Seventeen lessons across structure and reactivity: particles, the mole, bonding, energetics, rates, equilibrium, acids, redox, organic reactions, titration and the scientific investigation.
A skills course on calculation and explanation across the structure and reactivity themes of the first-assessment-2025 syllabus. It does not replace the syllabus. Higher-level-only content (entropy and spontaneity, rate equations, advanced organic mechanisms and spectroscopy) is not covered.
Algebra, logarithms and basic atomic structure.
Course outline
The particulate model: states, mixtures and separation
Explain properties of solids, liquids and gases and choose a separation method from particle behaviour.
The nuclear atom: isotopes and relative atomic mass
Use proton, neutron and electron counts and isotope abundances to find atomic mass.
Electron configurations and periodic trends
Write electron configurations and explain trends in atomic radius and ionisation energy.
The mole as a conversion factor
Convert between mass, amount and particle numbers using a single reliable method.
Limiting reagent, yield and the balanced equation
Use mole ratios from a balanced equation to find limiting reagent, theoretical yield and percentage yield.
Ideal gases: keep the units consistent
Apply pV = nRT and molar volume without unit mismatches.
Bonding models: ionic, covalent and metallic
Decide bonding type from the elements and link bonding to properties.
Shapes and polarity: VSEPR and intermolecular forces
Predict molecular shape and polarity and rank boiling points by intermolecular forces.
Functional groups and naming organic compounds
Identify functional groups and name simple organic compounds with correct prefixes and suffixes.
Energy from temperature change: q = m c delta T
Calculate enthalpy changes from calorimetry data with the right sign and units.
Hess's law and energy cycles
Combine known enthalpy changes to find an unknown value and sketch an energy cycle.
Rates of reaction: collisions, gradients and activation energy
Explain rate changes using collision theory and read rates from graphs.
Equilibrium: products over reactants
Write Kc expressions, interpret size, and use Le Chatelier reasoning.
Acids, bases and pH: small steps hide big changes
Use the pH scale, strength versus concentration and buffer ideas.
Redox and electrochemical cells
Assign oxidation states, balance half-equations and predict the direction in cells.
Organic reaction types: substitution, addition and oxidation
Identify the type of organic reaction and give reagents and conditions.
Titration, uncertainty and the scientific investigation
Turn titre data into concentration, combine uncertainties and plan an investigation.
Sources and curriculum note
Reviewed October 5, 2026 against the IB subject brief (first assessment 2025). The brief lists Paper 1 at 36 percent, Paper 2 at 44 percent and the internal assessment at 20 percent of the final grade overall with external at 80, and a 3,000-word maximum for the scientific investigation report. Constants (R = 8.31, Avogadro 6.02 x 10^23, specific heat capacity of water 4.18 J g-1 K-1, Kw 1.0 x 10^-14 at 298 K) are rounded teaching values; use the values in your own current IB data booklet. Check the official IB page for changes.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. The particulate model: states, mixtures and separation
Learning goal: Explain properties of solids, liquids and gases and choose a separation method from particle behaviour.
Matter is made of particles that are in constant motion and attract each other. The particle model explains the properties of the three common states. In a solid, particles vibrate about fixed positions in a regular or irregular arrangement, so a solid keeps its shape and volume. In a liquid, particles are close but can slide past each other, so a liquid has a fixed volume and takes the shape of its container. In a gas, particles are far apart and move freely, so a gas fills its container and can be compressed.
Changes of state are explained by energy and attractions. Heating a solid increases the vibration until the attractions between particles are partly overcome and it melts. Boiling and evaporation both turn liquid into gas, but boiling occurs throughout the liquid at the boiling point, while evaporation occurs at the surface at any temperature. During a change of state the temperature stays constant because the energy supplied goes into overcoming attractions, not into raising the average kinetic energy.
Pure substances have a fixed composition; elements contain one kind of atom and compounds contain two or more elements chemically bonded in a fixed ratio. A mixture contains two or more substances that are not chemically combined, with variable composition, and each substance keeps its own properties. A homogeneous mixture has a uniform composition throughout, like salt dissolved in water. A heterogeneous mixture has visibly different parts, like sand in water. Naming the type of mixture helps you choose a method of separation.
Separation methods use differences in physical properties. Filtration separates an insoluble solid from a liquid by particle size. Evaporation recovers a dissolved solid. Distillation separates liquids with different boiling points, and fractional distillation does this with a column for closer boiling points. Paper chromatography separates dissolved substances by their relative attraction to the paper and the solvent, and the retardation factor Rf is the distance moved by the substance divided by the distance moved by the solvent front. Rf has no units and is at most one.
Worked example
A chromatogram shows a spot 3.0 cm from the baseline when the solvent front is 8.0 cm from the baseline. Find the Rf value.
Rf = distance moved by spot / distance moved by solvent front.
Rf = 3.0 / 8.0.
Rf = 0.375.
Rf has no units; to two significant figures it is 0.38.
Practice problem and solution
A spot moves 4.5 cm and the solvent front moves 6.0 cm. Find the Rf value as a decimal.
Rf = 4.5 / 6.0 = 0.75.
Mental model: Choose the separation method from the physical property that differs.
Common trap: Measuring both distances from different starting lines.
2. The nuclear atom: isotopes and relative atomic mass
Learning goal: Use proton, neutron and electron counts and isotope abundances to find atomic mass.
