CHAPTER 1 - MATTER

CHAPTER 1
MATTER

CHEMISTRY UNIT
KOLEJ MATRIKULASI MELAKA

SHARED BY: DALINA BINTI DAUD

Chapter 1 : MATTER

1.1 Atoms and Molecules
1.2 Mole Concept
1.3 Stoichiometry

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1.1 Atoms and Molecules

2

Learning outcomes

At the end of this topic, students should be able to:
i. Write isotope notation
ii. Interpret mass spectrum.
iii. Calculate the average atomic mass of an element

given of a given relative abundances of isotopes
or a mass spectrum.

3

Introduction

• Chemistry is the study of matter and the
change it undergoes.

• Matter is anything that occupies space
and has mass.
e.g: air, water, animals, trees, atoms, …..

• The three states of matter are solid,
liquid and gas.

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1.1 Atoms and Molecules

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
atom

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Modern Model of the Atom

• Electrons move around the region of the atom.

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Subatomic Particles

Subatomic Mass Charge Charge
Particle (gram) (Coulomb) (units)

Electron (e) 9.1 x 10-28 -1.6 x 10-19 1-

Proton (p) 1.67 x 10-24 +1.6 x 10-19 1+

Neutron (n) 1.67 x 10-24 0 0
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Proton number and Nucleon number

• The proton number, Z (known as atomic
number), of an atom is the number of
protons in the nucleus of the atom of
an element.

• The nucleon number,A (known as mass
number) is the total number of protons
and neutrons in the nucleus of the
atom of an element.

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Nuclide Notation

An atom/ion can be represented by an
nuclide notation ( atomic symbol )

* nuclide – a type of atom as characterised by its atomic number and its
nucleon number

X = Element symbol
Z = Proton Number of X

=p
A = Nucleon Number of X

= p+n

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Nucleon number Total charge
of mercury, on the ion
A = 202

Proton number The number of neutrons
of mercury, =A–Z
Z = 80 = 202 – 80
= 122

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EXAMPLE

Sodium

23Na

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Number of protons = 11
Number of electrons = 11
Number of neutrons = 23 – 11

= 12

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• Atom

has the number of protons equals to the
number of electrons

• CATION (Ion with a positive charge)

has the number of electrons less than the
number of protons (atom lost electrons)

• ANION (Ion with a negative charge)

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

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EXAMPLE 13

• Magnesium ion

2142Mg2+

Number of protons = 12
Number of electrons = 12 – 2

= 10
Number of neutrons = 24 – 12

= 12

Isotopes

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

• Example:

- Hydrogen has 3 isotopes while uranium has 2.

11H 21H (D) H (T)200Hg

hydrogen deuterium 1380

U235 U238 tritium
92 92
-
uranium-235 uranium-238
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• Isotopes of an element have similar
chemical properties but have slightly
different physical properties.

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Exercise 1

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

Symbol Number of : Charge

Proton Neutron Electron

200 Hg
80

63 Cu
29

O17 2
8

59 Co3 
27

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Answer

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

Symbol Number of : Charge

200 Hg Proton Neutron Electron 0
80 80 120 80 0
29 34 29 2-
63 Cu 8 9 10 3+
29 27 32 24

O17 2
8

59 Co3 
27

* Ionic charge has the sign AFTER the number , e.g 2+, 3+,
2- etc.

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Exercise 2

• Write the appropriate notation for each of the
following element :

Number of :

Species Notation

A Proton Neutron Electron 4 A
B 2 22 2
C 1 20
D 1 11 3 B
7 7 10 1

21 C

147D3

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Molecule

• A molecule consists of a small number of atoms
joined together by covalent bonds.

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• A diatomic molecule contains only two
atoms
Example :
H2, N2, O2, Br2, HCl, CO

• A polyatomic molecule contains more
than two atoms
Example :
O3, H2O, NH3, CH4

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Mass Spectrometer

• An atom is very light and its mass cannot be
measured directly.

• A mass spectrometer is an instrument used to
measure the precise masses and relative
quantity of atoms and molecules

• A mass spectrometer is used to determine the:

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 21

A Mass Spectrometer

22

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Simplified diagram of a Mass Spectrometer

Ionisation
Chamber

Vaporisation Accelaration Magnetic
Chamber Chamber
+ Chamber

--

Heated Vacuum Ion Beam
Filament Pump Ion Detector

AMPLIFIER

24 Recorder

Beam of 35Cl+ and 37Cl+
37Cl+

35Cl+

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• The mass and intensity of the ions are recorded
in the form of a graph called the mass spectrum

• Example : A mass spectrum of Mg

Relative 63
abundance
8.1 9.1
m/e (amu)
24 25 26

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Relative abundance Mass Spectrum of Magnesium

• The mass spectrum of Mg
shows that Mg consists of
three isotopes: 24Mg, 25Mg
and 26Mg.

