SK015

lecture notes & questions

Name
Class

lecture notes & questions

SPECIAL THANKS TO:
AZURAH BINTI DAPOR
ZAINAB BINTI BASRI

MARZIAH BINTI MOHAMAD
BIBI ZARITA BINTI SHER JANG KHAN

ARYATI BINTI BUSERO
NORSALWANA BINTI NGASRI

MURNI BINTI SAMSUDDIN
SUHADA BINTI ALIAS

MOHD RUSYDI BIN MOHD ROSLI

lecture notes & questions

CHAPTER 1: MATTER 1
CHAPTER 2: ATOMIC STRUCTURE 43
CHAPTER 3: PERIODIC TABLE 75
CHAPTER 4: CHEMICAL BONDING 96
CHAPTER 5: STATES OF MATTER 154
CHAPTER 6: CHEMICAL EQUILIBRIUM 190
CHAPTER 7: IONIC EQUILIBRIA 225
TABLE OF RELATIVE ATOMIC MASSES
AND LIST OF CONSTANTS 267

CHAPTER 1: MATTER

LECTURE NOTE

lecture notes & questions

1

Chapter 1 : MATTER 1.1 Atoms and Molecules

1.1 Atoms and Molecules 2
1.2 Mole Concept
1.3 Stoichiometry

1

Learning outcomes Introduction

At the end of this topic, students should be able to: • Chemistry is the study of matter and the
change it undergoes.
i. Write isotopic notation
ii. Interpret mass spectrum. • Matter is anything that occupies space
iii. Calculate the average atomic mass of an element and has mass.

given the relative abundances of isotopes or a e.g: air, water, animals, trees, atoms, …..
mass spectrum.
• The three states of matter are solid,
3 liquid and gas.

4

1.1 Atoms and Molecules Modern Model of the Atom

Atoms

• An atom is the smallest unit of a chemical
element/compound.

• An atom is composed of three subatomic
particles:

- Proton (p)
- Neutron (n) Packed in a small nucleus

- Electron (e) Move rapidly around the nucleus of an • Electrons move around the region of the atom.
atom
6
5

2

Subatomic Particles Proton number and Nucleon number

Subatomic Mass Charge Charge • The proton number, Z (known as atomic
Particle (gram) (Coulomb) (units) number), of an atom is the number of
protons in the nucleus of the atom of
Electron (e) 9.10 x 10-28 -1.6 x 10-19 1- an element.

Proton (p) 1.67 x 10-24 +1.6 x 10-19 1+ • The nucleon number,A (known as mass
number) is the total number of protons
Neutron (n) 1.67 x 10-24 0 0 and neutrons in the nucleus of the
7
atom of an element.

8

Nuclide Notation Nucleon number Total charge
of mercury, on the ion
An atom/ion can be represented by an A = 202
nuclide notation ( atomic symbol )
Proton number The number of neutrons
* nuclide – a type of atom as characterised by its atomic number and its of mercury, =A–Z
nucleon number Z = 80 = 202 – 80
= 122
X = Element symbol
10
A = Nucleon Number/Mass
number of X (p + n)

Z = Proton number/Atomic
number of X (p)

9

EXAMPLE 1 • Atom

Sodium has the number of protons equals to the
number of electrons
2131Na
• CATION (Ion with a positive charge)
Number of protons = 11
Number of electrons = 11 has the number of electrons less than the
Number of neutrons = 23 – 11 number of protons (atom lost electrons)

= 12 • ANION (Ion with a negative charge)

11 has the number of electrons more than the
number of protons (atom gained electrons)

12

3

EXAMPLE 2 EXAMPLE 3

13 14

Isotopes • Isotopes of an element have similar
chemical properties but have slightly
• Isotopes are two or more atoms of the same different physical properties.
element that have the same number of protons
but different number of neutrons in their nucleus. 16
.

• Example:

- Hydrogen has 3 isotopes while uranium has 2.

11H 21H (D) H (T)320800 Hg
hydrogen
deuterium 1
U235 U238
92 tritium
92
uranium-235 -
uranium-238
15

Exercise 1 Answer

• Give the number of protons, neutrons,electrons • Give the number of protons, neutrons,electrons
and charge in each of the following species: and charge in each of the following species:

Symbol Number of : Charge

Symbol Number of : Charge Proton Neutron Electron

Proton Neutron Electron 200 Hg 80 120 80 0
80

200 Hg 63 Cu 29 34 29 0
80 29

63 Cu O17 2 8 9 10 2-
29 8

O17 2 Co59 3 27 32 24 3+
8
27
Co59 3
* Ionic charge has the sign AFTER the number , e.g 2+, 3+,
27 2- etc.

17 18

4

Exercise 2 Molecule

• Write the appropriate notation for each of the • A molecule consists of a small number of atoms
following element : joined together by covalent bonds.

Species Number of : Notation 20
Proton Neutron Electron

A2 2 2 4 A
B1 2
C1
D7 2 0 31B

1 1 21C

7 10 174D3

19

• A diatomic molecule contains only two Mass Spectrometer
atoms
Example : • An atom is very light and its mass cannot be
H2, N2, O2, Br2, HCl, CO measured directly.

• A polyatomic molecule contains more • A mass spectrometer is an instrument used to
than two atoms measure the precise masses and relative
Example : quantity of atoms and molecules
O3, H2O, NH3, CH4
• A mass spectrometer is used to determine the:
21
i. relative atomic mass of an element

ii. relative molecular mass of a compound

iii. types of isotopes, abundance and relative
isotopic mass

iv. structural formula of the compound in an

unknown sample 22

A Mass Spectrometer

23 24

5

Simplified diagram of a Mass Spectrometer

Ionisation
Chamber

Vaporisation Accelaration Magnetic Beam of 35Cl+ and 37Cl+
Chamber Chamber 37Cl+
+ Chamber
-- 35Cl+

Heated Vacuum Ion Beam 26
Filament Pump Ion Detector

AMPLIFIER

Reco2r5der

• The mass and intensity of the ions are recorded Mass Spectrum of chlorine
in the form of a graph called the mass spectrum
• The mass spectrum of Cl
• Example : A mass spectrum of Cl shows that Cl consists of
Relative two isotopes: 35Cl and
abundance 0.76 37Cl.

Relative abundance0.240.76 • The height of each line is
0.24 proportional to the
35 37 m/e (amu) abundance of each
isotope.
27
35 37 m/e (amu)

• 35Cl is the most abundant of
the two isotopes

28

How to calculate the relative atomic mass 2. Calculate the relative atomic mass.
from a mass spectrum?

1. Calculate the atomic mass. Relative atomic mass, Ar  Mass of one atom of element
112 X Mass of one atom of 12C
The atomic mass of an element is the average
mass of the mixture of isotopes.

 Relative x Isotopic mass of
abundance of each isotope

Atomic each isotope
mass of an =
 Relative
element abundance

29 30

6

Example 1: Solution

Calculate the relative atomic mass of chlorine from Average atomic mass of chlorine
the mass spectrum.
Average atomic mass of chlorine is 35.48 amu
Relative 0.76 Relative atomic mass of chlorine
abundance 0.24
= 35.48
35 37 m/e (amu)
The relative atomic mass of chlorine is 35.48
31 32

Example 2: Solution:

Calculate the relative atomic mass of neon Average atomic mass of neon
from the mass spectrum.

