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IB® · HL/SL

IB® Chemistry HL/SL

First assessment 2025: complete syllabus-aligned review of the Structure and Reactivity themes, with worked problems.

Start Unit 1 free. Structure 1: Particulate Nature of Matter & The Mole is open to everyone, no account needed. Other topics are locked.

Structure 1: Particulate Nature of Matter & The Mole

THE BIG PICTURE. Structure 1 covers the PARTICULATE NATURE OF MATTER: what stuff is made of and how to count it. Sub-topics: S1.1 Introduction to the particulate nature, S1.2 The nuclear atom, S1.3 Electron configurations, S1.4 Counting particles by mass (the mole), S1.5 Ideal gases. Together these give you the language of chemistry: atoms, isotopes, electrons, moles, gases: and the math to handle quantitative problems.

KINETIC MOLECULAR THEORY (S1.1)

Matter consists of PARTICLES (atoms, molecules, ions) in constant motion. Key statements:

  • Particles have KINETIC ENERGY that depends on temperature.
  • The Kelvin temperature TT is a measure of the AVERAGE KINETIC ENERGY of particles.
  • In the SI system, TT is measured in Kelvin (K). Conversion: T(K)=θ(°C)+273.15T(\text{K}) = \theta(°\text{C}) + 273.15.
  • At absolute zero (0 K), particles have the minimum possible thermal kinetic energy (zero-point motion remains: quantum mechanically, motion is not literally zero).

STATES OF MATTER

  • SOLID: particles vibrate in fixed positions, definite shape and volume.
  • LIQUID: particles slide past each other, definite volume but no fixed shape.
  • GAS: particles move freely, no fixed shape or volume.

PHASE CHANGES: melting, freezing, evaporation, condensation, sublimation, deposition. Each involves energy absorbed/released without temperature change at the transition.

Particulate diagrams Count the kinds of particle and check whether atoms are bonded together. One kind of atom means an element, identical units containing different atoms mean a compound, and two or more kinds of unit mean a mixture. The lower row shows how spacing and freedom of movement change from solid to liquid to gas.

PRACTICE: CLASSIFY THE PARTICULATE DIAGRAM

What the box showsClassificationWhy
Identical single atoms, widely spacedElement (gas, e.g. He)one kind of atom, not bonded to others
Identical pairs of the same atomElement (e.g. O₂, Cl₂)molecules, but only one element
Identical units, each with two kinds of atomCompoundatoms of different elements bonded in a fixed ratio
Single atoms plus diatomic moleculesMixture of elementstwo kinds of particle, no fixed ratio
Two kinds of compound unitMixture of compoundsseparable by physical methods
Regular rows, touchingSolidparticles vibrate about fixed positions

WORKED EXAMPLE (PAPER 1A STYLE): KELVIN AND KINETIC ENERGY

A sample of neon is at 27 °C. To which temperature must it be heated to double the average kinetic energy of its atoms? A. 54 °C B. 300 °C C. 327 °C D. 600 °C

  • Average kinetic energy is proportional to the Kelvin temperature, never to °C.
  • T1=27+273=300T_1 = 27 + 273 = 300 K, so doubling needs T2=600T_2 = 600 K.
  • Convert back: 600−273=327600 - 273 = 327 °C. Answer C. Option A is the classic trap (doubling the Celsius value).

ELEMENTS, COMPOUNDS, MIXTURES

Antoine Lavoisier (1743–1794), often called the father of modern chemistry. He established the law of conservation of mass and helped put the study of elements and reactions on a quantitative footing.

Unknown (opens in new tab), Public domain
  • ELEMENT: pure substance that cannot be chemically broken down (e.g., Fe, O, Au).
  • COMPOUND: chemically combined elements in fixed ratios (e.g., H2O\text{H}_2\text{O}, NaCl\text{NaCl}).
  • MIXTURE: physical combination; components retain identity (e.g., air, salt water).
    • HOMOGENEOUS: uniform composition (solutions).
    • HETEROGENEOUS: non-uniform (sand in water).

Separation techniques: filtration, distillation, chromatography, evaporation.

THE NUCLEAR ATOM (S1.2)

Marie Curie (1867–1934), the first person to win Nobel Prizes in two sciences. Her work on radioactivity showed that atoms are not indivisible and helped open the modern study of atomic structure.

