complete modul sem 1 2021 2022

FOUNDATION OF MARA 2021/2022

SCI 1054 CHEMISTRY
SEMESTER 1

EDITED BY: | CHEMISTRY UNIT
MARA COLLEGE OF KUALA NERANG & MARA COLLEGE OF KULIM

SCI 1054 CHEMISTRY SEMESTER 1

1. Name of

Course/Module CHEMISTRY I

2. Course Code SCI 1054

3. Synopsis This course is designed to provide the essential foundations of chemistry to prepare

students for higher studies where chemistry or chemistry-related subjects are taught. Students

will be exposed to a thorough introduction to chemistry, scientific methods and development

of skills relevant to the safe practice of science. Coverage of this course includes

stoichiometry, atomic structure, the periodic table, chemical bonding, thermochemistry,

hydrocarbon and halogenalkanes, chemical equilibrium and reaction kinetics.

4. Name(s) of 1. Siti Khadijah Bt Mat Daud (KMKN)

academic staff 2. Nurul Farehan Bt Rosli (KMKN)

3. Nur Hayati Bt Che Zan (KMKN)

4. Dr Mohd Shahir Bin Mohd Sunar (KMKN)

5. Nurul Suhadah Bt Murzahan (KMKN)

6. Nursafura Bt Abd Manap (KMK)

7. Nurul Elyani Bt Eleas (KMK)

8. Siti Zubaidah Bt Azizan (KMK)

9. Yuzaila bt Muhamad Yunus (KMK)

5. Semester offered Semester 1, Foundation in Science (KMKN & KMK)

(Program) Semester 1, Foundation in Technology (KMKN)

6. Credit Value 4

7. Prerequisite/co- -

requisite (if any)

8. Course Learning Upon successful completion of the course students will be able to:

Outcomes

1. Describe the concepts of physical and inorganic chemistry theories related

to definitions, laws/principles, chemical bonding and reactions. (C2, PLO1)

2. Solve problems with analytical and critical thinking by applying chemistry facts

and principles.( C4, PLO1,PLO2)

3. Apply some techniques used in chemistry experiments.(C3, P3, PLO2, PLO3, PLO4)

9. Mapping of the Course Programme Learning Outcomes (PLO) Teaching Assessment
Course Learning Learning strategies
Outcomes to the Outcomes PLO PLO PLO PLO PLO PLO Test, Quizzes, Final
Programme 123456 Examination
Learning (CLO)
Outcomes,
Teaching CLO 1 / Lectures,
Tutorials

1

SCI 1054 CHEMISTRY SEMESTER 1

Methods and CLO 2 // Lectures, Test, Quizzes, Final
Assessment: CLO 3 /// Tutorials, Examination,

Laboratory Assignment (Essay )
Practicals
Lab reports,
Presentations

-
10. Transferable Skills Critical Thinking and Problem Solving

(if applicable):

11. Distribution of Student Learning Time (SLT):

Teaching and Learning Activities Total
SLT
CLO* Guided Guided 20
Learning (F2F) Learning
Course Content Outline Independent 14
L TPO (NF2F) Learning
e.g., e- (NF2F)
Learning

1. Stoichiometry

1.1 Structure of atoms 1,2,3 622 10

1.2 Proton number, nucleon

number, isotopes, Avogadro

number

1.3 Relative atomic masses

1.4 Composition (%) of

compounds with the empirical

and molecular formula

1.5 Concentration of solution

1.6 Limiting reagent and

application to stoichiometric

calculations

2. Atomic structure

2.1 Nucleus of atom 1,2,3 4 1 2 7

2.2 Bohr atomic model

2.3 Bohr atom and Rydberg

equation calculations

2.4 Atomic orbital & quantum

numbers

2.5 Electronic configurations of

elements

2.6 Rules of the electronic

configuration

3. Periodic Table 1,2 3 1 48

3.1 Introduction
3.2 Classification
3.3 Periodicity properties

2

SCI 1054 CHEMISTRY SEMESTER 1

4. Chemical Bonding 9 18
8 15
4.1 Lewis dot symbol 1,2,3 5 2 2
4.2 Three types of chemical 8 16
9 17
bonding: ionic, covalent and

metallic bonding

4.3 Intermolecular bonding

4.4 Lewis dot structures

4.5 Molecular shape and polarity

4.6 Orbital overlap and

hybridisation

5. Chemical Equilibrium

5.1 Equilibrium and the 1, 2,3 3 2 2
equilibrium
constant

5.2 Writing expressions of the
equilibrium constant

5.3 Direction of reaction
5.4 Le Chatelier’s principle and its

application

5.5 Factors affecting chemical
equilibrium

6. Thermochemistry

6.1 Introduction to 1,2,3 4 2 2
thermochemistry

6.2 Enthalpy of reaction

6.3 Specific heat capacity and

heat capacity

6.4 Hess Law

6.5 Born-Haber cycle

7. Introduction to Organic 1,2 62
Chemistry

7.1 Functional groups and
nomenclature

7.2 Nucleophiles, electrophiles,
free radicals, homolytic and
heterolytic cleavage of bonds

7.3 Isomerism: structural (chain,
positional, functional group)
and stereoisomerism
(geometrical, cis-trans)

3

SCI 1054 CHEMISTRY SEMESTER 1

8.Hydrocarbons & Halogenalkanes

8.1 Alkanes

8.2 Alkenes

8.3 Arenes: resonance structure, 10 19
nomenclature, 1,2,3 5 2 2 65 127
IUPAC

physical properties, reaction,

mechanism

8.4 Classification of alkyl halides

8.5 Hydrolysis, formation of

nitriles primary amines and

elimination reaction (Grignard

reagents)

Total 36 14 12

Laboratory Practical (12 hours):
1. Introduction to Lab Techniques and Apparatus
2. Determination of Formula Unit of a Compound
3. Properties of Ionic and Covalent Bonds
4. Chemical Equilibria and Le Chatelier’s Principle
5. Determination of Heat & Reaction
6. Chemical properties of Alkanes and Alkenes

Assessment Percentage (%) F2F NF2F Total
SLT
Lab reports 15 3 6
Test 20 2 6 9
Assignments 15 2 4 8
Final Examination 50% 2.5 7.5
6
10

GRAND TOTAL SLT 160
L = Lecture, T = Tutorial, P = Practical, O = Others, F2F = Face to Face, NF2F = Non Face to Face

*Indicates the CLO based on the CLO’s numbering in Item 8.

12. Identify special requirements or resources to

deliver the course (e.g., software, nursery,

computer lab, simulation room):

13. Recommended text/reading: Prof. Madya Dr. Norbani Abdullah et al. (2018). Comprehensive

Note: HEPs to update and ensure the college chemistry.

latest edition/publication. SAP Education.

Nivaldo J. Tro (2013). Principles of chemistry: A molecular approach
(2nd ed.).

Pearson.

