Foreword Academic success can be measured in many different ways, and I often tell my students that scoring high marks in exams is only one of the rewards from diligent study. The true measures of academic success should be the enjoyment of learning and the sense of accomplishment always held a special place for me as what could be more important than understanding the composition, structure, and properties of all the matter that surrounds us? The International Baccalaureate (IB) Diploma Programme is a challenging two-year curriculum for exam success as they continue to pursue secondary and college education. As a leading of students reach their academic goals. Whether they are in primary, secondary, undergraduate, programs and expertise in all major international curricula and exams. enjoy the learning process as well. T.K. Ng Founder and Managing Director iii
The IBDP Chemistry Syllabus and Examination to study extension materials that are more demanding. Core Syllabus Topic 1 Topic 3 Topic 6 Topic 7 Topic 8 Topic 9 Topic 10 Topic 11 Topic 13 Topic 16 Topic 17 Topic 18 Topic 19 Atomic structure Periodicity Bonding Energetics Kinetics Acids and bases Oxidation and reduction Organic chemistry Measurement and data processing Additional Higher Level Material Atomic structure Periodicity Bonding Energetics Kinetics Acids and bases Oxidation and reduction Organic chemistry
Options Option A Option B Option D Option E Option F Modern analytical chemistry Medicines and drugs Food chemistry Further organic chemistry material Level Paper Overall weighting Duration (minutes) Paper 1 Paper 3 60 Paper 1 60 Paper 3 Format* calculators allowed each of the two options studied calculators allowed each of the two options studied
HL and / or a shaded background. Table of Contents Chapter 1 Quantitative Chemistry...1 Mole concept... 1... 1 Percentage by mass... 3 Empirical formula... Molecular formula... 10...... Reacting masses... Percentage yield... 16 Properties of gases... 17... 18... The ideal gas law...... Electrolytes... Preparation of solutions... Dilution of solutions...... Chapter 2 Atomic Structure...27 The structure of an atom... Properties of subatomic particles... Atoms and ions... Atomic symbols... Isotopes... 30 The mass spectrometer... 31... 33 The electromagnetic spectrum...... 36... 36 The Bohr model... 37 Interpretation of the hydrogen emission spectrum... 39 Electron arrangement...
HL... HL... HL... Electron spin and the Pauli exclusion principle HL... HL... Exceptions to the Aufbau principle HL... Chapter 3 Periodicity...53 The periodic table... Electron arrangements in the periodic table... HL... The periodic trends... Explanations for the periodic trends in physical properties... 61 Properties of some main group elements... 67 Properties of oxides across period 3... 70 Properties of chlorides across period 3 HL... HL... 73 Transition metals HL... 73 Chapter 4 Bonding...77 Introduction... 77 Ionic bonding... 78 Monatomic ions... 79 Polyatomic ions... 81 Formation of ionic compounds.................. 86 Valence shell electron pair repulsion model...87 ) and pi ( ) bonds HL... 90 HL... 91 HL... 93 Bond polarity... Polarity of molecules... Intermolecular forces... 96 Effect of intermolecular forces on boiling points...98 Metallic bonding...99... 100
Allotropes of carbon...... Chapter 5 Energetics... 105 Enthalpy and enthalpy change...... 106 Exothermic and endothermic reactions... 106... 107... 107... Bond enthalpies... HL... 117 HL... 119 HL... HL... HL... HL... Entropy HL... Entropy change HL... Free energy change HL... HL... Chapter 6 Kinetics...133 Rate of reaction... 133 Experiments to determine reaction rate......... Rate expression HL... HL... Determination of the rate expression HL... Elementary steps HL... Reaction mechanism HL... HL... Temperature dependence of rate constants HL... Chapter 7 Introduction to Equilibrium... 161... 161... 161... 161