An atom has a small dense nucleus of protons and neutrons surrounded by electrons. The atomic number Z is the number of protons and defines the element. The mass number A is the number of protons plus neutrons. In a neutral atom the number of electrons equals the number of protons. An ion forms when electrons are gained or lost; a positive ion has fewer electrons than protons, and a negative ion has more. The nuclide notation writes A as a top left number and Z as a bottom left number.
Isotopes are atoms of the same element with different numbers of neutrons, so they have the same atomic number but different mass numbers. They have the same chemical properties, because chemical behaviour depends on the electron arrangement, but slightly different physical properties such as density. Naming an isotope uses the element and its mass number, for example carbon-13, which has six protons and seven neutrons.
The relative atomic mass of an element is the weighted mean of the masses of its isotopes, relative to one twelfth of the mass of a carbon-12 atom. To find it, multiply each isotope mass by its fractional abundance and add the results. Because it is an average, it is usually not a whole number, and no individual atom has exactly that mass. The relative atomic mass has no units since it is a ratio.
A mass spectrum shows the abundance of each isotope as a bar at its mass-to-charge ratio. Read the peak positions as the isotope masses and the peak heights, or the percentages given, as the abundances. If the abundances are given as percentages, divide the sum of mass times percentage by 100. Check that the answer lies between the lightest and heaviest isotope, and nearer the more abundant one. Chlorine, with isotopes 35 and 37 in roughly a three to one ratio, gives an average near 35.5.
Worked example
An element has two isotopes: 10.0 u with 20.0 percent abundance and 11.0 u with 80.0 percent. Find the relative atomic mass.
Multiply mass by percentage: 10.0 x 20.0 = 200.
11.0 x 80.0 = 880.
Add: 200 + 880 = 1080.
Divide by 100: 10.8.
Practice problem and solution
Boron has two isotopes: 10.0 u (19.9 percent) and 11.0 u (80.1 percent). Find the relative atomic mass to three significant figures.
(10.0 x 19.9 + 11.0 x 80.1) / 100 = (199 + 881.1) / 100 = 10.8.
Mental model: Chemistry depends on electrons, mass depends on neutrons.
Common trap: Taking the simple mean of the isotope masses.
3. Electron configurations and periodic trends
Learning goal: Write electron configurations and explain trends in atomic radius and ionisation energy.
Electrons occupy energy levels (shells) and, within them, subshells labelled s, p, d and f. The s subshell holds up to two electrons, p up to six, d up to ten and f up to fourteen. Subshells fill in order of increasing energy. A common order is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, which is the pattern you read across the periodic table. Write the configuration as subshell labels with superscripts for the electron counts, for example 1s2 2s2 2p6 3s1 for sodium.
Orbital rules refine this. The Aufbau principle fills the lowest available energy first. The Pauli exclusion principle says an orbital holds at most two electrons, with opposite spin. Hund's rule says that electrons occupy orbitals of the same energy singly before pairing. Two exceptions are chromium and copper, where a half-filled or filled 3d subshell makes the 4s1 3d5 and 4s1 3d10 arrangements lower in energy than the expected patterns. Positive transition metal ions lose the 4s electrons first.
The periodic table is arranged by atomic number. Elements in a group have the same number of outer-shell electrons and similar chemistry. The period number is the number of the highest occupied shell. Across a period, from left to right, the nuclear charge increases while the electrons enter the same shell, so atomic radius decreases and first ionisation energy generally increases. Down a group, the outer electrons are in shells further from the nucleus and are shielded by inner electrons, so atomic radius increases and ionisation energy decreases.
First ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms. Two small dips across period 3 are worth explaining. Aluminium is lower than magnesium because its outer electron is in a 3p subshell, further out and higher in energy than 3s. Sulfur is lower than phosphorus because the paired electron in a 3p orbital experiences repulsion from its partner. Always give the reason in terms of nuclear charge, distance, shielding or repulsion, not just the trend.
Worked example
Write the electron configuration of a neutral phosphorus atom (Z = 15).
Fill in order: 1s2 (2 electrons).
2s2 2p6 adds 8, total 10.
3s2 adds 2, total 12.
3p3 adds 3, total 15: 1s2 2s2 2p6 3s2 3p3.
Practice problem and solution
How many electrons are in the 3d subshell of a neutral chromium atom? Enter the number.
Chromium is 4s1 3d5, because a half-filled 3d subshell is lower in energy than 4s2 3d4.
Mental model: Give the reason in terms of nuclear charge, distance, shielding or repulsion.
Common trap: Describing the trend without a reason.
4. The mole as a conversion factor
Learning goal: Convert between mass, amount and particle numbers using a single reliable method.
The mole is the amount of substance containing as many elementary entities as there are atoms in 12 g of carbon-12. The number of entities in one mole is the Avogadro constant, about 6.02 x 10^23 per mole. The molar mass M is the mass of one mole in grams per mole, and numerically equals the relative atomic or formula mass. The core equation is n = m / M, where n is the amount in moles and m is mass in grams. Keep units with every number so mistakes stand out.
Think of the mole as a bridge. To go from mass to particles, convert grams to moles by dividing by M, then moles to particles by multiplying by the Avogadro constant. To go from particles to mass, reverse the steps. For a compound, the number of atoms of a given element in one mole of the compound comes from the formula, so one mole of calcium chloride, CaCl2, contains one mole of calcium ions and two moles of chloride ions. Write the formula before counting.