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• The height of each line is

8.1 9.1 proportional to the
abundance of each

24 25 26 m/e (amu) isotope.

• 24Mg is the most abundant
of the three isotopes

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How to calculate the relative atomic mass
from a mass spectrum?

1. Calculate the atomic mass.

The atomic mass of an element is the average
mass of the mixture of isotopes.

 Relative Isotopic mass of

abundance of x each isotope
each isotope
Atomic
 Relative
mass of an = abundance
element

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2. Calculate the relative atomic mass.

Re lative atomic mass, Ar  Mass of one atom of element
1 X Mass of one atom of 12 C
12

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Example 1:
Calculate the relative atomic mass of neon
from the mass spectrum.

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Solution:

Average atomic =  (% abundanceisotopicmass)

% abundanec

mass of Ne = (90.5  20 u)  (0.3  21u)  (9.2  22 u)

(90.5  0.3  9.2)

= 20.2 u

Ar of Ne  20.2 amu
1 x 12 amu
2

 20.2

Relative atomic mass Ne = 20.2 31

Example 2

Relative 18
abundance

7

85 87 m/e (amu)

Figure above shows the mass spectrum of
the element rubidium, Rb;

a. What are the isotopes of Rb?

85Rb and 87Rb

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Example 2 (cont…)

b.What is the percentage abundance of each
isotope?

% abundance 85Rb
= 18 x 100
25
= 72 %

% abundance 87Rb
= 7 x 100
25
= 28 %

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Example 2 (cont…)

c. Calculate the relative atomic mass of Rb.

Atomic mass of Rb =(18 x 85 amu) + ( 7x 87amu)
(18+ 7)

= 85.56amu

A or f Rb = 85.56amu
112 x12.00amu

= 85.56

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Example 3

The relative atomic mass of 6Li and 7Li are 6.01
and 7.02. What is the percentage abundance of
each isotope if the relative atomic mass of Li is
6.94

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Assume that, X%
% abundance of 6Li = (100 - x) %
% abundance of 7Li =

atomic mass Li = X (6.01) + (100 – X) 7.02
6.94 = X + (100 – X)
X=
6.01 X + 702 – 7.02 X

100

7.92

 % abundance of 6Li = 7.92 % 36
% abundance of 7Li = 92.08 %

Exercise 1

Chlorine has two naturally occuring isotopes, 35Cl
(isotopic mass 34.9689 amu) and 37Cl (isotopic mass
36.9659 amu). If chlorine has an atomic mass of
35.4527 amu, what is the percentage abundance of
each isotope?

Answer: 75.774 % 35Cl and 24.226 % 37Cl

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Assume that, X%
% abundance of 35Cl = (100 - x) %
% abundance of 37Cl =

atomis mass Cl =X (34.9689) + (100 – X) 36.9659
X + (100 – X)

35.4527 = 34.9689 X + 3696.59 – 36.9659 X

100

X = 75.774

 % abundance of 35Cl = 75.774 %
% abundance of 37Cl = 24.226 %

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Exercise 2

Naturally occuring iridium, Ir is composed of 2
isotopes 191Ir and 193Ir in the ratio of 5:8. The relative
mass of 191Ir and 193Ir are 191.021 and 193.025
respectively. Calculate the relative atomic mass of
iridium.
(Ans : 192.254)

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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

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Success is the sum of small efforts,
repeated day in and day out.

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1.2 MOLE CONCEPT

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Learning outcomes…

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, molality, mole fraction, percentage by
mass and percentage by volume.

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Empirical and Molecular Formula

• Empirical formula is a chemical formula
that shows the simplest ratio of all
elements in a molecule.

• Molecular formula is a formula that
indicates the actual number of atoms of
each element in a molecule.

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• The empirical formula and molecular
formula of a compound could be the same.
e.g: NH3 and CO2

• There are molecules that have different
molecular formulae but the same empirical
formula.
e.g: C2H4 and C3H6

Empirical formula and molecular formula can
be related as

molecular formula = n(empirical formula)

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Example 1

Ascorbic acid cures scurvy and may help to
prevent the common cold. It is composed of
40.92% carbon, 4.58% hydrogen and 54.50%
oxygen by mass. The molar mass of ascorbic
acid is 176 gmol-1.
Determine its empirical formula and
molecular formula.

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Answer:

Element C H O

Mass (g) 40.92 4.58 54.50
54.50
Number 40.92 4.58 16.00
of moles 12.01 1.01 =3.41
=3.41 =4.53
(mol) 1.33 1
1
Ratio of 4 3
moles 3

Simplest
ratio

∴ Empirical fomula = C3H4O3

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Molecular formula = n (C3H4O3)
Mr n (C3H4O3) = 176

n[3(12.01)+4(1.01)+3(16.00)] =176
88.07n = 176
n=2

∴ Molecular formula = C6H8O6

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