Unknown (1920s photograph) (opens in new tab), Public domain

An atom has:

  • A dense, positively charged NUCLEUS containing PROTONS (charge +1+1, mass ~1 u) and NEUTRONS (charge 0, mass ~1 u).
  • Electrons (charge −1-1, mass ~1/1836 u) in regions of probability around the nucleus.
  • ATOMIC NUMBER (ZZ) = number of protons. Defines the element.
  • MASS NUMBER (AA) = protons + neutrons.
  • NEUTRON NUMBER (NN) = A−ZA - Z.
  • In a neutral atom: number of electrons = ZZ.

ISOTOPES: atoms of the same element with the same ZZ but different AA (different numbers of neutrons). Isotopes have

  • Same chemical properties (same electron configuration).
  • Different physical properties (mass affects density, rates of diffusion, melting point slightly).
  • Examples: 12C^{12}\text{C}, 13C^{13}\text{C}, 14C^{14}\text{C} all carbon, but with 6, 7, 8 neutrons respectively.

RELATIVE ATOMIC MASS: SL/HL (S1.2.2)

RELATIVE ATOMIC MASS (ArA_r): the weighted average mass of an atom of an element on the carbon-12 scale. Ar=∑(isotope mass×fractional abundance)A_r = \sum (\text{isotope mass} \times \text{fractional abundance}) Example: chlorine has two stable isotopes: 35Cl^{35}\text{Cl} (75.77%) and 37Cl^{37}\text{Cl} (24.23%). Find ArA_r. Ar=(0.7577×35)+(0.2423×37)=26.52+8.97=35.5A_r = (0.7577 \times 35) + (0.2423 \times 37) = 26.52 + 8.97 = 35.5 This is why the periodic table shows Cl as ~35.5: not a whole number, because of isotope mixture. At SL you need to calculate ArA_r from given abundance data; specific isotopes need not be memorised.

4b. MASS SPECTRUM INTERPRETATION: HL ONLY (S1.2.3).

A mass spectrometer separates ions by mass-to-charge ratio (m/zm/z). The output is a mass spectrum: vertical lines on an m/zm/z axis whose heights show relative abundance of each isotope (or fragment).

  • HL students must interpret a mass spectrum: identify each isotope, read its abundance, and calculate ArA_r.
  • Operational details of the instrument are not assessed.
  • Example: a mass spectrum with peaks at m/z=35m/z = 35 (75.77%) and m/z=37m/z = 37 (24.23%) → Ar=35.5A_r = 35.5 → element is chlorine.

Mass spectrum of magnesium (HL) Each line is one isotope; its height is the relative abundance. Weighting each mass by its abundance gives Ar=24.32A_r = 24.32, matching the data booklet value of 24.31. Abundances: NIST isotopic compositions.

WORKED EXAMPLE (HL ONLY, PAPER 1B STYLE): RELATIVE ATOMIC MASS FROM A SPECTRUM

Use the magnesium spectrum above to calculate ArA_r of magnesium to two decimal places.

  • Read each line: m/zm/z 24 (78.99%), 25 (10.00%), 26 (11.01%). The abundances add to 100.00%, so no isotope is missing.
  • Weight each mass by its fractional abundance:

Ar=(24×0.7899)+(25×0.1000)+(26×0.1101)=18.96+2.50+2.86=24.32A_r = (24 \times 0.7899) + (25 \times 0.1000) + (26 \times 0.1101) = 18.96 + 2.50 + 2.86 = 24.32

  • The data booklet gives 24.31. The tiny difference comes from using whole-number mass numbers instead of exact isotopic masses.
  • Exam habit: the answer must lie between the lightest and heaviest isotope and sit closest to the most abundant one. 24.32 passes both checks.

ELECTRON CONFIGURATIONS (S1.3)

Electrons occupy ENERGY LEVELS (shells) labeled n=1,2,3,…n = 1, 2, 3, \ldots. Each shell contains SUBSHELLS (s, p, d, f) and each subshell has ORBITALS (regions of probability).

  • s subshell: 1 orbital, holds 2 electrons.
  • p subshell: 3 orbitals, holds 6 electrons.
  • d subshell: 5 orbitals, holds 10 electrons.
  • f subshell: 7 orbitals, holds 14 electrons.

