1.1 atoms and molecules-sk016 lecturer (1)
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SEMESTER 1OVERVIEW
7CHAPTERS:
1 : Matter ( 7 hrs )2 : States of Matters ( 8 hrs )3 : Atomic Structure ( 10 hrs )4 : Periodic Table ( 6 hrs )5 : Chemical Bonding ( 16 hrs )
6 : Chemical Equilibrium (6 hrs)7 : Ionic Equilibria ( 14 hrs )
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ASSESMENT
Components Paper Code Format Time Marks Percentage
MID-SEM TEST 1 SK016 Structured andessay
1 hour - 10%
CONTINUOUS
ASSESMENT
- - Practical
Throughoutthe
semester
- 10%
- - Assignment/Quiz
- 10%
FINALEXAMINATION
Paper1
SK016/1 Multiple ChoiceQuestions
1 hour 30
70%
Paper2
SK016/2 Structured
(Part A) 2½ hours
40
Essay (Part B) 60
TOTAL 100%
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CONSULTATION
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Chapter 1 : MATTER
1.1 Atoms and Molecules
1.2 Mole Concept
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1.1 Atoms andMolecules
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LEARNING OUTCOMES
At the end of this topic, I should be able to:
(a) Identify and describe proton, electron andneutron as subatomic particle.
(b) Define proton number, Z, nucleon number,A and isotope.
(c) Write isotope notation.
(.
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LEARNING OUTCOMES…(continue)
(d) Define relative atomic mass, Ar andrelative molecular mass, Mr based onthe C-12 scale
(e) Calculate the average atomic mass of anelement given the relative abundance ofisotopes or a mass spectrum
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• Matter
Anything that occupies space and has mass.
Examples:
air, water, animals, trees, atoms, …..
• Matter may consists of atoms, molecules or ions.
Introduction
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SOLID LIQUID GAS
Three states of Matter
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Classification of Matter
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A substance is a form of matter that has adefinite composition and distinct properties.
Example : water, ammonia, sucrose, gold, oxygen
Substance
Water (H2O)
Oxygen (O2)
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A mixture is a combination of two or more substances inwhich the substances retain their distinct identities.
1. Homogenous mixture – composition of the mixtureis the same throughout.
2. Heterogeneous mixture – composition is not uniform
throughout.
soft drink, milk, solder
cement,iron filings in sand
1.4
Mixture
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A compound is a substance composed of atoms of twoor more elements chemically united in fixedproportions.
Compounds can only be separated into their purecomponents (elements) by chemical reactions.
Water (H2O)
Glucose (C6H12O6)
Ammonia (NH3)
1.4
Compound
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An element is a substance that cannot be separatedinto simpler substances by chemical means .
An element is composed of atoms of only one kind.
• 115 elements have been identified
• 83 elements occur naturally on Earth(gold, aluminum, lead, oxygen, carbon)
• 32 elements have been created byscientists (technetium, americium, seaborgium)
1.4
Element
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1.1.1 Atoms
• An atom is the smallest unit of a chemical
element/compound.• In an atom, there are three subatomic particles:
- Proton (p)
- Neutron (n)
- Electron (e)
Packed in a small nucleus
Move rapidly around the nucleusof an atom
Atoms and Molecules
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Nucleus } proton + neutron
Shell
Electron
An atom contains: Proton
Neutron and
Electron
Modern Atomic Model
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The size of the nucleus is about100,000 times smaller than that of
the whole atom, and therefore mostof the atom consist of empty spaceand almost 99.99% of the mass of anatom is concentrated within the tinynucleus.
Do You Know?
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Particle Mass
(gram)
Charge
(Coulomb)
Charge
(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
Subatomic Particles
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The total sum of protons and neutrons foundin the nucleus of the atom. (also known as massnumber)
Thetotal sum of protons
found in the nucleusof the atom. (also known as atomic number)
Nucleon Number, A = protons (Z) + neutrons (n)
Nucleon Number, A
Proton Number, Z
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Isotopes are two or more atoms of the sameelement that have the same number of protons in their nucleus but different number ofneutrons.
Have the same chemical properties butdifferent physical properties.
Examples:
(D)H2
1
U235
92U238
92
(T)H3
1H1
1
Isotopes
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• An atom can be represented by an isotope notation( atomic symbol )
Element Symbol
Proton Number
Nucleon Number
= Proton number(Z) +
Number of
neutrons
Isotopes Notation
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Total charge onthe ion
proton number ofmercury, Z = 80
Nucleon number ofmercury, A = 202
The numberOf neutrons= A – Z= 202 – 80= 122
Example
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• A molecule consists of a small number ofatoms joined together by bonds.
