WELCOME TO SKO16 CHEMISTRYCHEMISTRY CHEMISTRY CHEMISTRY SK016 Chapter Topics Hours 1.0 2.0 3.0 4.0 5.0 6.0 7.0 Matter Atomic Structure Periodic Table Chemical Bonding State of Matter Chemical Equilibrium Ionic Equilibria Total matter 7 7 4 2 7 5 12 54 2 08/16/11 CHEMISTRY SK026 Chapter Topic Hours 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 08/16/11 Thermochemistry Electrochemistry Reaction Kinetics Intro To Organic Chemistry Hydrocarbons Aromatic Compounds Haloalkanes (Alkyl halides) Hydroxy compounds matter 4 6 7 4 8 3 4 3 3 0 17.0 Carbonyl Carboxylic acids & Derivatives Amines Amino acids and Proteins Polymers 4 4 5 2 1 08/16/11 matter 4 .CHEMISTRY SK026 Chapter Topic Hour 16.0 20.0 18.0 19. FINAL EXAMINATION (70%) Paper 1 (30 multiple choice questions) . MID-SEMESTER EXAMINATION .30% Paper 2 (Part A-structured) (Part B-long structured) -100% 08/16/11 matter 5 .10% 2.10% 3.10% Practical work .ASSESSMENT 1. COURSEWORK (20%) Continuous evaluation (tutorial/test/quiz) . McGraw Hill CHEMISTRY – The Central Science.REFERENCE BOOKS CHEMISTRY . 6th Ed. 9th Ed. Theodore L. Pearson Education GENERAL CHEMISTRY – Principle & Structure. John Wiley and Sons. Bruce E Bursten.– Martin Silberberg. McGraw-Hill CHEMISTRY –The Molecular Nature of Matter and Change. James E Brady.Eugene LeMay. 08/16/11 matter 6 .Jr. H.9th Ed. 3rd Ed. – Raymond Chang.Brown. GENERAL CHEMISTRY – Principle and Modern Applications. 6th Ed – John McMurry.G. Prentice Hall ORGANIC CHEMISTRY. Jr.Fryhle. 7th Ed – T. Wade.W. Harwood.Graham Solomon. William S. Prentice-Hall ORGANIC CHEMISTRY. Thompson – Brooks/Cole 08/16/11 matter 7 . John Wiley and Sons ORGANIC CHEMISTRY. 4th Ed – L. Petrucci. Ralph H.Craig B. 8th Ed. Chapter 1 : MATTER 1.1 Atoms and Molecules 1.2 Mole Concept 1.3 Stoichiometry 08/16/11 matter 8 . 1 Atoms and Molecules 08/16/11 matter 9 .1. students should be able to: (a) Describe proton.Learning Outcome At the end of this topic. (b) Define proton number. A and isotope. (c) Write isotope notation. nucleon number. electron and neutron in terms of the relative mass and relative charge. 08/16/11 matter 10 . Z. g: air.Introduction Matter Anything that occupies space and has mass. water. etc Matter may consists of atoms. atoms. molecules or ions. trees. animals. e. 08/16/11 matter 11 . Classifying Matter 08/16/11 matter 12 . Example : air. A substance is a form of matter that has a definite or constant composition and distinct properties. NH3. milk. O2 A mixture is a combination of two or more substances in which the substances retain their identity. cement 08/16/11 matter 13 . Example: H2O. Example : CO2. Al. Example : Na. K. H2O.Fe A compound is a substance composed of atoms of two or more elements chemically united in fixed proportion.An element is a substance that cannot be separated into simpler substances by chemical means. CuO . Three States of Matter SOLID 08/16/11 LIQUID matter GAS 15 . 