Chemistry Malaysian Matriculation Full Notes & Slides for Semester 1 and 2

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WELCOME TO SKO16

CHEMISTRY

CHEMISTRY

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CHEMISTRY SK016

1.0 Matter 7 2.0 Atomic Structure 7 3.0 Periodic Table 4 4.0 Chemical Bonding 2 5.0 State of Matter 7 6.0 Chemical Equilibrium 5 7.0 Ionic Equilibria 12 Total 54

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CHEMISTRY SK026

8.0 Thermochemistry 4 9.0 Electrochemistry 6 10.0 Reaction Kinetics 7 11.0 Intro To Organic Chemistry 4 12.0 Hydrocarbons 8 13.0 Aromatic Compounds 3 14.0 Haloalkanes (Alkyl halides) 4 15.0 Hydroxy compounds 3

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CHEMISTRY SK026

16.0 Carbonyl 4

17.0 Carboxylic acids & Derivatives 4

18.0 Amines 5

19.0 Amino acids and Proteins 2

20.0 Polymers 1

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ASSESSMENT

1. COURSEWORK (20%)

Continuous evaluation (tutorial/test/quiz) - 10%

Practical work - 10%

2. MID-SEMESTER EXAMINATION - 10%

3. FINAL EXAMINATION (70%)

Paper 1 (30 multiple choice questions) - 30%Paper 2 (Part

A-structured) (Part

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REFERENCE BOOKS

CHEMISTRY ,9th Ed. – Raymond Chang, McGraw-Hill  CHEMISTRY –The Molecular Nature of Matter and

Change, 3rd Ed.– Martin Silberberg, McGraw Hill

 CHEMISTRY – The Central Science, 9th Ed. Theodore

L.Brown, H.Eugene LeMay,Jr, Bruce E Bursten,

Pearson Education

 GENERAL CHEMISTRY – Principle & Structure, 6th Ed.

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 GENERAL CHEMISTRY – Principle and Modern

Applications, 8th Ed. Ralph H. Petrucci, William

S. Harwood, Prentice-Hall

 ORGANIC CHEMISTRY, 7th Ed – T.W.Graham

Solomon,Craig B.Fryhle, John Wiley and Sons

 ORGANIC CHEMISTRY, 4th Ed – L.G. Wade, Jr,

Prentice Hall

 ORGANIC CHEMISTRY, 6th Ed – John

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Chapter 1 : MATTER

1.1 Atoms and Molecules 1.2 Mole Concept

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Learning Outcome

At the end of this topic, students 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.

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Introduction

Matter

 Anything that occupies space and has mass.

e.g: air, water, animals, trees, atoms, etc

 Matter may consists of atoms, molecules or

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Classifying Matter

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 A substance is a form of matter that has

a definite or constant composition and distinct properties.

Example: H2O, NH3, O2

 A mixture is a combination of two or

more substances in which the substances retain their identity. Example : air, milk, cement

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An element is a substance that cannot be separated into simpler substances by

chemical means.

Example : Na, K, Al,Fe

A compound is a substance composed of

atoms of two or more elements chemically united in fixed proportion.

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Three States of Matter

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a) Atoms

 An atom is the smallest unit of a chemical

element/compound.

 In an atom, there are 3 subatomic particles:

- Proton (p) - Neutron (n) - Electron (e)

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Modern Model of the Atom

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 All neutral atoms can be identified by the

number of protons and neutrons they contain.

Proton number (Z)  is the number of

protons in the nucleus of the atom of an element (which is equal to the number of

electrons). Protons number is also known as

atomic number.

Nucleon number (A)  is the total number

of protons and neutrons present in the nucleus of the atom of an element. Also known as mass number.

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Subatomic 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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Isotope

 Isotopes are two or more atoms of the same

element that have the same proton number in their nucleus but different nucleon number.

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Examples:

H

1 1

H

(D)

2 1

H

(T)

3 1

U

235 92

U

238 92

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Isotope Notation

X = Element symbol Z = Proton number of X (p) A = Nucleon number of X = p + n

An atom can be represented by an isotope notation ( atomic symbol )

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08/16/11 08/16/11 mattermatter 2323 Total charge on 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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 In a neutral atom:

 number of protons equals number of electrons

 In a positive ion:

 number of protons is more than number of

electrons

 In a negative ion:

 number of protons is less than number of

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Exercise 1

Symbol Number of : Charge

Proton Neutron Electron

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

Hg 200 80 Cu 63 29 2 17 8O 3 59 27 Co

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Exercise 2

Species Number of : Notation

for nuclide Proton 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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A molecule consists of a small number of atoms joined together by bonds.

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A diatomic molecule

 Contains only two atoms

Ex : H2, N2, O2, Br2, HCl, CO

A polyatomic molecule

 Contains more than two atoms

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Learning Outcomes

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

(a) Define relative atomic mass, Ar and relative molecular mass, Mr based on the C-12 scale.

(b) Calculate the average atomic mass of an element given the relative abundance of isotopes or a mass spectrum.

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Relative 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 amu

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

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 Mass of an atom is often expressed in atomic

mass unit, amu (or u).