FILLING RULES

  • AUFBAU principle: electrons fill lowest-energy orbitals first.
  • PAULI exclusion principle: max 2 electrons per orbital with opposite spins.
  • HUND's rule: degenerate orbitals fill singly first (parallel spins) before pairing.

ORDER: 1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,…1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, \ldots (note 4s before 3d).

EXAMPLES

  • H (1 electron): 1s11s^1.
  • C (6 electrons): 1s22s22p21s^2 2s^2 2p^2.
  • Fe (26 electrons): 1s22s22p63s23p64s23d61s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6 or [Ar]4s23d6\text{[Ar]} 4s^2 3d^6.
  • EXCEPTIONS: Cr and Cu (S1.3.5): the observed configurations differ from a strict Aufbau prediction; treat them as observed facts (S1.3.5 asks you to know these two exceptions; avoid explaining them by "special stability" of half-filled or filled sublevels).
    • Cr: observed [Ar]4s13d5\text{[Ar]} 4s^1 3d^5 (not 4s23d44s^2 3d^4).
    • Cu: observed [Ar]4s13d10\text{[Ar]} 4s^1 3d^{10} (not 4s23d94s^2 3d^9).
  • IONS up to Z=36Z = 36: remove from the 4s4s subshell before 3d3d. Examples: Fe2+:[Ar]3d6\text{Fe}^{2+}: \text{[Ar]} 3d^6; Fe3+:[Ar]3d5\text{Fe}^{3+}: \text{[Ar]} 3d^5; Ca2+:[Ar]\text{Ca}^{2+}: \text{[Ar]}; Br−:[Kr]\text{Br}^-: \text{[Kr]}.

PRACTICE: WRITE THE CONFIGURATION

SpeciesFull or condensed configurationCheck
S (16 e⁻)1s22s22p63s23p41s^2 2s^2 2p^6 3s^2 3p^4two unpaired 3p electrons (Hund's rule)
S²⁻ (18 e⁻)[Ne]3s23p6\text{[Ne]} 3s^2 3p^6isoelectronic with Ar
Cr (24 e⁻)[Ar]4s13d5\text{[Ar]} 4s^1 3d^5observed exception
Cu²⁺ (27 e⁻)[Ar]3d9\text{[Ar]} 3d^9both 4s electrons lost first, then one 3d
Zn²⁺ (28 e⁻)[Ar]3d10\text{[Ar]} 3d^{10}4s emptied before 3d
Fe³⁺ (23 e⁻)[Ar]3d5\text{[Ar]} 3d^5five unpaired d electrons

6a. EMISSION SPECTRA & QUANTISED ENERGY: SL/HL (S1.3.1–S1.3.2).

When excited atoms return to lower energy states, they emit photons of specific energies. Each transition corresponds to a specific wavelength of light → discrete lines in the emission spectrum (contrast a continuous spectrum from a hot solid/incandescent source).

  • EVIDENCE FOR QUANTISATION: discrete lines (not a continuum) prove electrons can only occupy specific energies: energy is quantised.
  • The hydrogen emission spectrum shows lines that converge at higher energy: evidence for discrete levels whose spacing decreases as nn increases.
  • Energy of a photon: E=hf=hc/λE = hf = hc/\lambda (with hh = Planck's constant, ff = frequency, λ\lambda = wavelength: values and equations in the data booklet).
  • Spectra are atomic fingerprints: used in astronomy, forensics, manufacturing.
  • Hydrogen transitions ending at n=1n = 1 lie in the UV; ending at n=2n = 2 in the visible; ending at n=3n = 3 in the IR. The names of these series (Lyman, Balmer, Paschen, etc.) are enrichment only: the IB syllabus states they will not be assessed.

Energy levels and the line spectrum of hydrogen Level energies are En=−1312/n2E_n = -1312/n^2 kJ mol−1^{-1} (1312 kJ mol−1^{-1} is the first ionization energy of hydrogen in the data booklet). Falls to n=1n = 1 release the most energy (UV); falls to n=2n = 2 give the four visible lines; falls to n=3n = 3 give IR.