Relative atomic mass of neon 34

Relative atomic mass of neon is 20.2
33

Example 3 Example 3 (cont…)

Relative 18 b.What is the percentage abundance of each
abundance isotope?
7
% abundance 85Rb
85 87 m/e (amu) = 18 x 100
25
Figure above shows the mass spectrum of = 72 %
the element rubidium, Rb;
% abundance 87Rb
a. What are the isotopes of Rb? = 7 x 100
25
85Rb and 87Rb = 28 %

35 36

7

Example 3 (cont…) Example 4

c. Calculate the relative atomic mass of Rb. The relative atomic mass of 6Li and 7Li are 6.01
and 7.02. What is the percentage abundance of
Atomic mass of Rb =(18 x 85 amu) + ( 7x 87amu) each isotope if the relative atomic mass of Li is
(18+ 7) 6.94

= 85.56amu 38

Ar of Rb = 85.56amu
112 x12.00amu

= 85.56

37

Assume that, = X% Exercise
% abundance of 6Li = (100 - x) %
% abundance of 7Li Naturally occuring iridium, Ir is composed of 2
isotopes 191Ir and 193Ir in the ratio of 5:8. The relative
atomic mass Li = X (6.01) + (100 – X) 7.02 mass of 191Ir and 193Ir are 191.021 and 193.025
6.94 = X + (100 – X) respectively. Calculate the relative atomic mass of
X= iridium.
6.01 X + 702 – 7.02 X (Ans : 192.254)
100
40
7.92

 % abundance of 6Li = 7.92 % 39
% abundance of 7Li = 92.08 %

Answer

Atomic mass Ir = 5 (191.021) + 8(193.025)
13

= 192.254 amu

Ar of Ir = 192.254 amu
112x12.00 amu

= 192.254

41

8

1.2 MOLE CONCEPT Learning outcomes…

42 At the end of today’s lesson, students should be able to:
i. Define the terms empirical and molecular formulae.
ii. Determine empirical and molecular formulae from

mass composition or combustion data
iii. Define and perform calculations for the following

concentration measurements: molarity (M), molality
(m), mole fraction (X), percentage by mass (%w/w) and
percentage by volume (%v/v).

43

Empirical and Molecular Formula • The empirical formula and molecular
formula of a compound could be the same.
• Empirical formula is a chemical formula e.g: NH3 and CO2
that shows the simplest ratio of all
elements in a molecule. • There are molecules that have different
molecular formulae but the same empirical
• Molecular formula is a formula that formula.
indicates the actual number of atoms of e.g: C2H4 and C3H6
each element in a molecule.
Empirical formula and molecular formula can
44 be related as

molecular formula = n(empirical formula)

45

Example 1 Answer:

Ascorbic acid cures scurvy and may help to Element C H O
prevent the common cold. It is composed of
40.92% carbon, 4.58% hydrogen and 54.50% Mass (g) 40.92 4.58 54.50
oxygen by mass. The molar mass of ascorbic 54.50
acid is 176 gmol-1. Number 40.92 4.58 16.00
Determine its empirical formula and of moles 12.01 1.01 =3.41
molecular formula. =3.41 =4.53
(mol) 1
46 1 1.33
Ratio of 3
moles 3 4

Simplest
ratio

∴ Empirical fomula = C3H4O3 47

9

Molecular formula = n (C3H4O3) Example 2
Mr n (C3H4O3) = 176
1.00 g sample of compound A was burnt
n[3(12.01)+4(1.01)+3(16.00)] =176 in excess oxygen yield 2.52 g of CO2
88.07n = 176 and 0.443 g of H2O. Determine the
n=2 empirical formula of the compound.

∴ Molecular formula = C6H8O6 49

48

Answer: Answer:

Element C H O

Element C H O Mass (g) 0.6877 0.04922 0.2631

Mass 12.01 x 2.52 2.02 x 0.443 = 1- 0.6877 - 0.04922 Number 0.6877 0.04922 0.2631
(g) = 0.2631 of moles 12.01 1.01 16.00
44.01 18.02 =0.0573 =0.01644
(mol) =0.04873
=0.6877 =0.04922 3.5 1
Ratio of 3
moles 7 2
6
Simplest
ratio

50 ∴ Empirical formula = C7H6O2 51

Exercise Answer:

A combustion of 0.202 g of an organic sample that Element C H O
contains carbon, hydrogen and oxygen produce 0.361g
carbon dioxide and 0.147 g water. If the relative Mass 12.01 x 0.361 2.02 x 0.147 = 0.202- 0.0985 -
molecular mass of the sample is 148, what is the (g) 0.01648
molecular formula. 44.01 18.02
= 0.08702
Ans : C6H12O4 =0.0985 =0.01648

52 53

10

Answer: Molecular formula = n (C3H6O2)
Mr n (C3H6O2) = 148
Element C H O
n[3(12.01)+6(1.01)+2(16.00)] =148
Mass (g) 0.0985 0.01648 0.08702 74.09n = 176
n=2
Number 0.0985 0.01648 0.08702
of moles 12.01 1.01 16.00 ∴ Molecular formula = C6H12O4
=0.008201
(mol) =0.01632 =0.005439 55
1.5
Ratio of 3 1
moles 3
6 2
Simplest
ratio

∴ Empirical formula = C3H6O2 54

Concentration of Solutions • Concentration of a solution can be
expressed in various ways:
• A solution is a homogenous mixture of two
or more substances. A. Molarity (M)
B. Molality (m)
• Example: 56 C. Mole Fraction (X)
sugar + water = solution D. Percentage by Mass (%w/w)
sugar = solute E. Percentage by Volume (%v/v)
water = solvent
57
• The concentration of solution is the
amount of solute present in a given
quantity of solvent or solution.

A. Molarity (M) Example 1

• The number of moles of solute per cubic Determine the molarity of 85.0 ml of
decimetre (dm3) or litre (L) of solution. ethanol solution contains 1.77 g of
ethanol, C2H5OH.
Molarity, M  moles of solute
volume of solution (dm3) 59

The unit of molarity : molL-1 or moldm-3 or M

58

11

Answer: Example 2
A solution is prepared by dissolving 0.586 g
= 0.0384 mol of sodium carbonate, Na2CO3 in 250 cm3 of
water. Calculate its molarity.

60 61

Answer: Exercise: Molarity

= 5.53 X 10-3 mol 1. Calculate the molarity of a solution of
1.71 g sucrose, C12H22O11 dissolved in
= 0.022 M 500 ml of water.
[0.01 M ]

2. How many gram of potassium
dichromate, K2Cr2O7 required to prepare
a solution of 250 ml with concentration of
2.16 M

[158.9 g] 63

62

Answer: Answer:

2. Molarity  mol solute (L)
volume of solution
1. nsucrose 1.71
 342.34 2.16 mol solute
0.25
 4.995 x103 mol 

 0.54 mol

M  4.995 x 103 mol mass K 2Cr2 O7
0.5 L 294.2
sucrose 0.54 

 0.01molL1 mass K 2Cr2O7 158.9 g

64 65

12

B. Molality (m) Example 1

• Molality is the number of moles of • Calculate the molality of a sulphuric acid
solute dissolved in a given of solvent solution containing 16.8 g of sulphuric acid
in kg. in 200 g of water.

Molality, m  moles of solute 67
mass of solvent (kg)

• The unit for molality is m, molkg-1 or
molal.

66

Answer: Example 2

Molality, m  moles of solute Calculate the mass of NaOH to be added to
mass of solvent (kg) 150 g of water for the preparation of a 1.2 m
of NaOH solution.

68 69

Answer: Exercise: Molality

Molality, m  moles of solute 1. What is the molal concentration of a
mass of solvent (kg) solution when 0.30 mol of CuCl2 is
dissolved in 40.0 mol of water?
molality of NaOH  n NaOH [0.416 m]
150

1000

1.2  n NaOH 2. What is the molality of a solution made by
0.15 dissolving 36.5 g of naphthalene, C10H8 in
425 g of toluene, C7H8?

n NaOH  0.18 mol [0.670 m]

70 71

13

Answer: Answer:

1. mass of H2O40.0 x18.02 2. mass of toloune (solvent)  0.425 kg
 720.8 g
 0.7208 kg mol napthalene  36.5 g 1
128.08 gmol

 0.28498 mol

molality of CuCl2  0.30 molality of C10H8  0.28498 mol
0.7208 0.425

 0.416 m  0.67 m

72 73

C. Mole Fraction (X)  Mol fraction is always smaller than 1

• Mole fraction is the ratio of the number of  The total mol fraction in a mixture
moles of one component to the total (solution) is equal to one.
number of moles of all components XA + XB + XC + X….. = 1
present.
 Mole fraction has no unit (dimensionless)
mole fraction of A, XA = total moles of A since it is a ratio of two similar quantities.
moles of all components
75
XA = nA
n total