Atomic mass unit, amu is defined to be one

twelfth of the mass of 12C atom

 Mass of a 12C atom is given a value of exactly 12

amu

1 u = 1.660538710-24 g

 The relative isotopic mass is the mass of an

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Example 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

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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 12.000amu

C of atom one of Mass x 12 1 molecule one of Mass Mr mass, molecular Relative 12 

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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 2

Calculate the relative molecular mass of C5H5N,

Ar C = 12.01 Ar H = 1.01 Ar N = 14.01

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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

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Mass Spectrum of Monoatomic Elements

 Modern mass

spectrum converts the abundance into percent abundance

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Mass Spectrum of Magnesium

 The mass spectrum of Mg

shows that Mg consists of 3 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 3 isotopes Re la ti v e a b u n d a n c e 63 8.1 9.1 24 25 26 m/e (amu)

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Learning Outcomes

At the end of this topic, student should be able

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

(a) Calculate the average atomic mass of an element given the relative abundances of isotopes or a mass spectrum.

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How to calculate the relative atomic

mass, A

r

from mass spectrum?

    abundance ) abundance mass (isotopic mass atomic Average

Ar is calculated using data from the mass

spectrum.

 The average of atomic masses of the entire

element’s isotope as found in a particular

environment is the relative atomic mass, Ar of the atom.

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Example 1:

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

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Solution:

Average atomic = mass of Ne

=

= 20.2 u

Relative atomic mass Ne = 20.2

   e % abundanc mass) isotopic abundance (% ) 2 . 9 3 . 0 5 . 90 ( ) u 22 2 . 9 ( ) u 21 3 . 0 ( ) u 20 5 . 90 (       

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Example 2:

Copper occurs naturally as mixture of

69.09% of 63Cu and 30.91% of 65Cu. The

isotopic masses of 63Cu and 65Cu are 62.93 u and 64.93 u respectively. Calculate the

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Solution:

Average atomic = mass of Cu

=

= 63.55 u

Relative atomic mass Cu = 63.55

 e % abundanc mass) isotopic abundance (% ) 91 . 30 09 . 69 ( ) u 93 . 64 91 . 30 ( ) u 93 . 62 09 . 69 (    

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Example 3:

Naturally occurring iridium, Ir is composed of two isotopes, 191Ir and 193Ir in the ratio of 5:8. The relative isotopic mass of 191Ir and 193Ir are 191.021 u and 193.025 u

respectively. Calculate the relative atomic mass of Iridium

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08/16/11 08/16/11 4747 Solution: Average atomic = mass of Ir = = 192.254 u

Relative atomic mass Ir = 192.254

   abundance mass isotopic abundance ) 8 5 ( ) u 025 . 193 8 ( ) u 021 . 191 5 (    

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Mass Spectrum of Molecular

Elements

A sample of chlorine which contains 2 isotopes with nucleon number 35 and 37 is analyzed in a mass spectrometer. How many peaks would be expected in the mass spectrum of chlorine?

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MASS SPECTROMETER

+ _ _ Cl2 + e  Cl2+ + 2e Cl2 Cl2 + e  2Cl+ + 2e 35Cl-35Cl 35Cl-37Cl 37Cl-37Cl 35Cl-35Cl+ 35Cl-37Cl+ 37Cl-37Cl+ 35Cl+ 37Cl+

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Mass Spectrum of Diatomic

Elements

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Exercise:

How many peaks would be expected in a mass spectrum of X2 which consists of 3 isotopes?

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MATTER

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Learning Outcome

At the end of this topic, students should be

able to:

a)

Define mole in terms of mass of

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Avogadro’s Number, NA

 Atoms and molecules are so small – impossible

to count

 A unit called mole (abbreviated mol) is devised

to count chemical substances by weighing them

 A mole is the amount of matter that contains

as many objects as the number of atoms in exactly 12.00 g of carbon-12 isotope

 The number of atoms in 12 g of 12C is called

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Example:

1 mol of Cu contains Cu atoms

1 mol of O2 contains O2 molecules O atoms

1 mol of NH3 contains NH3 molecules N atoms H atoms 6.02 6.02  10102323 3 3  6.02 6.02  10102323 2 2  6.02 6.02  10102323 6.02 6.02  10102323 6.02 6.02  10102323 6.02 6.02  10102323

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1 mol of CuCl2 contains Cu2+ ions

Cl- ions

6.02  1023

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Mole and Mass

Example:

Relative atomic mass for carbon, C = 12.01 Mass of 1 C atom = 12.01 amu

Mass of 1 mol C atoms = 12.01 g

Mass of 1 mol C atoms consists of 6.02 x 1023 C atoms = 12.01 g  Mass of 1 C atom = = 1.995 x 10-23 g 23 10 x 6.02 g 01 . 12

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12.01 amu = 1.995 x 10-23 g 1 amu = = 1.66 x 10-23 g amu 12.01 g 10 x .995 1 23

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Example:

From the periodic table, Ar of nitrogen, N is The mass of 1 N atom =

The mass of 1 mol of N atoms = The molar mass of N atom =

The molar mass of nitrogen gas =  The nucleon number of N =

14.01 14.01 14.01 amu 14.01 amu 14.01 g 14.01 g 14.01 g mol 14.01 g mol11 28.02 g mol 28.02 g mol11 14 14

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Mr of CH4 is

The mass of 1 CH4 molecule =

The mass of 1 mol of CH4 molecules = The molar mass of CH4 molecule =

16.05 16.05 16.05 amu 16.05 amu 16.05 g 16.05 g 16.05 g mol 16.05 g mol11

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Learning Outcome

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

(a) Interconvert between moles, mass,

number of particles, molar volume of gas at STP and room temperature.

(b) Define the terms empirical & molecular formulae

(c) Determine empirical and molecular formulae from mass composition or from mass composition or

combustion data.

