WELCOME TO SKO16 CHEMISTRY

CHEMISTRY SK016 1.0 Matter 7 2.0 Atomic Structure 7 Chapter Topics Hours 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 08/16/11 matter 2

CHEMISTRY SK026 8.0 Thermochemistry 4 9.0 Electrochemistry 6 Chapter Topic Hours 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 08/16/11 matter 3

CHEMISTRY SK026 16.0 Carbonyl 4 17.0 Carboxylic acids & Derivatives 4 Chapter Topic Hour 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 08/16/11 matter 4

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 B-long structured) -100% 08/16/11 matter 5 5

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. James E Brady, John Wiley and Sons. 08/16/11 matter 6 6

ORGANIC CHEMISTRY, 4th Ed – L.G. Wade, Jr, Prentice Hall 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 McMurry, Thompson – Brooks/Cole 08/16/11 matter 7 7

Chapter 1 : MATTER 1.1 Atoms and Molecules 1.2 Mole Concept 1.3 Stoichiometry 08/16/11 matter 8 8

1.1 Atoms and Molecules 08/16/11 matter 9 9

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. (c) Write isotope notation. 08/16/11 matter 10 10

Introduction Matter  Anything that occupies space and has mass. e.g: air, water, animals, trees, atoms, etc Matter may consists of atoms, molecules or ions. 08/16/11 matter 11 11

Classifying Matter 08/16/11 matter 12

Example : air, milk, cement 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 08/16/11 matter 13

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. Example : CO2, H2O, CuO

Three States of Matter SOLID LIQUID GAS 08/16/11 matter 15 15

1.1 Atoms and Molecules 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) 08/16/11 matter 16 16

Modern Model of the Atom Electrons move around the region of the atom. 08/16/11 matter 17 17

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.

Subatomic Particles Electron (e) 9.1 x 10-28 -1.6 x 10-19 -1 Mass (gram) Charge (Coulomb) (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) 08/16/11 matter 19 19

Isotope Isotopes are two or more atoms of the same element that have the same proton number in their nucleus but different nucleon number. 08/16/11 matter 20 20

Examples:

Isotope Notation An atom can be represented by an isotope notation ( atomic symbol ) X = Element symbol Z = Proton number of X (p) A = Nucleon number of X = p + n 08/16/11 matter 22 22

Nucleon number of mercury, A = 202 Total charge on the ion The number of neutrons = A – Z = 202 – 80 = 122 Proton number of mercury, Z = 80 08/16/11 matter 23 23

In a neutral atom:  number of protons equals number of electrons In a positive ion:  number of protons is more than number of In a negative ion:  number of protons is less than number of

Exercise 1 Give the number of protons, neutrons, electrons and charge in each of the following species: Symbol Number of : Charge Proton Neutron Electron 08/16/11 matter 25 25

Exercise 2 Write the appropriate notation for each of the following nuclide : Species Number of : Notation for nuclide Proton Neutron Electron A 2 B 1 C D 7 10 08/16/11 matter 26 26

b) Molecules A molecule consists of a small number of atoms joined together by bonds. 08/16/11 matter 27 27

Contains only two atoms Ex : H2, N2, O2, Br2, HCl, CO A diatomic molecule Contains only two atoms Ex : H2, N2, O2, Br2, HCl, CO A polyatomic molecule Contains more than two atoms Ex : O3, H2O, NH3, CH4 08/16/11 matter 28 28

Learning Outcomes At the end of this topic, student should be able (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. 08/16/11 matter 29 29

Relative Mass 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 08/16/11 matter 30 30

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 atom, scaled with 12C. 08/16/11 31

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 ANSWER: 08/16/11 matter 32 32

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 08/16/11 matter 33 33

The relative molecular mass of a compound is the summation of the relative atomic masses of all atoms in a molecular formula.