74

Example: Solution: 200.0 g
A sample of ethanol, C2H5OH contains a) n ethanol
200.0 g of ethanol and 150.0 g of water. -1
Calculate the mole fraction of n water
(a) ethanol 45.03 gmol
(b) water X ethanol  4.4375 mol
in the solution.
150.0 g
76 18.02 gmol  1
 8.324 mol

4.4375 mol
(4.4375  8.324) mol
 0.3477

77

14

b) X water = 1 - 0.3477 Exercise: Mole Fraction
= 0.6523
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?
[Ar C=12.01, H=1.01, Br=79.9]

78 79

Solution: D. Percentage by Mass (%w/w)

n C7H8 925.515 • Percentage by mass is defined as the
percentage of the mass of solute per mass
 0.5969 mol of solution.

n C6H5Br 15575.01

 0.3503 mol %ww  mass of solute x100
mass of solution
n C7H8
X C7H8  n  n C6H5Br

C7H8 note: mass of solution  mass of solute mass of solvent
0.5969 % w/w has no unit.
 0.5969  0.3503
80 81
 0.63

X C6H5Br  1  0.63
 0.37

Example 1: Example 2:
A sample of 0.892 g of potassium chloride,
KCl is dissolved in 54.362 g of water. A solution is made by dissolving 4.2 g of
What is the percent by mass of KCl in the sodium chloride, NaCl in 100.00 mL of
solution? water. Calculate the mass percent of
sodium chloride in the solution.
Solution:
%00 mmaassss 0.08.98290g2.80g9.52849g.52346g.23g62x1g00 10000% Solution:
 1.6100
mass NaCl  4.2 g
82
mass H2O 100 g

00 mass of NaCl  4.2 g x 100 0 0
g 100
4.2 g

 4.03 00 83

15

Exercise: % w/w Solution:

1. How many grams of NaOH and water are needed 1. mass NaOH  x g
to prepare 250.0 g of 1.00% NaOH solution?
Ans : 2.50g; 247.5 g mass H2O  y g
mass solution  x  y  250 g
2. Hydrochloric acid can be purchased as a solution
of 37% HCl. What is the mass of this solution contains 00 mass of NaOH  xg x 100 0 0  1.00%
7.5 g of HCl? 250 g
Ans : 20.27 g
mass NaOH  x  2.5 g
84
mass H2O  250 g  2.5 g
 247.5 g

85

Solution: E. Percentage By Volume (%V / V)

2. 00 mass of HCl  7.5 g x 100 0 0  37% • Percentage by volume is defined as the
xg percentage of volume of solute in milliliter
per volume of solution in milliliter.

mass of solution  x  20.27 g %VV  volume of solute (mL) x 100
volume of solution (mL)

note : of solution  mass of solution
Density volume of solution

86 87

Example 1: Solution:

A sample of 250.0 mL ethanol is labeled 00 volume of ethanol  Vethanol x100 00
as 35.5% (v/v) ethanol. How many Vsolution
milliliters of ethanol does the solution VethVaentohal nol 35.53050.1x502%050100.020050%0m.0L0 mL
contain?

 88.8 mL

88 89

16

Conversion of Concentration Units Solution:

Example 1: (a) M = nNaOH
A 6.25 m of sodium hydroxide, NaOH Vsolution
solution has has a density of 1.33 g mL-1
at 20 ºC. Calculate the concentration 6.25 m of NaOH
NaOH in:  there is 6.25 mol of NaOH in 1 kg of water
(a) molarity
(b) mole fraction for a solution consists of 6.25 mol of NaOH and 1
(c) percent by mass kg of water;

90 Vsolution = mass solution
solution

91

masssolution = massNaOH + mass water MNaOH = 6.25 mol
massNaOH = nNaOH  molar mass of NaOH
 1250  10 3 L 
= 6.25 mol  (22.99 + 16.00 + 1.01) g mol 1  1.33 
= 250 g
= 6.65 mol L1

masssolution = 250 g + 1000 g

= 1250 g

Vsolution = 1250 g
1.33 g mL1
92 93

(b) XNaOH = nNaOH
nNaOH  nwater
massNaOH
1 kg of water contains 6.25 mol of NaOH (c) %(w/w) of NaOH = massNaOH  mass water  100%

nwater = mass water = 250 250 g g  100%
molar mass of water g  1000

= 1000 g = 20.0%
 16.00)
(2(1.01) g mol1

XNaOH = 6.25 mol

 6.25 mol  1000 mol 
 18.02 

= 0.101 94 95

17

Example 2: Solution:

An 8.00%(w/w) aqueous solution of a) % by mass of NH3  Mass of NH3 x 100%  8.00 %
ammonia has a density of 0.9651 g mL-1. Mass of solution
Calculate the
assume that ,
(a) molality
mass of NH3  8 g,
(b) molarity mass of solution 100g

(c) mole fraction mass solvent  100 g  8 g

of the NH3 solution  92 g  0.092 kg

mole of NH3  8g 1
17.03 gmol

Answer: a) 5.10 m  0.4698 mol
b) 4.53 mol L-1
c) 0.0842 molality NH3  0.4698 mol
0.092 kg

96  5.11m 97

Solution: Solution: 8g
17.03 gmol1
b) mass of solution 100g c) mole of NH3 

mole of NH3  8g  0.4698 mol
17.03 gmol1

 0.4698 mol mass of solvent  92 g

mass solution mole of solvent  92 g
Vsolution 18.02 gmol1
from density of NH3   0.9651
 5.1054 mol

V solution  100 g
0.9651
XNH3 0.4698
103.61ml  0.10361L  5.1054  0.4698

molarity NH3  0.4698 mol 0.0843
0.1036 L
98 99
 4.53 M

Example 3: Solution:

A solution of hydrochloric acid contains 36% HCl a) %by mass of HCl  mass HCl x 100 %  36%
by mass. mass solution
a) Calculate the mole fraction of HCl in the
assume that,
solution.
mass HCl  36 g
answer : 0.2173
mass solution 100g
b) Calculate the molality of HCl in the
solution. nHCl  36 g
36.46gmol 1
answer : 15.4069 m
 0.9874 mol
100
nsolvent 64 g
 18.02 gmol1

 3.5516 mol

XHCl  0.9874
0.9874  3.5516
101
 0.2175

18

Solution: Example 4:
A 16 M commercial aqueous nitric acid,
b ) mass HCl  36 g HNO3 has a density of 1.42 g mL-1.
mass solution  100 g Calculate the percentage HNO3 by mass
mass solvent  100  36 in solution.
 64 g  0 .064 kg answer: 71%

n HCl  36 g 1 102 103
36 .46 gmol

 0 .9874 mol

molality HCl  0 .9874
0 .064 kg

 15 .42 molkg 1

Solution Solution

Molarity of HNO3  V of nHNO3 16 M density of HNO3  mass solution  1.42 gmL1
solution (L) V solution

assume that, mass solution 1.42 x1000 1420 g

V of solution 1L 1000 mL

nHNO3 16 mol % by mass of HNO3  mass solute x 100%
mass of solution

mass HNO3  63.02 gmol1 x 16 mol  1008.32 x 100
1008.32 g 1420

 71.01%

104 105

Exercise 3. Calculate the molality of a 5.86 M ethanol (C2H5OH)
solution which has a density of 0.927 gmL-1.
1. A solution contains 35% HBr by mass and has a (answer: 8.88 m)
density of 1.30 g mL-1. Calculate the molality and
molarity of this solution. The molar mass for HBr is 4. The density of an aqueous solution containing 10.0%
80.91 g mol-1. of ethanol (C2H5OH) by mass is 0.984 g mL-1.
(answer: molarity =5.63 M; molality=6.66 m)
a) Calculate its molarity.
2. A sulphuric acid solution contains 66% H2SO4 by b) Calculate the molality of this solution.
weight and has the density of 1.58 gmL-1. How c) What volume of the solution would contains
many moles of the acid are present in 1.00 L of the
solution. Molar mass for H2SO4 is 98.09 gmol-1. 0.125 mole of ethanol?
(answer: 10.63 mol) [a) 2.14 M; b) 2.41 m; c)0.0587 L]

106 107

19

1.3 Stoichiometry Learning outcomes…

At the end of today’s lesson, students should be
able to:
i. Write and balance chemical equation by inspection
method and by ion-electron method (redox
equation).
ii. define limiting reactant and percent yield
iii. perform stoichiometric calculations using mole
concept including limiting reactant and percent yield

108 109

Balancing Chemical Equation Example: yB zC + wD

• A chemical equation shows a chemical reaction xA + Products
using symbols for the reactants and products.
Reactants
• The formulae of the reactants are written on the left
side of the equation while the products are on the • The total number of atoms of each element is the
right. same on both sides in a balanced equation.

