1.0 Matter

CHAPTER 1

MATTER

1.1 Atoms & Molecules
1.2 Mole Concept
1.3 Stoichiometry

1

MATTER

Anything that occupies space and has mass.
Examples:
air, water, animals, trees, atoms, …..

Matter may consists of atoms, molecules or ions.
Can be in form of solid, liquid and gas.

2

STATES OF MATTER

SOLID LIQUID GAS

3

SUBSTANCES

Matter that has a specific composition and
specific properties.

Compound Element

 Two or more different  Composed of atoms of
chemical element only one type of atoms

 Can be separated into  Cannot be separated
simpler substance by into simpler substances by
chemical reactions. chemical reactions.

Eg: water,glucose, ammonia Eg: gold, tin, oxygen

4

MIXTURE

Different substances mixed together but
not chemically combined.

Homogeneous mixture Heterogeneous mixture

Uniform in composition and Not uniform in composition
cannot be separated
physically.

Eg: water,milk Eg: Water with cement

5

6

1.1 ATOMS AND
MOLECULES

77

LEARNING OUTCOMES

At the end of this topic, I should be able to:

(a) Describe proton, electron and neutron in terms of the relative
mass and relative charge.

(b) Define proton number, Z, nucleon number, A and isotope.
(c) Write isotope notation.
(d) Define relative atomic mass, Ar and relative molecular mass, Mr

based on the C-12 scale.
(e) Calculate the average atomic mass of an element given the relative

abundances of isotopes or a mass spectrum.

8

ATOMS

The smallest unit of a chemical element/compound

Consist of three subatomic particles:
- Proton (p)
- Neutron (n) Packed in a small nucleus

- Electron (e) Move rapidly around the nucleus
of an atom

9

MODERN ATOMIC MODEL

Nucleus } proton + neutron

Shell

Electron

10

SUBATOMIC PARTICLES

Particle Mass Charge Charge Mass
(gram) (Coulomb) (units) (a.m.u)

Electron (e) 9.1 x 10-28 -1.6 x 10-19 -1 0

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

Neutron (n) 1.67 x 10-24 0 01

11

PROTON NUMBER/ ATOMIC NUMBER, Z

Number of protons found in the nucleus of an atom.

Eg: Na Proton 11
Proton number 11

Neutral charge atoms :

number of proton = number of electron
Na

Proton number = number of electron = 11

12

NUCLEON NUMBER/ MASS NUMBER, A

Total number of protons and neutrons found in the
nucleus of an atom.

Nucleon Number, A = protons (Z ) + neutrons (n)

 Eg: Na

Proton 11

Neutron 12

Nucleon number 23

13

ISOTOPES

Two or more atoms of the same element with

same number of protons but different

number of neutrons.

Same chemical properties but different physical
properties.

Eg:
11H 21H(D) 31H(T)

U235 U238
92 92

14

ISOTOPE NOTATION

Nucleon Number Element Symbol
Proton Number

Eg:

15

Example: Total charge on
the ion
Nucleon number of
mercury, A = 202 16

The number
of neutrons
=A–Z
= 202 – 80
= 122

proton number of
mercury, Z = 80

QUESTION 1

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

Symbol Proton Number of : Charge
7 Neutron Electron -3
D14 33- 7 10

7

63 Cu 29 34 29 0
29

O17 2 8 9 10 -2
8

Co59 3 27 32 24 +3

27

17

QUESTION 2

Write the appropriate symbol for each of the following
isotopes:

(a) Z = 11, A = 23
(b) Z = 28, A = 64
(c) Z = 74, A = 186
(d) Z = 80, A = 201

18

MOLECULES

Atoms joined together by bonds.

Diatomic Molecule Polyatomic Molecule

Contains only two atoms Contains more than two
atoms
Example :
H2, N2, O2, Br2, HCl Example :
O3, H2O, NH3, CH4

19

EXAMPLE:

20

IONS

Cation Anion

a positive charge ion a negative charge ion
formed when a neutral formed when a neutral
atom loses an electron(s). atom gains an electron(s).

Na Na+ Cl Cl-

11 protons 11 protons 17 protons 17 protons
11 electrons 10 electrons 17 electrons 18 electrons

21

CATIONS / ANIONS

A monoatomic ions A polyatomic ions
Consist of one atom
Consist two or more
Example : atoms
Fe3+ : Iron (III) ion
S2- : Sulfide ion Example :
O2- : Oxide ion H30+ : Hydroxonium
ion
CN- : Cyanide ion
I3- : tri-iodide ion

22

RELATIVE ATOMIC MASS, Ar

A mass of one atom of an element compared to 1/12
mass of one atom of 12C with the mass 12.000 (a.m.u)

Relative Atomic Mass, Ar= mass of one atom of an element (a.m.u)
1/12 X mass of one atom of 12C (a.m.u)

Ar has no unit

23

QUESTION 3

Determine the relative atomic mass of Y atom, if the
mass of one atom Y is 6.00 a.m.u ?