Example 2 Calculate the relative molecular mass of C5H5N, Ar C = 12.01 Ar H = 1.01 Ar N = 14.01 08/16/11 matter 35 35

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 quantity of atoms and molecules 08/16/11 36 36

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Mass Spectrum of Monoatomic Elements Modern mass spectrum converts the abundance into percent abundance 08/16/11 38

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 63 8.1 9.1 24 25 26 m/e (amu) Relative abundance 08/16/11 matter 39 39

Learning Outcomes At the end of this topic, student should be able (a) Calculate the average atomic mass of an element given the relative abundances of isotopes or a mass spectrum. 08/16/11 matter 40 40

How to calculate the relative atomic mass, Ar from mass spectrum? 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. 08/16/11 matter 41 41

Calculate the relative atomic mass of neon from the mass spectrum. Example 1: Calculate the relative atomic mass of neon from the mass spectrum. 08/16/11 42

Solution: Average atomic = mass of Ne = = 20.2 u Relative atomic mass Ne = 20.2

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 relative atomic mass of copper. 08/16/11 44

Solution: Average atomic = mass of Cu = = 63.55 u Relative atomic mass Cu = 63.55

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 08/16/11 46

Relative atomic mass Ir = 192.254 Solution: Average atomic = mass of Ir = = 192.254 u Relative atomic mass Ir = 192.254 08/16/11 47

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?

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

Mass Spectrum of Diatomic Elements

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

MATTER 1.2 Mole Concept

Learning Outcome At the end of this topic, students should be able to: a) Define mole in terms of mass of carbon-12 and Avogadro’s constant, NA

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 Avogadro’s number, NA = 6.02 x 1023

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  1023 6.02  1023 2  6.02  1023 6.02  1023 6.02  1023 3  6.02  1023

1 mol of CuCl2 contains Cu2+ ions Cl- ions 6.02  1023 2  6.02  1023

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

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

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 amu 14.01 g 14.01 g mol1 28.02 g mol1 14

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 amu 16.05 g 16.05 g mol1

Learning Outcome At the end of this topic, students should be able to: Interconvert between moles, mass, number of particles, molar volume of gas at STP and room temperature. Define the terms empirical & molecular formulae Determine empirical and molecular formulae from mass composition or combustion data. 08/16/11 MATTER 61 61

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 O2  3.011 x 1023 molecules of O2  = 0.5000 mol of O2 molecules 1 mol of O2 molecules

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 N2   2 mol of N atoms 1.204 x 1023 molecules of N2  = 0.4000 mol of N atoms 1 mol of N2 molecules

Calculate the mass of 0.25 mol of chlorine gas. 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  35.45 g

Calculate the mass of 7.528 x 1023 molecules of methane, CH4 Example 4: Calculate the mass of 7.528 x 1023 molecules of methane, CH4 Solution: 6.02 x 1023 CH4 molecules  (12.01 + 4(1.01)) g 7.528 x 1023 CH4 molecules  = 20.06 g

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

Standard Molar Volume 08/16/11 MATTER 67

At room conditions (1 atm, 25 C), the molar volume of a gas = 24 L mol-1

Example 1: Calculate the volume occupied by 1.60 mol of Cl2 gas at STP. Solution: At STP, 1 mol Cl2 occupies 1.60 mol Cl2 occupies = 35.8 L 22.4 L 08/16/11 MATTER 69

Example 2: Calculate the volume occupied by 19.61 g of N2 at STP Solution: 1 mol of N2 occupies 22.4 L of N2 occupies = 15.7 L 08/16/11 MATTER 70

Example 3: 0.50 mol methane, CH4 gas is kept in a cylinder at STP. Calculate: (a) The mass of the gas (b) The volume of the cylinder (c) The number of hydrogen atoms in the cylinder Solution: (a) Mass of 1 mol CH4 = Mass of 0.50 mol CH4 = = 8.0 g 16.05 g 08/16/11 MATTER 71