• Reactants are the starting substances in a chemical • The numbers x, y, z and w (show the relative
reaction. number of molecules reacting) are called the
stoichiometric coefficients.
• Products are the substances formed as a result of a
chemical reaction. 111
110

The methods to balance a chemical A) Inspection Method
equation are:
a. Write down the unbalanced equation, with
A) Inspection method the correct formulae of the reactants and
B) Ion-electron method products.

112 b. Balance the metallic atoms, followed by
non-metallic atoms.

c. Balance the hydrogen and oxygen atoms.

d. Check to ensure that the total number of
atoms of each element is the same on
both sides of equation.

113

20

Example: Exercise
Balance the chemical equation by applying the
inspection method. Balance the chemical equation below by
applying inspection method.
NH3 + CuO → Cu + N2 + H2O a. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O
b. C6H6 + O2 → CO2 + H2O
Answer:
2NH3 + 3CuO → 3Cu + N2 + 3H2O 115

114

Answer: B) Ion-electron Method

a. 2Fe(OH)3 + 3H2SO4 → Fe2(SO4)3 + 6H2O • Mainly for redox reaction.
b. C6H6 + 15/2O2 → 6CO2 + 3H2O
• Redox reaction is a reaction that involves
both reduction and oxidation.

116 117

Oxidation : Reduction:
 A process of electron loss .  A process of electron gain .
 When a species undergo oxidation,  When a species undergo reduction,

- its oxidation number increase - its oxidation number decrease
- it loses one or more electron(s) - it gains one or more electron(s)
- act as a reducing agent - act as an oxidation agent
e.g e.g :

Mg Mg2+ + 2e Br2 + 2e → 2Br-
Sn4+ + 2e → Sn2+
118
119

21

Rules for Assigning Oxidation Numbers 2. For monoatomic ions, the oxidation
1. The oxidation number of an atom or a number is equal to the charge on the ion.

molecule in its elementary form (or free Example:
elements) is zero.
Ion Oxidation
Example: number
Na, Mg, O2, Br2, H2
Na+ +1
120 Al3+ +3

I- -1
S2- -2

121

Exception:

a. Hydrogen bonded to metal, i.e metal hydrides 3. In a neutral compound, the total
(e.g: NaH, HMgisH-21) the oxidation number for oxidation number of all atoms that made
hydrogen, up the molecule is zero.

1. b. Oxygen Example: 0
0
- In peroxides, its oxidation number is -1 Oxidation number of H2O = 0
Oxidation number of HCl =
(e.g: H2O2 = -1).
- when combine with fluorine, posses a Oxidation number of KMnO4 =

positive oxidation number (e.g: in OF2 = +2)

c. Halogen bonded to oxygen (e.g: Cl2O7), the 123
oxidation number for halogen = +ve

122

4. For polyatomic ion, the total oxidation Example
number of all atoms that made up the ion is
equal to the net charge of the ion. Assign the oxidation number of Cr in Cr2O72-
.
Example:
answer
Oxidation number of MnO4- = -1 2Cr  (7 x  2)   2
Oxidation number of Cr2O72- = -2
Oxidation number of NO3- = -1 2Cr   12
Cr   6
124
125

22

Exercise Answer

1. Assign the oxidation number of Mn in the 1. i. MnO2 ii. MnO4-
following chemical compounds. Mn  2(2)  0 Mn  4(2)   1
Mn   4 Mn   7
i. MnO2 ii. MnO4-
2. i. KClO3 ii. Cl2O72-
2. Assign the oxidation number of Cl in the
following chemical compounds.  1 Cl  3(2)  0 2Cl  7(2)   2
Cl   5 2Cl   12
i. KClO3 ii. Cl2O72- Cl  6

3. Assign the oxidation number of following:
i. Cr in K2Cr2O7
ii. U in UO22+
iii. C in C2O42-

126 127

Answer Answer

3. i. Cr in K2Cr2O7 3. iii. C in C2O42-
2C  (4 x  2)   2
2(1)  2Cr  (7 x  2)  0 2C   6
2Cr   12 C3
Cr   6
128
ii. U in UO22+

U  (2 x  2)   2
U 6

129

Balancing Redox Reaction using ion- Steps for balancing redox equation
electron method
For reactions in acidic solution:
 Redox reaction may occur in acidic and basic
solutions. Example:

 Follow the steps systematically so that MnO4- + C2O42-  Mn2+ + CO2
equations become easier to balance.
1. Divide the equation into two half-reactions:
reduction reaction and oxidation reaction

oxidation: C2O42-  CO2
reduction: MnO4-  Mn2+

2. Balance atoms other than O and H in each half
reaction.

04/29/12 MATTER 131030 oxidation: C2O42-  2CO2 131
reduction: MnO4-  Mn2+

23

3. BanadlatnhceehtyhderoogxyegneantoamtombybaydaddindginHg+H2O 5. Multiply each half-reaction with a coefficient to
equalize the number of electrons
oxidation: C2O42-  2CO2
reduction: MnO4- + 8H+  Mn2+ + 4H2O oxidation: (C2O42-  2CO2 + 2e) x 5
reduction: ( MnO4- + 8H+ + 5e  Mn2+ + 4H2O)x2
4. Balance the charge by adding electrons
to the side with the greater overall charge 6. Combine the two half-reactions and cancel out
(more tve side). species that appear on both sides of the equation

oxidation: C2O42-  2CO2 + 2e oxidation: 5C2O42-  10CO2 + 10e
reduction: 2 MnO4- + 16H+ + 10e  2Mn2+ + 8H2O
reduction: MnO4- + 8H+ + 5e  Mn2+ + 2MnO4- + 5C2O42- + 16H+ 2Mn2+ + 10CO2 + 8H2O
4H2O
133
132

7. Check to make sure that there are same number For reactions in basic solution
of atoms of each element and the same total
charge on both sides. balance the equation as in acidic form, then
continue,

2MnO4- + 5C2O42- + 16H+ 2Mn2+ + 10CO2 + 8H2O 8. Add the same number of OH- as H+ on both
sides of the equation.

Total charge reactant Total charge product (OH- and H+ on the same side of the equation
can be combined to form H2O.)
= 2(-1) + 5(-2) +16(+1) = 2(+2) + 10(0) + 8(0)
= -2 – 10 + 16 = +4 + 0 + 0 2M1n0OC4O- +2 5C2O42- + 16H+ 2Mn2+ + 8H2O +
= +4 = +4
+ 16OH- +16OH-

2MnO4- + 5C2O42- + 8H2O  2Mn2+ + 10CO2 + 16OH-

134 135

Example 2 Answer

Balance the equation below as in acidic 6Cl  3Cl2  6e
medium 6e 14H  Cr2O72  2Cr 3  7H2O
6Cl  14H  Cr2O72  2Cr 3  7H2O  3Cl2
Cr2O72- + Cl-  Cr3+ + Cl2

136 137

24

Example 3 Solution:

Balance the equation below as in basic 3H2O  3CN  3CNO  6e  6H
medium
6e  8H  2 MnO   2MnO 2  4H2O
CN- + MnO4-  CNO- + MnO2 4

2OH  3CN  2H  2 MnO   2MnO 2  H2O  3CNO  2OH
4

3CN  H2O  2 MnO   2MnO 2  3CNO  2OH
4

138 139

Stoichiometry • Example:
2CO(g) + O2(g)  2CO2(g)
• Stoichiometry is a quantitative study of
reactants and products in a chemical • It shows that 2 moles of CO reacts with 1 mole
reaction. of O2 to produce 2 moles of CO2.

