Atoms, Ions and Molecules The Building Blocks of Matter Chapter 2.
MATTER (1.1 Atoms and Molecules)2
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Chapter 1 : MATTER
1.1 Atoms and Molecules
1.2 Mole Concept
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1.1 Atoms and Molecules
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Learning OutcomeLearning Outcome At the end of this topic, students should be able :At the end of this topic, students should be able :
(a) (a) Identify and describe proton, electron and
neutron as subatomic particle. as subatomic particle.
(b) (b) Define proton number, Z, nucleon number, A proton number, Z, nucleon number, A
and isotope. and isotope. Write isotope notation.
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(c) (c) Define relative atomic mass, Ar and
relative molecular mass, Mr based on based on
the C-12 scale.the C-12 scale.
(d) (d) Sketch and and explain the following the following main
components of a simple of a simple mass
spectrometer.
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(e) (e) Analyse mass spectrum of an element. of an element.
Calculate the average atomic mass of Calculate the average atomic mass of
an element given the relative an element given the relative
abundance of isotopes or a mass abundance of isotopes or a mass
spectrum.spectrum.
(f) Name (f) Name cation, anions and salt according according
to the to the IUPAC nomenclature. .
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IntroductionIntroduction
Matter
Anything that occupies space and has mass.
e.g
air, water, animals, trees, atoms, …..
Matter may consists of atoms, molecules or ions.
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Three States of MatterThree States of Matter
SOLID LIQUID GAS
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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)
1.1 Atoms and Molecules1.1 Atoms and Molecules
Packed in a small nucleus
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.Electrons move around the region of the atom.
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Subatomic ParticlesSubatomic Particles
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
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ElementsElements
A substance that A substance that cannot be separated into cannot be separated into simpler substancessimpler substances by chemical reactions. by chemical reactions.
An element is An element is composed of atoms of only composed of atoms of only one kind.one kind.
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IsotopeIsotope
Isotopes are two or more atoms of the same element that have the same number of protons in their nucleus but different number of neutrons.
Examples:
Hg20080
Hg20080
(D) H21
U23592 U238
92
(T) H31H11
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Isotope Notation Isotope Notation
X = element symbolZ = Proton Number of
X = pA = Nucleon Number of X
= Z + n
• An atom can be represented by an isotope notation
( atomic symbol )
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Total charge on the ion
Number of atoms that formed the
ion proton number of mercury,
Z = 80
Nucleon number of mercury, A = 202
The number of neutrons= A – Z= 202 – 80= 122
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Exercise 1Exercise 1
+3243227
-21098
0293429
08012080
ElectronNeutronProton
ChargeNumber of :Symbol
Give the number of protons, neutrons,electrons and charge in each of the following species:
Hg20080
Cu6329
2178O
35927Co
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Exercise 2Exercise 2
Species Number of : Notation for nuclideProton Neutron Electron
A 2 2 2
B 1 2 0
C 1 1 1
D 7 7 10
Write the appropriate notation for each of the following nuclide :
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1.1.5 Ion 1.1.5 Ion
Cation
a positive charge ion formed when a neutral atom loses an electron(s).
11 protons 11 protons
11 electrons 10 electrons
Two types of ions : a) cation b) anion
Na Na+
Anion
a negative charge ion formed when a neutral atom gains an electron(s).
17 protons 17 protons
17 electrons 18 electrons
Cl Cl-
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MoleculeMolecule A molecule consists of a small number of atoms A molecule consists of a small number of atoms
joined together by bonds.joined together by bonds.
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A diatomic molecule Contains only two atomsExample :
H2, N2, O2, Br2, HCl, CO
A polyatomic moleculeContains more than two atomsExample :
O3, H2O, NH3, CH4
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Relative MassRelative Mass
i. 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
Cof atom one of Mass X 121
element of atom one of Mass Ar mass, atomic lativeRe
12
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Example 1Example 1 Determine the relative atomic mass of an element Y if
the ratio of the atomic mass of Y to carbon-12 atom is 0.45
ANSWER:
Ar (Y) = Mass of one atom of Y____ 1/12 x Mass of one atom of C-12
= 0.45 x 12
= 5.4
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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
C12 of atom one of Mass X 121
compound a of
molecule one of Mass
Mr mass, molecular Relative
The relative molecular mass of a compound is the summation of the relative atomic masses of all atoms in a molecular formula.
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Example 2Example 2 Calculate the relative molecular
mass of C5H5N,
Ar C = 12.01
Ar H = 1.01
Ar N = 14.01ANSWER:
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
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Yesterday, we have learned about:Yesterday, we have learned about:MatterMatter
AtomsAtomsElementsElementsMoleculeMoleculeArArMrMr
Today, we are going to learn about:Today, we are going to learn about:Mass spectrometerMass spectrometer
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Mass SpectrometerMass Spectrometer
A mass spectrometer is used to determine:
i. Ar of an element
ii. Mr of a compound
iii. Types of isotopes, the abundance and its relative isotopic mass
iv. Recognize the structure of the compound in an
unknown sample
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+
AMPLIFIER
- -
Accelaration Chamber
VacuumPump
HeatedFilament
VaporisationChamber
IonisationChamber
Magnetic Chamber
Ion Detector
Recorder
A Mass Spectrometer
Ion Beam
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Vaporisation ChamberVaporisation Chamber
- sample of the element is vaporised into gaseous atom
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Ionisation ChamberIonisation Chamber
- A gaseous sample is bombarded by a stream of high-energy electrons that are emitted from a hot filament.