WORKED EXAMPLE (PAPER 1A STYLE): WHICH TRANSITION EMITS THE HIGHEST-ENERGY PHOTON?

Which electron transition in a hydrogen atom emits the photon with the shortest wavelength? A. n=6→n=2n = 6 \to n = 2 B. n=3→n=2n = 3 \to n = 2 C. n=2→n=1n = 2 \to n = 1 D. n=4→n=3n = 4 \to n = 3

  • Shortest wavelength means highest frequency, which means the largest energy gap (E=hc/λE = hc/\lambda).
  • Gaps shrink as nn rises, so any fall to n=1n = 1 beats every fall to n=2n = 2 or n=3n = 3.
  • Check with En=−1312/n2E_n = -1312/n^2: C gives 1312(1−14)=9841312(1 - \tfrac{1}{4}) = 984 kJ mol−1^{-1}; A gives 1312(14−136)=2921312(\tfrac{1}{4} - \tfrac{1}{36}) = 292 kJ mol−1^{-1}. Answer C (a UV line).

CASE STUDY: AN ELEMENT FOUND IN SUNLIGHT FIRST (HELIUM, 1868)

During the total solar eclipse of 18 August 1868, the French astronomer Jules Janssen observed a bright yellow emission line (about 588 nm) from the Sun's chromosphere that sat close to, but not exactly on, the known sodium lines. Norman Lockyer independently saw the same line in October 1868, concluded that it came from an element unknown on Earth and, with the chemist Edward Frankland, named it helium after the Greek word for the Sun. In 1895 William Ramsay released a gas from the uranium mineral cleveite and found that its spectrum showed the same yellow line: helium existed on Earth too.

  • Concept: every element has its own set of energy levels, so its line spectrum is a fingerprint (S1.3.1).
  • What it shows: an element was identified from light alone, 27 years before anyone held a sample.
  • Exam link: explain why an emission spectrum is a set of lines rather than a continuum, and why no two elements share the same pattern.

6b. CONVERGENCE LIMIT & IE FROM SPECTRA: HL ONLY (S1.3.6).

The limit of convergence at the highest-frequency end of an emission series corresponds to ionisation: the electron is removed completely (n=∞n = \infty).

  • Calculate the first ionisation energy (IE1IE_1) from spectral data:

IE1=NA⋅hflimit=NA⋅hcλlimitIE_1 = N_A \cdot h f_\text{limit} = N_A \cdot \dfrac{hc}{\lambda_\text{limit}}

  • Multiply by NAN_A to convert from energy per atom (J) to energy per mole (J mol−1^{-1}); divide by 10001000 for kJ mol−1^{-1}.

WORKED EXAMPLE (HL ONLY, PAPER 2 STYLE): IONIZATION ENERGY FROM THE CONVERGENCE LIMIT

The lines of the hydrogen series ending at n=1n = 1 converge at a wavelength of 91.291.2 nm. Calculate the first ionization energy of hydrogen in kJ mol−1^{-1}. (h=6.63×10−34h = 6.63 \times 10^{-34} J s, c=3.00×108c = 3.00 \times 10^{8} m s−1^{-1}, NA=6.02×1023N_A = 6.02 \times 10^{23} mol−1^{-1}, all from the data booklet.)

Step 1, energy of one photon. Convert nm to m first: λ=91.2×10−9\lambda = 91.2 \times 10^{-9} m. E=hcλ=6.63×10−34×3.00×10891.2×10−9=2.18×10−18 JE = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{91.2 \times 10^{-9}} = 2.18 \times 10^{-18} \text{ J}

Step 2, scale to one mole. IE1=2.18×10−18×6.02×1023=1.31×106 J mol−1=1.31×103 kJ mol−1IE_1 = 2.18 \times 10^{-18} \times 6.02 \times 10^{23} = 1.31 \times 10^{6} \text{ J mol}^{-1} = 1.31 \times 10^{3} \text{ kJ mol}^{-1}

Three significant figures, matching the data. The data booklet value is 1312 kJ mol−1^{-1}, so the answer is sensible. The common slip is forgetting NAN_A, which leaves an answer in joules per atom.