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Example 1:

Calculate the number of moles of molecules for 3.011 x 1023 molecules of oxygen gas.

Solution: 6.02 x 1023 molecules of O 2  3.011 x 1023 molecules of O 2  = 0.5000 mol of O2 molecules 1 mol of O

1 mol of O22 moleculesmolecules

molecules 10 6.02 mol 1 molecules 23   1023 011 . 3

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Example 2:

Calculate the number of moles of atoms for 1.204 x 1023 molecules of nitrogen gas.

Solution: 6.02 x 1023 molecules of N 2   2 mol of N atoms 1.204 x 1023 molecules of N 2  = 0.4000 mol of N atoms 1 mol of N

1 mol of N22 moleculesmolecules

molecules mol 2 molecules 23 23 10 02 . 6 10 204 . 1   

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Example 3:

Calculate the mass of 0.25 mol of chlorine gas.

Solution: 1 mol Cl2

0.25 mol Cl2

18 g

or

mass = mol x molar mass

= 0.25 mol x (2 x 35.45 g mol-1) = 18 g 2 2  35.45 g35.45 g mol 1 mol 0.25 g 35.45 2  

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Example 4:

Calculate the mass of 7.528 x 1023 molecules of

methane, CH4 Solution: 6.02 x 1023 CH 4 molecules  (12.01 + 4(1.01)) g 7.528 x 1023 CH 4 molecules  = 20.06 g 23 23 10 02 . 6 10 7.528 g 05 . 16   

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Molar Volume of Gases

 Avogadro (1811) stated that equal volumes of gases

at the same temperature and pressure contain

equal number of molecules

 Molar volume is a volume occupied by 1 mol of gas  At standard temperature and pressure (STP), the

molar volume of an ideal gas is 22.4 L mol 1 Standard Temperature and Pressure

273.15 K 1 atm 760 mmHg

0 C 101325 N m-2 101325 Pa

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 At room conditions (1 atm, 25 C), the molar

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Example 1:

Calculate the volume occupied by 1.60 mol of

Calculate the volume occupied by 1.60 mol of

Cl Cl22 gas at STP.gas at STP. Solution: At STP, At STP, 1 mol Cl

1 mol Cl22 occupiesoccupies

1.60 mol Cl

1.60 mol Cl22 occupies occupies

= 35.8 L = 35.8 L 22.4 L

mol

1

L

4

.

22

mol

60

.

1

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08/16/11 MATTERMATTER 7070

Example 2:

Calculate the volume occupied by 19.61 g of

Calculate the volume occupied by 19.61 g of

N

N22 at STPat STP

Solution:

1 mol of N

1 mol of N22 occupiesoccupies 22.4 L22.4 L

of N of N22 occupies occupies = 15.7 L = 15.7 L mol 1 L 22.4 mol ) 01 . 14 ( 2 61 . 19        1 mol g 2(14.01) g 61 . 19 

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08/16/11 MATTERMATTER 7171

Example 3:

0.50 mol methane, CH

0.50 mol methane, CH44 gas is kept in a cylinder at gas is kept in a cylinder at STP. Calculate:

STP. Calculate:

(a)

(a) The mass of the gasThe mass of the gas (b)

(b) The volume of the cylinderThe volume of the cylinder (c)

(c) The number of hydrogen atoms in the cylinderThe number of hydrogen atoms in the cylinder

Solution:

(a)

(a) Mass of 1 mol CHMass of 1 mol CH44 ==

Mass of 0.50 mol CH Mass of 0.50 mol CH44 == = 8.0 g = 8.0 g 16.05 g mol 1 mol 0.50 g 05 . 16 

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(b)

(b)At STP;At STP; 1 mol CH1 mol CH44 gas gas occupiesoccupies

0.50 mol CH

0.50 mol CH44 gas gas occupies occupies

= 11 L

= 11 L (c)

(c) 1 mol of CH1 mol of CH44 molecules molecules  4 mol of H atoms4 mol of H atoms

0.50 mol of CH

0.50 mol of CH44 moleculesmolecules  2 mol of H atoms2 mol of H atoms

1 mol of H atoms 1 mol of H atoms  2 mol of H atoms 2 mol of H atoms  1.2 x 101.2 x 102424 atomsatoms 22.4 L 6.02 x 1023 atoms 2 x 6.02 x 1023 atoms mol 1 mol 0.50 L 4 . 22 

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08/16/11 MATTERMATTER 7373

Exercise

A sample of CO2 has a volume of 56 cm3 at STP.

Calculate:

a) The number of moles of gas molecules

(0.0025 mol)

a) The number of CO2 molecules

(1.506 x 1021 molecules)

a) The number of oxygen atoms in the sample

(3.011x1021atoms)

Notes: 1 dm3 = 1000 cm3

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Empirical And Molecular Formulae

- Empirical formula => chemical formula =>

that shows the simplest ratio of all elements in a molecule.