- Collisions between the electrons and the gaseous sample produce positive ions
M M +
M s + e -f M +
s + e -s + e -
f
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Vacuum PumpVacuum Pump
A pump maintains a vacuum inside the mass spectrometer to avoid any small particle that would block the movement.
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Acceleration ChamberAcceleration Chamber
- the positive ions are accelerated by an electric
field towards the two oppositely charge plates
- the electric field is produced by a high voltage
between the two plates
- the emerging ions are of high and constant
velocity.
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Magnetic FieldMagnetic Field
- The positive ions are separated and deflected into a circular path by a magnet according to its mass / charge (m/e) ratio.
- Positive ions with small m/e ratio are deflected most Ions with large m/e ratio are deflected least.
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Beam of Beam of 3535ClCl++ and and 3737ClCl++
3535ClCl++3737ClCl++
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Ion DetectorIon Detector The numbers of ions and types of isotopes are recorded
as a mass spectrum. Example : A mass spectrum of Mg
63
8.1 9.1
24 25 26
Re
lativ
e a
bu
nda
nce
m/e (amu)
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Mass Spectrum of MagnesiumMass Spectrum of Magnesium The mass spectrum of Mg
shows that Mg consists of three isotopes: 24Mg, 25Mg and 26Mg.
The height of each line is proportional to the abundance of each isotope.
24Mg is the most abundant of the three isotopes
63
8.1 9.1
24 25 26
Re
lativ
e a
bu
nda
nce
m/e (amu)
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How to calculate the relative atomic How to calculate the relative atomic mass from mass spectrum?mass from mass spectrum?
i
ii
QMQ
Ar
i
ii
QMQ
Ar
Q = the relative abundance / percentage abundance
of an isotope of the element
M = the relative isotopic mass of the element
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Example 1Example 11. 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 of each isotope?
18
7
85 87
Re
lativ
e a
bu
nda
nce
m/e (amu)
85Rb and 87Rb
% abundance 85Rb= 18 x 100
25= 72 %
% abundance 87Rb= 7 x 100
25= 28 %
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Example 1 (cont…)Example 1 (cont…)
c. Calculate the relative atomic mass of Rb.
85.56
amu x12.00121
amu 85.56Rb of A
amu 85.5625
)87x7()85x18(
Qi Rb of mass Average
r
QiMi
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Example 2Example 2
6.94? is Li of mass atomic relative
theif isotope each of abundance percentage theis What
. 7.02 and 6.01 are Li73 and Li6
3 of mass atomic relative The
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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) 7.02 X + 100 – X
6.94 = 6.01 X + 702 – 7.02 X 100
694 - 702 = -1.01 X+8 = +1.01 X X = 7.92 %
So, % abundance of 6Li = 7.92 %And % abundance of 7Li = 92.08 %
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Exercise 1Exercise 1
35.45) :(Ans
chlorine. ofAr theCalculate 36.9659.Cl37 and
34.9689Cl35 of mass atomic relative thescale, 12-carbon theon Based
3.127 Cl37Cl35
: followas is isotopes
chlorine of occuring naturally of abundance relative of ratio The
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We have learned about:We have learned about:Mass spectrometerMass spectrometer
Today, lets learn about:Today, lets learn about: IUPAC Nomenclature of IonsIUPAC Nomenclature of IonsMole conceptsMole concepts
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Exercise 2Exercise 2
192.254) :(Ans
iridium. of
mass atomic relative theCalculate ly.respective 193.025 and 191.021
areIr 193 andIr 191 of mass relative The . 8:5 of ratio theinIr 193
andIr 191 isotopes 2 of composed isIr iridium, occuring Naturally
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IUPAC Nomenclature of IonsA) Cations
i) For the metals of group 1, 2 and 13 :
Name the metals followed by the word ‘ ions ‘
e.g : Na+ : sodium ion, Al3+ : aluminium ion
ii) For the metal with more oxidation states, Roman
numerals are used to indicate the oxidation state.
e.g : Cu2+ : copper(II) ion, Fe3+ : iron(III) ion
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B. Anions
• Monoatomic ions have names that ended with ‘ide’
e.g : F- : fluoride ion, O2- : oxide ion
• Other polyatomic anions have their own names
e.g : CO3 : carbonate ion, SO42- : sulphate ion,
Cr2O72- : dichromate ion
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• When a metal combines with a nonmetal element, the metal is named before the nonmetal
Example : Fe2(SO4)3 - Iron(III) sulphate
FeCl3 - Iron(III) chloride
CuCl2 - copper(II) chloride