IONISATION ENERGIES: SL/HL trends + HL extension (S3.1.3 SL/HL; S1.3.6–S1.3.7 HL)

The FIRST IONISATION ENERGY (IE1IE_1) is the energy to remove one mole of electrons from one mole of gaseous atoms. X(g)→X+(g)+e−(ΔH=IE1)X(g) \rightarrow X^+(g) + e^- \quad (\Delta H = IE_1)

TRENDS

  • Across a period: IEIE generally INCREASES (more nuclear charge, similar shielding).
  • Down a group: IEIE DECREASES (outer electron further from nucleus, more shielding).

ANOMALIES IN PERIOD 3 (HL ONLY: S1.3.6, S3.1.7)

  • Mg → Al: IE1IE_1 DROPS (Al's outer electron is in higher-energy 3p; Mg's is in 3s).
  • P → S: IE1IE_1 DROPS (S's paired 3p electron has electron-electron repulsion).

SUCCESSIVE IONISATION ENERGIES (HL ONLY: S1.3.7) reveal electron configuration

  • Large jump after removing all valence electrons → reveals group number.
  • Example: a sequence with a big jump after the 2nd IE → element is in Group 2.

Successive ionization energies of aluminium (HL) Plotting log10_{10} IE makes the pattern visible: three electrons in the outer shell (n=3n = 3), eight in n=2n = 2 and two in n=1n = 1. The big jump after the third electron places aluminium in group 13. Values: NIST Atomic Spectra Database, converted to kJ mol−1^{-1}.

WORKED EXAMPLE (HL ONLY, PAPER 1B STYLE): FIND THE GROUP FROM SUCCESSIVE IEs

The first five ionization energies of a period 3 element X are 578, 1817, 2745, 11577 and 14842 kJ mol−1^{-1}. Deduce the group of X and identify it.

StepIE2/IE1IE_2/IE_1IE3/IE2IE_3/IE_2IE4/IE3IE_4/IE_3IE5/IE4IE_5/IE_4
Ratio3.141.514.221.28
  • Every IE is larger than the last (each electron leaves an increasingly positive ion), so look for the step that is out of proportion.
  • The biggest ratio is between the 3rd and 4th: the 4th electron comes from the full inner shell (n=2n = 2), much closer to the nucleus and less shielded.
  • So X has 3 valence electrons: group 13. In period 3 that is aluminium (the booklet's IE1IE_1 for Al is 578 kJ mol−1^{-1}).

THE MOLE (S1.4)

Chemistry connects atomic-scale events to macroscopic measurements via the MOLE.

  • 1 mole = 6.02×10236.02 \times 10^{23} particles (the AVOGADRO CONSTANT, NAN_A).
  • MOLAR MASS (MM, g mol−1^{-1}): mass of 1 mole, numerically equal to relative atomic/molecular mass.
  • Relationship between mass and moles:

n=mMn = \dfrac{m}{M} Example: how many moles of CO2\text{CO}_2 in 22.0 g? M(CO2)=12+2(16)=44 g mol−1M(\text{CO}_2) = 12 + 2(16) = 44 \text{ g mol}^{-1} n=22.044=0.500 moln = \dfrac{22.0}{44} = 0.500 \text{ mol} This contains 0.500×6.02×1023=3.01×10230.500 \times 6.02 \times 10^{23} = 3.01 \times 10^{23} molecules.

WORKED EXAMPLE (PAPER 2 STYLE): GAS VOLUMES WITH AVOGADRO'S LAW

50.050.0 cm3^3 of propane is burned in 300.0300.0 cm3^3 of oxygen: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l). All volumes are measured at the same temperature and pressure, with water condensed to a liquid. Calculate the total volume of gas remaining.

  • Equal volumes contain equal numbers of molecules (S1.4.6), so volume ratios = mole ratios for gases. No molar volume is needed.
  • Oxygen needed: 5×50.0=250.05 \times 50.0 = 250.0 cm3^3. Oxygen is in excess by 300.0−250.0=50.0300.0 - 250.0 = 50.0 cm3^3; propane is limiting.
  • CO₂ formed: 3×50.0=150.03 \times 50.0 = 150.0 cm3^3. Water is a liquid, so it contributes no gas volume.
  • Gas remaining: 150.0+50.0=200.0150.0 + 50.0 = 200.0 cm3^3 (CO₂ plus unreacted O₂).

