Chapter 1 - Principle of Chemistry(1)

Centre for Foundation Studies, UTAR

FHSC1114 Physical Chemistry 1

Unit code: FHSC1114

Unit title: PHYSICAL CHEMISTRY

Course of study:Foundation in Science

Credit Hour: 4 credit hours

Contact Hours:2 lecture contact hours per week for the duration of 14 weeks.1 tutorial contact hour per week for theduration of 12 weeks.2 practical contact hours per week for the duration of 10 weeks.

Mode of delivery: Lecture, tutorial & experiment

Attendance for lectures, tutorials and experiments are compulsory for every student.

Main Text: Kotz, J. C., Treichel, P. M., & Townsend, J.

R. (2012). Chemistry & chemical reactivity (8th ed.). Belmont, CA: Thomson Brooks/Cole.

Method of AssessmentNo.No. Method of AssessmentMethod of Assessment TotalTotal

1.1. CourseworkCoursework

a) Test 1 & 2 (15% each) a) Test 1 & 2 (15% each) b) Experiment (20%)b) Experiment (20%)

50%50%

2.2. Final ExaminationFinal Examination 50%50%

Grand totalGrand total 100%100%

FHSC1114PHYSICAL CHEMISTRY

CHAPTER 1Principle of Chemistry

Chapter Scopes Relative Atomic Masses of atoms &

molecules Mass number & atomic number Atomic structure (neutrons, protons &

electrons) Mole concept & conversion Avogadros concept Empirical & molecular formulae Isotopes



Objectives To define relative atomic masses of atoms

& molecules To define & determine mass no. & atomic

no.

To determine no. of neutrons, protons & electrons

To understand mole concept & Avogadros concept

To determine the empirical & molecular formulae

Atomic Composition3 subatomic particles made up all atoms: Electrically positive protons Electrically neutral neutrons Electrically negative electrons

Structure of An Atom Table: Properties & Location of Protons, Neutrons & Electrons In Atom

Subatomic Particle

Symbol Relative electrical charge

Mass (g)

Mass (amu)

Location

Proton p+ + 1 1.6726 x 10-24

1 In the nucleus

Electron e- - 1 9.1094 x 10-28

0.0005 Outside the nucleus

Neutron n0 0 1.6749 x 10-24

1 In the nucleus

1 atomic mass unit (amu) = 1.6605 x 10-24 g

ATOMIC NUMBER (Z) Number of protons in the nucleus of an atom Atomic number (Z) = Number of protons

MASS NUMBER (A) Sum of the number of protons and neutrons

in the nucleus of an atom Mass number (A)

= Number of protons + Number of neutrons= Atomic Number (Z) + Number of neutrons

Mass numberMass numberElement symbolElement symbolAtomic numberAtomic number

A A XX

ZZ

Example :What is the atomic number and the mass number of the element FLUORINE that contains 9 protons and 10 neutrons ? Write the element symbol. Atomic number = Mass number =



Isotopes Atoms of an element with the same atomic

number but different mass numbers (with different number of neutrons)

e.g. Hydrogen-1, Hydrogen-2, Hydrogen-3

H11 H21 H

31

Protium 1 p & 0 n

Deuterium (D) 1 p & 1 n

Tritium (T)1 p & 2 n

radioactive

ExerciseHow many neutrons are in each isotope of oxygen? Write the symbol of each isotope.Oxygen (atomic no.) = 8 a) Oxygen-16 b) Oxygen-17 c) Oxygen-18

Relative Atomic Masses Masses of fundamental atomic particles

often expressed in atomic mass units (amu)

The mass of an atom is measured relative to the mass of an atomic standard, Carbon-12

1 carbon atom has a mass of 12.000 amu Atomic mass of an element the average

relative mass of the isotopes of that element compared to atomic mass of carbon-12 (12 amu)

1 amu 1/12 of the mass of an atom of carbon with 6 protons and 6 neutrons (a carbon-12 atom)

1 amu = 1.66054 10 -24 g Average atomic mass a weighted

average of the masses of all the isotopespresent in the sample

Average atomic mass =[(% abundance isotope 1)/100] (Isotopic mass 1) + [(% abundance isotope 2)/100] (Isotopic mass 2) + .

Example :Isotope Isotope mass (amu) Abundance

(%)63

29Cu 62.9298 69.0965

29Cu 64.9278 30.91

Average atomic mass for Cu=

=

=

Relative Molecular Masses The mass of one MOLECULE of the

substance compared to 1/12 the mass of one ATOM of carbon-12 isotope. Its symbol is Mr.

Mr is calculated by adding together the relative atomic masses of all the atomspresent in the molecular formula of the substance.



Mole is a chemical unit used inquantitative measurement of particles involved in chemical reactions

A mole is the amount of a substance that contains as many elementary entities (atoms, molecules, ions or other particles) as there are atoms in exactly 12 g of the carbon-12 isotope.

Mole Concept Avogadros Number One mole always contains the same

number of particles, no matter what the substance is.

1 mole = 6.0221415 x 1023 particles This value is known as Avogadros

number in honour of Amedeo Avogadro, an Italian lawyer and physicist (1776-1856)

Empirical Formula Simplest formula Gives the smallest whole-number ratio of

atoms present in a compound

Molecular Formula True formula Total number of atoms of each element

present in one molecule of a compound Knowing the relative numbers of atoms of

each element in a molecule

Molecular Formula

Molecular formula = [Empirical formula]nwhere n should be integers (n = 1, 2, 3)

To determine molecular formula from empirical formula, the molar mass must be obtained from experiment

Empirical Formula from % CompositionFor a compound composed of atoms of A & B,

% A

% B

g A

g B

x mol A

x mol B

x mol A

y mol BAxBy

Ratio gives formula

Find mole ratio

Convert weight % to mass (g) Convert mass (g)

to moles (mol)

Example 1:Finding Empirical Formula

25.00g of orange compound, contains 6.64g of potassium, 8.84g of chromium and 9.52g of oxygen

K, Cr, OGiven the molar massK = 39.40 g/mol Cr = 52.00 g/mol O = 16.00 g/mol



Atom K Cr OMass (g) 6.64 8.84 9.52 Amount(mole)

Mol Ratio

Whole-number

Mol RatioEmpirical formula

Example 2 : Finding Empirical formula & Molecular formula

Eugenol is the major component in oil of cloves. It has a molar mass of 164.2 g/mol and is 73.14 % C and 7.37 % H, the remainder is oxygen. What are the empirical and molecular formulas of eugenol? (Given the molar mass of C = 12.011 g/mol; H = 1.008 g/mol; O = 15.999 g/mol)[ Assumption : Mass % mass (g)]The mass of O in a 100.0 g sample :73.14 g C + 7.37 g H + mass of O = 100.00 g Mass of O = 19.49 g O

Atom C H OMass (%) 73.14 7.37 19.49 Mass (g)Amount(mole)

Mol Ratio

Empirical formula

The molecular mass of eugenol= 164.2 g/mol

[C5H6O]n = 164.2 g/mol

n =

Molecular formula ==

Learning Outcomes Determine the empirical and molecular

formulae of compounds and calculate their relative formula/molecular mass

Use atomic number and mass number to differentiate different element and different isotopes of the same element

Chapter 1 - Principle of Chemistry(1)

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