• The most important thing in stoichiometric • The relationship can be simplified using symbol
calculation is a balanced equation. ‘≡’ which means ‘stoichiometrically equivalent
to’.
• The stoichiometric coefficient in a balanced 2 moles of CO ≡ 1 mole of O2 ≡ 2 moles of CO2
equation shows the ratio of amounts (in
moles) of reactants and products. 141

140

• The calculation based on the chemical equation Example 1
can be solved by 4 simple steps:
Excess barium nitrate was added to
1. write a balance equation. aluminium sulphate and 6.260g barium
2. calculate the number of mol (convert any sulphate was recovered. Calculate the mass
of aluminium sulphate in the original solution.
given quantities to mols)
3. use the equation (mol ratio) to determine 143

the mole of unknown.
4. convert mol calculated in 3. to the required

units.

142

25

Solution 3. use the equation (mol ratio) to determine the
mole of unknown.
1. write a balance equation.
from equation:
3Ba(NO3)2 + Al2(SO4)3  3BaSO4 + 2Al(NO3)3 1 mol Al(SO 4 )3  3 mol BaSO 4

? (g) 6.260 g

2. calculate the number of mol mol of Al2 (SO4 )3 = 2.682 x 10-2mol BaSO4 x  1 mol Al2 (SO4 )3 
 3 mol BaSO4 
mol of BaSO4 = 6.260 g  
233.36 g mol-
= 8.940 x 10-3mol

=2.682x10-2 mol

144 145

4. convert mol calculated in step 3 to the Example 2
required units.
Propane (C3H8) is a common fuel used for
mass of Al2 (SO 4 )3 = 8.940 x 10-3mol x 342.14 g mol-1 cooking and home heating. What mass of
=3.059 g O2 is consumed in the combustion of 1.00
g of propane?

146 147

Solution from equation:
1 mol C3H 8  5 mol O 2
C3H8 + 5O2  3CO2 + 4H2O
5 mol O2
1.00 g ? (g) mol of O2 = 2.27 x 10-2mol C3H8 x  1 mol C3H8 
 
1.00 g  
44.11 g mol-
mol of C3H8 = = 0.114 mol O2

=2.27x10-2 mol mass of O2 = 0.114 mol x 32.0 g mol-1
=3.64 g

148 149

26

Example 3 Solution:

What volume of 0.812 M HCl, in mililiters, is HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
required to titrate 1.45 g of NaOH to the
equivalence point? 0.812 M 1.45 g

Note: V= ?

the pH at which the number of moles of OH- mole of NaOH = 1.45 g
ions added to a solution is equal 40.0 g mol-1
stoichiometrically to the number of moles of
H+ ions originally present. = 0.0363mol

150 151

from equation: Exercise 1
1 mole of HCl  1 mole of NaOH
A 16.50 tmionLoaxn0ida.1icz3ied2i27c0Mm.0e0KdmMiuLnmOo.f4WashoFaleutStiiosOnt4hise
 mole of HCl= mole of NaOH needed
= 0.0363 mol solution

volume of HCl = mole of HCl cnoent cioennitcraetqiounatoiof nthies:FeSO4 solution? The
molarity of HCl

= 0.0363 mol 5Fe2+ + MnO4- + 8H+  Mn2+ + 5Fe3+ + 4H2O
0.812 mol L-1 Answer : 0.5474 M

= 0.0447 L = 44.7 mL

152 153

Solution 3. use the equation (mol ratio) to determine the
mole of unknown.
1. check/ write a balance equation.
5Fe 2+ + MnO4- + 8H+  Mn 2+ + 5Fe 3+ + 4H2O 1mol MnO   5 mol Fe2
4
20.00 mL 16.50 mL
?(M) 0.1327M 2.1896 x103 mol MnO   2.1896 x 103 mol MnO  x 5 mol Fe2
4 4

1mol MnO 
4

2. calculate the number of mol  0.01095 mol

mol MnO   0.1327 x16.5
4 1000

 2.1896 x 103 mol

154 155

27

4. convert mol calculated in step 3 to the Exercise 2
required units.
How many mililitres of 0.112 M HCl will
 Fe2  FeSO4  0.01095 react exactly with 21.2 mL ,0.150 M
0.02L sodium carbonate (Na2CO3)?
Answer : 56.8 mL
 0.5474M
157
156

Solution

1. check/ write a balance equation. 3. use the equation (mol ratio) to determine the
mole of unknown.

2HCl + Na2CO3  2NaCl + H2CO3 1 mol Na2CO3  2 mol HCl

0.112M 21.2 mL 3.18x 103 mol Na2CO3  3.18 x103 mol x2 mol
? (M) 0.150 M 1mol

2. calculate the number of mol  6.36 x 103 mol

mol Na2CO3  0.150 x 21.2
1000

 3.18 x 103 mol

158 159

4. convert mol calculated in step 3 to the Learning Outcome
required units.
At the end of this topic, students should be
Volume of HCl  6.36 x103 mol x1000 able to:
0.112 a) define the limiting reactant and

 56.79 mL percentage yield
b) perfome stoichiometric calculations
160
using mole concept including limiting
reactant and percentage yield.

161

28

Limiting reactant Limiting reactant & excess reactant

• In practice, the reactants used usually not • Limiting reactant
present in exact stoichiometric amounts. A reactant that is completely consumed
in a reaction and limits the amount of
– Consequently, some reactant will be product formed.
used up while others will be left over
at the end of the reaction. • Excess reactant
Reactant that remains after reaction,
162 presents at greater quantity compared
to the limiting reactant.

163

Example Example

Consider a bicycle: 1 frame • Consider the reaction:
requires
2 tyres Zn + 2HCl ZnCl2 + H2

Given: + • From equation:
1 mol Zn reacts completely with 2 mol HCl
2 frames Five tyres
• If 2 mole of Zn were added to 3 mole of HCl
How many bicycles will be formed ? - Limiting reactant : HCl
What limits the product ? The frame (limiting reactant) - Reactant in excess: Zn

164 165

Steps to determine limiting reactant Example1

i) Calculate no of moles available for each reactant A 10.0 g of hydrochloric acid was reacted
with 10.0 g of calcium carbonate to
ii) Stoichiometrically, calculate no of moles of produce carbon dioxide gas. Determine
reactant a) the limiting reactant
b) the mass of CO2 gas produced
iii) Compare no of mol calculated (needed) in (ii) to c) the mass of unused reactant
no of moles available in (i)
167
mol needed > mol available limiting reactant

mol needed < mol available excess reactant

iv) Apply no of moles of limiting reactant to
determine the product.

166

29

Solution: from equation: 2 mol HCl  1 mol CaCO3

2HCl + C10a.0CgO3  CaCl2 + H2O + CO2 The number of moles of CaCO3 needed to react with
all HCl is
a) 10.0 g

mol of HCl = 10.0 g mol of CaCO3 = 0.274 mol HCl x 1 mol CaCO3
36.46 g mol-1 2 mol HCl

= 0.274 mol = 0.137 mol CaCO3

mol of CaCO3 = 10.0 g Only 0.0999 mol CaCO3 available. Thus, not
100.09 g mol-1 enough CaCO3 to react with all HCl.

= 0.0999 mol ∴ CaCO3 is the limiting reagent.