1 Atoms and Molecules a) Atoms An atom is the smallest unit of a chemical element/compound. In an atom.Neutron (n) .Electron (e) 08/16/11 matter 16 .1. there are 3 subatomic particles: .Proton (p) . Modern Model of the Atom Electrons move around the region of the atom. 08/16/11 matter 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 Particle Mass (gram) Charge (Coulomb) Charge (units) Electron (e) Proton (p) Neutron (n) 9.1 x 10-28 1.67 x 10-24 1.67 x 10-24 -1.6 x 10-19 +1.6 x 10-19 0 -1 +1 0 08/16/11 matter 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 Examples: 1 1 H 2 1 H(D) 3 1 H(T) 235 92 U 238 92 U . Isotope Notation An atom can be represented by an isotope notation ( atomic symbol ) X= Z = A= = 08/16/11 matter Element symbol Proton number of X (p) Nucleon number of X p+n 22 . A = 202 Total charge on the ion Proton number of mercury.Nucleon number of mercury. Z = 80 The number of neutrons =A–Z = 202 – 80 = 122 matter 23 08/16/11 . In a neutral atom: number of protons equals number of electrons a positive ion: number of protons is more than number of electrons In In a negative ion: number of protons is less than number of electrons . Exercise 1 Give the number of protons. neutrons. electrons and charge in each of the following species: Symbol Proton 200 80 63 29 Number of : Neutron Electron Charge Hg Cu 17 8 59 27 08/16/11 O2− Co 3+ matter 25 . Exercise 2 Write the appropriate notation for each of the following nuclide : Species Number of : Notation for Proton Neutron Electron nuclide 2 1 1 7 2 2 1 7 matter A B C D 08/16/11 2 0 1 10 26 . 08/16/11 matter 27 .b) Molecules A molecule consists of a small number of atoms joined together by bonds. N2. Br2. CO A polyatomic molecule Contains more than two atoms Ex : O3.A diatomic molecule Contains only two atoms Ex : H2. HCl. CH4 08/16/11 matter 28 . NH3. H2O. O2. Ar and relative molecular mass. Mr based on the C-12 scale. student should be able to : (a) Define relative atomic mass. (b) Calculate the average atomic mass of an element given the relative abundance of isotopes or a mass spectrum.Learning Outcomes At the end of this topic. 08/16/11 matter 29 . Relative Atomic Mass. A r = 1 X Mass of one atom of 12 C 12 08/16/11 matter 30 .Relative Mass i. 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 Mass of one atom of element Re lative atomic mass. Atomic mass unit. amu (or u). 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. Mass of an atom is often expressed in atomic mass unit. scaled with 12C. 08/16/11 31 . 45 ANSWER: 08/16/11 matter 32 .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. Mr Mass of one molecule x Mass of one atom of 12 C 08/16/11 matter 33 .000amu Relative = 1 12 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.ii) Relative Molecular Mass. The relative molecular mass of a compound is the summation of the relative atomic masses of all atoms in a molecular formula. . 01 08/16/11 matter 35 .01 Ar N = 14.01 Ar H = 1. Ar C = 12.Example 2 Calculate the relative molecular mass of C5H5N. 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 . 08/16/11 matter 37 . Mass Spectrum of Monoatomic Elements Modern mass spectrum converts the abundance into percent abundance 08/16/11 38 . 25Mg and 26 Mg. The height of each line is proportional to the abundance of each isotope.Mass Spectrum of Magnesium Relative abundance 63 8. m/e 24 (amu) Mg is the most abundant of the 3 isotopes 08/16/11 matter 39 .1 24 25 26 The mass spectrum of Mg shows that Mg consists of 3 isotopes: 24Mg.1 9. student should be able to : (a) Calculate the average atomic mass of an element given the relative abundances of isotopes or a mass spectrum. 08/16/11 matter 40 .Learning Outcomes At the end of this topic. Ar of the atom. Average atomic mass = 08/16/11 ∑ (isotopic mass × abundance ) ∑ abundance matter 41 .How to calculate the relative atomic mass. The average of atomic masses of the entire element’s isotope as found in a particular environment is the relative atomic mass. Ar from mass spectrum? Ar is calculated using data from the mass spectrum. 