplied by 10.
The 1 g then changes to 10 g.
g / 100 g (%) = Y g /1000 g
Y = 10
Thus,
g / kg = % x 10
139
Show that g/kg = mg/g.
When working with solid-liquid solutions,
ppm = μg/mL or mg/L
The density of water is 1 g/cm3 or 1g/mL
1 cm3 = 1 mL
The density of water being 1 g/cm3 means
that 1 cm3 or 1 mL of water weighs 1 g.
If 1cm3 or 1 mL of water = 1g
Then μg/mL = ug/g = ppm
140
Also 1000 mL (cm3) = 1L
If 1 mL or 1 cm3 of water weighs 1 g
The 1000 mL or 1L weighs 1000 g = 1 kg
Therefore mg/L = mg/kg = ppm
Assignment
Show that 1 dm3 = 1000 cm3 = 1L
The density of water is 1g/cm3 = Mg/m3
Show that g/cm3 = Mg/m3
141
Example 6
A 0.002 g NaCl pellet is weighed into a flask
and topped up to 100 mL with distilled water.
Calculate the concentration in
%
ppm
The final volume of the solution prepared is
100 mL. But
1 mL of water = 1g
Therefore, 100 mL = 100 g
Concentration is 0.002 g / 100 mL
=0.00 2g / 100 g
= 0.002%
142
We have proved that ppm = mg/kg = mg/L
0.002 g NaCl = 2 mg NaCl
1000 mg = 1 g
Thus 0.002 g NaCl = 2 mg NaCl
If 100 mL water = 2 mg NaCl
Then 1000mL i.e. 1 L (1kg) =
(1000mL/100mL) x 2 mg = 20 mg
Concentration is therefore 20 mg /L = 20 mg/kg
= 20 ppm
143
Exercise
Calculate the concentration of 4% KCl solution
in;
g/L
ppm
μg/mL
A soil has 4 ppm P. Calculate the level of P in
ppm
ppb
mg/kg
144
Can the concentration of the P in the soil be
expressed in
mg/L
μg/kg
Explain the following concentrations
1. 3 ppm sugar solution
2. 4% blood sugar
145
PREPARATION OF SOLUTIONS
• Definitions:Solution
A homogenous mixture of a solute and solvent.
Solute
It is the substance that dissolves in a solvent.
Solvent
It is the substance that dissolves the solute and it is usually a liquid. The
solute can be a solid, liquid or a gas.
A solution whose solvent is water, is called an aqueous solution.
146
• If the solute is a liquid, then in the solution, the substance with the smaller
volume is the solute.
• Generally, the quantity of a solute in a unit quantity of solution is called
concentration of solute.
Concentration can be expressed in four main ways:
1. Amount concentration (moldm-3)
2. Mass concentration (gdm-3)
3. Molal concentration (molkg-1)
147
• 4. Mole fraction
• 1. Amount concentration (mol dm-3)
It is the amount of substance or number of moles of a substance
in 1 dm3 of a solution.
Unit is mol dm-3 or mol L-1or M (meaning molar).
For example, 0.5 mol of NaOH dissolved in 1 dm3 of solution is dm3
dissolved
expressed as 0.5 mol dm-3 or 0.5 mol L-1 or 0.5 M
148
• NB: Concentration (mol dm-3) =
n (mol)
=
n (mol)
v (dm3)
m (g)
M (g mol−1)
• 2. Mass concentration is the mass of a substance dissolved in 1 dm3 of
solution.
Mass concentration is sometimes called mass density and can be
expressed as g cm-3 .
149
• g dm-3 = mass of solute in g in 1 dm3 of solution.
Question
Calculate
i. the concentration of a solution prepared by dissolving 5 g of NaCl to make 500
cm3 of solution.