The empirical formula is the simplest whole-number ratio of atoms in a compound. To find it from percentage composition or masses, convert each mass to moles, divide all by the smallest value, and round to a whole-number ratio. If a ratio such as 1 to 1.5 appears, multiply through by two. The molecular formula is a whole-number multiple of the empirical formula; find the multiple by dividing the molar mass of the compound by the empirical formula mass.
Concentration in a solution is c = n / V, usually in mol dm-3, where the volume V is in cubic decimetres. Convert cm^3 to dm^3 by dividing by 1000. For a dilution, the amount of solute stays the same, so c1 V1 = c2 V2. Percentage yield is the actual yield divided by the theoretical yield, multiplied by 100. A frequent error is rounding too early. Keep extra digits in intermediate steps and round only the final answer to the same significant figures as the least precise data.
Worked example
A compound is 40.0 percent carbon, 6.7 percent hydrogen and 53.3 percent oxygen by mass. Find the empirical formula (C 12.0, H 1.0, O 16.0).
Take 100 g: C 40.0 g, H 6.7 g, O 53.3 g.
Moles: C 3.33, H 6.7, O 3.33.
Divide by 3.33: C 1, H 2, O 1.
Empirical formula CH2O.
Practice problem and solution
Find the amount in moles in 5.85 g of sodium chloride, NaCl (M = 58.5 g mol-1).
n = m / M = 5.85 / 58.5 = 0.100 mol.
Mental model: The mole links mass, particles and concentration.
Common trap: Using cm^3 instead of dm^3 in a concentration.
5. Limiting reagent, yield and the balanced equation
Learning goal: Use mole ratios from a balanced equation to find limiting reagent, theoretical yield and percentage yield.
A balanced equation shows the ratio in which substances react. The coefficients are mole ratios, not mass ratios. For example, in 2H2 + O2 gives 2H2O, two moles of hydrogen react with one mole of oxygen and form two moles of water. To use an equation, convert any quantity given in grams or volume into moles first, apply the ratio, then convert back into the units the question wants.
When reactants are not supplied in the exact ratio, one runs out first. This is the limiting reagent, and it decides how much product forms. The other is in excess. To find the limiting reagent, divide the moles of each reactant by its coefficient. The smaller value belongs to the limiting reagent. Then calculate the product amount from the limiting reagent only, never from the excess one. The excess amount left over is the initial amount minus the amount that reacted.
The theoretical yield is the maximum mass of product that could form from the limiting reagent if the reaction is complete. The actual yield is what is obtained in practice, usually lower because of incomplete reactions, side reactions, or losses during separation and transfer. The percentage yield is the actual yield divided by the theoretical yield, multiplied by 100. A percentage yield above 100 suggests a problem such as an impure or wet product, not a successful reaction.
Atom economy measures how much of the mass of the reactants ends up in the desired product, calculated as the molar mass of the desired product divided by the total molar mass of all products, multiplied by 100 percent. It is a property of the equation, not of the experiment, so it does not depend on how well the reaction goes. A high atom economy means little waste by mass, which is an aim of green chemistry. Do not confuse atom economy with percentage yield; one is about the equation, the other about the experiment.
Worked example
Mg + 2HCl gives MgCl2 + H2. 0.20 mol Mg reacts with 0.30 mol HCl. Which is limiting, and how many moles of H2 form?
Mg: 0.20 / 1 = 0.20. HCl: 0.30 / 2 = 0.15.
The smaller value is HCl, so HCl is limiting.
Ratio HCl to H2 is 2 to 1.
Moles H2 = 0.30 / 2 = 0.15 mol.
Practice problem and solution
2Al + 3Cl2 gives 2AlCl3. 0.40 mol Al reacts with excess chlorine. How many moles of AlCl3 form?
The ratio of Al to AlCl3 is 2 to 2, so 0.40 mol of AlCl3 forms.
Mental model: Convert to moles, find the limiting reagent, then calculate from it.
Common trap: Using the amount of the excess reagent.
6. Ideal gases: keep the units consistent
Learning goal: Apply pV = nRT and molar volume without unit mismatches.
An ideal gas is a model in which particles have negligible volume and no attractions between them. Real gases behave nearly ideally at high temperature and low pressure, where particles are far apart and moving quickly. They deviate at low temperature and high pressure, where the volume of the particles and the attractions between them matter. Polar gases and gases with large molecules deviate more because their attractions are stronger.
The ideal gas equation is pV = nRT. With R = 8.31 J K-1 mol-1, the pressure must be in pascals, the volume in cubic metres and the temperature in kelvin. Convert degrees Celsius to kelvin by adding 273. Common conversions are 1 kPa = 1000 Pa, 1 dm^3 = 10^-3 m^3 and 1 cm^3 = 10^-6 m^3. Write every value in these units before substituting, because the units of the answer follow from the units of the inputs.
Molar volume is the volume occupied by one mole of an ideal gas at a stated temperature and pressure. At standard temperature and pressure used in the IB course the value is given in the data booklet, so check it for your cohort. Equal volumes of different ideal gases at the same temperature and pressure contain the same number of particles, which is Avogadro's law. This lets you use volume ratios in the same way as mole ratios for reactions between gases.
For two states of a fixed amount of gas, p1 V1 / T1 = p2 V2 / T2. Units may stay consistent but temperature must be in kelvin. To find molar mass from gas data, find n from pV = nRT, then M = m / n. Also, in an investigation the gas laws predict the shape of graphs, for instance volume against temperature in kelvin is a straight line through the origin. If your experimental graph does not pass through the origin, discuss systematic errors such as a zero error or a trapped air pocket.