... 163 Kc HL...... 166... HL... 173 Chapter 8 Acids and Bases...175...... 176 Monoprotic and polyprotic acids... 177 Amphoteric substances... 178... 179... 180... 181... 183 HL... 186 HL... 188 HL... 189 HL... 190 Ka, Kb, pka and pkb HL... 191 HL... 193 HL... Relationship between Ka and Kb of conjugate acids and bases HL... 197 Buffer solutions HL... 198 Preparation of buffer solutions HL... 199 HL... HL... Acid-base titrations HL... Indicators HL... Chapter 9 Oxidation and Reduction...215 Introduction to redox reactions... Assignments of oxidation numbers... Nomenclature... Recognition of oxidation, reduction and redox reactions... Balancing redox reactions......... Voltaic cells... ix
HL... HL... HL... Introduction to electrolytic cells... HL... HL... HL... Chapter 10 Organic Chemistry...249... Representation of organic molecules... Naming esters HL............ Alkanes... Alkenes... Alcohols......... Reaction pathways... HL... Chapter 11 Measurement and Data Processing... 283 Uncertainty and error in measurement... Precision and accuracy... Absolute error and percentage error... Random uncertainties due to instruments...... Propagation of uncertainty...... Plotting appropriate graphs... Practice Examination (Standard Level)...291 Practice Examination (Higher Level) HL...321 Practice Examination (Solutions)...363 x Index...387
Chapter 1 Quantitative Chemistry Mole concept The mole, abbreviated as mol, is a unit used to count substances such as atoms, molecules, ions and electrons. To convert between the amount of a substance and the number of moles, we use the following formula. Amount of substance Moles of substance L where L is known as Avogadro s constant, which, in IB examinations, has an accepted value of 6.02 10 23 mol 1. Example 1.1 Q : What is the number of oxygen atoms in 015 moles of carbon dioxide (CO2) molecules? A : Number of CO2 molecules 015 mol (6.02 10 23 mol 1 ) 9.0 10 21 Each CO2 molecule contains two oxygen atoms. Number of oxygen atoms in 015 moles of CO2 molecules (9.0 10 21 ) 2 1.8 10 22 Throughout this study guide, the intermediate calculated values are reported in accordance with the propagation of uncertainty. However, the final answers are obtained using calculator displayed values and rounded off with an appropriate uncertainty. Relative atomic mass and relative molecular mass The relative atomic mass of an element, denoted by Ar, is the average mass of all naturally occurring atoms of an element relative to carbon-12. It has no unit and the value for each element is commonly listed in the periodic table. The molar mass of an atom is numerically equivalent to the relative atomic mass. Its unit is g mol 1 and is the conversion factor between the mass and the number of moles of a substance, as shown in the following formula. Mass Moles Molar Mass 1
IBDP Chemistry Study Guide Example 1.2 Q : What is the mass of 3.01 10 23 sodium atoms? A : Moles of sodium atoms 23 301. 10 602. 10 mol 23 1 500 mol Referring to the periodic table, the relative atomic mass of sodium is 22.99. Mass of sodium atoms Moles Molar mass 500 mol 22.99 g mol 1 11.5 g For molecules, the relative molecular mass, denoted by Mr, is the average mass of a molecule relative to carbon-12. It is often approximated as the sum of the relative atomic masses of the molecule s constituent atoms. Example 1.3 Q : What is the relative molecular mass of water (H2O)? A : Mr of H2O (Ar of H 2) Ar of O (1.01 2) 16.00 18.02 Similar to the relative atomic mass, the relative molecular mass is also numerically equivalent to the molar mass of a molecule. Example 1.4 Q : What is the number of moles of molecules in 8.5 g of water (H2O)? A : From Example 1.3, the relative molecular mass of H2O is 18.02. Moles of H2O Mass Molar mass 85. g 47 mol 1 18. 02 gmol For a substance that is not made up of molecules but which can still be represented by a formula, the molar mass of the substance is numerically the same as the relative formula mass, also denoted by Mr, which is often approximated as the sum of relative atomic masses. Example 1.5 Q : What is the relative formula mass of magnesium nitrate (Mg(NO3)2)? A : Mr[Mg(NO3)2] Ar of Mg [Ar of N (Ar of O 3)] 2 24.31 [14.01 (16.00 3)] 2 148.33 2