14. Other additional information:

4

SCI 1054 CHEMISTRY SEMESTER 1

List of constant

Ionization constant for water at 250C Kw = 1.00 x 10-14 mol2dm-16
Molar Volume of gases Vm = 22.4 dm3mol-1 at stp
24 dm3mol-1 (at room
Speed of light =
Specific heat of water, Temperature)
c = 3.0 x 108ms-1
Avogadro‟s number c = 4.18 kJ kg-1K-1
Faraday constant = 4.18 Jg-1K-1
Planck‟s constant NA = 4.18 Jg-1C-1
Rydberg constant F = 6.02 x 1023 mol-1
h = 9.65 x 104Cmol-1
Gas constant RH = 6.6256 x 10-34 Js
R 1.097 x 107m-1
Density of water at 25 0C ρ = 2.18 x 10-18 J
Freezing point of water PH2O = 8.314 JK-1mol-1
Vapor pressure of water = 0.0821 L atm mol-1 K-1
= 1 g cm-3
= 0oC
= 23.8 torr
=

Unit and conversion factor

Volume 1 liter = 1 dm3
Energy 1 mL = 1cm3
1 J = 1 kgm2s-2 = 1 Nm = 107 erg
Pressure 1 calorie = 4.184 J
Others 1 eV/molecular = 96.7 kJ mol-1
1 atm = 760 mm Hg = 760 torr = 101.325 kPa= 101325 nm-2
1 m = 1 x 109 nm
1 F = 96500 C
1 N = 1 kg ms-2

5

SCI 1054 CHEMISTRY SEMESTER 1

Element Symbol Proton Number Molar mass
Aluminium Al 13 27
Argentum Ag 47 108
Argon Ar 18 40
Arsenik As 33 75
Aurum Au 79 197
Barium Ba 56 137
Berilium Be 4 9
Boron B 5 11
Bromine Br 35 80
Cadmium Cd 48
Calcium Ca 20 112.4
Carbon C 6 40
Cesium Cs 55 12
Chlorine Cl 17 133
Chromium Cr 24 35.5
Cobalt Co 27 52
Copper Cu 29 59
Ferum Fe 26 63.5
Flourine F 9 56
Helium He 2 19
Hydrogen H 1 4
Iodine I 53 1
Krypton Kr 36 127
Litium Li 3 84
Magnesium Mg 12 7
Manganese Mn 25 24
Mercury Hg 80 55
Natrium Na 11
Neon Ne 10 200.6
Nikel Ni 28 23
Nitrogen N 7 20
Oxygen O 8 59
Phosphorus P 15 14
Platinum Pt 78 16
Plumbum Pb 82 31
Potassium K 19 195
Rubidium Rb 37 207
Selenium Se 34 39
Silicon Si 14 85.5
Stanum Sn 50 79
Stibium Sb 51 28
Strontium Sr 38
Sulphur S 16 118.7
Zinc Zn 30 121.8
87.6

32
63.4

6

SCI 1054 CHEMISTRY SEMESTER 1

CHAPTER 1 STOICHIOMETRY

7

SCI 1054 CHEMISTRY SEMESTER 1

1.0 MATTER

Anything that occupies space and possesses mass is called matter. Example: air, earth,
animals, trees, atoms and others. The three states of matter are solid, liquid and gas.

Matter

Mixtures Separation by physical method Pure
Substances

Heterogeneous Homogeneous Compound Separation Elements
Mixtures Mixtures s by chemical

method

Figure 1.1 Classification of matter

1.1 Structure of Atoms

1.1.1 Atom

electron nucleus
cloud

Figure 1.2 Structure of an atom

Atom consist of positively charged nucleus located at the centre, surrounding by negatively
charged electron cloud.
In an atom, there are three fundamental/subatomic particles:

I. proton (p)
II. neutron (n)
III. electron (e)
The protons and neutrons of an atom are densely packed in an extremely small nucleus. They
also known as nucleons.

8

SCI 1054 CHEMISTRY SEMESTER 1

The relative mass of an atom is almost the same as its nucleon number (p + n). Almost all the
mass of atom is concentrated in the nucleus. The nucleon number is sometimes used as the
approximate relative mass in calculations.
The number of proton it contains determines the positive charge of the nucleus, meanwhile
neutrons are neutral.
The electrons move rapidly around the nucleus of an atom, and the space they occupy as they
move defines the volume of the atom – electrons cloud. Large volume of an atom is actually
made up of empty space which is occupied by this electron cloud.
The number of protons = the number of electrons

Particle Charge Relative Mass
p +1 1.0 amu
n 0 1.0 amu
e -1 »0.0

Table 1.1 Properties of subatomic particles

1.2 Proton number, nucleon number, isotopes and isotope
denotation.

1.2.1 Atomic (Nuclide) Symbol or Notation for Nuclides

XA charge X = element symbol
A = Nucleon Number of the nuclide X= Z + n
Z
Z = Proton Number of the nuclide X = p

9

SCI 1054 CHEMISTRY SEMESTER 1

Example 1.2.1 :

Hg202 2
80 2

 the right superscript shows that the total charge on the ion is 2+
 the right subscript shows that the ion is formed from two mercury atoms.
 the left superscript: nucleon number of mercury, A = 202
Þ the left subscript: proton number of mercury, Z = 80

the number of neutrons in a nucleus of mercury = 202 – 80 = 122

Example 1.2.2:
Give the number of protons, neutrons and electrons in each of the following species:

(a) 200 Hg (b) 63 Cu (c) O17 2 (d) 59 Co 3
80 29 8 27

Symbol Number of Charge

200 Hg proton neutron electron 0
80 80 120 80 0
29 34 29 -2
63 Cu 9 +3
29 8 32 10
27 24
O17 2
8

59 Co 3
27

Example 1.2.3 :
Write the appropriate notation for each of the following nuclide:

Species Number of Notation for
proton neutron electron nuclides
A 4A
2 22
B 2
C 1 20
D 3 B
1 11
1
7 7 10
2C

1

14D 3

7

10

SCI 1054 CHEMISTRY SEMESTER 1

Exercise 1.2.1 :

1) Atom 39 K consists of ___ neutrons and ___ electrons.
19

2) An element Y contains 22 neutrons and 18 electrons. What is the proton number
and nucleon number of the element? Write the symbol of Y.

3) An ion X 2- has 34 neutrons and 30 protons. Determine the nucleon number and the
number of electrons in the ion.

4) Determine the number of electrons and neutrons in the following species:

M55 n2 , F19 

23 9

1. The proton number, Z, is the nuclear charge and also the number of
electrons in a neutral atom of the element.

2. no. p = no. e  atom (neutral)
3. no. p > no. e  positively charged – cation (atom lost electrons)
4. no. p < no. e  negatively charged – anion (atom gained electrons)

1.2.2 Isotope
Isotopes are termed as two or more atoms of the same element that have the same number of
protons in their nucleus but different number of neutrons.