Worked example
Find the volume of 0.050 mol of an ideal gas at 300 K and 100 kPa. Use R = 8.31.
p = 100000 Pa.
V = nRT / p.
V = 0.050 x 8.31 x 300 / 100000.
V = 1.25 x 10^-3 m^3 = 1.25 dm^3.
Practice problem and solution
Find the amount of gas, in mol, in a volume of 2.00 x 10^-3 m^3 at 200 kPa and 300 K. Use R = 8.31 and give three decimal places.
n = pV / RT = 200000 x 2.00 x 10^-3 / (8.31 x 300) = 400 / 2493 = 0.160 mol.
Mental model: Convert units first, then substitute.
Common trap: Using Celsius temperatures or cm^3 with R in joules.
7. Bonding models: ionic, covalent and metallic
Learning goal: Decide bonding type from the elements and link bonding to properties.
Ionic bonding is the electrostatic attraction between oppositely charged ions in a lattice. It forms typically between a metal, which loses electrons to form a cation, and a non-metal, which gains electrons to form an anion. Ionic compounds have high melting points because many strong attractions must be overcome, they are brittle because layers of ions shift and like charges meet, and they conduct electricity when molten or dissolved because the ions are free to move, but not as solids.
Covalent bonding is the electrostatic attraction between a shared pair of electrons and the two nuclei. It occurs between non-metal atoms. A single bond shares one pair, a double bond two pairs and a triple bond three pairs, with higher bond order meaning a shorter and stronger bond. A coordinate (dative) covalent bond is a shared pair where both electrons come from one atom, as in the ammonium ion. In Lewis structures, count the valence electrons, join the atoms, place lone pairs and check that atoms have a full outer shell where the octet rule applies.
Metallic bonding is the attraction between a lattice of positive metal ions and a sea of delocalised electrons. It explains electrical conductivity, because the electrons are free to move, and malleability, because layers of ions can slide without breaking the bonding. Melting points are generally higher for metals with more delocalised electrons per ion and smaller ions, because the attraction is stronger. For example, magnesium has a higher melting point than sodium because each ion has a charge of 2+ and gives two electrons to the sea.
Electronegativity measures the ability of an atom to attract the bonding electrons. A large difference between two atoms gives a bond that is largely ionic, a small difference gives a polar covalent bond with a dipole, and no difference gives a non-polar covalent bond. This is a continuum, not a sharp line, so describe the bond type with the evidence given. Network covalent solids such as diamond and silicon dioxide have very high melting points because the whole lattice is held together by covalent bonds, unlike simple molecular substances.
Worked example
Explain why magnesium oxide has a much higher melting point than sodium chloride.
Both are ionic lattices.
Mg2+ and O2- have charges of 2+ and 2-; Na+ and Cl- have 1+ and 1-.
Attractions scale with the product of the charges and are also stronger for smaller ions.
More energy is needed to separate the ions in magnesium oxide.
Practice problem and solution
How many shared electron pairs are in a double bond between two atoms? Enter the number.
A double bond consists of two shared pairs of electrons, four electrons in total.
Mental model: Link properties to the particles and forces involved.
Common trap: Saying that molecules melt by breaking covalent bonds.
8. Shapes and polarity: VSEPR and intermolecular forces
Learning goal: Predict molecular shape and polarity and rank boiling points by intermolecular forces.
The VSEPR model says that regions of electron density around a central atom repel each other and so spread as far apart as possible. A region is a bonding pair, a lone pair or a multiple bond counted as one region. Two regions give a linear shape with 180 degrees, three give trigonal planar at 120 degrees, and four give tetrahedral at about 109.5 degrees. Lone pairs repel more strongly than bonding pairs, so the bond angles in molecules with lone pairs are slightly smaller.
To predict a shape, draw the Lewis structure, count the regions, then name the shape by the positions of atoms only. Water has four regions, two bonding and two lone pairs, so the electron geometry is tetrahedral but the molecular shape is bent. Ammonia has four regions with one lone pair, so it is trigonal pyramidal. Methane has four bonding pairs and is tetrahedral. Carbon dioxide has two regions and is linear. Always separate the electron-pair arrangement from the molecular shape.
A molecule is polar if it has polar bonds and the dipoles do not cancel. Carbon dioxide has two polar bonds, but they point in opposite directions, so the molecule is non-polar. Water is bent, the dipoles do not cancel, and the molecule is polar. Tetrahedral molecules with identical bonds, such as tetrachloromethane, are non-polar, while those with different atoms, such as chloromethane, are polar. State both ideas: bond polarity and molecular symmetry.
Intermolecular forces are weaker than covalent bonds. London dispersion forces exist between all molecules and increase with the number of electrons, so larger molecules have stronger forces. Dipole-dipole forces occur between polar molecules. Hydrogen bonding is a strong dipole-dipole attraction between a hydrogen atom bonded to nitrogen, oxygen or fluorine and a lone pair on another such atom. To rank boiling points, compare the type of strongest force, then the size of molecules within the same type. This explains why water boils at a much higher temperature than hydrogen sulfide.
Worked example
Predict the shape of ammonia, NH3, and state whether it is polar.
Nitrogen has 5 valence electrons: 3 bonding pairs and 1 lone pair.
Four regions, one a lone pair: trigonal pyramidal.
The N-H dipoles do not cancel.
The molecule is polar.
Practice problem and solution
How many regions of electron density surround the central atom in methane, CH4? Enter the number.