- Molecular formula => formula that show => the actual number of atoms of each

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- The relationship between empirical formula and molecular formula is :

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Example:

A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its

molar mass is 56. Determine the empirical formula and molecular formula of the

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= == = = = = = Mass Number of moles Simplest ratio 85.72 1.984 14.1584 1 7.1357 14.3 01 . 12 7 . 85 01 . 1 3 . 14 C H Solution: Empirical formula = Empirical formula = CH2

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14.03 56 n  4 3.99   8 4 2 H C formula Molecular ) n(CH formula Molecular   

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08/16/11 MATTERMATTER 7979

Exercise:

A combustion of 0.202 g of an organic sample that contains carbon, hydrogen and oxygen

produce 0.361g carbon dioxide and 0.147 g water. If the relative molecular mass of the sample is

148, what is the molecular formula of the sample? Answer : C6H12O4

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08/16/11 MATTERMATTER 8080

At the end of this topic, students should be

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

(a)

(a) Define and perform calculation for each of for each of the

the following concentration measurements :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

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Concentration of Solutions

 A solution is a homogeneous mixture of

two or more substances:

solvent + solute(s)

e.g: sugar + water – solution

sugar – solute water – solvent

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08/16/11 MATTERMATTER 8383

Concentration of a solution can be expressed in various ways : a) molarity b) molality c) mole fraction d) percentage by mass e) percentage by volume

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a) Molarity

 Molarity is the number of moles of

solute in 1 litre of solution

 Units of molarity: mol L-1

mol dm-3 M (L) solution of volume (mol) solute of moles M molarity, 

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Example 1:

Determine the molarity of a solution

containing 29.22 g of sodium chloride, NaCl in a 2.00 L solution.

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Solution: solution NaCl NaCl

V

n

M

L

00

.

2

mol

)

45

.

35

99

.

22

(

22

.

29

= 0.250 mol L-1

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Example 2:

How many grams of calcium chloride, CaCl2 should be used to prepare 250.00 mL

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Solution:

solution CaCl CaCl

M

x V

n

2 2

= 0.500 mol L1  250.00  103 L

mass

molar

x

n

CaCl

of

mass

2 CaCl 2

= (0.500  250.00  103) mol  (40.08 + 2(35.45)) g mol1 = 13.9 g

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b) Molality

 Molality is the number of moles of

solute dissolved in 1 kg of solvent

 Units of molality:mol kg-1

molal m (kg) solvent of mass (mol) solute of moles m molality, 

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Example:

What is the molality of a solution

prepared by dissolving 32.0 g of CaCl2 in 271 g of water?

(91)

Solution: 1 -CaCl

mol

g

2(35.45)

08

.

40

g

0

.

32

n

2

kg 10 271 mol 98 . 110 0 . 32 CaCl of Molality 2 3         1 kg mol 1.06  

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Exercise:

Calculate the molality of a solution

prepared by dissolving 24.52 g of sulphuric acid in 200.00 mL of distilled water.

(Density of water = 1 g mL-1)

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c) Mole Fraction (X)

Mole fraction is the ratio of number of

moles of one component to the total

number of moles of all component present.

For a solution containing A, B and C:

T A C B A A A n n n n n n X of A, fraction Mol    

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08/16/11 MATTERMATTER 9494

 Mol fraction is always smaller than 1

 The total mol fraction in a mixture

(solution) is equal to one.

XA + XB + XC + X….. = 1

 Mole fraction has Mole fraction has no unit (dimensionless)

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Example:

A sample of ethanol, C2H5OH contains

200.0 g of ethanol and 150.0 g of water. Calculate the mole fraction of

(a) ethanol (b) water

(96)

Solution: nethanol = 1 mol g 16.00) 5(1.01) (2(12.01) g 0 . 200    nwater = 1 mol g 16.00) (2(1.01) g 0 . 150   Xethanol =                    mol 02 . 18 0 . 150 mol 07 . 45 0 . 200 mol 07 . 45 0 . 200 = 0.3477

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Xwater = 1  0.3477 = 0.6523

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08/16/11 MATTERMATTER 9898

d) Percentage by Mass (%w/w)

Percentage by mass is defined as the

percentage of the mass of solute per mass of solution.

Note:

Mass of solution = mass of solute + mass of solvent

100

x

solution

of

mass

solute

of

mass

w

w

%

(99)

Example:

A sample of 0.892 g of potassium

chloride, KCl is dissolved in 54.362 g of water. What is the percent by mass of KCl in the solution? Solution: % 100 g 54.362 g 0.892 g 892 . 0 mass %    = 1.61%

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Exercise:

A solution is made by dissolving 4.2 g of sodium chloride, NaCl in 100.00 mL of water. Calculate the mass percent of sodium chloride in the solution.

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08/16/11 MATTERMATTER 101101 e) Percentage by Volume (%V / V)

Percentage by volume is defined as the

percentage of volume of solute in milliliter per volume of solution in milliliter.

Note: solution of volume solution of mass solution of Density  100 x solution of volume solute of volume v v %  x 100 solution of volume solute of volume v v % 

(102)

Example 1:

25 mL of benzene is mixed with 125 mL

of acetone. Calculate the volume percent of benzene solution. Solution: 100% mL 125 mL 25 mL 25 volume %   

=

17%

(103)

Example 2:

A sample of 250.00 mL ethanol is labeled as 35.5% (v/v) ethanol. How many

milliliters of ethanol does the solution contain?