Figure 1.3 : Isotopes of Li

11

SCI 1054 CHEMISTRY SEMESTER 1

Protium Deuterium Tritium

Isotopes of 1 H 2 H 3 H
hydrogen 1 1 1

Proton number 1 1 1

Number of neutrons 0 12

Table 1.2 Isotopes of hydrogen

Uranium- Uranium-
235 238

Isotopes of U2 3 5 U2 3 8
uranium 92 92

Proton number 92 92

Number of 143 146
neutrons

Table 1.3 Isotopes of uranium

Example 1.2.4:

For the following nuclides:

163T , 14 R and 15 Z
6
7

a) Write the nuclides that are isotopes.
and (reason: same proton number but different nucleon number)
163T 14 R
6

b) State the atoms that have the same number of neutrons.
(no. neutrons = 14 - 6 = 8) and 15 Z (no. neutrons =15 - 7 = 8)
14 R
6 7

Isotopes of an element have the SAME,
(a) number of protons (proton number)
(b) charge of nucleus of the atoms (ionization energy; electron affinity; size of the
atom; electronegativity are the same)
(c) number of electrons in a neutral atom
(d) electronic configuration (the number of valence electrons)
(e) chemical properties

12

SCI 1054 CHEMISTRY SEMESTER 1

Isotopes of an element have different,
(a) number of neutrons (nucleon number) in the nucleus of the atoms
(b) relative isotopic mass
(c) physical properties (e.g boiling point/melting point, density, effusion rate)

1.2.3 Molecule
A molecule consists of a small number of atoms joined together by covalent bond.

 Diatomic molecule: contains two atoms (example: H2, Cl2, HCl, CO,)
 Polyatomic molecule: contains more than two atoms (example: H2O, NH3)

1.2.4 Ion

An ion is a charged species formed from a neutral atom or molecule when electrons are gained
or lost as the result of a chemical reaction.

 Cation: a positively charged ion (number e < number p)(example: Mg2+, NH  )
4

 Anion: a negatively charged ion (number e > number p) (example: Cl-, OH-)

 Monatomic ion: ion contains only one nucleus (example: Fe3+, S2-)

 Polyatomic ion: ion contains more than one nucleus (example: H3O+, CN-)

1.2.5 Mole concept

Mole: The amount of substance that contains as many elementary particles
(atoms/molecules/ions) as there are atoms in exactly 12.000 g of carbon-12.

One mole of carbon-12 atoms has a mass of exactly 12.000 grams and contains 6.023x1023
atoms.

1 mole of atoms of an element = atomic mass of the element = 6.023x1023

To find mole(n),

n= mass, m
Molar mass, M

Molar Mass: The mass (in grams) of 1 mole of units (atoms/molecules/electrons/ions) of a
substance.

13

SCI 1054 CHEMISTRY SEMESTER 1

Avogadro Constant (symbol: L @ NA)
NA = 6.023x1023 mol-1 elementary particles/entities/units

Example 1.2.5:

1.0 mol of chlorine atoms = 6.023x1023 chlorine atoms
= 35.5 g Cl (in gram or a.m.u)

1.0 mol of chlorine molecules = 6.023x1023 chlorine molecules
= 2(35.5) g = 71.0 g Cl2
= 6.023x1023 x 2 chlorine atoms (Cl)

1.0 mol of calcium bromide = 6.023x1023 units of CaBr2
= 200 g CaBr2
= 6.023x1023 calcium ions
= 6.023x1023 x 2 bromide ions

1.0 mol of ammonia, NH3 contains 1.0 mol of nitrogen atoms and 3.0 mol of hydrogen
atoms.

1.0 mol of phosphorus, P4 contains 4.0 mol of phosphorus atoms.
1.0 mol of Na2SO4.10H2O contains 2.0 mol of natrium ions, 1.0 mol of sulphur atoms, 4.0
mol of oxygen atoms and 10.0 mol of water molecules.

1.2.6 Relative Mass

a) Relative Atomic Mass, Ar
Relative atomic mass, Ar of an element is defined as the ratio of the mass of one atom of
the element to 1/12 of the mass of a carbon-12 atom.

Ar = average mass of one atom of the element

1 x mass of one atom of 12C
12

The mass of an atom is mostly concentrated in the nucleus. So, the relative atomic
mass of an element can be considered to be the same as its nucleon number.
If the elements consist of a mixture of isotopes, then the relative isotopic mass of the
atoms must be used for calculation.

14

SCI 1054 CHEMISTRY SEMESTER 1

b) Relative Molecular Mass, Mr
Relative molecular mass,Mr of an element or compound is defined as the ratio of the mass of
one molecule of the substance to 1/12 of the mass of a carbon-12 atom.

Mr = average mass of one molecule of the substance

1 x mass of one atom of 12C
12

Or

Mr = sum of the relative atomic masses of all the atoms shown in the molecular
formula.

Example 1.2.6:
Calculate relative molecular mass of water, H2O. Given relative atomic mass of H = 1.008 and
O = 15.999.

Solution 1.2.5 :
Mr of H2O = 2 (Ar of H) + Ar of O

= 2(1.008) + 15.999 = 18.15
Relative molecular mass of CaCl2 = 40 + 2(35.5) = 111
Relative molecular mass of SO42- = 32 + 4(16) = 96

 The word relative means dimensionless. Ar and Mr has no unit.
 Molecular mass, atomic mass – atomic mass unit (amu)

Exercise 1.2.2:

1) Relative atomic mass of chlorine atom is 35.5. Explain why this value is not a
whole number.

2) Relative atomic mass of bromine atom is 80. What is the mass of bromine atom
compared to carbon-12 atom?

3) Calculate the relative molecular mass of C5H5N.

15

SCI 1054 CHEMISTRY SEMESTER 1

1.2.7 Relationship between mole, mass, Ar @ Mr and amount of particles

In general, if the relative molecular mass of SO2 = 32 + 2(16) = 64
Þ the mass of 1.0 mol of SO2 = 64 g
Þ the mass of 6.023x1023 SO2 molecules = 64 g
the mass of one SO2 molecule = ________ g
Þ the molar mass of SO2 = 64 g/mol

Notes:

The amount of A (mol) = Mass A (g)
Molar mass A (g/mol)

= Number of particles A
NA

1 MOLE SUBSTANCE:
= 6.023x1023 elementary entities/units/particles

(atoms/ions/molecules/electrons)
= Ar @ Mr

= 22.4 L at STP where T= 273.15K
= 24.0 L @ dm3 at room temperature where T= 298 K

Example 1.2.7:
Calculate the number of atoms in 0.20 mol of magnesium.

Solution 1.2.7:
1 mol Mg = 6.023x1023 atom.
So, if we have 0.2 mol = ? atom

16

SCI 1054 CHEMISTRY SEMESTER 1

The number of Mg atoms = 0.20 mol Mg x 6.023 x 1023 atoms
1.0 mol Mg

= 1.2 x 1023 atoms.

Example 1.2.8 :
Calculate the mass of (NH4)2CO3 that contains

a) 0.300 mol NH4+ b) 6.023 x 1023 hydrogen atoms

Solution 1.2.8:

(a) The mass of (NH4)2CO3 = 0.300 mol NH4+ x 96 g (NH ) CO
42 3

2.0 mol NH
4

= 14.4 g

(b) The mass of (NH4)2CO3

= 6.023x1023 H atoms x 96 g (NH4 )2 CO3
8 x 6.023 x 1023 H atoms

= 12.0 g

Exercise 1.2.3 :

1) Calculate the number of mol of oxygen atoms in 80 g of CuSO4.
2) How many atoms are there in 17 g of NH3 gas?
3) How many moles of Fe2O3 are there in 1.00 kg of rust?
4) What is the number of atoms in 2.5 g of phosphorus, P4?