Four bonding pairs and no lone pairs: four regions, giving a tetrahedral shape.
Mental model: Shape depends on regions, polarity on shape and bonds.
Common trap: Naming the electron-pair geometry instead of the molecular shape.
9. Functional groups and naming organic compounds
Learning goal: Identify functional groups and name simple organic compounds with correct prefixes and suffixes.
Organic compounds are built around carbon chains. A homologous series is a family of compounds with the same functional group and general formula, in which successive members differ by a CH2 unit. Members show a gradual change in physical properties, such as boiling point rising with chain length, and similar chemical properties because of the shared functional group. Alkanes have the general formula CnH2n+2, alkenes CnH2n with a carbon-carbon double bond, and alcohols CnH2n+1OH.
A functional group is the atom or group of atoms that determines most of a molecule's chemical behaviour. Key groups are the hydroxyl group in alcohols, the carbonyl group in aldehydes and ketones, the carboxyl group in carboxylic acids, the ester group, the amine group, the amide group and the halogens in halogenoalkanes. In aldehydes the carbonyl carbon is at the end of the chain, while in ketones it is inside the chain, so an aldehyde ends in -al and a ketone in -one.
IUPAC naming follows a routine. Find the longest carbon chain containing the principal functional group; this gives the stem (meth, eth, prop, but, pent, hex). Choose the suffix for the principal group, for example -ol for alcohols and -oic acid for carboxylic acids. Number the chain from the end that gives the principal group the lowest number, then name the side groups as prefixes in alphabetical order with their position numbers, using di or tri for repeats. A name such as 2-methylpropan-1-ol shows each part clearly.
Isomers are compounds with the same molecular formula but different structures. Structural isomers differ in the connectivity of the atoms, which may be the chain, the position of the group or the type of group. Butan-1-ol and butan-2-ol are position isomers. Alcohols are classified as primary, secondary or tertiary by the number of carbon atoms attached to the carbon bearing the hydroxyl group, and this classification determines how they react in oxidation. Always draw the structure, count carbons and hydrogens, then check the formula.
Worked example
Name the compound CH3CH(CH3)CH2OH.
The longest chain containing the OH group has three carbons: propane.
The -OH group is at carbon 1: propan-1-ol.
A methyl group is on carbon 2.
The name is 2-methylpropan-1-ol.
Practice problem and solution
How many hydrogen atoms are in an alkane with 6 carbon atoms? Enter the number.
CnH2n+2 with n = 6 gives 2 x 6 + 2 = 14 hydrogens.
Mental model: Find the functional group and the longest chain.
Common trap: Numbering the chain from the wrong end.
10. Energy from temperature change: q = m c delta T
Learning goal: Calculate enthalpy changes from calorimetry data with the right sign and units.
Enthalpy change is the heat energy transferred at constant pressure. In an exothermic reaction the system releases energy and the enthalpy change is negative, and the surroundings warm up. In an endothermic reaction the system absorbs energy, the enthalpy change is positive and the surroundings cool. The key idea is that the temperature change is measured in the surroundings, usually in the water, and the sign of the enthalpy change is opposite to the sign of the temperature change of the water.
The heat absorbed by the water is q = m c delta T, where m is the mass of water in grams, c is the specific heat capacity (about 4.18 J g-1 K-1 for water) and delta T is the temperature change in kelvin or degrees Celsius. Then divide q by the amount in moles of the limiting reactant to get the enthalpy change per mole. Convert from joules to kilojoules by dividing by 1000. The final enthalpy change for an exothermic reaction is negative, with the unit kJ mol-1.
A common assumption in school calorimetry is that the solution has the same density and specific heat capacity as water, so one cm^3 has a mass of one gram. Heat losses to the surroundings make the temperature rise smaller than the true value, so the measured enthalpy change is less exothermic than the real one. Improvements include insulating the cup, using a lid, and extrapolating a cooling curve back to the time of mixing rather than reading a single final temperature.
The standard enthalpy change of combustion is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions. In a flame calorimeter, heat from burning fuel warms water in a can. The result is usually smaller in size than the data book value because of incomplete combustion and heat loss to the can and air. Describe such sources of error in terms of their direction, which means saying whether they make the result too high or too low.
Worked example
50.0 g of water rises in temperature by 6.5 C when 0.0050 mol of a salt dissolves. Use c = 4.18 J g-1 K-1. Find the enthalpy change per mole in kJ mol-1.
q = m c delta T = 50.0 x 4.18 x 6.5 = 1358.5 J.
In kJ: 1.3585 kJ.
Per mole: 1.3585 / 0.0050 = 271.7 kJ mol-1.
The water warmed, so the reaction is exothermic: enthalpy change = -270 kJ mol-1 to two significant figures.
Practice problem and solution
100 g of water rises by 3.0 C. Use c = 4.18 J g-1 K-1. Find the heat absorbed by the water in joules.
q = 100 x 4.18 x 3.0 = 1254 J.
Mental model: The sign comes from the direction of energy flow, not from the sign of q.
Common trap: Dividing by the mass of the reactant rather than the amount in moles.
11. Hess's law and energy cycles
Learning goal: Combine known enthalpy changes to find an unknown value and sketch an energy cycle.
Hess's law states that the enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same. This follows from conservation of energy. It allows you to find an enthalpy change that is hard to measure directly, such as the formation of carbon monoxide from carbon and oxygen, from changes that are easy to measure, such as combustion. Draw a cycle showing the direct route and the alternative route, then equate the two.