(104)

Solution: 100% mL 00 . 250 % 5 . 35 Vethanol   % 100 V V ethanol of volume % solution ethanol  = 88.8 mL

(105)

Example 3:

A 6.25 m of sodium hydroxide, NaOH

solution has has a density of 1.33 g mL-1 at 20 ºC. Calculate the concentration NaOH in:

(a) molarity

(b) mole fraction

(106)

Solution:

(a) M = solution NaOH V n 6.25 m of NaOH

 there is 6.25 mol of NaOH in 1 kg of water

for a solution consists of 6.25 mol of NaOH and 1 kg of water; Vsolution = solution solution mass 

(107)

masssolution = massNaOH + masswater

massNaOH = nNaOH  molar mass of NaOH

= 6.25 mol  (22.99 + 16.00 + 1.01) g mol1 = 250 g masssolution = 250 g + 1000 g = 1250 g Vsolution = 1 mL g 1.33 g 1250 

(108)

MNaOH =       10 L 33 . 1 1250 mol 25 . 6 3 = 6.65 mol L1

(109)

(b) XNaOH = water NaOH NaOH n n n 

1 kg of water contains 6.25 mol of NaOH

nwater = water of mass molar masswater = 1 mol g 16.00) (2(1.01) g 1000   XNaOH =        mol 02 . 18 1000 mol 25 . 6 mol 25 . 6 = 0.101

(110)

(c) %(w/w) of NaOH = water NaOH NaOH mass mass mass   100% = g 1000 g 250 g 250   100% = 20.0%

(111)

Exercise:

An 8.00%(w/w) aqueous solution of

ammonia has a density of 0.9651 g mL-1. Calculate the (a) molality (b) molarity (c) mole fraction of the NH3 solution Answer: a) 5.10 mol kg-1 b) 4.53 mol L-1 c) 0.0842

(112)

MATTER

(113)

Learning Outcome

At the end of the lesson, students should be able to:

a) Determine the oxidation number of an element in a chemical formula.

b) Write and balance :

i) Chemical equation by inspection method ii) redox equation by ion-electron method

(114)

08/16/11

08/16/11 MATTERMATTER 114114

Balancing Chemical Equation

 A chemical equation shows a chemical

reaction using symbols for the reactants and products.

The formulae of the reactants are written on

the left side of the equation while the

(115)

08/16/11

08/16/11 MATTERMATTER 115115

Example:

x A + y B z C + w D

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08/16/11

08/16/11 MATTERMATTER 116116

 A chemical equation must have an equal number

of atoms of each element on each side of the arrow

 The number x, y, z and w, showing the relative

number of molecules reacting, are called the

stoichiometric coefficients.

 A balanced equation should contain the smallest

possible whole-number coefficients

 The methods to balance an equation:

a) Inspection method

(117)

08/16/11

08/16/11 MATTERMATTER 117117

Inspection Method

1. Write down the unbalanced equation. Write the

correct formulae for the reactants and products.

1. Balance the metallic atom, followed by non-metallic atoms.

1. Balance the hydrogen and oxygen atoms.

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

(118)

08/16/11

08/16/11 MATTERMATTER 118118

Example:

Balance the chemical equation by applying the inspection method.

(119)

08/16/11

08/16/11 MATTERMATTER 119119

Exercise

Balance the chemical equation below by applying inspection method.

1. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O

2. C6H6 + O2 → CO2 + H2O

3. N2H4 + H2O2 → HNO3 + H2O

(120)

08/16/11

08/16/11 MATTERMATTER 120120

Redox Reaction

 Mainly for redox (reduction-oxidation)

(121)

08/16/11

08/16/11 MATTERMATTER 121121

Oxidation is defined as a process of electron loss.  The substance undergoes oxidation

loses one or more electrons.  increase in oxidation number

 act as an reducing agent (electron donor)

Half equation representing oxidation:

Mg  Mg2+ 2e Fe2+  Fe3+ + e 2Cl-  Cl2 + 2e

(122)

Reduction is defined as a process of electron gain.  The substance undergoes reduction

gains one or more electrons.  decrease in oxidation number

 act as an oxidizing agent (electron acceptor)

Half equation representing reduction:

Br2 + 2e → Br

-Sn4+ + 2e → Sn2+ Al3+ + 3e → Al

(123)

08/16/11

08/16/11 MATTERMATTER 123123

Oxidation numbers of any atoms can be determined by applying the following rules:

1. For monoatomic ions,

oxidation number = the charge on the ion

e.g: ion oxidation number

Na+ +1

Cl- -1

Al3+ +3

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08/16/11

08/16/11 MATTERMATTER 124124

2. For free elements, e.g: Na, Fe, O2, Br2, P4, S8 oxidation number on each atom = 0

1. For most cases, oxidation number for O = -2

H = +1 Halogens = -1

(125)

08/16/11

08/16/11 MATTERMATTER 125125

Exception:

1. H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1

1. Halogen bonded to oxygen (e.g: Cl2O7) oxidation number for halogen = +ve

1. In a neutral compound (e.g: H2O, KMnO4) the total of oxidation number of every atoms that made up the molecule = 0

1. In a polyatomic ion (e.g: MnO4-, NO3-) the total oxidation number of every atoms that made up the

(126)

08/16/11

08/16/11 MATTERMATTER 126126

Exercise

1. Assign the oxidation number of Mn in the following

chemical compounds.

i. MnO2 ii. MnO4

-1. Assign the oxidation number of Cl in the following

chemical compounds.

i. KClO3 ii. Cl2O7

2-1. Assign the oxidation number of following:

i. Cr in K2Cr2O7 ii. U in UO22+

(127)

2-08/16/11

08/16/11 MATTERMATTER 127127

Balancing Redox Reaction

Redox reaction may occur in acidic and basic

solutions.