17

SCI 1054 CHEMISTRY SEMESTER 1

1.2.8 Mole Concept of Gases

Molar Volume at STP
Molar volume of any gas at STP, Vm = 22.4 dm3mol-1
STP = standard temperature and pressure
Where T = 273.15 K P = 1 atm @ 760 mmHg

Amount of gas at STP, n = V, Volume of gas (L)
Vm (22.4 dm3 mol-1)

Example: 1.2.9

How many moles are there in 6.5 L oxygen at STP?

Solution 1.2.9

n = 6.5 L O2 x 1 mol O 2 = 0.29 mol

O 22.4 L

2

Molar volume 22.4 L can only be used when the temperature and pressure is 273 K and 1 atm.

Molar Volume at Room Temperature.

At room temperature, the molar volume of any gas = 24 dm3
STP =standard temperature and pressure

Where T= 273.15 K , P = 1 atm @ 760 mmHg

18

SCI 1054 CHEMISTRY SEMESTER 1

Exercise 1.2.4 :
1) A balloon is filled with hydrogen gas at STP. If the volume of the balloon is 2.24 L,
calculate the amount of hydrogen gas (in mmol).

2) Calculate the volume of 24x1023 molecules of gas at STP.

3) A sample of carbon dioxide has a volume of 56 cm3 at STP. Calculate
(a) The number of moles of gas molecules,

(b) The number of molecules, and

(c) The number of oxygen atoms in the sample.

1.3 Relative Atomic Mass based on Isotopic Abundance

Most of the elements exist as mixtures of two or more naturally occurring isotopes. For
example, chlorine can exist as the mixture of 75.5% chlorine-35 and 24.5% chlorine-37. The
respective symbol is 35Cl and 37Cl.

Mass spectrometry is an instrument used to determine the presence of isotopes together with
their abundance and isotopic mass. These data can help us to calculate the relative atomic
mass of the respective atom. The data is usually recorded in the form of mass spectrum as in
the figure below.

Figure 1.4 Mass spectrum of chlorine
19

SCI 1054 CHEMISTRY SEMESTER 1

To calculate relative atomic mass, Ar of an element that has isotopes,
Step 1,

Average atomic mass =

Where mi = isotopic mass, Qi = abundance (percentage@ratio)

Step 2,

Relative atomic mass, Ar =

Example 1.3.1:
The element europium (Eu) has two naturally occurring isotopes. The first isotope has a mass
of 150.920 a.m.u and its abundance is 47.820%. The second isotope has a mass of 152.921
a.m.u with abundance 52.180%. Calculate its relative atomic mass.

Solution 1.3.1:

Average atomic mass =
= 151.964 a.m.u

Relative atomic mass =

= 151.964

20

SCI 1054 CHEMISTRY SEMESTER 1

1.4 Empirical and molecular formula

Definition:
Empirical formula is the simplest formula that shows the relative numbers of the different
kinds of atoms in a molecule.
Molecular formula is a formula that states the actual number of each kind of atom found in
the molecule. Example : C2H4 , C2H4O2

Example 1.4.1:
A sample of hydrocarbon contains 85.7 % carbon and 14.3 % hydrogen by mass. Its molar
mass is 56. Determine the empirical formula and molecular formula of the compound.

Solution 1.4.1 :

Element C H
Mass (g) 85.7 14.3
Moles (n) 85.7 14.3
12.0 = 7.142 1.0 = 14.3

Smallest ratio 7.142 14.3

7.142 7.142
=1 =2

Empirical formula = CH2
n ( empirical molar mass) = molar mass
n ( 12 x 1 + 1 x 2 ) = 56
4
n= C4 H8
Molecular formula =

21

SCI 1054 CHEMISTRY SEMESTER 1

Exercise 1.4.1:
1. A complete combustion of 10.0 g of compound X, CxHyOz forms 12.0 g H2O and 22.0 g

CO2. Its molar mass is 60.
i. Determine the percentage composition of C, H and O.
ii. Determine the molecular formula of compound X.

2. A complete combustion of 0.6 g of hydrocarbon, CxH12 forms 1.98 g CO2. Determine the
percentage of C and H in the hydrocarbon and write the molecular formula.

1.5 Concentration Units

The concentration of solutions is the quantity of dissolved substance per unit quantity of
solvent in a solution. Concentration is measured in various ways as below:

Molarity, M Mole fraction, X

Molality, m

Percent by mass, Volume percent, Parts per million,
% w/w % v/v ppm

1.5.1. Molarity, M

The number of mole of solute dissolved per unit volume.
(unit: mol L-1 @ mol dm-3 @ M)

Molarity, M = n of solute 1 L = 1 dm3
V of solution 1 mL = 1 cm3
1 dm3 = 1000 cm3

22

SCI 1054 CHEMISTRY SEMESTER 1

Example 1.5.1:

A matriculation student prepared a solution by dissolving 0.586 g of sodium carbonate, Na2CO3
in 250.0 cm3 of water. Calculate its molarity.
Solution 1.5.1:

M (Na2CO3(aq)) = 0.586 g Na CO 1mol Na CO x 1000 mL
23 x 23 1.0 L
250.0 mL
106 g Na CO
23

= 0.0221 mol L-1.

Exercise 1.5.1:
1) What is the molarity of 85.0 mL ethanol solution that contains 1.77 g of ethanol,

C2H5OH?
[Ar C = 12.0, H = 1.0, O = 16.0]
2) Calculate the molarity of a solution of 1.71 g sucrose (C12H22O11) dissolved in ½ litre
of water.
[Ar H = 1.0, C = 12.0, O = 16.0]
3) How many grams of potassium dichromate, K2Cr2O7 required to prepare a solution of
250 mL with 2.16 M?
4) Calculate the amount of moles of solute in 200 cm3 of ammonia solution, having a
concentration of 0.125 mol dm .
5) A mixture is prepared from 45.0 g of benzene, C6H6 and 80.0 g of toluene, C7H8.
Calculate the molarity of the solution. The density of the solution is 1.0 g/ml.
[Ar C=12 ; H=1]

1.5.2. Molality, m
Molality of a solution is defined as the number of moles of solute per kg of solvent.
Symbol: m
(Unit: molkg-1 or molal, m)

Molality, m = n of solute

23

SCI 1054 CHEMISTRY SEMESTER 1

m solvent (kg)

Example 1.5.2 :

Calculate the molal concentration of ethylene glycol (C2H6O2) solution containing 8.40 g of
ethylene glycol in 200 g of water. The molar mass of ethylene glycol is 62 g/mol.

Solution 1.5.2 :

Mol of C2H6O2 = mass / Mr
= 8.40 g / 62 gmol-1
= 0.135 mol

Mass of H2O in kg = 200 x 10-3 kg

Molal = mol of solute (mol) .
mass of solvent (kg)

Molal = 0.135 mol / 200 x 10-3 kg
= 0.675 molkg-1

1.5.3. Mole Fraction, X
The mole fraction of component A is given by

n n total = nA + nB + nC + ..….
XA = n A

total

where nA is the number of mole of one component in a mixture, A (for a given entity)
and ntotal is the total number of mole of all substances present in the mixture (for the

same entity).