08/16/11 42 .Example 1: Calculate the relative atomic mass of neon from the mass spectrum. 05 × 2 0 ) + ( 0 . 3 × 2 u1 ) + ( 9 . 3 + 9 . 2 ) 2) u = 20.2 .05 + 0 .2 u ∴ Relative atomic mass Ne = 20. 2 × 2 u ( 9 .Solution: Average atomic mass of Ne = ∑ (% abundance × isotopic mass) = ∑% abundanc e ( 9 . Example 2: Copper occurs naturally as mixture of 69. 08/16/11 44 . The isotopic masses of 63Cu and 65Cu are 62. Calculate the relative atomic mass of copper.93 u and 64.91% of 65Cu.09% of 63Cu and 30.93 u respectively. Solution: Average atomic mass of Cu ( 6 .90 ∑ (% abundance× isotopic mass) = ∑ % abundance × 6 .29 3 ) + ( 3 .09 ×1 6 .49 9 u ( 6 .90 9+ 3 .09 ) 1 3) u = = 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 193 Ir are 191.021 u and 193.025 u respectively. Calculate the relative atomic mass of Iridium 08/16/11 46 Solution: Average atomic = ∑ ( abundance × isotopic mass ) ∑ abundance mass of Ir = = (5 × 191 .021 u) + (8 × 193 .025 u) (5 + 8 ) 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? . Cl-35Cl 35 Cl-37Cl 37 Cl-37Cl 35 MASS SPECTROMETER + _ _ Cl2 Cl2 + e Cl2+ + 2e → Cl2 + e → 2Cl+ + 2e Cl-35Cl+ 35 Cl-37Cl+ 37 Cl-37Cl+ 35 Cl+ 37 Cl+ 35 . Mass Spectrum of Diatomic Elements . Exercise: How many peaks would be expected in a mass spectrum of X2 which consists of 3 isotopes? . 2 Mole Concept .MATTER 1. students should be able to: a) Define mole in terms of mass of carbon-12 and Avogadro’s constant.Learning Outcome At the end of this topic. NA . NA = 6.02 x 1023 . 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.Avogadro’s Number.00 g of carbon-12 isotope The number of atoms in 12 g of 12C is called Avogadro’s number. Example: 1 mol of Cu contains 6.02 × 1023 H atoms .02 × 1023 Cu atoms 1 mol of O2 contains 6.02 × 1023 N atoms 3 × 6.02 × 1023 NH3 molecules 6.02 × 1023 O2 molecules 2 × 6.02 × 1023 O atoms 1 mol of NH3 contains 6. 02 × 1023 2+ ions Cu 2 × 6.02 × 1023 Cl.1 mol of CuCl2 contains 6.ions . 01 Mass of 1 C atom = 12. C = 12.Mole and Mass Example: Relative atomic mass for carbon. 20 g1 ∴ Mass of 1 C atom = 6 .01 g Mass of 1 mol C atoms consists of 6. 0 x1 2 20 3 = 1.01 g 1 .995 x 10-23 g .01 amu Mass of 1 mol C atoms = 12.02 x 1023 C atoms = 12. 12.66 x 10-23 g . a 0 m1 u = 1.01 amu = 1. 9 9 x 1 5 20 3 g 1 2 .995 x 10-23 g 1 amu = 1 . 02 g mol−1 = The nucleon number of N = 14 . N is 14.01 amu = 14.01 g mol−1 The molar mass of N atom = The molar mass of nitrogen gas 28.01 g The mass of 1 mol of N atoms = 14.Example: From the periodic table. Ar of nitrogen.01 The mass of 1 N atom 14. 05 The mass of 1 CH4 molecule = 16.Mr of CH4 is 16.05 amu The mass of 1 mol of CH4 molecules = 16.05 g mol−1 .05 g The molar mass of CH4 molecule = 16. molar volume of gas at STP and room temperature. 08/16/11 MATTER 61 . mass.Learning Outcome At the end of this topic. students should be able to: (a) Interconvert between moles. (b) Define the terms empirical & molecular formulae (c) Determine empirical and molecular formulae from mass composition or combustion data. number of particles. 011 x 1023 molecules of O2 ≡ 1 × 1 1 2 0 3 m o l e ×c 1 m l oe u 6 .02 x 1023 molecules of O2 ≡ 3 .011 x 1023 molecules of oxygen gas.Example 1: Calculate the number of moles of molecules for 3. 