168 169

b) The mass of CO2 produced depends on c) HCl is the excess reactant
the amount of the limiting reactant.
from equation: 1 mol CaCO3  2 mol HCl
from equation: 1 mol CaCO3  1 mol CO2
mole of CO2 = mole of CaCO3 mole of HCl used = 0.0999 mol CaCO3x1 2 mol HCl
= 0.0999 mol mol CaCO3

mass of CO2 = 0.0999 mol x 44.01 g mol-1 = 0.200 mol
= 4.40 g
mass of HCl used = 0.200 mol x 36.46 g mol-1
170 = 7.29 g

171

Mass of HCl available = 10.0 g Example 2
∴ Mass of HCl unused = 10.0 – 7.29 g
A 3.200 g sample of NaOH is added to a
= 2.71 g solution containing 1.125 g of H2SO4.
Determine
172 a) the limiting reactant
b) the mass of sodium sulphate formed
c) the mass of unused reactant

173

30

Solution: from equation: 2 mol NaOH  1 mol H2SO4
Moles of H2SO4 needed to react with all NaOH,
2NaOH + 1H.122S5Og 4  Na2SO4 + 2H2O

3.200 g
a)

mol of NaOH = 3.200 g mol of H2SO4 = 0.08000 mol NaOH x1 mol H2SO4
40.00 g mol-1 2 mol NaOH

= 0.08000 mol = 0.04000 mol

mol of H2SO4 = 1.125 g Only 0.01147 mol H2SO4 available. Thus, not
98.08 g mol-1 enough H2SO4 to react with all NaOH.

= 0.01147 mol ∴ H2SO4 is the limiting reactant.

174 175

b) c) The excess reactant = NaOH

from equation: 1 mol H2SO4  1 mol Na2SO4 from equation: 1 mol H2SO4  2 mol NaOH

mole of Na2SO4 = mole of H2SO4 mole of NaOH used = 0.01147 mol H2SO4 x 2 mol NaOH
= 0.01147 mol 1 mol H2SO4

mass of Na2SO4 = 0.01147 mol x 142.05 g mol-1 = 0.02294 mol
= 1.629 g
mass of NaOH used = 0.02294 mol x 40.00 g mol-1
176 = 0.9176 g

177

Mass of NaOH available = 3.200 g Percent yield
∴ Mass of NaOH unused = 3.200 – 0.9176 g
• Theoretical yield
= 2.282 g The amount of product predicted to form
based on the balanced equation.
178 (based on stoichiometric calculation)

• Actual yield
The amount of product actually obtained
from a reaction in an experiment.

179

31

• Percent yield Example

% yield = actual yield x 100 Titanium is prepared by the reaction of
theoretical yield titanium(IV) choride with molten magnesium
between 950oC and 1150oC.
180
TiCl4(g) + 2Mg(l) → Ti(s) + 2MgCl2(l)

In a certain industrial operation 3.54 X 107 g
of TiCl4 are reacted with 1.13 x 107g of Mg.

a) Calculate the theoretical yield of Ti in
grams

b) Calculate the percent yield if 7.91 X 106 g
of Ti are actually obtained.

181

Solution From equation: 1 mol TiCl4  2 mol Mg

a) 3.54 x 107g
189.7 g mol-1
n TiCl4 = mol of Mg = 1.87 x 105mol TiCl4 x 2 mol Mg
1 mol TiCl4
= 1.87 x 105 mol
= 3.74 x 105 mol

n Mg = 1.13 x 107 g 4.65 x 105 mol of Mg available. Thus, Mg is the
24.31 g mol-1 excess reactant.

= 4.65x105 mol ∴ TiCl4 is the limiting reactant.

182 183

From equation: 1 mol TiCl4  1 mol Ti b)

mole of Ti = mole of TiCl4 % Yield = Actual yield x 100%
= 1.87x 105 mol
Theoretical yield

The theoretical = 1.87 x 105 mol x 47.88g/mol = 7.91 x 106 g x 100%
mass of Ti formed = 8.95 X 106 g 8.93 x 106 g

= 88.6 %

184 185

32

Exercise Solution:

Consider the reaction: n Sb  1.20 n I2  2.40
2Sb (s) + 3I2 (s)  2SbI3 (s) 121.8 126.9x2

Determine the limiting reactant and the theoretical  9.8522 x103 mol  9.4563 x 103 mol
yield when 1.20g of Sb and 2.40g I2 are mixed.
What mass of excess reactant is left when the (given) (given)
reaction is complete.
( Sb=121.8, I= 126.9) 2 mol Sb  3 molI2

186 9.8522 x 103 mol Sb  9.8522 x 10 3 x3
2

 0.01479 molI2 (needed)

Mole needed I2 > mole given I2
Thus, I2 is limiting reactant

187

Solution: 3mol I2  2 mol Sb

3 mol I2  2 mol SbI3 9.4563 x 103 molI2 9.4563 x 103 x2
3

9.4563 x 103 mol I2  9.4563 x 10 3 x2 mole needed Sb  6.3042 x103 mol Sb
3
mole Sb unreacted  9.8522 x103  6.3042 x103
theoretical yield  6.3042 x103 mol SbI3  3.548 x103 mol

mass Sb unreacted  3.548 x103 mol x121.8 gmol1
 0.4321g

188 189

33

CHAPTER 1: MATTER

QUESTIONS

TUTORIAL.MEKA.KUMBE

lecture notes & questions

34

CHEMISTRY
TUTORIAL TOPIC 1: MATTER

1.1: Atoms and Molecule

1. a) Define:
i. proton number
ii. nucleon number
iii. isotopes

b) Write the isotope notation for each of the following species:

Species protons Number of electrons
neutrons

P 56 82 56

Q 13 14 10

R 14 14 14

S 16 16 18

c) State the number of sub-atomic particles for each of the following
elements:
i. chromium-52
ii. fluorine-19

2. Natural occurring magnesium has the following isotopic abundances:

Isotope Abundance Atomic mass (amu)
24Mg 78.9% 23.98504
25Mg 10.0% 24.98584
26Mg 11.1% 25.98259

a) Calculate the relative atomic mass of magnesium.
b) Sketch the mass spectrum of magnesium.

3. Two natural occurring isotopes of copper are 63Cu and 65Cu respectively. Given that
the relative atomic mass of copper is 63.5. What is the percentage abundance of each
isotope?

35

CHEMISTRY
TUTORIAL TOPIC 1: MATTER

1.2: Mole Concept

1. The hydrated copper (II) sulphate, CuSO4.5H2O is the compound used to prepare
CuSO4 solution. Given a sample of 10.0 g CuSO4.5H2O, calculate the:
a) number of moles of CuSO4.5H2O.
b) number of moles of copper atoms.
c) mass of anhydrous copper (II) sulphate.

2.
a) Define empirical formula and molecular formula.
b) Peroxyacylnitrate (PAN) is one of the components of smog containing C, H, N
and O. Analysis of 1.00 g of PAN was found to contain 0.198 g C, 0.025 g H
and 0.116 g N. Determine the molecular formula of PAN.
Given the molar mass of PAN is 121 gmol-1.

3. Isopropyl alcohol, sold as rubbing alcohol is composed of C, H and O. Combustion of
0.255 g of isopropyl alcohol produces 0.561 g CO2 and 0.306 g H2O. Determine the
empirical formula for isopropyl alcohol.

4. a) Define molality.
b) Given the density of 2.5 M aqueous solution of methanol CH3OH is 0.954
g/mL. Calculate the molality of solution.
c) A solution of hydrochloric acid with 37.8 % by mass of HCl has a density of
1.19 gml-1. Calculate the molarity of hydrochloric acid.
d) A solution is made by dissolving 16.0 g CaCl2 in 64.0 g of water. Calculate the
mole fraction of CaCl2 and percentage by mass (%w/w).

1.3: Stoichiometry

1. Balance the equations below:

i. C5H6O (l) + 6O2 (g) 5CO2 (g) + 3H2O (g)

ii. 8I-(aq) + SO42-(aq) + 10H+ (aq) 4I2(s) + H2S (aq) + 4H2O(l)

iii. 4Cl2 (aq) + 8OH- (aq) ClO4- (aq) + 7Cl- (aq) + 4H2O (l)

2. An impure sample of zinc weighing 50.0 g reacts with excess hydrochloric acid to
produce 18.0 L hydrogen gas at room condition.
a) Write a balanced equation for the reaction.
b) Calculate the mole of H2 gas produced.
c) Determine the percentage of zinc in the impure sample.

36

CHEMISTRY
TUTORIAL TOPIC 1: MATTER

3. Menthol CH3OH is an excellent fuel which can be made by a reaction of carbon
monoxide and hydrogen.

CO (g) + 2H2(g) CH3OH (l)

If 50.0 g of carbon monoxide react with 0.80 moles of hydrogen gas, determine:
a) the limiting reactant
b) the percentage yield if the actual yield of methanol is 11.9 g.