24

SCI 1054 CHEMISTRY SEMESTER 1

Exercise 1.5.2:

1. What is the mole fraction of CuCl2 in a solution prepared by dissolving 0.30 mol of
CuCl2 in 40.0 mol of H2O?

2. A solution is prepared by mixing 55 g of toluene, C7H8 and 55 g of bromobenzene,
C6H5Br. What is the mole fraction of each component?

3. A mixture containing benzene and toluene has 18.4 g of toluene and its percentage
composition is 30%. Calculate the number of moles of benzene in this solution.
[Mr benzene = 78 and toluene = 92]

1.5.4. Weight Percent (% w/w) @ Percent by Mass

% w/w = Weight solute × 100%
Weight solution

10% w/w NaOH
10 g of NaOH dissolved in 100 g of solution
10 g of NaOH dissolved in 90 g of solvent (water)

mass of solution = mass of solute + mass of solvent

Example: 1.5.4

A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.3 g of water. Calculate the
percent by mass of KCl in this solution?

Percent by mass of KCl = 0.892 g x 100% = 1.61%
0.892 g  54.2 g

25

SCI 1054 CHEMISTRY SEMESTER 1

Exercise 1.5.3:
1. Calculate the percent by mass of the solute in a aqueous solution containing 5.50 g of
NaBr in 78.2 g of solution.
2. Calculate the amount of water (in grams) that must be added to 5.00 g of urea,
(NH2)2CO in the preparation of a 16.2 percent by mass solution.
3. How many grams of NaOH and water are needed to prepare 250.0 g of 1.00% NaOH
solution?
4. Hydrochloric acid can be purchased as a solution of 37% HCl. What mass of this solution
contains 7.5 g of HCl?

1.5.5. Weight/volume percent (% v/v)

% v/v = volume of solute × 100%
Volume of solution

Example 1.5.5 :

A 200 ml of perfume contains 28ml of alcohol. What is the concentration of alcohol by volume
in this solution?

Solution 1.5.5 :
% v/v = volume of solute ___ x 100%

volume of solution
% v/v = 28 x 100%

200
= 14%
This indicates that there are 14 cm3 of alcohol in 100 cm3 of perfume.

26

SCI 1054 CHEMISTRY SEMESTER 1

1.6 Chemical Equation and Stoichiometry

A chemical reaction is a process in which one set of substances called reactants is converted
to a new set of substances called products.

Chemical equation is a way of denoting a chemical reaction using the symbols for the
participating particles (atoms, molecules, ions, etc.); formulae of the reactants are written on
the left side of the equation and formulae of the products, on the right.

xA + yB zC + wD

Types of arrow:

( ) single way : used for an irreversible reaction
( ) double way : are used for reversible reactions.

When reactions involve different phases it is usual to put the phase in brackets after the symbol
(s = solid, l = liquid; g = gas; aq = aqueous).

Because atoms can neither be created nor destroyed in a chemical reaction, the total number of
atoms of each element is the same on both sides in a balanced equation.

The numbers x, y, z, and w, showing the relative numbers of molecules reacting, are called the
stoichiometric coefficients.

Stoichiometry is the quantitative study of reactants and products in a chemical reaction.
→ x moles of A react with y moles of B to yield z moles of C and w moles of D.

Example 1.6.1:

CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

The chemicals reacting together and the chemicals produced.

Reactants: CaCO3 and HCl Products: CaCl2; CO2; and H2O

27

SCI 1054 CHEMISTRY SEMESTER 1

The physical states of reactants and products.

The coefficients in a chemical equation provide the atom ratios or the mole ratios by which
moles of one substance react with or form moles of another.

1 mole of CaCO3 reacts with 2 moles of HCl to yield 1 mole of CaCl2, 1 mole of CO2 and 1 mole

of H2O. or

1 molecule CaCO3(s) + 2 molecules HCl(aq) 1 molecule CaCl2(aq) + 1 molecule

CO2(g) + 1 molecule H2O(l)

Symbol : irreversible reaction.

1.6.1 Stoichiometric Calculations, Limiting Reactant, Reaction Yield And
Percentage Yield

Stoichiometric calculations and limiting reactant

When all the reactants are completely and simultaneously consumed in a chemical reaction,
they are exactly in stoichiometric proportions/amounts (the proportions indicated by the
balanced equation).

However, some reactions proceed better when one reactant is in stoichiometric excess.
Therefore, in many situations a chemist deliberately mixes reactants in a mole ratio that does
not agree with the coefficients of the equation. In this sense, the reactant that is completely
consumed first (the limiting reagent) determines the amount of products that form.

(before the reaction)

(after the reaction)

Excess Product Balance
Reactant d

Limiting
reactant

28

SCI 1054 CHEMISTRY SEMESTER 1

Excess Reagents are the reactants present in quantities greater than necessary to react with
the quantity of the limiting reagent.

The concept of the limiting reagent is analogous to the relationship between men and women in
a dance contest at a club. If there are fourteen men and only nine women, then only nine
female/male pairs can compete. Five men will be left without partners. The number of women
thus limits the number of men that can dance in the contest, and there is an excess of men.

In stoichiometric calculations involving limiting reagents, the first step is to decide / identify
which reactant is the limiting reagent. Then, calculate the amount of product formed based on
the amount of the limiting reagent available.

1) The first step in a stoichiometric calculation is to write a balanced equation. Either
this must be given or you must be able to supply your own.

2) The coefficients in the balanced equation tell you only the molar ratios in which
the species combine or are formed.

3) Always base the calculation of the maximum yield of product on the
stoichiometric equivalency (molar ratio) between it and the limiting
reagent.

4) To identify the limiting reagent:

The limiting reagent is the reactant → The number of moles

with the smallest ratio of The stoichiometric coefficient

Example 1.6.2:

Urea [(NH2)2CO] is prepared through equation:
2NH3(g) + CO2(g) ¾® (NH2)2CO(aq) + H2O(l)

In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2.
a) Which of the two reactants is the limiting reagent?
b) Calculate the mass of (NH2)2CO formed.
c) How much of the excess reagent (in grams) is left at the end of the reaction?

29

SCI 1054 CHEMISTRY SEMESTER 1

Solution 1.6.2 :
Part (a)

a) Since we cannot tell by inspection which of the two reactants is the limiting reagent, we
have to proceed by first converting their masses into numbers of moles.

Moles of NH3 = 637.2 g NH3 x 1 mol NH = 37.42 mol NH3
3
17.03 g 3
NH

Moles of CO2 = 1142 g CO2 x 1 mol CO 2 = 25.95 mol CO2

44.01g CO 2

Method 1 (Assumption)

Let assume CO2 as limiting reactant.
2NH3(g) + CO2(g) ¾® (NH2)2CO(aq) + H2O(l)

The balanced equation shows that 2 mol NH3 @ 1 mol CO2; number of mole of CO2 is 25.95
mol (as calculated above- the available amount) .Therefore, the number of moles of NH3
needed to react with 25.95 mol CO2 is given by

25.95 mol CO2 x 2 mol NH 3 = 51.90 mol NH3

1 mol CO 2

As we compare amount available and amount needed for NH3,

Available mole Needed mole
37.42 mole 51.90

The amount mol NH3 of needed for this reaction to occur is MORE than available (present), so
not enough NH3 to react completely with the CO2. So, the assumption can be ignored.