ii. the amount (moles) of NaOH in 1 dm of solution of concentration 0.1 mol dm-3 [
Na = 23; Cl= 35.5 ; O = 16 ; H = 1 ]
Solution:
Concentration (mol dm-3) =
also n =
n (mol)
v (dm3)
m (g)
M (gmol−1 )
150
Molar Mass of NaCl = 23 + 35.5 = 58.5 g mol – 1
Volume of solution = 500 = 0.5 dm-3
m (g)
n =
M (gmol−1)
=
5 (g)
(58.5gmol−1)
=
Concentration (mol
0.085mol
dm-3)
=
=
n (mol)
v (dm3)
0.085mol
0.5dm3
151
= 0.0171 mol dm-3
ii. Concentration (mol dm-3) =
n (mol)
v (dm3)
n (mol) = concentration (mol dm3) x volume (dm3)
= 0.1 mol dm-3 x 1 dm3
= 0.1 mol
152
In an experiment involving the hydrated salt Na2CO3 .XH2O,
where X = number of molecules of H2O water of crystallization, the mass
Concentration of its solution is 5.72 g dm-3 and from the experiment, the
concentration was found to be 0.02 mol dm-3
Calculate
i. The molar mass of Na2CO3. xH2O in terms of x
ii. The molar mass of Na2CO3. xH2O
iii. The number of molecules of H2O of crystallization x of the hydrated salt
and hence it chemical formula.
[ Na = 23; O = 16 ; C = 12 ]
153
PREPARATION OF STANDARD SOLUTIONS
A: Using the salt
A standard solution is a solution whose concentration is
accurately known.
In preparing standard solutions the following glassware and
materials are used
• volumetric flasks
• beaker
• distilled water
• electronic balance
154
• Stirrer
Procedure
1. The amount of the salt (moles) to be prepared is calculated in terms of
mass.
2. The mass is then weighed accurately with a weighing balance in a
weighing boat or in a beaker
3. The salt is then dissolved in the beaker with distilled water.
4. The content of the beaker is then poured into the standard volumetric
flask of 1L or 1000 cm3
5. The beaker is rinsed at least 3x to ensure quantitative transfer of all the
substance
155
6. More distilled water is added to reach the maximum mark of the
volumetric flask
7. The solution is then labelled by indicating the concentration and the name
of the reagent or the salt.
156
Questions
How would you prepare a solution of 500 ml of 0.5 mol dm-3 of
NaOH solution? [ NaOH = 40 g mol-1
1. Calculate the mass of NaOH required.
C = n v = 500 ml = 0.5 dm3
v
0.5 mol dm -3 = n
0.5 dm3
n = 0.5 mol dm-3 x 0.5 dm3
= 0.25 mol.
157
• moles of NaOH = mass of NaOH
molar mass
mass of NaOH = 0.25 mol x 40 g mol-1
= 10 g of NaOH
2. weigh 10g of NaOH using a weighing boat on an electronic
balance into a beaker.
158
3. Add distilled water to the NaOH in the beaker to dissolve.
4. Transfer the solution into a 500 ml volumetric flask. Rinse at least 3x to
quantitatively transfer all the NaOH into the volumetric flask.
5. Add distilled water to the mark.
6. Labelled the solution by indicating the concentration and the name of
the reagent or substance
159
Preparation of standard solution using the Assay on a reagent bottle of a
concentrated solution:
Definition:
An Assay: is a full list of the experimentally determined purity of a chemical
including its impurities.
For example the following label specifications on an HCl solution container
is called an assey:
Mass per cm3 (density or specific gravity) = 1.18g cm3 ; percentage purity =
36 %; Molecular mass = 36.5 g mol-1
160
from the information we can calculate the concentration of the pure
HCl solution.
Solution:
Density of 1.18g cm-3 means that, for every 1 cm3 of the concentrated
solution, there is 1.18 g of inpure HCl
36% w/v means that, for the 1.18 g of impure HCl in the 1 cm3 of
solution, there is only 36 /100 x 1.18 g
= 0.4248 g of pure HCl
Hence, 1000 cm3 of the concentrated solution will contain
161
1000 /1 x 0.4248
= 424.8 g of pure HCl.
Mass concentration of pure HCl
Concentration of HCl = g dm-3
Molar Mass
= 424.8 g dm3
36.5 gmol= 11.64 M
162
Dilution of Solutions:
 Standard solutions can be diluted to smaller concentrations by adding water. This
process is called dilution.
 The concentrated solution from which the diluted solution is prepared is called a
stock solution.
 During dilution, it is always the volume of the solution that changes but the number
of moles or amount of substance in the solution remain constant.