(b) At STP; 1 mol CH4 gas occupies 0.50 mol CH4 gas occupies = 11 L (c) 1 mol of CH4 molecules  4 mol of H atoms 0.50 mol of CH4 molecules  2 mol of H atoms 1 mol of H atoms  2 mol of H atoms   1.2 x 1024 atoms 22.4 L 6.02 x 1023 atoms 2 x 6.02 x 1023 atoms 08/16/11 MATTER 72

Exercise A sample of CO2 has a volume of 56 cm3 at STP. Calculate: The number of moles of gas molecules (0.0025 mol) The number of CO2 molecules (1.506 x 1021 molecules) The number of oxygen atoms in the sample (3.011x1021atoms) Notes: 1 dm3 = 1000 cm3 1 dm3 = 1 L 08/16/11 MATTER 73 73

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 element in a molecule. 08/16/11 MATTER 74 74

The relationship between empirical formula and molecular formula is : Molecular formula = n ( empirical formula ) 08/16/11 MATTER 75 75

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 compound. 08/16/11 MATTER 76 76

Solution: 77

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 08/16/11 MATTER 79 79

Learning Outcome At the end of this topic, students should be able to: (a) Define and perform calculation for each of the 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 08/16/11 MATTER 80 80

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 08/16/11 MATTER 81 81

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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 08/16/11 MATTER 83 83

Molarity is the number of moles of solute in 1 litre of solution 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 08/16/11 MATTER 84 84

Example 1: Determine the molarity of a solution containing 29.22 g of sodium chloride, NaCl in a 2.00 L solution.

Solution: = 0.250 mol L-1

Example 2: How many grams of calcium chloride, CaCl2 should be used to prepare 250.00 mL solution with a concentration of 0.500 M

Solution:

b) Molality Molality is the number of moles of solute dissolved in 1 kg of solvent Units of molality: mol kg-1 molal m 08/16/11 MATTER 89 89

Example: What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?

Solution:

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) Ans = 1.250 mol kg-1

c) Mole Fraction (X) For a solution containing A, B and C: 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: 08/16/11 MATTER 93 93

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 no unit (dimensionless) since it is a ratio of two similar quantities. 08/16/11 MATTER 94 94

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 in the solution.

Solution:

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 08/16/11 MATTER 98 98

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:

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. Answer = 4.0%

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. 08/16/11 MATTER 101 101

Example 1: 25 mL of benzene is mixed with 125 mL of acetone. Calculate the volume percent of benzene solution. Solution:

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?

Solution:

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 (c) percent by mass

Solution:

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

MATTER 1.3 Stoichiometry

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

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 products are on the right. 08/16/11 MATTER 114 114

Example: x A + y B z C + w D Reactants Products 115 08/16/11 MATTER

The methods to balance an equation: a) Inspection method 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 b) Ion-electron method 08/16/11 MATTER 116 116

Inspection Method Write down the unbalanced equation. Write the correct formulae for the reactants and products. Balance the metallic atom, followed by non- metallic atoms. Balance the hydrogen and oxygen atoms. Check to ensure that the total number of atoms of each element is the same on both sides of equation. 08/16/11 MATTER 117 117

Balance the chemical equation by applying the inspection method. Example: Balance the chemical equation by applying the inspection method. NH3 + CuO → Cu + N2 + H2O 08/16/11 MATTER 118 118

Exercise Balance the chemical equation below by applying inspection method. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O C6H6 + O2 → CO2 + H2O N2H4 + H2O2 → HNO3 + H2O ClO2 + H2O → HClO3 + HCl 08/16/11 MATTER 119 119

Redox Reaction Mainly for redox (reduction-oxidation) reaction 120 08/16/11 MATTER 120 120

Half equation representing oxidation: 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 08/16/11 MATTER 121 121

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

oxidation number = the charge on the ion e.g: ion oxidation number Oxidation numbers of any atoms can be determined by applying the following rules: For monoatomic ions, oxidation number = the charge on the ion e.g: ion oxidation number Na+ +1 Cl- -1 Al3+ +3 S2- -2 08/16/11 MATTER 123 123