Hence, NH3 must be the limiting reactant and CO2 the excess reactant.

Method 2 (Ratio)

After finding number of moles for each reactant as above,

Ratio = Number of moles

Stoichiometric coefficient

30

SCI 1054 CHEMISTRY SEMESTER 1

Ratio for NH3 = 34.72 Ratio for CO2 = 25.95
2
1

= 18.71 = 25.95

Since the NH3 has the smallest ratio, NH3 is the limiting reagent.
Part (b)

To calculate the maximum yield of (NH2)2CO formed, it is based on the stoichiometric
proportions between the product and the limiting reagent, NH3.

The balanced equation shows that 2 mol NH3 @ 1 mol (NH2)2CO. This means 2 mole of NH3 will
produce 1 mole of (NH2)2CO.

Number of mole (NH2)2CO = 34.72 mol NH3 x 1 mol (NH2)2CO
2 mol NH3

= 17.36 mol

Mass of (NH2)2CO = 17.36 mol x 60.06 g/mol
= 1124 g (NH2)2CO

Part (c)
Since NH3 is the limiting reagent, CO2 must be the reactants left over after the reaction is
completed/finished. The difference between the amount of CO2 available and the amount
consumed is the amount left over.

Number of mole CO2 consumed = 34.72 mol NH3 x 1 mol CO2
2 mol NH3

= 17.36 mol

N of CO2 leftover = 25.95 mol – 17.36 mol
= 7.24 mol

Mass of CO2 leftover = 7.24 mol CO2 x 44.01 g/mol
= 319 g CO2

31

SCI 1054 CHEMISTRY SEMESTER 1

Exercise 1.6.1 :

1. The reaction between aluminium and iron (III) oxide can generate temperatures

approaching 3000 oC and is used in welding metals:

2Al + Fe2O3 Al2O3 + 2Fe

In one process 124 g of Al are reacted with 601 g of Fe2O3.

a) Identify the limiting reagent.

b) Calculate the mass (in grams) of Al2O3 formed.
c) How much of the excess reagent (in grams) is left over at the end of the reaction?

2. Aluminium, Al reacts with sulphuric acid, H2SO4, which is the acid in automobile
batteries, according to the equation:

2Al + 3H2SO4 2(SO4)3 + 3H2

If 20.0 g of Al is put into a solution containing 115 g of H2SO4,

a) Which reactant will be used up first?

b) Calculate the number of moles of Al2(SO4)3 will be produced?
c) How many grams of Al2(SO4)3 can be formed?
d) Calculate the mass (in grams) of the excess reagent remaining at the end of the

reaction.

3. Hydrazine, N2H4 and hydrogen peroxide, H2O2 are used as rocket reagent in which the
mixture of these two substances will produce nitric acid, HNO3 and water.

a) Write a balanced chemical equation for this reaction.

b) How many moles of HNO3 will be formed from 3.5 g of N2H4?
c) Calculate the number of moles of H2O2 required to react with 22.0 g of N2H4?
d) How many moles of HNO3 are produced when 30.5 g H2O2 is allowed to react with

25.6 g of N2H4? Which compound is the limiting reagent?

1.6.2 Reaction yield and percentage yield

The amount of limiting reagent present at the start of a reaction determines the theoretical
yield of the reaction, that is, the amount of product that would result if the entire limiting
reagent reacted.

The theoretical yield, then, is the maximum obtainable yield, predicted by the balanced
equation. This value can be obtained from calculation by applying the concept of limiting
reactant together.

32

SCI 1054 CHEMISTRY SEMESTER 1

Actual yield is the amount of product actually obtained from a reaction. This yield is obtained
from the experiment.
There are several reasons for the difference between actual and theoretical yield:

1. Side reaction may occur, that is other reactions in addition to the principle one
may take place.

2. Procedures may be necessary to separate the product from the reaction
mixture and obtain it in a pure state. Product may be lost during the separation
and purification processes.

3. The reaction may not go to completion.

4. There may have been impurities in one or more of the reactants.

Percentage yield is used by chemist to determine how efficient a given reaction in percent,
which describes the proportion of the actual yield to the theoretical yield.

% yield = Actual yield x 100%
Theoretical yield

33

SCI 1054 CHEMISTRY SEMESTER 1

Exercise 1.6.2:
Chlorofluorocarbons, commonly known as Freons, have been implicated in the gradual
destruction of the earth‟s ozone shield. One of these, Freon-12 (CCl2F2), is a gas that is
used as a refrigerant and is prepared by the reaction:

3CCl4 + 2SbF3 2F2 + 2SbCl3

In a certain experiment, 14.6 g of SbF3 was allowed to react with an excess of CCl4. After
the reaction was finished, 8.62 g of CCl2F2 was isolated from the reaction mixture.

a) Calculate the theoretical yield of CCl2F2 in this experiment in grams?
b) What was the percentage yield of CCl2F2 in this experiment?

“Today, I will do what others won’t, so tomorrow I can do what
others can’t” _Piccolor

34

SCI 1054 CHEMISTRY SEMESTER 1

ANSWER

Exercise 1.2.1
1. Neutron: 20 ; electron:19
2.

3. Nucleon number: 64 ; electron: 32
4. Mn2+ : electron : 21; neutron:32

F- : electron : 10; neutron:10

Exercise 1.2.2
1) the value is not a whole number because of relative atomic mass is average mass of one
atom of the element
2)

Relative atomic mass = average mass of one atom of the element
x mass of one atom of 12C
80
=x
3) 79 g/mol x x 12.000 a.m.u

= 80 a.m.u

Exercise 1.2.3
1) mole CuSo4 = 80/162

= 0.494 mole

1 mole of CuSo4 = 4 mole of O atom
0.494 mole = 1.975 mol O

2) mole NH3 = 17/17
= 1 mole

1 mole of NH3 = 4 mole of atom
= 4 x NA
= 2.24492 x 1024 atoms

3) mole Fe2O3 = 1000/94
= 10.64 mole Fe2O3

4) mole P4 = 2.5/124
1 mole of P4 = 0.0202 mole
No. atom P = 4 mole of atom P
= 0.0202 x 4 x NA = 4.857x 1022 atom P

35

SCI 1054 CHEMISTRY SEMESTER 1

Exercise 1.2.4
1) at STP

22.4 dm3 = 1 mole of hydrogen gas
2.24 dm3 = x
x = 0.1 mole
= 1x10-4 mmol
2) 1 mol of = 6.023 x 1023
molecules
x = 24 x 1023 molecules
x = 3.985 mol x 22.4
= 89.26 dm3
3)
22.4 dm3 = 1 mole of hydrogen gas
0.056 dm3
=x
x = 0.0025 mole CO2
a) 0.0025 mole CO2
b) 0.0025 mole CO2 x NA= 1.506 x1021 molecules

c) 0.0025 mole CO2 x NA x 2= 3.0115 x1021 atom O

Exercise 1.4.1 C H O
1. CO2 = 22.0g H2O = 12.0g ?
i) 10.0g of CxHyOz Mass of oxygen
Mass of C Mass of H = 10-6-1.333
Mass = x 22= 6g = x 12= 1.333g = 2.666 g
percentage O
percentage i) percentage C percentage H
= x 100= 60% = x 100= = x 100= 26.67%