Chapter 1 Quantitative Chemistry Percentage by mass The composition of a compound can be expressed in terms of the percentage of an element by mass, which can be determined by the following formula. Percentage of an element by mass 100% Example 1.6 Q : 2.00 g of ammonium nitrate contains 70 g of nitrogen, 1.20 g of oxygen and 10 g of hydrogen. What is the percentage of each element by mass? Determine the mass of nitrogen contained in 32.5 g of ammonium nitrate. A : Percentage of nitrogen by mass 070. g 200. g 100% 35% Percentage of oxygen by mass 120. g 200. g 100% 60% Percentage of hydrogen by mass 010. g 200. g 100% 5.0% In 32.5 g of ammonium nitrate, the percentage of nitrogen by mass is 35%. Therefore, Mass of nitrogen Mass of ammonium nitrate Percentage of nitrogen by mass 32.5 g 35% 11 g Note that the sum of the percentages of elements by mass is always 100%. 3
IBDP Chemistry Study Guide Empirical formula An empirical formula shows the simplest whole-number ratio of the constituent elements in a compound. The general method to determine the empirical formula of a compound which contains three elements, denoted by the letters X, Y and Z, is shown below. Mass of X Mass of Y Mass of Z Ar of X Ar of Y Ar of Z Moles of X Moles of Y Moles of Z Mole ratio between X, Y and Z Expand to whole numbers if necessary Empirical formula Rounding off values in the early stages of a calculation can lead to a wrong empirical formula. Exam Tip Example 1.7 Q : What is the empirical formula of a compound if 13.00 g of this compound contains 5.20 g of carbon, 87 g of hydrogen and 6.93 g of oxygen? A : Carbon Hydrogen Oxygen Mass (g) 5.20 87 6.93 Moles (mol) 520. 12. 01 433 087. 101. 86 693. 16. 00 433 Mole ratio 433 433 1 086. 433 2 433 433 1 The empirical formula of the compound is CH2O. 4
Chapter 1 Quantitative Chemistry To deduce the empirical formula of a compound by the percentages of different elements by mass, we can assume that there is 100 g of the compound. The masses of the elements are then numerically equivalent to their percentages by mass. Example 1.8 Q : What is the empirical formula of a compound, made of hydrogen and carbon only, which is 88.1% carbon? A : Assume that there is 100 g of the compound. Mass of carbon 100 g 88.1% 88.1 g Mass of hydrogen 100 g 88.1 g 11.9 g Carbon Hydrogen Mass (g) 88.1 11.9 Moles (mol) 88. 1 12. 01 7.34 11. 9 101. 11.8 Mole ratio 734. 734. 1 11. 8 734. 1.61 Whole number ratio 1 5 5 1.61 5 8 The empirical formula of the compound is C5H8. Empirical formulas of binary compounds The empirical formula of a binary compound is often determined through thermal decomposition, combustion or reduction. Usually, either the reactant or the product is a pure element. The difference between the reactant and the product reflects the mass of another element. Example 1.9 Q : A sample of iron is placed in a crucible. The sample is then burnt in excess air, producing an oxide of iron. The experimental data is shown below. Mass of the crucible 11.23 g Mass of the crucible and iron 14.58 g Mass of the crucible and the oxide of iron 15.86 g What is the empirical formula of the oxide of iron? 5