That is, initial moles before dilution = final moles after dilution
Initial moles = Initial concentration (ci) x volume (vf) x 10-3
Final moles = Final concentration (cf) x volume (vf) x 10-3
163
Therefore after dilution,
ci x vi = cf x vf x 10-3
ci x vi = cf x vf
ci = initial concentration
vi = initial volume
cf = final concentration after dilution
vf = final volume after dilution
164
For a 10 ml of 0.5 M NaOH solution,
0.5M means 0.5 mol dm-3 solution. That is 0.5 moles of the NaOH is in 1000
cm3
Therefore moles of NaOH in 10 cm3 = 0.5 x 10 moles
1000
= 0.5 x 10 x 10-3 moles
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The number of times the volume or concentration of the stock solution is
diluted is called the dilution factor (Df).
Dilution factor (Df) = ci = vf
cf
vi
ci = initial concentration of solution
cf = final concentration of diluted solution,
vi = initial volume of the concentrated solution
vf = final volume of diluted solution
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Example
10 cm3 of 2M solution is diluted to 200 cm3 . What is the concentration of the
diluted solution?
Answer:
Initial moles of the solution before dilution = final moles of the solution after
dilution.
ci vi = cf x vf
where ci, cf are the initial and final concentrations respectively
vi and vf are the initial and final volumes respectively.
167
ci = 2M
vi = 10 cm3
vf = 200 cm3
cf = ?
ci vi = cf x vf
cf
= ci x vi = 20 x 10
vf
200
= 0.1 M
168
The dilution factor,
Df = ci = 2 M
cf 0.1M
= 20
169
AMOUNT OF SUBSTANCE AND THE MOLE
CONCEPT.
The mole is a term from Latin which means “to heap” or “pile up”.
It is the chemists’ way of counting. It is used to count electrons,
atoms, molecules and other small particles.
Definition:
• The Mole is the amount of substance which contains as many
particulate entities as there are in exactly 12 g of carbon - 12
isotope.
• The particles which are elementary units or particulate entities
of matter can be atoms, molecules, ions, formula units or
electrons.
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• The molar mass is the mass of 1 mole of an element and is
given the unit ‘ g/mol’. The molar mass of H being 1 g/mol
means 1 mol of H weighs 1 gramme.
• Oxygen has a molar mass of 16 g/mol. This , therefore,
means that 1 mole of the element weighs 16 g.
Since the mole is the chemist way of counting, it implies that it is
a quantity.
Thus quantity of any element, ion, atom or substance in mass
can also be expressed in mole or vice versa.
171
Avogadro’s Constant, L
It is defined as the number of particles in a mole of a
substance.
Since a mole of substance contains particles as many as there
are atoms in 12 g of C-12,
Avogadro’s constant can also be defined as the number of
atoms in 12 g of C-12
Particles in 1 mole ≡ Avogadro’s constant ≡ number of atoms in
12 g of c-12
Avogadro’s constant, L = 6.02 x 1023 entities per mole.
172
i ) Therefore 1 mole of a substance contains 6.02 x 1023
entities.
ii ) The number of atoms in 12 g of C- 12 = 6.02 x 1023
173
Note
i ) 1 mole of oxygen gas (O2) contains 6.02 x 1023 oxygen
molecules
ii ) 1 mole of oxygen atoms (O) contains 6.02 x 1023 oxygen
atoms
m
Mole of a substance = n =
M
n = amount of substance (mol); m = mass of substance (g); M
= molecular weight/ Molar mass (g/mol)
N=nxL
Where
n = amount of substance ( in mole)
L = Avogadro’s constant
N = Number of entities
The molar mass is the mass of 1 mole of a substance (12 g of
carbon -12) it is measured in g/mol
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The molar mass is the mass of 1 mole of a substance (12 g of
carbon -12) it is measured in g/mol
Molar volume, Vm is the volume occupied by 1 mol of a gas at
standard temperature (273 K ; 0 oC ) and pressure (100 kPa ; 1
bar ; )
1 mole of a gas occupies 22.4 dm3 volume at STP (i.e Vm = 22.4
dm3)
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Question 1
Calculate the number of moles and the number of Al atoms in
10.0 g sample.
[ Molar mass of Al = 36.98 g mol- , L = 6.022 x 1023 mol-1 ]
Solution:
Given mass of Al = 10.00 g and M = 26.98 g/mol
n=
moles of Al =
𝑚
𝑀
10.00 g
26.98 g /mol
= 0.3706 mol Al
Hence, there are 0.3706 mol of Al in 10.00g
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To calculate for the number of atoms of Al, we have to use the
Avogadro’s number.
