1.1 Atoms and Molecules-SK016 Lecturer (1)

8/4/2019 1.1 Atoms and Molecules-SK016 Lecturer (1)

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1

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 )



2

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%



3

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

63

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

63

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

1.1 Atoms and Molecules-SK016 Lecturer (1)

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