Ar = Mass of one atom of an element
1/12 X mass of carbon-12 atom

= 6 a.m.u
1/12 X 12 a.m.u

=6

24

RELATIVE MOLECULAR MASS, Mr

A mass of one molecule of a compound compared to 1/12 mass
of one atom of 12C with the mass 12.000 (a.m.u)

Relative Molecular Mass, Mr = mass of one molecule of a compound
1/12 X mass of one atom of 12C

Mr has no unit

Mr of a compound = ∑ Ar of all atoms in a molecular formula.

25

QUESTION 4

Calculate the relative molecular mass of C5H5N and
CaSO4.2H2O

[Ar C = 12.0 ; Ar H = 1.0 ; Ar O = 16 ; Ar N = 14.0 ; Ar S = 32.0 ; Ar Ca = 40.0]

Mr C5H5N = [ 5(12.0) + 5(1.0) + (1)14.0 ]
= 79.0

Mr CaSO4.2H2O = [ 1(40.0) + 1(32.0) + 4(16.0) + 2(18.0) ]
= 172.0

26

MASS SPECTROMETER

A mass spectrometer is used to :
i. Determine relative atomic mass of an element
ii. Determine relative molecular mass of a compound
Determine types of isotopes, the abundance
iii. and its relative isotopic mass
iv. Recognize the structure of the compound in an
unknown sample

27

MASS SPECTROMETER

Ionisation
Chamber

Vaporisation Accelaration Magnetic
Chamber Chamber
+ Chamber
--

Ion Beam

Heated Vacuum Ion
Filament Pump Detector

AMPLIFIER
Recorde2r8

MASS SPECTRUM OF MAGNESIUM

Relative abundance 63

9.1 m/e
8.1
24 25 26 29

How to calculate the relative
atomic mass from mass spectrum?

Q mAverage
atomic
ii

mass Qi

Q = the relative abundance / percentage abundance
of an isotope of the element

m = the isotopic mass of the element

30

QUESTION 5 Relative abundance 18
7
Fig shows the mass spectrum of the
element rubidium, Rb; 85 87 m/e

a. What isotopes are present in Rb? 31

b. What is the percentage abundance
of each isotope?

c. Calculate the relative atomic
mass of Rb

a. 85Rb and 87Rb

b. % abundance of 85Rb = 18 × 100 % = 72 %
18+7

% abundance of 87Rb = 7 × 100 % = 28 %
18+7

c. Average atomic mass = σ
σ

= 18 ×85 +(7 ×87)
7+18

= 85.56 a.m.u

∴ = 85.56 . .

1 ×12.000 . .
12

= 85.56

QUESTION 6

Naturally occurring 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)

Ratio of 191 = 5
193 8

Average atomic mass = σ
σ
191.021 ×5 +(193.025 ×8)
= 191 5+8

= 192.254 a.m.u

∴ = 192.254 . .

1 ×12.000 . .
12

= 192.254

33

QUESTION 7

A naturally occurring element Q, composes of two
isotopes 25Q and 26Q. If the relative atomic mass of Q
is 25.3, determine the percentage abundance of each
isotope.

Let % abundance of 25Q = x %

% abundance of 26Q = (100 – x) %

Average atomic mass = σ
σ
25 + 26(100 − )
25.3 = 100

x = 70 %

% abundance of 25Q = 70%

% abundance of 26Q = 30%

34

1.2 MOLE CONCEPT

35

LEARNING OUTCOMES 36

At the end of this topic, I should be able to:

(a) Define mole in terms of mass of carbon-12 and Avogadro’s
constant, NA

(b) Interconvert between moles, mass, number of particles, molar
volume of gas at s.t.p. and room temperature.

(c) Define the terms empirical and molecular formulae.
(d) Determine empirical and molecular formulae from mass

composition or combustion data.
(e) Define and perform calculations for each of the following

concentration measurements :
i) molarity (M)
ii) molality (m)
iii) mole fraction (X)
iv) percentage by mass (% w/w)
v) percentage by volume (% v/v)

ONE MOLE

The amount of substance contains equal number of
particles(atoms /molecules /ions)
as in 12.0 g of carbon-12

One mole of carbon-12 atom has a mass of 12.0 g
and contains 6.02 x 1023 atoms.