0× 1 2 20 3 m o l e c u = 0.0 1 mol of O2 molecules 3.5000 mol of O2 molecules . Solution: 6. Example 2: Calculate the number of moles of atoms for x 1023 molecules of nitrogen gas. Solution: 6.204 x 1023 molecules of N2 ≡ = 0. 0 ×2 1 2 0 3 m o l e c u l e 1.4000 mol of N atoms .2 0 × 1 4 2 0 3 m o l e ×c 2 um l eo sl 6 .204 ≡ 1 mol of N2 molecules 2 mol of N atoms 1 .02 x 1023 molecules of N2 ≡ 1. Solution: 1 mol Cl2 ≡ 0. g 4 × 05 .25 mol of chlorine gas.45 g ≡ 2 × 3 5 . 2m 5 o l 1 m o l 18 g or mass = mol x molar mass = 0.25 mol Cl2 ≡ 2 × 35.25 mol x (2 x 35.Example 3: Calculate the mass of 0.45 g mol-1) = 18 g . 528 x 1023 molecules of methane.6 0 5g × 7 .06 g . 0 2× 1 20 3 = 20.528 x 1023 CH4 molecules ≡ 6 .Example 4: Calculate the mass of 7.02 x 1023 CH4 molecules ≡ (12.01)) g 1 . CH4 Solution: 6. 5 × 21 820 3 7.01 + 4(1. 15 K 0 °C .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 mol1 Standard Temperature and Pressure 1 atm 101325 N m-2 101325 Pa 760 mmHg 273. Standard Molar Volume 08/16/11 MATTER 67 . At room conditions (1 atm. the molar volume of a gas = 24 L mol-1 . 25 °C). 6 0m × 2 l .60 mol Cl2 occupies 1 .Example 1: Calculate the volume occupied by 1. 1 mol Cl2 22.8 L 69 08/16/11 MATTER .4 L occupies 1. Solution: At STP.24 L o 1m o l = 35.60 mol of Cl2 gas at STP. 9 6 1 m 2 ( 1 . o o l = 15.7 L 08/16/11 MATTER 70 .Example 2: Calculate the volume occupied by 19.4 L 1 .61 g of N2 at STP Solution: 1 mol of N2 occupies 22.4 0 ) 1 of N2 occupies 1m × 2 l 2 L. m 0 1−o1 ) l 1 .9 6 g1 2 ( 1 4 g . 5m 0 o l Mass of 0.50 mol CH4 = 1m o l 08/16/11 MATTER = 8. 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: 16.Example 3: 0.05 g (a) Mass of 1 mol CH4 = 1 .0 g 71 .50 mol methane.6 0 5g × 0 . 5m 0 o l 0.4 L 2 .2 4 L × 0 . 1 mol CH4 gas occupies 22.50 mol CH4 gas occupies 1m o l = 11 L (c) 1 mol of CH4 molecules ≡ 4 mol of H atoms 0.02 x 1023 atoms 1.02 x 1023 atoms 2 x 6.2 x 1024 atoms 72 .(b) At STP.50 mol of CH4 molecules ≡ 2 mol of H atoms 1 mol of H atoms ≡ 2 mol of H atoms ≡ ≡ 08/16/11 MATTER 6. 011x1021atoms) Notes: 08/16/11 1 dm3 1 dm3 = 1000 cm3 =1L MATTER 73 .506 x 1021 molecules) a) The number of oxygen atoms in the sample (3.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. 08/16/11 MATTER 74 . .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. The relationship between empirical formula and molecular formula is : Molecular formula = n ( empirical formula ) 08/16/11 MATTER 75 .. 08/16/11 MATTER 76 . Its molar mass is 56.3% hydrogen by mass.Example: A sample of hydrocarbon contains 85. Determine the empirical formula and molecular formula of the compound.7% carbon and 14. 9 8 4 ≈ 2 u = 7 .43 1 .57 b e r o l e s 1 . 7 H 1 4 .Solution: C M a s s 8 5 . 1 5 8 4 s t 1r a m t i o p i r i c a l C f 2H r m o . 3 N u m o f m S i m p 8 .0 1 1 . 1 3 5 = 71 4 .20 1 l e E 1 . n = 5 1 4 6 . 9 ≈ 4 M ∴M o o l e f c o u r lm a = rnu (l aC 2 ) H l e f c o u r lm a = rC 4 Hl a8 u . 0 9 3 = 3 . Exercise: A combustion of 0. what is the molecular formula of the sample? Answer : C6H12O4 08/16/11 MATTER 79 .147 g water.202 g of an organic sample that contains carbon. hydrogen and oxygen produce 0.361g carbon dioxide and 0. If the relative molecular mass of the sample is 148. % w/w v) percentage by volume. 