4. A reaction between 7.0 g of copper (II) oxide, CuO and 50 mL of 0.20 M nitric acid,
HNO3 produces copper (II) nitrate, Cu(NO3)2 and water.
a) Define limiting reactant.
b) Write a balanced chemical equation for the above reaction.
c) Determine the limiting reactant.
d) Calculate the expected mass of copper (II) nitrate formed.
e) Determine the percent yield if the actual mass of copper (II) nitrate obtained
from the reaction is 0.85 g.

37

CHEMISTRY
MEKA TOPIC 1: MATTER

1.1: Atoms and Molecule

1. The mass spectrum of magnesium, Mg obtained from a spectrometry analysis is
shown below.

Relative Abundance 78.99

x 11.01

m/e

24 25 26

a) List the isotopes found in the spectrum by using isotope notation.
b) Which is the most abundant isotope of Mg?
c) How are the isotopes different from one another in terms of sub-atomic

particles?
d) Determine the value of x if the average atomic mass for Mg is 24.32 amu.

2. A sample of element M was vapourised and injected into a mass spectrometer. The
mass spectrum show 60.10% of the metal is 69M and 39.90% is 71M. The atomic
masses for 69M and 71M are 68.93 amu and 70.92 amu respectively.
a) Sketch the mass spectrum of element M.
b) What is the average atomic mass of M?

1.2: Mole Concept

1. Calculate the amount of each of the following entities
a) mass in grams of 0.64 mol MnSO4
b) moles of 15.8 g of Fe(ClO4)3 compound
c) number of nitrogen atoms in 92.6 g of NH4NO2

2. Find the empirical formula of the following compounds:
a) 0.063 mol chlorine atoms combine with 0.220 mol of oxygen atoms
b) 2.45 g of silicon combine with 12.4 g of chlorine

3. An oxide of nitrogen contains 30.45% by mass of nitrogen,
a) what is the empirical formula of the oxide?
b) if the molar mass is 90 gmol-1, what is its molecular formula?

38

CHEMISTRY
MEKA TOPIC 1: MATTER

4. Calculate each of the following quantities,
a) mass of solute in 185.8 mL solution of 0.267 M calcium acetate,
Ca(CH3COO−)2.
b) molarity of 500 mL of solution containing 21.1 g of potassium iodide.
c) moles of solute in 145.6 L of 0.850 M sodium cyanide solution.
d) molarity of a 150 mL solution prepared by diluting 37.00 mL of 0.25 M
potassium cyanide solution.
e) volume (L) of 1.63 M calcium chloride that must be diluted with water to
prepare
a 350 mL of a 2.86×10−3 M chloride ion solution.

1.3: Stoichiometry

1. Balance the equations below: (acidic)
a) Cu(s) + S8(s) → Cu2S(s) (basic)

b) H3PO4(aq) + NaOH(aq) → Na2HPO4(aq) + H2O(l)
c) Co3+(aq) + Mn2+(aq) → Co2+(aq) + MnO2(s)
d) MnO4-(aq) + ClO3-(aq) → MnO2(s) + ClO4-(aq)

2. Calculate mass of each product formed when 33.61 g of diborane (B2H6) react with

excess water.
B2H6(g) + H2O(l) → H3BO3(s) + H2(g)

3. Iron reacts with steam according to the equation:
3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)

In an experiment, 24.5 g of iron were reacted with 12.6 g of steam. Calculate:
a) the mass of Fe3O4 produced.
b) the volume of H2 formed at STP.

4. In an experiment, 2.80 g of oxygen gas was added to 4.20 g of calcium to form
calcium oxide
2Ca(s) + O2(g) → 2CaO(s)
a) What is the mole of CaO that can be produced by the given mass of Ca?
b) What is the mole of CaO can be produced by the given mass of O2?

c) Which is the limiting reactant?
d) What is the expected mass of CaO produced in the experiment?

39

CHEMISTRY
KUMBE TOPIC 1: MATTER

1.1: Atoms and Molecule

1. The element Ne consists of three isotopes which are 20Ne, 21Ne and 22Ne. The relative
atomic mass of Ne is 20.2.
a) Define relative atomic mass
b) Which is the most abundant isotope of Ne?
b) State the number of protons and the number of neutrons of this isotope.

2. The table below lists the relative atomic mass of three types of carbon.

Substance Carbon-12 Carbon-13 Diamond

Relative mass 12.000 13.003 12.011

a) How many times is one carbon-13 atom heavier than one atom of carbon-12?

b) Calculate the mass of one 12C atom. [1.993×10−23g]

c) Diamond consists of the 12C and 13C isotopes. Calculate the percentage of 12C

in diamond. [98.9%]

1.2: Mole Concept

1. Quinine C20H24N2O2, is a compound extracted from cinchona tree which is

traditionally used to treat malaria. If given a 1.08 g of quinine sample, calculate:

a) the molecular mass of quinine. [324 g mol−1]

b) the number of moles of quinine. [3.33×10−3 mol]

c) the number of molecules of quinine. [2.01×1021 molecules]

d) the number of hydrogen atoms. [4.82×1022 atoms]

e) the mass of carbon atoms in gram. [0.800 g]

2. A hydrate of potassium carbonate has the formula K2CO3.xH2O. From 10.00 g of the

hydrate, 7.95 g of anhydrous salt was left after heating. Determine the value of x in

the formula. [2]

3. The combustion of 0.202 g of an organic compound produces 0.361 g of CO2 and

0.147 g of water. Determine the molecular formula of the compound if it consists of
C, H and O and has the molar mass of 148 g mol−1.

4. An alcohol solution contain 2.5% ethanol, C2H5OH. By assuming the density of the

solution is 1.00 g mL−1, determine the

a) ethanol concentration in mol L−1. [0.543 mol L−1]

b) mass of ethanol in a 200 mL glass of this beer. [5.00 g]

5. Calculate the molality of 6.5 M aqueous solution of an acid, HA with a density of

0.888 g cm−3. Given the molar mass of acid is 98.0 g mol−1. [26 molal]

40

CHEMISTRY
KUMBE TOPIC 1: MATTER

6. A sample of concentrated ethanoic acid, CH3COOH with 90% by mass and density of
1.23 g mL−1 is used to prepare 1.8 L of 4.0 M dilute ethanoic acid. Calculate the

a) molarity of the concentrated ethanoic acid. [18.5 M]

b) volume of concentrated ethanoic acid required to prepare the dilute ethanoic

acid. [0.39 L]

7. A sulphuric acid solution has a density of 0.182 g cm−3 and contains 96.0% H2SO4 by

mass.

a) Determine the molarity of the H2SO4 solution. [1.78 M]

b) Calculate the volume of H2SO4 required reacting with zinc in order to liberate

201.5 cm3 of hydrogen gas at room conditions. [4.71 mL]

8. Muriatic acid is an industrial grade of concentrated HCl, used to clean masonry and
cement. The concentration of stock solution is 10.0 M.
a) Write steps on how to dilute the stock solution to make 250 mL of 0.2 M acid
solution.
b) How many milliliters of the diluted muriatic acid solution contain 0.15 g HCl?
[20.5 mL]

1.3: Stoichiometry

1. A redox reaction between oxalate ions, C2O42− and permanganate ions, MnO4− in an
acidic solution produces carbon dioxide, CO2 and manganese(II) ions, Mn2+. Write a
balanced equation for the above reaction.