2. For free elements, e.g: Na, Fe, O2, Br2, P4, S8 oxidation number on each atom = 0 For most cases, oxidation number for O = -2 H = +1 Halogens = -1 08/16/11 MATTER 124 124

H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1 Exception: H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1 Halogen bonded to oxygen (e.g: Cl2O7) oxidation number for halogen = +ve In a neutral compound (e.g: H2O, KMnO4) the total of oxidation number of every atoms that made up the molecule = 0 In a polyatomic ion (e.g: MnO4-, NO3-) the total oxidation number of every atoms that made up the molecule = net charge on the ion 08/16/11 MATTER 125 125

Exercise Assign the oxidation number of Mn in the following chemical compounds. i. MnO2 ii. MnO4- Assign the oxidation number of Cl in the following chemical compounds. i. KClO3 ii. Cl2O72- Assign the oxidation number of following: i. Cr in K2Cr2O7 ii. U in UO22+ iii. C in C2O42- 08/16/11 MATTER 126 126

Balancing Redox Reaction Redox reaction may occur in acidic and basic solutions. Follow the steps systematically so that equations become easier to balance. 08/16/11 MATTER 127 127

Balancing Redox Reaction In Acidic Solution Fe2+ + MnO4- → Fe3+ + Mn2+ Separate the equation into two half- reactions: reduction reaction and oxidation reaction i. Fe2+ → Fe3+ ii. MnO4- → Mn2+ 08/16/11 MATTER 128 128

Balance atoms other than O and H in each half-reaction separately i. Fe2+ → Fe3+ ii. MnO4- → Mn2+ 08/16/11 MATTER 129 129

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+ + 4H2O 8H+ 4H2O 1 e 5 e 130 08/16/11 MATTER 130

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+ + 4H2O 08/16/11 MATTER 131 131

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+ + 4H2O ___________________________________ 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O ___________________________________

Exercise: In Acidic Solution C2O42- + MnO4- + H+ → CO2 + Mn2+ + H2O 08/16/11 MATTER 134 134

Balancing Redox Reaction In Basic Solution Firstly balance the equation as in acidic solution. Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O. The number of OH- added is equal to the number of H+ in the equation. 08/16/11 MATTER 135 135

Example: In Basic Solution Cr(OH)3 + IO3- + OH- → CrO32- + I- + H2O 08/16/11 MATTER 136 136

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 4. Cl2  ClO3- + Cl- (basic medium) 08/16/11 MATTER 137 137

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. 08/16/11 MATTER 138 138

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 O2 to produce 3(6.02 x 1023) molecules of CO2 and 4(6.02 x 1023) molecules of H2O

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 produce 15 moles of CO2 and 20 moles of H2O

At room condition, 25 ºC and 1 atm pressure; 22.4 dm3 of C3H8 reacts with 5(22.4 dm3) of O2 to produce 3(22.4 dm3) of CO2

Example 1: How many grams of water are produced in the oxidation of 0.125 mol of glucose? C6H12O6(s) + O2(g)  CO2(g) + H2O(l)

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

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?

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

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 reactant and percentage yield.

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 08/16/11 MATTER 148 148

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

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 H2 produce

or 1 mol N2 react with 3 mol H2 6 mol N2 react with mol NH3

3 mol H2 produce 2 mol NH3 6 mol N2 produce mol NH3 = 4 mol NH3

Consider the reaction: 2 Al(s) + 3Cl2(g)  2 AlCl3(s) 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? 08/16/11 MATTER 154 154

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 3. There may be impurities in the reactants 08/16/11 MATTER 155 155

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 reaction is the actual yield 08/16/11 MATTER 156 156

Percentage yield is the percent of the actual yield of a product to its theoretical yield 08/16/11 MATTER 157 157

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?