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.Learning Outcome At the end of this topic. X iv) percentage by mass. %v/v 08/16/11 MATTER 80 . Concentration of Solutions A solution is a homogeneous mixture of two or more substances: solvent + solute(s) e.g: sugar + water – sugar – water – 08/16/11 MATTER solution solute solvent 81 . 08/16/11 MATTER 82 . 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 . a) Molarity Molarity is the number of moles of solute in 1 litre of solution m o m o lo e f s s o l u( mt e o l ) l a Mr = t y . i v o l u o m f s e o l u ( t Li o ) n Units of molarity: mol L-1 mol dm-3 M 08/16/11 MATTER 84 . NaCl in a 2. .00 L solution.22 g of sodium chloride.Example 1: Determine the molarity of a solution containing 29. 92 2 m ( 2 .54 )5 2 .0 0 L 0.29 9+ 3 .250 mol L-1 o l .Solution: M N a C =l nN V s o a lu C l t io n = = 2 . 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 . 9 g . 0 8 + 2 ( 31 5 . 4 = 1 3 . 10 −003 ) m o l − × ( 4 0 .Solution: n C a 2C =l M = C a 2C l x s o l u t i o n V 0 . l 10 L−003 L m a o s Cs a 2 C= ln f x m o m a ar s s l = ( 0 . 5 C 0 − 1 0× 2 m5 a 2C l 0o × . 5 × 02 05 0 ×. b) Molality Molality is the number of moles of solute dissolved in 1 kg of solvent m o m o lo e f s s o l am = i t y . l m a os fs s o l u( mt e o l ) l v ( e k ng t ) Units of molality: molal m 08/16/11 MATTER mol kg-1 89 . 0 g of CaCl2 in 271 g of water? .Example: What is the molality of a solution prepared by dissolving 32. 2 0 m 8 o l a o l Ci t a y 2 C= l 1 1 . 0m 6 ok −g 1 l . 9 0 − 3 f 2 7× 1 0 k = 1 . m 4 5o ) l o l g M 3 .0 0 8+ 2 ( 3 5 g .Solution: nC a 3 .2 0 g C= l -1 2 4 . (Density of water = 1 g mL-1) Ans = 1.00 mL of distilled water.250 mol kg-1 .Exercise: Calculate the molality of a solution prepared by dissolving 24.52 g of sulphuric acid in 200. B and C: M of rl a c ot if o AnA =. X nA nA = nT nA + nB + nC 08/16/11 MATTER 93 . For a solution containing A.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. XA + XB + XC + X…. 08/16/11 MATTER 94 .Mol fraction is always smaller than 1 The total mol fraction in a mixture (solution) is equal to one. = 1 Mole fraction has no unit (dimensionless) since it is a ratio of two similar quantities.. 0 g of water. C2H5OH contains 200.Example: A sample of ethanol. .0 g of ethanol and 150. Calculate the mole fraction of (a) ethanol (b) water in the solution. 5 0 7 1 . 0 0 m o + l 1 5 . 3 4 7 7 X e t h a n =o l o l = . ) g0 m 0 )o l .Solution: n e t h a n o= 2 0 .5 0 ( 11 ). 00 g −1 ( 2 ( 1 + 10 61 .5 0 7 2 0 . 0 0 m 4 . + 01 16 ) .8 0 2 0 . g0 m 0 )−o1 l n w a t e =r 1 5 . 2 0 . 0 0 m o l 4 . 00 g l ( 2 ( 1 2 + . 6 5 2 3 . 3 4 7 7 = 0 .X w a t e =r −1 0 . % w w = m m a o s fs s o l u t e x1 0 0 a o s fs s o l u t i o n Note: Mass of solution = mass of solute + mass of solvent 08/16/11 MATTER 98 .d) Percentage by Mass (%w/w) Percentage by mass is defined as the percentage of the mass of solute per mass of solution. 3g 6 2 = 1 .Example: A sample of 0.8 9 g 2 a =s s × 1 0% 0 0 .892 g of potassium chloride. 6 1 % . 8 g 9 + 52 4 . KCl is dissolved in 54. What is the percent by mass of KCl in the solution? Solution: % m 0 .362 g of water. Exercise: A solution is made by dissolving 4.0% .00 mL of water.2 g of sodium chloride. NaCl in 100. Answer = 4. Calculate the mass percent of sodium chloride in the solution. % v v o l u o m sf oe l u t e = x1 0 0 v v o l u o m sf oe l u t i o n N D 08/16/11 a os sf s o l u t i o e n os ifs t o y l u t= i o n v o l u om f s e o l u t i o MATTER 101 o t e : m .