Follow the steps systematically so that

(128)

08/16/11

08/16/11 MATTERMATTER 128128

Balancing Redox Reaction In Acidic

Solution

Fe2+ + MnO4- → Fe3+ + Mn2+

1. Separate the equation into two

half-reactions: reduction reaction and oxidation reaction

i. Fe2+ → Fe3+

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08/16/11

08/16/11 MATTERMATTER 129129

1. Balance atoms other than O and H in each half-reaction separately

i. Fe2+ → Fe3+

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08/16/11

08/16/11 MATTERMATTER

130

130

3. Add H2O to balance the O atoms

Add H+ to balance the H atoms

i. Fe2+ → Fe3+

ii. MnO4- + → Mn2+ +

4. Add electrons to balance the charges i. Fe2+ → Fe3+ + ii. MnO4- + 8H+ + → Mn2+ + 4H 2O 4H2O 8H+ 1 e 5 e

(131)

08/16/11

08/16/11 MATTERMATTER 131131

3.Multiply each half-reaction by an integer, so that number of electron lost in one half-reaction equals the number gained in the other.

i. 5 x (Fe2+ → Fe3+ + 1e)

5Fe2+ → 5Fe3+ + 5e

ii. MnO4- + 8H+ + 5e → Mn2+ + 4H 2O

(132)

1. Add the two half-reactions and simplify where

possible by canceling the species appearing on both sides of the equation.

i. 5Fe2+ → 5Fe3+ + 5e ii. MnO4- + 8H+ + 5e → Mn2+ + 4H 2O ___________________________________ 5Fe2+ + MnO 4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O ___________________________________

(133)

5. Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.

5Fe2+ + MnO

4- + 8H+ 5Fe3+ + Mn2+ + 4H2O

Total charge reactant

= 5(+2) + (-1) + 8(+1) = + 10 - 1 + 8

= +17

Total charge product

= 5(+3) + (+2) + 4(0) = + 15 + (+2)

= +17

5Fe2+ + MnO

4- + 8H+ 5Fe3+ + Mn2+ + 4H2O

Total charge reactant

= 5(+2) + (-1) + 8(+1) = + 10 - 1 + 8

= +17

Total charge product

= 5(+3) + (+2) + 4(0) = + 15 + (+2)

(134)

08/16/11

08/16/11 MATTERMATTER 134134

Exercise: In Acidic Solution

C2O42- + MnO4- + H+ → CO2 + Mn2+ + H2O

(135)

08/16/11

08/16/11 MATTERMATTER 135135

Balancing Redox Reaction In Basic Solution

1. Firstly balance the equation as in acidic solution.

1. Then, add OH- to both sides of the equation so

that it can be combined with H+ to form H

2O.

1. The number of OH- added is equal to the

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08/16/11

08/16/11 MATTERMATTER 136136

Example: In Basic Solution

(137)

08/16/11 08/16/11 MATTERMATTER 137137 Exercise: 1. H2O2 + MnO4- + H+  O2 + Mn2+ + H2O (acidic medium) 2. Zn + SO42- + H2O  Zn2+ + SO2 + 4OH -(basic medium) 3. MnO4- + C2O42- + H+  Mn2+ + CO2 + H2O (acidic medium) 4. Cl2  ClO3- + Cl- (basic medium)

(138)

08/16/11

08/16/11 MATTERMATTER 138138

Stoichiometry

Stoichiometry is the quantitative study of

reactants and products in a chemical reaction.

 A chemical equation can be interpreted in

terms of molecules, moles, mass or even volume.

(139)

C3H8 + 5O2 3CO2 + 4H2O

 1 molecule of C3H8 reacts with 5 molecules of O2

to produce 3 molecules of CO2 and 4 molecules of H2O

 6.02 x 1023 molecules of C3H8 reacts with 5(6.02 x

1023) molecules of O

2 to produce 3(6.02 x 1023) molecules of CO2 and 4(6.02 x 1023) molecules of H2O

(140)

C3H8 + 5O2 3CO2 + 4H2O

 1 mol of C3H8 reacts with 5 moles of O2 to

produce 3 moles of CO2 and 4 moles of H2O

 44.09 g of C3H8 reacts with 160.00 g of O2 to

produce 132.03 g of CO2 and 72.06 g of H2O

 5 moles of C3H8 reacts with 25 moles of O2 to

(141)

 At room condition, 25 ºC and 1 atm pressure;

22.4 dm3 of C

3H8 reacts with 5(22.4 dm3) of O2

to produce 3(22.4 dm3) of CO 2

(142)

Example 1:

How many grams of water are produced in the oxidation of 0.125 mol of glucose?

(143)

Solution:

From the balanced equation;

1 mol C6H12O6 produce 6 mol H2O

0.125 mol C6H12O6 produce H2O mass of H2O = (0.125 x 6) mol x (2.02 + 16.00) g mol-1 = 13.5 g mol 1 mol 6 mol 125 . 0 

(144)

Example 2:

Ethene, C2H4 burns in excess oxygen to form carbon dioxide gas and water vapour.

(a) Write a balance equation of the

reaction

(b) If 20.0 dm3 of carbon dioxide gas is

produced in the reaction at STP, how many grams of ethene are used?