The standard enthalpy change of formation is the enthalpy change when one mole of a compound forms from its elements in their standard states. By definition the standard enthalpy of formation of an element in its standard state is zero. For a reaction, the enthalpy change equals the sum of formation enthalpies of products minus the sum of formation enthalpies of reactants, each multiplied by its coefficient. Take care to multiply by the stoichiometric coefficients.
When manipulating equations, two rules apply. If you reverse an equation, change the sign of the enthalpy change. If you multiply an equation by a factor, multiply the enthalpy change by the same factor. Add the manipulated equations so that the species that are not in the target equation cancel. Check by confirming that the sum gives exactly the target equation with the right states. Write state symbols, because the enthalpy of water as a gas differs from water as a liquid.
Average bond enthalpies provide an estimate. The enthalpy change of a reaction is approximately the energy to break the bonds in the reactants minus the energy released in forming the bonds in the products. Bond breaking is endothermic and bond making is exothermic. Because bond enthalpies are averages over many compounds and apply to gases, they give less accurate answers than a Hess cycle using measured data. If a question says to use bond enthalpies, state that the result is approximate.
Worked example
Use formation data: C(s) + O2 gives CO2, -394 kJ mol-1; H2 + O2/2 gives H2O(l), -286 kJ mol-1; the formation of CH4 is -75 kJ mol-1. Find the enthalpy of combustion of CH4 (to CO2 and 2 H2O(l)).
Reaction enthalpy = sum of formation (products) minus sum of formation (reactants).
Products: -394 + 2 x (-286) = -966.
Reactants: CH4 -75, oxygen 0.
Enthalpy change = -966 - (-75) = -891 kJ mol-1.
Practice problem and solution
Reaction A has an enthalpy change of -100 kJ mol-1. The reverse reaction is multiplied by 2. What is the enthalpy change of the new equation, in kJ mol-1?
Reversing changes the sign (+100); doubling multiplies by two (+200).
Mental model: The route does not matter, only start and end states.
Common trap: Forgetting to change the sign when reversing.
12. Rates of reaction: collisions, gradients and activation energy
Learning goal: Explain rate changes using collision theory and read rates from graphs.
Collision theory says that a reaction occurs when particles collide with enough energy and the right orientation. The minimum energy needed is the activation energy. Only a fraction of collisions have this energy, and the rate depends on how many successful collisions occur per unit time. Increasing the concentration or pressure of a gas increases the number of particles per unit volume, so collisions are more frequent. Increasing the surface area of a solid exposes more particles, which also raises the frequency of collisions.
Raising the temperature increases the rate for two reasons. Particles move faster, so collisions are more frequent, but the main effect is that a larger fraction of particles have energy at or above the activation energy. A Maxwell-Boltzmann distribution curve shows this: the area to the right of the activation energy grows when the temperature rises. In explanations, state the second reason clearly, since saying only that collisions are more frequent gives an incomplete answer.
A catalyst increases the rate by providing an alternative pathway with a lower activation energy, so a greater fraction of collisions are successful. It is not used up and does not change the position of equilibrium or the enthalpy change of the reaction. On an energy profile diagram, the catalysed pathway has a lower peak, but the reactants and products are at the same levels. A frequent error is to say that a catalyst gives particles more energy, which it does not.
The rate of reaction is the change in concentration or amount of a reactant or product per unit time. To measure it from a graph of product against time, draw a tangent to the curve at the chosen time and find its gradient. The initial rate is the gradient at time zero and is often preferred because the concentrations are known. Units should reflect the quantity plotted, for example cm^3 s-1 for gas volume or mol dm-3 s-1 for concentration. The rate falls with time because the reactant concentration decreases.
Worked example
On a graph of volume of gas against time, a tangent at 20 s passes through (10 s, 12 cm^3) and (30 s, 36 cm^3). Find the rate at 20 s.
Gradient = change in volume / change in time.
Change in volume = 36 - 12 = 24 cm^3.
Change in time = 30 - 10 = 20 s.
Rate = 24 / 20 = 1.2 cm^3 s-1.
Practice problem and solution
On a rate graph, the tangent passes through (0 s, 0 cm^3) and (40 s, 80 cm^3). Find the initial rate in cm^3 s-1.
Gradient = 80 / 40 = 2.0 cm^3 s-1.
Mental model: Rate depends on the fraction of collisions above the activation energy.
Common trap: Saying that a catalyst changes the position of equilibrium.
13. Equilibrium: products over reactants
Learning goal: Write Kc expressions, interpret size, and use Le Chatelier reasoning.
In a reversible reaction in a closed system, equilibrium is reached when the rates of the forward and reverse reactions are equal. The reaction continues in both directions, so the equilibrium is dynamic, but the macroscopic properties such as concentration stay constant. Equilibrium can be reached from either direction. A closed system is needed, because if a gas escapes, equilibrium cannot be established.
The equilibrium constant Kc is written as the product of the concentrations of the products, each raised to its coefficient, divided by the product of the concentrations of the reactants raised to theirs. Pure solids and pure liquids are left out. A large Kc means the position lies far to the right, so products dominate; a small Kc means reactants dominate. Kc depends only on temperature and has a unit that depends on the equation, so check whether it cancels.
Le Chatelier's principle says that when a system at equilibrium is disturbed, the position shifts to partly oppose the change. Adding a reactant shifts it toward products. Increasing the pressure shifts it toward the side with fewer gas molecules. Raising the temperature favours the endothermic direction. A catalyst changes neither the position nor Kc; it only helps the system reach equilibrium sooner. Of these changes, only a temperature change alters the value of Kc.