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. Solution: 2 5m L % v o l u =m e ×1 0 0 % 2 5m +L 1 2 m 5 L = 1 7 % .Example 1: 25 mL of benzene is mixed with 125 mL of acetone. Calculate the volume percent of benzene solution. 5% (v/v) ethanol. How many milliliters of ethanol does the solution contain? .Example 2: A sample of 250.00 mL ethanol is labeled as 35. Solution: Ve t % v o l u o mfe t e h a = o l n Vs o Ve t h a h a n o l l u t i o n × 1 0% 0 3 . 0 0 0m = o l n 1 0 0 % L = 8 8 .m 8 L .5 5 % × 2 5 . Example 3: A 6. NaOH solution has has a density of 1. Calculate the concentration NaOH in: (a) molarity (b) mole fraction (c) percent by mass .25 m of sodium hydroxide.33 g mL-1 at 20 ºC. V s o l u c o o n Nf s a i s O t s H 1 o a f n 6 d =t i o m n a ρs s so o l u l su t i o n n t i o .Solution: ( a M) nN = Vs a o O l u H t i o n 6 . 2 5 H m o l o f N a O f o r a s o l u t i o n k g o f w a t e r . 2 5 ⇒ t h e m r e o f N a O i s 6 . 0− 1 0 m a s so l su =t = i o n2 5 2 0 −1 0 5 0 g g + 1 0 0 0 g 1 2 g 5 . 3 g m3 V s o l u =t i o 1 n 1 L . 2 × ( 2 2o. 9 5 m l 5 0 g .m m a s so l su =t i o N nm N a a a sO + sH o m l a w aa t se r s r 9 m + a 1 s s 6 o f N + a N s a sO = H = = 6 2 n O× m H a . 2 5 m o 2 5 0 −3 ×1 0 L .M N a O = H 1 1 6 . 6 5 m− 1 o l .3 3 l L = 6 . ( b X) N 1 n a O =H nN nN o a O+ nH w a O a H t e r k g =r f w a t 6 e . 10 61 .80 2 o ) l X N a O =H 6 .2 5m l = 0 . ) g0 m 0 6 . 1 0 1 . r2 c5 o m n r to a l i no sf N a O w a t e m = s a st e o ml a ar o s f ws w m a a 0 t e r −1 1 ( 2 0 g0 ( 1 +.2 5m o l 1 0 0 0 + l o m o 1 . 0 % .( c %) ( w o / wf N) a= O m = a N s a sO H a N s a sO+ mH a m H w s a st 0 × 1 e r 0 0 % 2 2 5g0 × 1 5 g 0+ 1 0 g0 0 0 % = 2 0 . 10 mol kg-1 b) 4.53 mol L-1 c) 0.0842 . Calculate the (a) molality (b) molarity (c) mole fraction of the NH3 solution Answer: a) 5.9651 g mL-1.Exercise: An 8.00%(w/w) aqueous solution of ammonia has a density of 0. MATTER 1.3 Stoichiometry . 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 .Learning Outcome At the end of the lesson. 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 .Balancing Chemical Equation A chemical equation shows a chemical reaction using symbols for the reactants and products. Example: xA + yB zC + wD Reactants Products 08/16/11 MATTER 115 . A chemical equation must have an equal number of atoms of each element on each side of the arrow The number x. 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 . z and w. showing the relative number of molecules reacting. y. are called the stoichiometric coefficients. 08/16/11 MATTER 117 . Balance the metallic atom. followed by nonmetallic atoms.Inspection Method 1. Check to ensure that the total number of atoms of each element is the same on both sides of equation. Write down the unbalanced equation. Write the correct formulae for the reactants and products. 1. Balance the hydrogen and oxygen atoms. 1. 1. Example: Balance the chemical equation by applying the inspection method. NH3 + CuO → Cu + N2 + H2O 08/16/11 MATTER 118 . 1. N2H4 + H2O2 → HNO3 + H2O 4.Exercise Balance the chemical equation below by applying inspection method. ClO2 + H2O → HClO3 + HCl 08/16/11 MATTER 119 . Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O 2. C6H6 + O2 → CO2 + H2O 3. Redox Reaction Mainly for redox (reduction-oxidation) reaction 08/16/11 MATTER 120 . − 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 .Oxidation is defined as a process of electron loss. The substance undergoes oxidation − loses one or more electrons. The substance undergoes reduction − − − gains one or more electrons.Reduction is defined as a process of electron gain. decrease in oxidation number act as an oxidizing agent (electron acceptor) Half equation representing reduction: Br2 + 2e → BrSn4+ + 2e → Sn2+ Al3+ + 3e → Al . Oxidation numbers of any atoms can be determined by applying the following rules: 1.g: ion oxidation number Na+ +1 Cl-1 Al3+ +3 S2-2 08/16/11 MATTER 123 . For monoatomic ions. oxidation number = the charge on the ion e. 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 08/16/11 MATTER 124 Exception: 1. H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1 1. Halogen bonded to oxygen (e.g: Cl2O7) number for halogen = +ve oxidation 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 molecule = net charge on the ion 08/16/11 MATTER 125 Exercise 1. Assign the oxidation number of Mn in the following chemical compounds. i. MnO2 ii. MnO41. Assign the oxidation number of Cl in the following chemical compounds. i. KClO3 ii. Cl2O721. Assign the oxidation number of following: i. Cr in K2Cr2O7 ii. U in UO22+ iii. C in C2O42- 08/16/11 MATTER 126 08/16/11 MATTER 127 .Balancing Redox Reaction Redox reaction may occur in acidic and basic solutions. Follow the steps systematically so that equations become easier to balance. MnO4.Balancing Redox Reaction In Acidic Solution Fe2+ + MnO4. Separate the equation into two halfreactions: reduction reaction and oxidation reaction i.→ Fe3+ + Mn2+ 1. Fe2+ → Fe3+ ii.→ Mn2+ 08/16/11 MATTER 128 . Balance atoms other than O and H in each half-reaction separately i. Fe2+ → Fe3+ MnO4.1.→ Mn2+ 08/16/11 MATTER 129 . ii. Fe2+ → Fe3+ ii.+ 8H+ + 1 e 5e 130 08/16/11 MATTER 4H2O → Mn2+ + 4H2O .+ → Mn2+ + 8H+ 4. Fe2+ → Fe3+ + ii. MnO4. MnO4. Add H2O to balance the O atoms Add H+ to balance the H atoms i.3. Add electrons to balance the charges i. + 8H+ + 5e → Mn2+ + 4H2O ii.Multiply each half-reaction by an integer. i.3. so that number of electron lost in one half-reaction equals the number gained in the other. 08/16/11 MATTER 131 . 5 x (Fe2+ → Fe3+ + 1e) 5Fe2+ → 5Fe3+ + 5e MnO4. MnO4.+ 8H+ → 5Fe3+ + Mn2+ + 4H2O ___________________________________ .1. 5Fe2+ → 5Fe3+ + 5e ii. i.+ 8H+ + 5e → Mn2+ + 4H2O ___________________________________ 5Fe2+ + MnO4. Add the two half-reactions and simplify where possible by canceling the species appearing on both sides of the equation. C h e c k t h e e q u a t i o n t o a r e s t a h me e n u m b oe f r e o a f c a h t a n d s ta h me e t o t ao ln c b h o a t r h g e s 5 F 2 e+ + M 4 -n + O 8+ H→ 5 F 3 e+ + M2 ++n 4 2O H T o t a l c r he aa r c g t ea Tn ot t a l c p h r ao r d g u e c = 5 ( + .1 0 + 8 = + 1 5 + ( + 2 ) = + 1 7 = + 1 7 .5 .21 )) ++ (8 ( + =1 ) 5 ( + 3 ) + ( + 2 ) = + . + MnO4.+ H+ → CO2 + Mn2+ + H2O Solution : 08/16/11 MATTER 134 .Exercise: In Acidic Solution C2O42. add OH. 1. 1.added is equal to the number of H+ in the equation. Firstly balance the equation as in acidic solution. The number of OH.to both sides of the equation so that it can be combined with H+ to form H2O. Then.Balancing Redox Reaction In Basic Solution 1. 08/16/11 MATTER 135 . + I.→ CrO32.+ H2O 08/16/11 MATTER 136 .+ OH.Example: In Basic Solution Cr(OH)3 + IO3. 3.+ Cl- 08/16/11 MATTER 137 .+ H+ → O2 + Mn2+ + H2O Zn + SO42(acidic medium) + H2O → Zn2+ + SO2 + 4OH(basic medium) + H+ → Mn2+ + CO2 + H2O (acidic medium) (basic medium) MnO4. 