EMPIRICAL & MOLECULAR FORMULAS

  • EMPIRICAL = simplest whole-number ratio of atoms (e.g., CH2O\text{CH}_2\text{O}).
  • MOLECULAR = actual number of atoms in a molecule (e.g., glucose C6H12O6=6×CH2O\text{C}_6\text{H}_{12}\text{O}_6 = 6 \times \text{CH}_2\text{O}).

METHOD to find empirical formula

1. Convert each percentage to grams (assume 100 g sample). 2. Divide each mass by atomic mass → moles. 3. Divide all by smallest → ratio. 4. Multiply to get whole numbers. Example: a compound is 40.0% C, 6.67% H, 53.3% O. Find the empirical formula. C:40.012=3.33H:6.671=6.67O:53.316=3.33\text{C}: \dfrac{40.0}{12} = 3.33 \quad \text{H}: \dfrac{6.67}{1} = 6.67 \quad \text{O}: \dfrac{53.3}{16} = 3.33 Divide by smallest (3.33): C = 1, H = 2, O = 1 → empirical formula CH2O\text{CH}_2\text{O}. If the molar mass is 180 g mol−1^{-1}: empirical mass = 30, ratio = 6 → molecular formula C6H12O6\text{C}_6\text{H}_{12}\text{O}_6 (glucose).

CONCENTRATION & SOLUTIONS

MOLAR CONCENTRATION (cc, mol dm−3^{-3})

c=nVc = \dfrac{n}{V} where VV is volume in dm3^3 (= L). 1 dm3^3 = 1000 cm3^3 = 10−310^{-3} m3^3. Example: 4.0 g of NaOH dissolved in 250 cm3^3 of solution. Find the concentration. M(NaOH)=23+16+1=40 g mol−1M(\text{NaOH}) = 23 + 16 + 1 = 40 \text{ g mol}^{-1} n=4.040=0.10 moln = \dfrac{4.0}{40} = 0.10 \text{ mol} c=0.100.250=0.40 mol dm−3c = \dfrac{0.10}{0.250} = 0.40 \text{ mol dm}^{-3}

THE IDEAL GAS LAW & COMBINED GAS LAW (S1.5)

The ideal gas equation ties pressure, volume, amount and temperature together: PV=nRTPV = nRT The combined gas law describes a fixed amount of gas under two sets of conditions (data booklet form): P1V1T1=P2V2T2\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2} Use the combined gas law whenever nn is constant: no need for RR if you keep your unit choices consistent on each side.

Special cases

  • Constant TT → Boyle's law: P1V1=P2V2P_1 V_1 = P_2 V_2.
  • Constant PP → Charles's law: V1/T1=V2/T2V_1/T_1 = V_2/T_2.
  • Constant VV → Gay-Lussac: P1/T1=P2/T2P_1/T_1 = P_2/T_2.
  • PP = pressure (Pa). 1 atm = 101,325 Pa = 1.013 × 105^5 Pa.
  • VV = volume (m3^3). 1 dm3^3 = 10−310^{-3} m3^3.
  • nn = amount (mol).
  • RR = 8.31 J mol−1^{-1} K−1^{-1} (gas constant: given in data booklet).
  • TT = absolute temperature (K). MUST be Kelvin.

At STANDARD TEMPERATURE AND PRESSURE (STP): T = 273.15 K, P = 100 kPa, molar volume of an ideal gas = 22.7 dm3^3 mol−1^{-1}. Example: at STP, what volume does 0.50 mol of nitrogen gas occupy? V=n×Vm=0.50×22.7=11.35 dm3V = n \times V_m = 0.50 \times 22.7 = 11.35 \text{ dm}^3 ASSUMPTIONS of ideal gas behaviour: particles have negligible volume; no intermolecular forces; collisions are elastic. Real gases deviate at HIGH PRESSURE (particle volume matters) or LOW TEMPERATURE (intermolecular forces matter).

WORKED EXAMPLE (PAPER 2 STYLE): COMBINED GAS LAW IN SI UNITS

A gas occupies 2.50×10−32.50 \times 10^{-3} m3^3 at 20.020.0 °C and 1.00×1051.00 \times 10^{5} Pa. It is heated to 80.080.0 °C and the pressure rises to 1.50×1051.50 \times 10^{5} Pa. Calculate the new volume.