6.02 x 1023 mol-1 = Avogadro’s Constant = NA

One moles gas = 22.4 L at stp
= 24 L at room temperature

37

1 mole of shulpur

32 g 6.02 x 1023 S atoms

1 mole of potassium chloride

74.6 g 6.02 x 1023 KCl molecules
6.02 x 1023 K atoms
6.02 x 1023 Cl atoms

38

1 mole of an element = 6.02 x 1023 atoms of that
element

In terms of molecule and ions,

1 mole of a molecule = 6.02 x 1023 molecules

1 mole of an ions = 6.02 x 1023 ions

39

1.0 mole of chlorine atom = 6.02 x 1023 chlorine atoms

= 35.5 g chlorine atom

1.0 mole of chlorine molecules = 6.02 x 1023 chlorine molecules
= 71.0 g Cl2
= 6.02 x 1023 x 2 chlorine atoms

1.0 mole of NH3 = 6.02x 1023 molecules
= 6.02 x 1023 x 4 atoms

= 6.02 x 1023 N atom

= 6.02 x 1023 x 3 H atoms

40

QUESTION 8 6.02 x 1023 atoms of oxygen
6.02 x 1023 molecules of oxygen
1.0 mole of oxygen atom 2 x 6.02 x 1023 atoms of oxygen
1.0 mole of oxygen gas
1.0 mole NH3 6.02 x 1023 molecules of NH3
1.0 mole of N
13.2 g of Fe2O3 3.0 mole of H

6.02 x 1023 atoms of N
3 x 6.02 x 1023 atoms of H
8.26 x 10-2 mole of Fe2O3
2 x 8.26 x 10-2 mole of Fe

3 x 8.26 x 10-2 mole of O
4.97 x 1022 molecules of Fe2O3

2 x 4.97 x 1022 atoms of Fe
3 x 4.97 x 1022 atoms of O

41

MOLAR MASS

Mass of one mole of an element/compound in g mol-1

QUESTION 9

Calculate the molar mass for this compound Ca3(PO4)2

molar mass of Ca3(PO4)2
= 3 ( 40.1) + 2(31.0) + 8(16.0) = 310.3 g/mol

42

NUMBER OF MOLE, n

For compound
n = Mass (g)
Molar mass (gmol-1)

QUESTION 10
What is the mass of 0.671 mol of Al2O3 ?

43

QUESTION 10
What is the mass of 0.671 mol of Al2O3 ?

Mr of Al2O3 = 102.0
Molar mass of Al2O3 = 102.0 g/mol

Mass Al2O3 = 0.671 mol x 102.0 g/mol
= 68.44 g

44

MOLE CONCEPT OF GASES

Molar volume of any gas at STP = 22.4 dm3 mol-1

STP = Standard Temperature and Pressure
Where, T = 273.15 K = 0 oC

P = 1 atm

1 mole of gas has a volume of 22.4 dm3 at s.t.p

1 mole of gas has a volume of 24.0 dm3 at room
temperature ( 298.15 K @ 25 oC)

1 L = 1 dm3 = 1000cm3

45

NUMBER OF MOLE, n

For gases
=

n = Vol ( L) at S.T.P
22.4 L/mol

n = Vol (L) at room T
24 L/mol

1 L = 1 dm3 = 1000 cm3 = 1000 mL

46

QUESTION 11
A gas X occupies 345.1 cm3 at s.t.p. Calculate the
number of gas molecules in the sample of X.

22.4 dm3 Ξ 1 mol gas X

345.1 cm3 Ξ 0.3451 dm3 x 1 mol gas X

1000 22.4 dm3

= 1.5406 x 10-2 mol gas X

1 mol gas X Ξ 6.02 x 1023 molecules of gas X
1.5406 x 10-2 mol gas X Ξ 1.5406 x 10-2 mol x 6.02 x 1023 molecules

1 mol

= 9.27 x 1021 molecules

47

QUESTION 12

(a) Calculate the molar mass of a compound if 0.372 mole of it has a
mass of 152 g.

(b) How many molecules of ethane, C2H6 are present in 0.334 g of
C2H6?

(c) Calculate the number of C, H and O atoms in 1.50 g of glucose,
C6H12O6.

(d) How many atoms are there in:
i. 5.10 moles of sulphur?
ii. 3.14 g of copper?

(e) How many moles of:
i. Ca atoms are in 77.4 g of Ca?
ii. Co atoms are there in 6.00 x 109 Co atoms?

48

a) 0.372 mol Ξ 152 g
1 mol Ξ 1 mol x 152 g
0.372 mol

= 408.60 g

Molar mass of compound = 408.60 g/mol

b) 1 mol C2H6 Ξ 30 g Ξ 6.02 X 1023 molecules
0.334 g Ξ 6.02 X 1023 X 0.334 g
30 g

= 6.70 X 1021 molecules of C2H6

c) n of C6H12O6 = mass
molar mass

= 1.50 g
180 g/mol

= 8.3333 x 10-3 mol

49

c) 1 mol C6H12O6 Ξ 6.02 X 1023 molecules of C6H12O6
8.3333×10−3
8.3333x10-3 mol C6H12O6 Ξ 1 mol mol x 6.02x1023 molecules

= 5.0167 x 1021 molecules of C6H12O6

number of C atoms = 6 x 5.0167 x 1021
= 3.01 x 1022 atoms

number of H atoms = 12 x 5.0167 x 1021
= 6.02 x 1022 atoms

number of O atoms = 6 x 5.0167 x 1021
= 3.01 x 1022 atoms