Molecules
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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
Example
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Cation
a positive charge ionformed when a neutralatom loses an electron(s).
11 protons 11 protons11 electrons 10 electrons
• Two types of ions : a) cation b) anion
Na Na+
Anion
a negative charge ion formedwhen a neutral atom gains anelectron(s).
17 protons 17 protons
17 electrons 18 electrons
Cl Cl-
Ions
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A monoatomic ions
• Contains only one nucleus• Example :
Fe3+ : Iron (III) ion or ferric ionS2- : Sulfide ion
A polyatomic ions • Contains more than one nucleus• Example :
H30+ : Hydronium ion or hydroxonium ionCN- : Cyanide ion
Example
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i. Relative Atomic Mass, Ar
A mass of one atom of an element comparedto 1/12 mass of one atom of 12C with the mass
12.000Relative Atomic Mass, Ar= mass of one atom of element (a.m.u)
1/12 X mass of one atom of 12C (a.m.u)
Ar has no unit
The mass of one atom of carbon-12 =12 atomic mass units (a.m.u.)
Relatives Mass
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Determine the relative atomic mass of an element
Y if the ratio of the atomic mass of Y to carbon-12 atom is 0.45Solution :
Ar of Y = mass of one atom of element (a.m.u)
1/12 X mass of one atom of 12C (a.m.u )
= 12 x mass of Y
mass of 12C
= 12 x 0.45
= 5.4
Example
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ii 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
The relative molecular mass of a compound is the.summation of the relative atomic masses of allatoms in a molecular formula.
Relative Molecular Mass, Mr = mass of one molecule of a compound
1/12 X mass of one atom of 12C
Mr has no unit
Relatives Mass
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• Calculate the relative molecular mass of C5H5N,Ar C = 12.01
Ar H = 1.01
Ar N = 14.01
ANSWER:
Mr = 5(Ar of C) + 5(Ar of H) + Ar of N
= 5(12.01) + 5(1.01) + 14.01
= 60.05 + 5.05 + 14.01
= 79.11
Example
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A mass spectrometer is used to determine:i. Relative atomic mass of an element
ii. Relative molecular mass of a compound
iii. Types of isotopes, the abundance and itsrelative isotopic mass
iv. Recognize the structure of the compound inan unknown sample
Mass Spectrometer
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+
AMPLIFIER
- -
AccelarationChamber
VacuumPump
HeatedFilament
VaporisationChamber
IonisationChamber
Magneticfield
Ion Detector
Recorder
Ion Beam
A Mass Spectrometer
+ +
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• The numbers of ions and types of isotopes arerecorded as a mass spectrum.
• Example : A mass spectrum of Mg
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8.1 9.1
24 25 26
R e l a t i v e a b u n d a n c e
m/e (amu)
Mass Spectrum
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Mass Spectrum of Magnesium
• The mass spectrum of Mgshows that Mg consists ofthree isotopes: 24Mg, 25Mgand 26Mg.
• The height of each line ispropartional to theabundance of each isotope.
• 24Mg is the most abundantof the three isotopes
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8.1 9.1
24 25 26
R e l a t i v e a b u
n d a n c e
m/e (amu)
Example
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Q = the relative abundance / percentage
abundance of an isotope of the elementM = the relative isotopic mass of the element
How to calculate the relative
atomic mass from mass spectrum?
i
ii
Q
MQ
Ar Average
atomicmass
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1. Fig 1.1 shows the mass spectrum of the
element rubidium, Rb;a. What isotopes are present in Rb?
b. What is the percentage abundance ofeach isotope?
18
7
85 87
R e l a t i v e a b u n d a n c e
m/e(amu)
85Rb and 87Rb
% abundance 85Rb= 18 x 100
25
= 72 %
% abundance 87Rb= 7 x 100
25= 28 %
Example 1
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c. Calculate the relative atomic mass of Rb.
85.56
amux12.00121
amu85.56Rbof A
amu85.5625
)87x7()85x18(
Qi Rbof massAverage
r
QiMi
Example 1
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6.94?is Li of mass atomic relative
theif isotope each of abundance percentage theis What
.7.02 and 6.01are Li73
and Li63
of mass atomic relative The
Example 2
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Assume that,
% abundance of 6Li = X %% abundance of 7Li = (100 - x) %
Ar Li = ∑QiMi ∑Qi
6.94 = X (6.01) + (100 – X) 7.02X + 100 – X
6.94 = 6.01 X + 702 – 7.02 X100
694 - 702 = -1.01 X
+8 = +1.01 XX = 7.92 %
So, % abundance of 6Li = 7.92 %And % abundance of 7Li = (100 – 7.92) %
= 92.08 %
Example 2