  • Temperatures must be in kelvin: T1=293.15T_1 = 293.15 K, T2=353.15T_2 = 353.15 K. Using °C here gives a nonsense answer.
  • Rearrange P1V1T1=P2V2T2\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} for V2V_2:

V2=P1V1T2T1P2=(1.00×105)(2.50×10−3)(353.15)(293.15)(1.50×105)=2.01×10−3 m3V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{(1.00 \times 10^{5})(2.50 \times 10^{-3})(353.15)}{(293.15)(1.50 \times 10^{5})} = 2.01 \times 10^{-3} \text{ m}^3

  • Sense check: heating alone would expand the gas, but the larger pressure squeezes it more, so a small net decrease is reasonable.

PRACTICE: WILL THIS GAS BEHAVE IDEALLY?

Conditions or gasCloser to ideal or further?Reason
He at 300 K, 100 kPaClose to idealtiny atoms, very weak London forces, particles far apart
N₂ at 50 000 kPaFurtherparticle volume is no longer negligible compared with the container
NH₃ near its boiling pointFurtherhydrogen bonding between molecules at low kinetic energy
CH₄ vs H₂O(g) at the same T and PH₂O deviates morepolar molecule with hydrogen bonding

EXAM CONNECTIONS. Be ready to: calculate ArA_r from isotope abundances; write electron configurations using Aufbau, Pauli, Hund; explain emission spectra as evidence for quantisation; convert between mass, moles, particles, concentration, and gas volume; identify groups from successive ionisation energy patterns. The mole concept is the gateway to all stoichiometry.

Key Terms

Atomic Number (ZZ)

The number of protons in an atom's nucleus. Defines the element. Identical for all atoms of the same element.

Mass Number (AA)

Total number of protons + neutrons (nucleons) in an atom's nucleus. A=Z+NA = Z + N.

Isotope

Atoms of the same element (same ZZ) with different numbers of neutrons (different AA). Same chemical properties; different physical properties.

Relative Atomic Mass (ArA_r)

The weighted average mass of an atom of an element relative to 112\tfrac{1}{12} the mass of 12C^{12}\text{C}. Takes isotope abundances into account.

Mole (mol)

The SI unit of amount. One mole contains 6.02×10236.02 \times 10^{23} elementary entities (the Avogadro constant, NAN_A, whose exact defined value is 6.02214076×10236.02214076 \times 10^{23} mol−1^{-1}).

Molar Mass (MM)

The mass of one mole of a substance in g mol−1\text{g mol}^{-1}. Numerically equal to the relative atomic or molecular mass.

Empirical Formula

The simplest whole-number ratio of atoms of each element in a compound. Found from mass percentage data.

Emission Spectrum

Discrete lines of light emitted when excited electrons fall to lower energy levels. Evidence for quantised energy levels in atoms.

Ionisation Energy

Energy required to remove one mole of electrons from one mole of gaseous atoms/ions. Successive IEs provide evidence for electron shell structure.

Exam Tips

  • HL: ionisation energy anomalies (S1.3.6, S3.1.7): always explain Mg→Al (3p higher energy than 3s) and P→S (paired 3p electron repulsion): these are classic short-answer traps.
  • Empirical formula calculation: write out the method systematically: ÷\div atomic mass →÷\to \div smallest →\to round to integers. Do NOT skip steps.
  • Ideal gas equation: TT MUST be in Kelvin. A common error is using °°C. Know the conversion: T(K)=θ(°C)+273T(\text{K}) = \theta(°\text{C}) + 273.
  • Successive IEs: a large jump after the nnth IE indicates the element is in group nn. The jump shows a new, inner shell is being ionised.
  • IB skills connection: Tools & Inquiry: Tool 1: gas collection over water (gas-law experiments) and gravimetric measurement (mass-to-mole conversions). Tool 2: databases and spreadsheets for isotopic abundance and ArA_r calculation; simulations of emission spectra. Tool 3: uncertainty in mass and volume measurements, propagation through n=m/Mn = m/M and PV=nRTPV = nRT. Inquiry 1–3: design and evaluate a gas-collection experiment; collect and process PP–VV–TT data; compare measured ArA_r with literature values.

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