13.33%

ii) molecular fomula

C H O
13.33 26.67
Mass 60 13.33/1=13.33 26.67/16=1.667
13.33/1.667 1.667/1.667
mole 60/12=5.000
8 1
Simplest 5.000/1.667 C3H8O

ratio (C3H8O)n= 60
n=1
3
C3H8O
Emperical
Formula

Molecular
Formula

36

SCI 1054 CHEMISTRY SEMESTER 1

2. 0.6g CxH12

CO2 = 1.98g

Mass of C
= x 1.98= 0.54g

i) Percentage C Percentage H
= x 100= 90% =100-90
=10%
molecular fomula C X3
90 H
Mass 90/12=7.5 10
mole 7.5/7.5 10/1=10
Simplest ratio 1 10/7.5
1.333
Emperical Formula C3H4
(C3H4)n= CxH12
Molecular Formula n=3
C9H12

Exercise 1.5.1 = mole of solute
1. Volume of solution,L
=(1.77/46)
molarity
0.085
=0.453 M

2. = mole of solute
molarity Volume of solution,L
=(1.71/342)

0.5
=0.01 M

3. = mole of solute
molarity Volume of solution,L
2.16 =mole
mole
mass 0.250
=0.54 Mole
= mole x Mr

37

SCI 1054 CHEMISTRY SEMESTER 1

4. = 0.54 x 294.185
molarity = 158.86 g
0.125 = mole of solute
mole Volume of solution,L
=mole

0.20
=0.0025 Mole

5. Density = mass/volume
molarity = mole of solute = 80/1.0
Volume of solution,L = 80 cm3
=(45/78)
(80/1000)
=7.21 M

Exercise 1.5.2 = mole of solute
1. nA + nB

Mole fraction, XA = 0.3
0.3 + 40
=7.44 x10-3 M

2. 55g of toluene (Mr=92), 55g of bromobenzene (Mr= 157)
Mole fraction, Xtoluene = (55/92)
(55/92) + (55/157)
= 0.5978
0.5978 + 0.350
=0.631
Mole fraction, Xbromobenzene = 1- 6.631
= 0.369

3. 18.4 g of toluene and its percentage composition is 30%.

Mr benzene = 78 and toluene
Mole, ntoluene = 92
= 18.4
0.3
x 92
=0.2
Mass benzene = 0.2
X + 0.2
= 0.467 mol
= 36. 43g

38

SCI 1054 CHEMISTRY SEMESTER 1

Exercise 1.5.3 = mass of solute x 100%
1. Mass of solution

Percentage by mass (%w/w) = 5.50 x 100%
78.2
2.
Percentage by mass (%w/w) =7.03 %
16.2
x = mass of solute x 100%
Mass of solution
3.
Percentage by mass (%w/w) = 5.00 x 100%
x

=30.86g

= mass of solute x 100%
Mass of solution

1.00 = x x 100%
x 250

Mass of NaOH = 2.5g =2.5g
Mass of H2O = 250-2.5 g

= 247.5 g

4.
Percentage by mass (%w/w) = mass of solute x 100%
Mass of solution
37 = 7.5 x 100%
x
x =20.27 g solution

Exercise 1.6.1
1.
a)
2Al + Fe2O3 Al2O3 + 2Fe
124g 601g

mole = 124 = 601

27 160

Ratio =4.593 =3.756

21

= 2.296 = 3.756

Since Al is the smallest ratio; Al is limiting reactant

39

SCI 1054 CHEMISTRY SEMESTER 1

b)
2 mole Al = 1 mole Al2O3

4.593 mole = x
x = 2.2965 mole Al2O3

mass = 2.2965 x 102
= 234.24g

c)
2 mole Al = 1 mole Fel2O3

4.593 mole = x
x = 2.2965 mole Fe2O3

Mole excess = 3.756 – 2.2965
= 1.4595

mass = 1.4595 x 160
= 233.52 g

2. 3H2SO4 Al2(SO4)3 + 3H2
2Al + 115g
20.0g
= 115
mole = 20 98
Ratio 27
=1.1735
=0.7407 3
2
= 0.3912
= 0.3704

a) Since Al is the smallest ratio; Al is used up first

b)
2 mole Al = 1 mole Al2(SO4)3

0.7407 mole = x
x = 0. 3704 mole Al2(SO4)3

c)
mass =mole x Mr
= 0. 3704 x 342
=126.677 g

d)
2 mole Al = 3 mole H2SO4

0.7407 mole = x
x = 1.111 mole H2SO4

mass = 1.111 x 98
= 108.89 g H2SO4

40

SCI 1054 CHEMISTRY SEMESTER 1

3. N2H4 + 7H2O2 2HNO3 + 8H2O
a) 7H2O2 2HNO3 + 8H2O
N2H4 + ?
b) 3.5g
mass
mole = 3.5
32

=0.1094

c) 22.0 g N2H4 1 mole N2H4 = 2 mole HNO3
0.1094 mol = 2(0.1094) mol

x = 0. 2188 mol HNO3
mole =mass / Mr

= 22.0/ 32
=0.6875 g

1 mole N2H4 = 7 mole H2O2
0.6875 mole = y

y = 4.8125 mol H2O2

d) N2H4 + 7H2O2 2HNO3 + 8H2O
mass 25.6 g 30.5g

mole = 25.6 = 30.5

32 34

Ratio =0.800 =0.8971

17

= 0.800 = 0.1282

Since mole H2O2 is the smallest ratio; H2O2 is limiting reactant

7 mole H2O2 = 2 mole HNO3
0.8971 mole = x

x = 0.2563 mole HNO3

41

SCI 1054 CHEMISTRY SEMESTER 1
3CCl2F2 +
Exercise 1.6.2 3CCl4 + 2SbF3
1. excess 2SbCl3
14.6 g 8.62 g
mole = 14.6

179
= 0.082

a) 2 mole SbF3 = 3 mole CCl2F2
0.082 mole = y
y = 0.1223 mole CCl2F2

mass =0.1223 x 121
= 14.80 g

b) Percentage yield = mass actual x 100%
Mass theorical
= 8.62 x 100%
14.80
= 58.23%

42

SCI 1054 CHEMISTRY SEMESTER 1

CHAPTER 2 ATOMIC
STRUCTURE

43

SCI 1054 CHEMISTRY SEMESTER 1

2.0 ATOMIC STRUCTURE
2.1 Atomic models

2.1.1 Electromagnetic Radiation

Electromagnetic radiation has properties associated with particles called photons. All
forms of electromagnetic radiation exhibit wavelike behaviour and they all can be
characterized by a wavelength, λ, and a frequency, v.

Different forms of electromagnetic radiation have different wavelength and frequencies,
but they all move with the same constant speed in a vacuum. This speed of light is
denoted as c (3.00 x 108 m s-1).