2. A 3.8 g sample of bronze was dissolved in sulphuric acid. The copper in the alloy

reacted with sulphuric acid as follow:
Cu(s) + 2H2SO4(aq) → CuSO4(aq) + SO2(g) + 2H2O(l)

The CuSO4 formed in the reaction was mixed with sufficient KI to produce solid CuI,
I3− and SO42− ions. The I3− ion formed in this second reaction was then titrated against
thiosulphate, S2O32− solution to form I− and S4O62− ions.

a) Write the ionic equations for the last two reactions.

b) Calculate the percentage by mass of copper in the bronze sample if 26.20 mL

of 1.05 M S2O32− was consumed in the titration. [46%]

3. Ammonia, NH3 gas can be prepared by the reaction of a metal oxide such as calcium

oxide, CaO with ammonium chloride, NH4Cl.
2NH4Cl(s) + CaO(s) → 2NH3(g) + H2O(l) + CaCl2(s)

If 224 g of NH4Cl and 112 g of CaO are mixed, calculate the mass of:

a) ammonia. [67.7 g]

b) excess reactant remains at the end of the reaction. [10.4 g]

41

CHEMISTRY
KUMBE TOPIC 1: MATTER

4. a) A sample of 1.5 g of a metal oxide, MO2 reacts with excess hydrochloric acid,

HCl to produce 386 mL chlorine gas at STP. The reaction is given by the

following equation: MCl2(aq) + Cl2(g) + H2O(l)
MO2(s) + HCl(aq) →

i. Balance the above equation.

ii. Determine the relative molecular mass of MO2 and the relative atomic

mass of M. [55]

b) In a separate experiment, 0.20 g of MO2 was added to 25 mL of 0.10 M HCl

solution.

i. Determine the limiting reactant.

ii. Calculate the mass of MCl2 produced in the reaction. [0.079 g]

iii. Calculate the percentage yield if the actual mass of MCl2 produced is

0.072 g. [91%]

5. Adipic acid, H2C6H8O4 is used to produce nylon. The acid is made commercially by a

controlled reaction between cyclohexane, C6H12 and O2

2C6H12(l) + 5O2(g) → 2H2C6H8O4(l) + 2H2O(g)

When 25.0 g of cyclohexane is reacted with excess of O2(g), 33.5 g of adipic acid is

produced. Calculate the percentage yield of adipic acid. [77.1%]

6. An unknown mass of KIO3 was added to excess of acidified KI to produced I2 and

H2O. The iodine liberated is then titrated with sodium thiosulphate and requires 53.7
cm3 of 0.20 M sodium thiosulphate for a complete reaction.

a) Write a balanced ionic equation for the reaction between KIO3 and KI.

b) The equation for the reaction between iodine and sodium thiosulphate is

I2(s) + 2S2O32− (aq) → 2I− (aq) + S4O62−(aq)

Calculate the mass of KIO3 used. [0.38 g]

7. Ammonia can be generated by heating together the solids NH4Cl and Ca(OH)2 to

produce CaCl2 and H2O. If a mixture containing 33.0 g each of NH4Cl and Ca(OH)2 is

heated:

a) calculate the mass of NH3 formed. [10.5 g]

b) what is the mass of reactant remains in excess? [10.1 g]

8. A side reaction in the manufacture of rayon from wood pulp is
3CS2 + 6NaOH → 2Na2CS3 + Na2CO3 + 3H2O

How many grams of Na2CS3 are produced in the reaction between 92.5 mL of liquid

CS2 and 2.78 mol NaOH? [143 g]

(density of CS2 = 1.26 g/mL)

42

CHAPTER 2: ATOMIC
STRUCTURE

LECTURE NOTE

lecture notes & questions

43

CHAPTER 2 2.1 Bohr’s Atomic
ATOMIC STRUCTURE Model

2.1 Bohr’s atomic model 2
2.2 Quantum mechanics
2.3 Electronic configuration

1

2.1 Bohr’s atomic model

34

i) calculate ionisation energy of hydrogen atom BOHR’S ATOMIC MODELS
from Lyman series
• In 1913, a young Dutch physicist,
j) state the limitation of Bohr’s atomic model. Niels Böhr proposed a theory of
k) state the dual nature of electron using de atom. He described an atom based
on his postulates that electrons
Broglie’s postulate and Heisenberg’s are circling in orbits a nucleus of
uncertainty principle an atom.

5 • These orbits can only occur at specifically
“permitted” levels, according to the energy levels
of the electrons which explain the lines in the
hydrogen spectrum.

6

44

BOHR’S ATOMIC MODELS 2. The moving electron in the permitted orbit has
a specific amount of energy (quantized
Bohr postulates (assumptions) are : energy). This explain the lines in the hydrogen
spectrum.
1. Electron moves in circular orbit about the
nucleus . It does not radiate or absorb energy • Orbit is a pathway of an electron travels around the
while moving around the nucleus. nucleus of an atom

• [ orbit = energy level ]

moving electron

H Nucleus orbit e
(proton)
11H n=1 nucleus
n=2
7 n=3 8
n=4

• The quantized energy of an electron at its orbit ( Energy of an electron is quantized , it has a specific
value
energy level) is RH
n2
En   If an electron occupies Electron orbiting the
n=4, it has the energy of: nucleus at n=1 has the
Where: energy of E1 = -RH
E4= -RH
n = the number of particular orbit / energy level/ quantum 42 12
number ,1, 2, 3, …∞

RH = Rydberg’s constant, 2.18 x 10-18 J e

nucleus

Note: n=1

• n identifies the orbit of electron n=2
n=3
• Energy is zero if an electron is located infinitely far from n=4
nucleus
9 10

Example Solution

1. Is it likely that there is an energy level, n for a 1. En   RH
hydrogen atom when En = -1.00 x 1028 J? n2

2. Calculate the energy of an electron at energy  1 .00 x 10 28  2 .18 x 10 18 J
n2
level

i) n = 2 ii) n = 5 2 . 18 x 10 18 J
 1 . 00 x 10 28 J
n2  

Answer : ii) -8.72 x 10-20 J n  1 . 48 x 10  23
1. no, n = 1.47 x 10-23
2. i) -5.45 x 10-19 J

11 12

45

Solution E2   RH 3. At ordinary conditions the electron is at the
n2 ground state (lowest level).
2 . i) If energy is supplied, the electron absorbed
2 . 18 x 10 18 J the energy and is promoted from a lower to a
2 . ii )   22 higher energy level. (Electron is excited)

  5 . 45 x 10 19 J 14

E5   RH
n2

2 . 18 x 10 18 J
  52

  8 . 72 x 10  20 J

13

3. A specific amount of energy is absorbed by an 4. Electron at its excited states is unstable. It will
electron fall back to a lower energy level and releases a
E = h = E3-E1 (+ve) specific amount of energy in the form of light
(radiation), called a photon.
• Electron is excited from lower, n=1 to higher, n=3
energy level. The energy of the photon equals to the energy
difference between the levels, ΔE.
energy is absorbed
photon = ΔE
electron at ground e
state 16
n =1 n = 2
n=3
n=4

15

4. Electron falls from higher energy level, n=3 to lower energy Fig. 7.10
level, n=1 and releases a photon.
E = h = E1-E3 (-ve)

electron at excited
state

e n =1 n=2
n=3
photon, h
emitted n=4

17 22 July 2022 Atomic structure 18

46

Energy level diagram for the hydrogen atom • The amount of energy emitted (released) when the
electron falls can be determine by

n= Energy released = the energy difference between
n=4 E two energy levels, Ef - Ei
n=3
n=2 where E = Ef - Ei

Potential energy n=1 Ef  RH 1  and Ei  RH  1 
nf2   ni2 

Energy Energy E  RH 1     RH 1  
absorbed released nf2   ni2  

19 20

Thus, R H  1 1  • The photon i.e the amount of energy released ,
ni2 nf 2 E by the electron transitions between two energy
E   levels has an appropriate frequency,  and

wavelength, 

Where: h (Planck’s constant) = 6.63 10-34 J s
 = frequency
ni = energy level at initial state E = h
nf= energy level at final state
RH=Rydberg constant, 2.18 x 10-18 J and   c c (speed of light) = 3.00108 ms-1


Thus,

ΔE  hc


21 22

Example Solution

One mole of electron drops from the forth electron drops from the forth energy level to the third
energy level to the third energy level. energy level.
Calculate:
i) ni = 4 nf = 3
a) the energy of the photon emitted
b) the frequency of this photon E  RH ( 1  1 )
ni2 nf2
23
= 2 .1 8 x 1 0 -1 8 J  1 - 1 
 42 32 

= -1 .0 6 x 1 0 -19 J

24

47