The reaction quotient Q has the same form as Kc but uses the current concentrations. If Q is less than Kc, the forward reaction is favoured until Q equals Kc. If Q is greater, the reverse reaction is favoured. For calculations, an ICE table of initial, change and equilibrium amounts helps. Use the change in the amount from the stoichiometry, then divide by the volume to get concentrations before substituting into Kc. If the volume cancels, as when the powers on each side match, you may use amounts directly.
Worked example
N2 + 3H2 reversibly forms 2NH3. At equilibrium [N2] = 0.10, [H2] = 0.20 and [NH3] = 0.040 mol dm-3. Find Kc.
Kc = [NH3]^2 / ([N2][H2]^3).
Numerator: 0.040^2 = 0.0016.
Denominator: 0.10 x 0.20^3 = 0.10 x 0.008 = 0.0008.
Kc = 2.0 (units mol-2 dm^6).
Practice problem and solution
For A + B reversibly forms C, equilibrium concentrations are [A] = 0.50, [B] = 0.20 and [C] = 0.30 mol dm-3. Find Kc.
Kc = [C] / ([A][B]) = 0.30 / (0.50 x 0.20) = 3.0.
Mental model: Only temperature changes Kc.
Common trap: Including pure solids in the expression.
14. Acids, bases and pH: small steps hide big changes
Learning goal: Use the pH scale, strength versus concentration and buffer ideas.
The pH scale is logarithmic: pH = -log10[H+]. Each unit change in pH is a tenfold change in the hydrogen ion concentration. A solution of pH 3 has ten times more H+ than one of pH 4 and one hundred times more than pH 5. To find the concentration from pH, use [H+] = 10^-pH. A frequent error is to treat pH as linear, saying a change from pH 5 to pH 3 doubles the acidity when it is actually a hundredfold increase.
Brønsted-Lowry theory says that an acid is a proton donor and a base is a proton acceptor. When an acid donates a proton it forms its conjugate base, and when a base accepts a proton it forms its conjugate acid. A strong acid is fully ionised in water, so the hydrogen ion concentration equals the acid concentration for a monoprotic acid. A weak acid ionises only partly, so its hydrogen ion concentration is much lower than the acid concentration. Strength is about the degree of ionisation, while concentration is about the amount per volume.
Water ionises slightly, with the ionic product Kw = [H+][OH-], which is 1.0 x 10^-14 at 298 K. Kw varies with temperature, so check the value in the question. In a neutral solution at 298 K, pH = 7 and pOH = 7. Because pH + pOH = 14 at that temperature, a solution with [OH-] = 0.010 has pOH 2 and pH 12. For a strong base, find [OH-] from the concentration, then pOH, then pH. At other temperatures the neutral pH is not 7.
A buffer resists changes in pH when small amounts of acid or base are added. It contains a weak acid and its conjugate base in comparable amounts, for example ethanoic acid and sodium ethanoate. Added H+ reacts with the conjugate base, and added OH- reacts with the weak acid, so the ratio of acid to base, and therefore the pH, changes only slightly. To explain a buffer, name the two components and give a reaction for each type of addition. Neutralisation reactions between an acid and a base give a salt and water.
Worked example
Find the pH of 0.0010 mol dm-3 hydrochloric acid, a strong monoprotic acid.
HCl is fully ionised, so [H+] = 0.0010 mol dm-3.
pH = -log10(0.0010).
log10(10^-3) = -3.
pH = 3.0.
Practice problem and solution
Find the pH of a 0.010 mol dm-3 solution of sodium hydroxide at 298 K (Kw = 1.0 x 10^-14).
[OH-] = 0.010, pOH = 2, pH = 14 - 2 = 12.
Mental model: Strength is about ionisation, not concentration.
Common trap: Treating pH as a linear scale.
15. Redox and electrochemical cells
Learning goal: Assign oxidation states, balance half-equations and predict the direction in cells.
Oxidation is the loss of electrons and reduction is the gain of electrons. Oxidation states help track the transfer. Elements have an oxidation state of zero, simple ions have a state equal to their charge, and in compounds the sum of the oxidation states equals the overall charge. Hydrogen is usually +1 and oxygen usually -2 in compounds, with exceptions such as peroxides. An increase in oxidation state means oxidation; a decrease means reduction. The oxidising agent is reduced, and the reducing agent is oxidised.
To balance a half-equation in acidic solution, balance the atoms other than hydrogen and oxygen, add water to balance oxygen, add H+ to balance hydrogen, then add electrons to balance the charge. Combine two half-equations by multiplying so that the electrons cancel. For example, the reduction of permanganate in acid uses eight H+ and five electrons per MnO4- to form Mn2+ and four water molecules. Always check the final charge and atom totals on each side.
A voltaic cell converts chemical energy to electrical energy. Oxidation occurs at the anode, which is the negative electrode, and reduction at the cathode, which is the positive electrode. Electrons flow in the external wire from anode to cathode, and ions move through a salt bridge to keep each half-cell electrically neutral. The standard electrode potentials in the data booklet are compared: the half-cell with the more positive potential is the site of reduction, and the cell potential is the difference between the two.
An electrolytic cell uses electrical energy to drive a non-spontaneous reaction. Here the anode is the positive electrode and the cathode the negative electrode, but oxidation still occurs at the anode and reduction at the cathode. In the electrolysis of an aqueous solution, water may compete with the ions in solution. Predict the products by comparing the species and consider concentration and electrode material. The mass deposited is proportional to the charge passed, which is current multiplied by time.