(145)

Solution:

(a) C2H4 + O2 CO2 + H2O

(b) 22.4 dm3 is the volume of 1 mol CO2

20.0 dm3 is the volume of CO2

2 mol CO2 produced by 1 mol C2H4 mol CO2 produced by C2H4 3 3 dm 4 . 22 mol 1 dm 0 . 20  4 . 22 0 . 20 mol 2 mol 1 mol 4 . 22 0 . 20       

(146)

massethane mol x [2(12.01) 4(1.01)] g mol-1 2 4 . 22 0 . 20         = 12.5 g

(147)

Learning Outcome

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

a) Define the limiting reactant and percentage yield

b) Perfome stoichiometric calculations using mole concept including limiting

(148)

08/16/11

08/16/11 MATTERMATTER 148148

Limiting Reactant/Reagent

Limiting reactant is the reactant that is

completely consumed in a reaction and limits the amount of product formed

Excess reactant is the reactant present in

quantity greater than necessary to react with the quantity of limiting reactant

(149)
(150)

Example:

3H2 + N2  2NH3

If 6 moles of hydrogen is mixed with 6 moles of nitrogen, how many moles of ammonia will be produced?

Solution:

3 mol H2 reacts with 1 mol N2 6 mol H2 reacts with

mol 3 mol 1 mol 6  = 2 mol N2

(151)

N2 is the excess reactant H2 is the limiting reactant

 limits the amount of products formed

3 mol H2 produce 2 mol NH3

6 mol H2produce mol 3 mol 2 mol 6  = 4 mol NH3

(152)

or

1 mol N2 react with 3 mol H2

6 mol N2 react with mol NH3

mol 1 mol 3 mol 6  = 18 mol H2 H2 is not enough

 H2 limits the amount of products formed

(153)

3 mol H2 produce 2 mol NH3

6 mol N2 produce mol NH3 = 4 mol NH3 mol 3 mol 2 mol 6 

(154)

08/16/11

08/16/11 MATTERMATTER 154154

Exercise:

Consider the reaction:

2 Al(s) + 3Cl2(g)  2 AlCl3(s)

A mixture of 2.75 moles of Al and 5.00 moles of Cl2 are allowed to react.

(a) What is the limiting reactant?

(b) How many moles of AlCl3 are formed?

(c) How many moles of the reactant remain at the end of the reaction?

(155)

08/16/11

08/16/11 MATTERMATTER 155155

PERCENTAGE YIELD

 The amount of product predicted by a balanced

equation is the theoretical yield

 The theoretical yield is never obtain because:

1. The reaction may undergo side reaction 2. Many reaction are reversible

(156)

08/16/11

08/16/11 MATTERMATTER 156156

4. The product formed may react further to form other product

5. It may be difficult to recover all of the product from the reaction medium

 The amount product actually obtained in a

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08/16/11

08/16/11 MATTERMATTER 157157

Percentage yield is the percent of the actual yield of a product to its theoretical yield

100 x yield yield yield % l theoretica actual

(158)

Example 1:

Benzene, C6H6 and bromine undergo reaction as follows:

C6H6 + Br2  C6H5Br + HBr

In an experiment, 15.0 g of benzene are mixed with excess bromine

(a) Calculate the mass of bromobenzene, C6H5Br that would be produced in the reaction.

(b) What is the percent yield if only 28.5 g of bromobenzene obtain from the experiment?

(159)

CHAPTER 2

(160)
(161)

At the end of this topic students should be able

to:-a) Describe the Bohr’s atomic model.

b) Explain the existence of electron energy levels in an atom.

c) Calculate the energy of electron at its level (orbit) using.

Learning Outcomes

E

n

= −

R

H

1

n

2

, R

H

=

2.18x10

− 18

J

(162)

Learning Outcomes

d) Describe the formation of line spectrum of hydrogen atom.

e) Calculate the energy change of an electron during transition.

f) Calculate the photon of energy emitted by an electron that produces a particular

wavelength during transition.

E= RH 1 n12− 1 n22 where RH= 2.18× 10 −18 J E= h where = c/

(163)

Learning Outcomes

g) Perform calculation involving the Rydberg equation for Lyman, Balmer, Paschen,

Brackett and Pfund series:

g) Calculate the ionisation energy of hydrogen atom from the Lyman series.

1 = RH 1 n12− 1 n22 where RH= 1.097× 10 7 m− 1 n1 n2

(164)

Learning Outcomes

i) State the weaknesses of Bohr’s atomic model.

j) State the dual nature of electron using de Broglie’s postulate and Heisenberg’s

(165)

Bohr’s Atomic Model

In 1913, a young

Dutch physicist, Niels Böhr proposed a

theory of atom that shook the scientific world.

The atomic model he described had

electrons circling a central nucleus that contains positively

(166)

Bohr’s Atomic Model

Böhr also proposed that these orbits can only occur at

specifically

“permitted” levels according to the

energy levels of the electron and explain successfully the lines in the hydrogen

(167)

Bohr’s Atomic Model

First Postulates

Electron moves in circular orbits about the nucleus.

While moving in the orbit, the electron does not radiate or absorb any energy.

(168)

[orbit = energy level=shell]

Orbit is a pathway where the electron is move around the nucleus.

(169)

Bohr’s Atomic Model

Second Postulate

The moving electron has a specific amount of energy; its energy is quantised.

(170)

The energy of an electron in its level is given by:

RH (Rydberg constant) = 2.18 x 10-18J.

n (principal quantum number) = 1, 2, 3 …. ∞ (integer) Note:

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

E

n

= −

R

H

1

n

2

(171)

Bohr’s Atomic Model

Third postulate

At ordinary conditions, the electron is at the ground state (lowest level).

If energy is supplied, electron absorbed the energy and is promoted from a lower energy

level to a higher ones. (Electron is excited)

(172)

Bohr’s Atomic Model

Fourth Postulate

Electron at its excited states is unstable. It will fall back to lower energy level and

released a specific amount of energy in the form of light (photon).

The energy of the photon equals the energy difference between levels.