IBDP Chemistry Study Guide A : First, we need to determine the masses of the two elements in the compound. Mass of iron (14.58 11.23) g 3.35 g Mass of the oxide of iron (15.86 11.23) g 4.63 g Since the oxide of iron is composed only of iron and oxygen, the difference between the oxide and iron equals the mass of oxygen. Mass of oxygen Mass of the oxide Mass of iron (4.63 3.35) g 1.28 g Iron Oxygen Mass (g) 3.35 1.28 Moles (mol) 335. 55. 85 0600 128. 16. 00 0800 Mole ratio 0600 0600 1 0800 0600 1.33 Whole number ratio 1 3 3 1.33 3 4 The empirical formula of the oxide is Fe3O4. Water of crystallization of a hydrated salt Some salts combine with water molecules and form hydrated salts. The number of water molecules present in the formula of a salt is known as the water of crystallization. For example, anhydrous copper(ii) sulfate (CuSO4) combines with water molecules to form hydrated copper(ii) sulfate (CuSO4 5H2O). The water of crystallization in CuSO4 is five. The water of crystallization can be obtained by the ratio between the number of moles of the anhydrous salt and the number of moles of water. 6
Chapter 1 Quantitative Chemistry Example 1.10 Q : A weighed sample of hydrated iron(ii) sulfate (FeSO4 xh2o) is strongly heated on an evaporating dish to remove all water of crystallization. The experimental data is shown below. Mass of the evaporating dish 28.85 g Mass of the dish and the hydrated salt 35.25 g Mass of the dish and the remaining solid 32.35 g Find the water of crystallization (i.e. the value of x). A : We need to determine the mass of the anhydrous salt and that of water. The water of crystallization can then be found by dividing the number of moles of water by that of the anhydrous salt, which makes up the remaining solid. The mass of water removed equals the difference in mass between the hydrated salt and the remaining solid. Mass of water removed (35.25 32.35) g 2.90 g The remaining solid is the anhydrous salt. Mass of the anhydrous salt (32.35 28.85) g 3.50 g FeSO4 H2O Mass (g) 3.50 2.90 Moles (mol) 350. 151. 91 0230 290. 18. 02 161 Mole ratio 0230 0230 1 161 0230 7.00 7 The value of x 7. The formula of the salt is FeSO4 7H2O. 7
IBDP Chemistry Study Guide Empirical formulas of organic compounds Upon complete combustion, organic compounds react with oxygen to form different oxides, depending on their composition. For example, when a mass made up of carbon and hydrogen combines with oxygen, carbon produces carbon dioxide and hydrogen produces water. By collecting and weighing the products separately, the amounts of the constituent elements can be determined. The method of calculation is shown in the following diagram. Step 1 Mass of CO2 Mass of H2O Mr of CO2 Mr of H2O Moles of CO2 Moles of H2O CO2 : C 1 : 1 H2O : H 1 : 2 Moles of C Moles of H Moles of O Ar of C Ar of H Ar of O Mass of C Mass of H Mass of O Mass of sample (Mass of C Mass of H) Step 2 Moles of C Moles of H Moles of O Mole ratio between C, H and O Expand to whole numbers if necessary Empirical formula 8
Chapter 1 Quantitative Chemistry Example 1.11 Q : 3.00 g of ethylene glycol, which contains only carbon, hydrogen and oxygen, was burnt in excess oxygen. 4.25 g of carbon dioxide and 2.61 g of water were collected. What is the empirical formula of ethylene glycol? A : Amount of carbon: Moles of CO2 425. g 0966 mol 1 [ 12. 01+ ( 16. 00 2)] gmol One CO2 molecule contains one carbon atom, therefore, Moles of C Moles of CO2 0966 mol Amount of hydrogen: Moles of H2O 261. g 145 mol 1 [( 1. 01 2) + 16. 00] gmol One H2O molecule contains two hydrogen atoms. Moles of H Moles of H2O 2 290 mol Amount of oxygen: The mass of oxygen is the difference between the mass of the compound and the masses of the other elements. Mass of C 0966 mol 12.01 g mol 1 1.16 g Mass of H 290 mol 1.01 g mol 1 293 g Mass of O Mass of the compound Mass of C Mass of H (3.00 1.16 293) g 1.55 g Moles of O 155. g 0967 mol 1 16. 00 gmol C H O Moles (mol) 0966 290 0967 Mole ratio 0966 0966 1 290 0966 0967 0966 1.00 1 The empirical formula of ethylene glycol is: CH3O 9