Worked example
Find the oxidation state of manganese in KMnO4.
Potassium is +1 and oxygen is -2.
Total for oxygen: 4 x (-2) = -8.
The compound is neutral: +1 + x - 8 = 0.
x = +7.
Practice problem and solution
What is the oxidation state of chromium in the dichromate ion, Cr2O7 with a 2- charge? Enter the number.
Oxygen contributes 7 x (-2) = -14. Two chromium atoms and total charge -2: 2x - 14 = -2, so x = +6.
Mental model: Oxidation is loss; the anode is where it occurs.
Common trap: Mixing up the polarity of the electrodes between cell types.
16. Organic reaction types: substitution, addition and oxidation
Learning goal: Identify the type of organic reaction and give reagents and conditions.
Organic reactions fall into types. A substitution reaction replaces one atom or group with another. An addition reaction joins a molecule across a double bond, so two reactants form one product. An elimination reaction removes a small molecule such as water or hydrogen halide and forms a double bond. Oxidation and reduction reactions change the oxidation state of carbon. Naming the type helps you recall the reagents and the expected product.
Alkanes are fairly unreactive and undergo free radical substitution with halogens in ultraviolet light. The mechanism has three stages: initiation, where the halogen bond breaks homolytically to form two radicals; propagation, where radicals react with the alkane and the halogen in a chain; and termination, where two radicals combine. A mixture of products forms, since further substitution can occur. Curly arrows with single barbs (fish hooks) show the movement of single electrons.
Alkenes undergo electrophilic addition. The electron-rich double bond attacks an electrophile such as bromine or a hydrogen halide. Bromine water changes from orange to colourless in the test for unsaturation. With an unsymmetrical alkene and a hydrogen halide, two products are possible, and the major one usually forms through the more stable carbocation, which is the one with more alkyl groups attached. State the mechanism with curly arrows for the movement of electron pairs and show the intermediate.
Halogenoalkanes undergo nucleophilic substitution, where a nucleophile with a lone pair attacks the carbon bonded to the halogen. Primary halogenoalkanes tend to react in one step, which is called SN2, and tertiary ones in two steps through a carbocation, called SN1. Alcohols are oxidised: primary alcohols give aldehydes and then carboxylic acids, secondary alcohols give ketones, and tertiary alcohols do not oxidise with acidified dichromate. To obtain the aldehyde, distil it off as it forms; to obtain the acid, heat under reflux with excess oxidising agent.
Worked example
State what is formed when ethanol is heated under reflux with excess acidified potassium dichromate.
Ethanol is a primary alcohol.
Reflux with excess oxidising agent keeps the aldehyde in the flask.
The aldehyde is oxidised further.
The product is ethanoic acid.
Practice problem and solution
How many different organic products can a tertiary alcohol give with acidified potassium dichromate under normal oxidation conditions? Enter a number.
Tertiary alcohols have no hydrogen on the carbon bearing the OH group and are not oxidised under these conditions, so no oxidation product forms.
Mental model: Classify the reaction first, then recall the conditions.
Common trap: Distilling when the acid is wanted instead of refluxing.
17. Titration, uncertainty and the scientific investigation
Learning goal: Turn titre data into concentration, combine uncertainties and plan an investigation.
A titration finds the concentration of an unknown solution by reacting it with a solution of known concentration. The titre is the volume of the solution added from the burette to reach the end point. Run a rough titration first to find the approximate end point, then repeat accurately until you have concordant results, usually within 0.10 cm^3 of each other. Average only the concordant titres. Discard the rough one and any outliers in the mean.
To calculate the concentration, write the balanced equation, find the amount in moles of the known solution from c x V (with V in dm^3), use the mole ratio to find the moles of the unknown, then divide by its volume. For an acid-base titration with 25.0 cm^3 of sodium hydroxide and a mean titre of 20.0 cm^3 of 0.100 mol dm-3 hydrochloric acid, the acid contains 0.00200 mol, so the base contains 0.00200 mol, and the concentration is 0.0800 mol dm-3. An indicator should change colour at the pH of the equivalence point.
Every measuring instrument has an uncertainty. A burette is typically read to half of its smallest division and a titre involves two readings, so add the uncertainties. The percentage uncertainty is the absolute uncertainty divided by the measured value, multiplied by 100. For a calculated result that involves products and quotients, add the percentage uncertainties of the measured values. Use the percentage uncertainty to judge which measurement limits the precision, and propose an improvement that targets it, such as a larger titre or a more precise pipette.
The scientific investigation is an internal assessment in which you pose your own research question, collect and process data and write a report of at most 3,000 words, per the IB subject brief. Good chemistry questions identify one independent variable, one dependent variable and controlled variables, and can be answered with measurements you can make safely. Evaluation should name specific limitations with their effect on the result and suggest improvements that deal with each. Distinguish random errors, which cause scatter, from systematic errors, which shift all results in one direction.
Worked example
25.0 cm^3 of NaOH is neutralised by a mean titre of 20.0 cm^3 of 0.100 mol dm-3 HCl. Find the NaOH concentration.
10.0 cm^3 of 0.100 mol dm-3 NaOH is neutralised by 20.0 cm^3 of sulfuric acid in a 2 to 1 ratio (2 NaOH to 1 H2SO4). Find the sulfuric acid concentration in mol dm-3.