2. H2O2 + MnO4.Exercise: 1.+ C2O42- Cl2 → ClO3. 4. Stoichiometry Stoichiometry is the quantitative study of reactants and products in a chemical reaction. moles. mass or even volume. A chemical equation can be interpreted in terms of molecules. 08/16/11 MATTER 138 . 02 x 1023) molecules of H2O .02 x 1023 molecules of C3H8 reacts with 5(6.02 x 1023) molecules of CO2 and 4(6.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 O2 to produce 3(6. 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 .03 g of CO2 and 72.09 g of C3H8 reacts with 160.00 g of O2 to produce 132.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. At room condition.4 dm3 of C3H8 reacts with 5(22.4 dm3) of CO2 . 22.4 dm3) of O2 to produce 3(22. 25 ºC and 1 atm pressure. 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) . 02 + 16. 1 mol C6H12O6 produce 6 mol H2O 0.125 x 6) mol x (2.125 mol C6H12O6 produce 0 .Solution: From the balanced equation.1 H 2 m 5 ×o 6 lm 2O o 1m o l l mass of H2O = (0.00) g mol-1 = 13.5 g . Example 2: Ethene. C2H4 burns in excess oxygen to form carbon dioxide gas and water vapour. how many grams of ethene are used? . (a) Write a balance equation of the reaction (b) If 20.0 dm3 of carbon dioxide gas is produced in the reaction at STP. 4 dm3 is the volume of 1 mol CO2 20.00 m 2 .400 2 .24 2 m o l .Solution: (a) C2H4 + (b) O2 → CO2 + H2O 22.00 d m CO2 m ×1 m o × 1l m o l 3 3 o l 2 mol CO2 produced by 1 mol C2H4 mol CO2 produced by 2 .24 d 2 C2H.0 dm3 is the volume of 2 .24 2 . g0 m 1 -)o ] = 1 2 .m a e st h s a 2 .4 0 ( 11 ).0 0 = e 2 . 5 g .2 4 m n 2 1 o xl [ 2 ( 1 2 + . . 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.Learning Outcome At the end of this topic. 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 . . 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 6 mol H2 reacts with 1 mol N2 6 m ×1m o l 3 m o m2 o o l l l = 2 N 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 6 m × 2 lm o 3 m o o l l = 4 m o 3l N H or 1 mol N2 6 mol N2 react with 3 mol H2 react with 6 m × 3 mol NH3 l o lm o 1m o l m 2 = 1 8 o l H H 2 i s n o t e n o u g h o f → l i m i t i n g r e a c t a n t → H2 l i m i t s t h e a m o u n t 3 mol H2 6 mol N2 produce produce 2 mol NH3 6 m × 2 lm mol NH3 o o l 3 m o l = 4 mol NH3 . 75 moles of Al and 5. (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 .00 moles of Cl2 are allowed to react.Exercise: Consider the reaction: 2 Al(s) + 3Cl2(g) → 2 AlCl3(s) A mixture of 2. PERCENTAGE YIELD The amount of product predicted by a balanced equation is the theoretical yield The theoretical yield is never obtain because: 1. There may be impurities in the reactants 08/16/11 MATTER 155 . The reaction may undergo side reaction 2. Many reaction are reversible 3. The product formed may react further to form other product 5.4. 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 . Percentage yield is the percent of the actual yield of a product to its theoretical yield actual yield % yield = x 100 theoretica l yield 08/16/11 MATTER 157 . 5 g of bromobenzene obtain from the experiment? . C6H5Br that would be produced in the reaction.Example 1: Benzene. (b) What is the percent yield if only 28. 15.0 g of benzene are mixed with excess bromine (a) Calculate the mass of bromobenzene. C6H6 and bromine undergo reaction as follows: C6H6 + Br2 → C6H5Br + HBr In an experiment.