In 1900, Max Plank proposed the electromagnetic radiation can be viewed as the
packets of energy called quanta. It is later confirmed that the frequency of
electromagnetic waves is directly proportioned to its energy.

The light emitted when electron moves from an orbit with a larger n value to one with
smaller n value is photon. The energy of a photon may be determined from the
following expression.

∆E = hv ……………….. (2.1)
(h = Plank’s constant, 6.63 x 10-34 Js)

The wavelength and frequency are inversely proportional
c
λv = ………………. (2.2)

Substitute equation 2.2 into 2.1

∴ ∆E = hc …………… (2.3)
λ

44

SCI 1054 CHEMISTRY SEMESTER 1

Example 2.1.1

What is the energy of photons of a monocromatic radiation with the wavelength of 250
nm?

Solution 2.1.1:

∆E = hv
= hc
λ
= 6.626 x 10-34 Js x 3.00 x 108 ms-1
2.50 x 10-7 m

= 7.95 x 10-19 J

Exercise 2.1.1:

1. Calculate the photon of energy emitted by an electron that produces a wavelength of

4.342x10-7 m. (Ans: 4.58 x 10-19 J)

2. A light emitted from a discharge tube has a frequency of 7.31 x 1014 Hz. What is the

wavelength of this light? (Ans : 410 nm)

2.1.2 Bohr’s Atomic Postulates

In 1913, a theoretical explanation of the emission spectrum of the hydrogen atom was
presented by the Danish physicist Niels Bohr. He assumed that:

1. Electrons move in circular orbits.
2. The energies associated with electron motion in the permitted orbits must be

fixed in value, or quantized.
3. The emission of radiation by an energized hydrogen atom when electron dropped

from a higher energy orbit to a lower one and give off a quantum of energy (a
photon) in the form of light.

En = -RH (1/n2)

45

SCI 1054 CHEMISTRY SEMESTER 1

where RH, the Rydberg constant, has the value 2.18 x 10-18 J. The number n is
an integer called the principal quantum number, it has the values n = 1,2,3, …
Notes:
- ground state, refers to the lowest energy state of an atom.
- excited state, has higher energy than the ground state.
- A hydrogen electron is said to be in an excited state when its electron is

in n > 1.

4. Radiant energy (in the form of a photon) is emitted when the electron moves

from a higher-energy state (Ei) to a lower-energy state (Ef). The difference in
energy between energy levels is

Where ΔE = Ef - Ei
Therefore Ef = – RH (1/nf2) and Ei = –RH (1/ni2)

ni = initial orbit
nf = final orbit

ΔE = –RH (1/nf2) – ( –RH (1/ni2))
= RH (1/ni2 – 1/nf2)

Because this transition results in the emission of a
photon
of frequency, v and energy, hv it can be written:

ΔE = hv = RH (1/ni2 – 1/nf2)

2.1.3 Line Spectra and Continuous Spectra

When white light from an incandescent lamp is passed through a prism, it produces a
continuous spectrum, or rainbow colours. The different colours of light represent different
wavelengths. All wavelengths are present in a continuous spectrum. White light is simply a
combination of all the various colours.

46

SCI 1054 CHEMISTRY SEMESTER 1

If the light from a gas discharge tube containing a particular element is passed through a prism,
only narrow coloured line is observed. Each line corresponds to light of a particular wavelength.
The pattern of emitted by an element is called its line spectrum.

The line spectrum of hydrogen is fairly simple;
it consists of four lines in the visible portion of
the electromagnetic spectrum.

The emission spectrum of hydrogen includes a
wide range of wavelengths from the infrared to
the ultraviolet. Table 2.1 lists the series of
transition in the hydrogen spectrum; they are
named after their discoverers. The Balmer series was particularly easy to study because a
number of its line falls in the visible range.

Figure 2.1 A continuous visible spectrum is produces when a narrow beam of white light is
passed through a prism. The white light could be sunlight or light from an incandescent lamp.

Figure 2.2 (a) An experimental
arrangement for studying the emission spectra
of atoms and molecules. The gas under study
is in a discharge tube containing two
electrodes. As electrons flow from the negative
electrode to the positive electrode, they collide
with the gas. The collision process eventually
leads to the emission of light by the atoms (or
molecules). The emitted light is separated into
its components by a prism. Each component
color is focused on a definite position, according to its wavelength, and forms a colored image
of the slit on the photographic plate. The colored images are called spectral lines. (b) The line
emission spectrum of hydrogen atoms.

Figure 2.3 Shows the

energy levels in the hydrogen

atom and the various emission

series. Each energy level

corresponds to the energy

associated with an allowed

energy state for an orbit, as

47

SCI 1054 CHEMISTRY SEMESTER 1

postulated by Bohr for the Balmer Series. The emission lines are labelled according to the
scheme in Table 2.1.

Series nf ni Spectrum Region
Lyman 1 2,3,4, .. Ultraviolet
Balmer 2 3,4,5, .. Visible
Paschen 3 4,5,6, .. Infrared
Brackett 4 5,6,7, .. Infrared
Pfund 5 6,7,8, .. Infrared

Table 2.1 The various series in atomic hydrogen emission spectrum.

2.1.4 Rydberg Equation

In 1885, J. J Balmer found a simple formula for the four wavelength of the hydrogen
emission spectra in visible range. Balmer‟s formula is usually written compactly as

(1 = RH )1 - 1 ……………… (2.4)
λ
22 n2

where,

RH = Rydberg constant ( 1.097 x 107 m-1)
n = integer > 2

J. R Rydberg measured the frequencies of the lines in visible spectrum and found that
they could be related by an equation that gives the energies, in joule, of the photon in
each line:

(1 = RH 1 - )1 …………… (2.5)
λ n21 n22

Where, n1 and n2 are positive integers, and n1 is smaller than n2

Niels Bohr used this concept as a basis of model of atom.

48

SCI 1054 CHEMISTRY SEMESTER 1

This equation solves problems based on transitions of the electrons in the Lyman to Pfund
Series.

1 /  = RH (1/n12 – 1/n22) where n1 < n2
RH = 1.097 x 107 m-1

Example 2.1.4 :

Calculate the wavelengths of the first line and the onset of the continuum limit for the Lyman
series for hydrogen.

First line : 1/ = RH (1/ni2 – 1/nf2)
= 1.097 x 107 (1/12 – ½2)

 = 1.21 x 10-7m

Onset of the continuum limit : 1/ = RH (1/ni2 – 1/nf2)
= 1.097 x 107 (1/12 – 1/)
= 1.097 x 107 (1/12 – 0)

 = 9.12 x 10-8 m

Exercise 2.1.4:
1. Calculate the energy liberated when an electron from the fifth energy level falls to the

second energy level in a hydrogen atom.
(Ans: -4.58x10-19J)
2. Calculate the wavelength of the fourth line in Balmer series of hydrogen atom.
(Ans: 410 nm)
3. Calculate the frequency of a line in the hydrogen spectrum corresponding to a
transition from n = 5 to n = 2.
(Ans: 6.91 x 1014 Hz)
2.1.5 IONIZATION ENERGY

49