(173)

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level.

A specific amount of energy is absorbed

∆E = hν = E3-E1

Electron falls from higher to lower energy level .

A photon is emitted.

∆E = hν = E1-E3

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level.

A specific amount of energy is absorbed

∆E = hν = E3-E1

Electron falls from higher to lower energy level .

A photon is emitted.

(174)

Radiant energy emitted when the electron

moves from higher-energy state to lower-energy state is given by:

Where: E= Ef − Ei                 − −         − = ∆ 2 i H 2 f H n 1 R n 1 R E        − = ∆ 2 f 2 i H n 1 n 1 R E Ef= − RH 1 nf2 Ei= − RH 1 ni2 Thus,

(175)

The amount of energy released by the electron is called a photon of energy.

A photon of energy is emitted in the form of radiation with appropriate frequency and

(176)

=

c

v

Where : c (speed of light) = 3.00 x 108 ms-1 Thus :

λ

hc

=

E

Where : h (Planck's constant) = 6.63 x 10-34 J s v = frequency (s-1)

(177)

Rydberg Equation

Wavelength emitted by the transition of electron between two energy levels is calculated using Rydberg equation:

1 = RH 1 n12 − 1 n22 where RH = 1.097× 10 7 m− 1 n1 n2

(178)

Example 1

Calculate the wavelength, in nanometers of the spectrum of hydrogen corresponding to n = 2 and n = 4 in the Rydberg equation.

(179)

Exercises:

1. Calculate the energy of hydrogen electron in the: (a) 1st orbit

(b) 3rd orbit (c) 8th orbit

1. Calculate the energy change (J), that occurs when an electron falls from n = 5 to n = 3 energy level in a

hydrogen atom.

2. Calculate the frequency and wavelength (nm) of the radiation emitted in question 2.

(180)

Emission Spectra

Emission Spectra Continuous Spectra Line Spectra

(181)

Continuous Spectrum

A spectrum consists of radiation distributed over all wavelength without any blank spot.

Example : electromagnetic spectrum, rainbow It is produced by white light (sunlight or

incandescent lamp) that passed through a prism

(182)
(183)
(184)

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous & discrete lines with specific wavelength.

It is composed when the light from a gas

discharge tube containing a particular element is passed through a prism.

(185)

Formation of Atomic / Line

Spectrum

The emitted light (photons) is then separated into its components by a prism.

Each component is focused at a definite

position, according to its wavelength and forms as an image on the photographic plate.

(186)

Formation of Atomic / Line

Spectrum

(187)

Formation of Atomic / Line

Spectrum

Example : The line emission spectrum of hydrogen atom

Line spectrum are composed a few

wavelengths giving a series of discrete line separated by blank areas

It means each line corresponds to a specific wavelength or frequency.

(188)

Formation of Line Spectrum

When electron absorbed radiant energy, they will move from lower energy level to higher

energy level (excited state).

This excited electron is unstable and it will fall back to lower energy level.

During the transition, electron will release energy in the form of light with specific

wavelength and can detected as a line spectrum.

(189)

Differences Between Line &

Continuous Spectra

Continuous Spectrum A spectrum that contains all wavelength without any blank spots.

Example: Rainbow.

Line Spectrum

A spectrum that contain only specific

wavelengths.

A spectrum of discrete lines with certain

wavelengths.

Example: Emission

(190)

Formation of Line Spectrum

(Lyman Series)

n = 1 n = 2 n = 3 n = 4 n = 5 Lyman Series Emission of photon Line spectrum λ E Energy n =

(191)

Formation of Line Spectrum

(Balmer Series)

n = 1 n = 2 n = 3 n = 4 n = 5 n = Lyman Series Emission of photon Line spectrum Balmer Series λ E Energy

(192)
(193)

Example

Series nf ni Spectrum region Lyman 2,3,4,… 2 3,4,5,… Paschen 4,5,6,… Infrared 4 5,6,7,… Infrared 5 6,7,8,… Infrared ultraviolet Visible/uv Balmer Brackett Pfund 1 3

(194)

The following diagram is the line spectrum of hydrogen atom. Line A is the first line of the Lyman series.

Specify the increasing order of the radiant energy, frequency and wavelength of the emitted photon. Which of the line that corresponds to

i) the shortest wavelength? ii) the lowest frequency?

Line spectrum E λ A B C D E v

Example

(195)

Example

The line spectrum of Balmer is given as below:

Describe the transitions of electrons that lead to the lines W, and Y, respectively.

Solution

W Y

For W: transition of electron is from n =4 to n = 2 For Y: electron shifts from n = 7 to n = 2

(196)

Example

Line spectrum A B C D E Paschen series

(a) Which of the line in the Paschen series corresponds to the longest wavelength of photon?

(b) Describe the transition that gives rise to the line.

Solution

Line A.

The electron moves from n=4 to n=3.

(197)

Example

With refer to the second line in the Balmer series of the hydrogen spectrum, Calculate;

a)the wavelength in nm b)the frequency

(198)

Example

Refer to last line of hydrogen spectrum in Lyman series, Calculate:

a) Wavelength b) Frequency

c) Wave number; where wave number =

1

For Lyman series; n1 = 1 & n2 = ∞

Ans:

i. 9.116 x10-8 m

ii. 3.29 x1015 s-1

(199)

Ionization Energy

Defination : Ionization energy is the minimum energy required to remove an electron in its ground state from an atom (or an ion) in

gaseous state.

Figure

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References

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