1 mol of Al contains 6.022 x 1023 atoms, therefore, 0.3706 mol
will contain
=
0.3706 mol x 6.022x 1023
1 mol
= 2.232 x 1023 atoms
hence, there are 2.232 x 1023 atoms of Al in 10.00g sample.
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Question 2
Calculate the mass of Americium (Am) in grams in a sample of
Americium containing 6 atoms. Molar mass of Am = 243 g/mol L =
6.022 x 1023 ]
Solution:
To calculate the mass, we must first know the amount of Americium
that contains 6 atoms. If we know the number of atoms present in the
sample, we can calculate the number of moles of Americium.
If 6.022 x 1023 atoms are contained 1 mole of Am
6 atoms x 1 mol
then 6 atoms will be contain in
= 9.9635 x 10-24 mol
23
6.022 x10 atoms
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hence, 6 atoms of Am contains 9.9635 x 10-24 mol.
We can calculate the mass of the sample since the molar mass
is known.
m
n=
M
m= nxM
= 9.9635 x 10-24 mol x 243 g/mol
= 2.4211 x 10-21 g.
179
Question 3
How many atoms are there in 0.3 mole of sodium ? [L = 6.022 x
1023 mol-1]
Solution
1 mole of Na contains 6.022 x 1023 Na atoms
0.3
∴ 0.3 mol. of Na contains =
x 6.022 x 1023
1
= 1.806 x 1023 Na atoms
Alternative solution
using the N = n x L
Where
n = amount of substance (in mole)
L = Avogadro’s constant
N = Number of entities
180
No. of atoms of Na = 0.3 mol x 6.022 x 1023 mol-1
= 1.86 x 1023 atoms.
Question 4
Calculate the number of molecules in 0.25 mol of oxygen [L =
6.022 x 1023 mol-1]
Solution
N=nxL
= 0.25 mol 6.022 x 1023 mol-1
= 1.505 x 1023 molecules
Alternative solution
1 mol of oxygen contains 6.022 x 1023 molecules
181
∴ 0.25 mol will contain
0.25
1
x 6.022 x 1023
= 1.505 x 1023 molecules
Question 5
Calculate the amount of oxygen gas in moles of 1.505 x 1023
molecules of the gas [L = 6.022 x 1023 mol-1]
Solution
6.022 x 1023 molecules contains 1 mol of oxygen
∴1.505 x 1023 molecules of the gas will be contained in
1.505 x1023
6.02 x1023
x 1 = 025 mole
182
Question 5
Calculate the number of atoms of oxygen in 0.20 mol. of
oxygen. [L = 6.022 x 1023]
Solution
NB: Oxygen exits naturally as diatomic molecule
∴ 0.2 mole of oxygen can also be read as 0.2 mole of
oxygen molecules.
1 mol of oxygen molecule (O2) = 2 moles of oxygen
atoms (O)
∴ 0.2 mole of oxygen molecules
0.2
=
1
x2
= 0.4 mol. of oxygen atoms
183
∴
number of atoms of oxygen
=nxL
= 0.4 x 6.022 x 1023
= 2.408 x 1023 oxygen atoms.
Question 7
Calculate the number of atoms of hydrogen in 0.2 mol of
methane gas (CH4) [L = 6.022 x 1023 ]
Solution
CH4 → C
1 mol
1 mol
1 mol of CH4
+
+
≡
4H
4 mol H
4 mol of H
184
∴ 0.2 mol
0.2 mol
=
x 4 mol
1 mol
= 0.8 mol H
N = nxL
= 0.8 x 6.02 x 1023
= 4.816 x 1023 atoms of H.
185
Molar Mass, M
1. It is the mass of one mole of a substance. It uint is g mol2.1 mole contains Avogadro’s constant, 6.02 x1023
∴ mass of 1 mole of a substance ( Molar mass ) also contains , 6.02 x1023
entities
3. Numerically, molar mass = relative atomic mass = relative molecular mass.
Note:
Mole
=
𝑚𝑎𝑠𝑠
Molar 𝑚𝑎𝑠𝑠
.
Quantitatively, the mole is mass of an element divided by its molar mass.
Mass / molar mass = g /mol= g/g x mol = mol
186
Question
Calculate the mass of 0.2 mol of magnesium atoms [Mg = 24]
Solution
1 mol of Mg ≡ 24
∴ 0.2 mol
0.2 x 24
=
1
= 4.8g
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189
190
191
192
PERIODIC CLASSIFICATION OF THE ELEMENTS
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