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IB Chemistry: Structure 1 · Models of the particulate nature of matter

Atoms, isotopes, electron configuration, the mole. 5 subtopics, 25 key terms, full free lesson below.

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In this topic: S1.1 · S1.2 · S1.3 · S1.4 · S1.5

S1.1Introduction to the particulate nature of matter

Key terms

Element
A substance made of only one type of atom; it cannot be broken down chemically.
Compound
Two or more elements chemically combined in a fixed ratio, with different properties from its elements.
Mixture
Two or more substances physically combined in no fixed ratio; each keeps its own properties.
Homogeneous mixture
Uniform composition throughout, in a single phase (e.g. salt solution, air).
Heterogeneous mixture
Non-uniform composition, with visibly different phases (e.g. sand in water).
Kinetic molecular theory
Matter is made of particles in constant random motion; their average kinetic energy is proportional to absolute temperature.
\[T(\text{K})=\theta(^{\circ}\text{C})+273\qquad R_F=\frac{\text{distance moved by spot}}{\text{distance moved by solvent front}}\]

Worked example

In paper chromatography a spot moves 3.6 cm while the solvent front moves 8.0 cm. Find the \(R_F\) value.

  1. \(R_F=\dfrac{3.6}{8.0}=\mathbf{0.45}\).
  2. \(R_F\) has no units and is always between 0 and 1. Comparing it with known values identifies the substance (same solvent and paper).

Exam tip: Name the physical property each separation technique relies on (solubility, boiling point, particle size): that is usually where the marks are.

S1.2The nuclear atom

Key terms

Atomic number (Z)
Number of protons in the nucleus; it defines the element.
Mass number (A)
Number of protons plus neutrons.
Isotopes
Atoms of the same element with different numbers of neutrons.
Relative atomic mass (Ar)
Weighted mean mass of an element's atoms compared with 1/12 of a carbon-12 atom.
Ion
A charged particle formed when an atom gains or loses electrons.
\[\text{neutrons}=A-Z\qquad {}^{A}_{Z}\text{X}^{\text{charge}}\]

HL A mass spectrometer separates isotopes by their mass-to-charge ratio (m/z). The heights of the peaks give relative abundances, from which \(A_r\) is calculated. Fragments in the spectra of compounds are used to identify molecules (Structure 3.2).

Worked example

Chlorine is 75.8% chlorine-35 (mass 34.97) and 24.2% chlorine-37 (mass 36.97). Calculate its relative atomic mass.

  1. \(A_r=\dfrac{(75.8\times34.97)+(24.2\times36.97)}{100}\)
  2. \(A_r=\dfrac{2650.7+894.7}{100}=\mathbf{35.45}\)

Exam tip: Check your Ar answer lies between the isotope masses and is closer to the more abundant one.

S1.3Electron configurations

Key terms

Emission spectrum
Lines of specific wavelengths emitted when excited electrons fall to lower energy levels.
Orbital
A region of space where there is a high probability of finding an electron; holds up to 2 electrons with opposite spins.
Aufbau principle
Electrons fill orbitals of lowest energy first.
Hund's rule
Electrons occupy orbitals of equal energy singly, with parallel spins, before pairing.
Pauli exclusion principle
No orbital holds more than two electrons, and they must have opposite spins.
Ionization energy
Minimum energy to remove one mole of electrons from one mole of gaseous atoms.
\[\Delta E=hf=\frac{hc}{\lambda}\qquad c=f\lambda\]

HL The ionization energy can be found from the convergence limit of the Lyman series: \(IE=\dfrac{hc}{\lambda}\times N_A\). Successive ionization energies jump when an electron is removed from a new, inner shell, revealing the group of an element. First ionization energies across a period show dips at groups 13 (new p sublevel) and 16 (first paired p electron, extra repulsion).

Worked example

Write the full electron configuration of iron (Z = 26) and the configuration of Fe³⁺.

  1. Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶, or [Ar] 3d⁶ 4s².
  2. Fe³⁺ loses the two 4s electrons first, then one 3d electron: [Ar] 3d⁵.

Exam tip: In orbital diagrams, fill each box of a sublevel with one up-arrow before adding any down-arrows (Hund's rule).

S1.4Counting particles by mass: the mole

Key terms

Mole
Amount of substance containing 6.02 × 10²³ particles (Avogadro's constant, N_A).
Molar mass (M)
Mass of one mole of a substance, in g mol⁻¹.
Empirical formula
Simplest whole-number ratio of atoms of each element in a compound.
Molecular formula
Actual number of atoms of each element in a molecule.
Concentration
Amount of solute per volume of solution, in mol dm⁻³ (or g dm⁻³).
\[n=\frac{m}{M}\qquad N=nN_A\qquad c=\frac{n}{V}\qquad 1\text{ dm}^{3}=1000\text{ cm}^{3}\]

Worked example

A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its \(M_r\) is 180. Find its empirical and molecular formulas.

  1. C: \(\dfrac{40.0}{12.01}=3.33\); H: \(\dfrac{6.7}{1.01}=6.63\); O: \(\dfrac{53.3}{16.00}=3.33\).
  2. Divide by 3.33: C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O (\(M_r=30\)).
  3. \(180\div30=6\), so the molecular formula is C₆H₁₂O₆.

Exam tip: Keep at least 3 significant figures in ratio steps: rounding too early turns 1.5 into 2 and gives the wrong formula.

S1.5Ideal gases

Key terms

Ideal gas
A model gas whose particles have negligible volume, collide elastically and have no intermolecular forces.
Molar volume
Volume occupied by one mole of gas: 22.7 dm³ mol⁻¹ at STP (273 K, 100 kPa).
STP
Standard temperature and pressure: 273 K and 100 kPa.
\[pV=nRT\qquad \frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}\qquad V=n\times22.7\text{ dm}^{3}\text{ (at STP)}\qquad R=8.31\text{ J K}^{-1}\text{ mol}^{-1}\]

Worked example

0.320 g of a gas occupies 0.250 dm³ at 100 kPa and 300 K. Find its molar mass and suggest its identity.

  1. \(n=\dfrac{pV}{RT}=\dfrac{1.00\times10^{5}\times2.50\times10^{-4}}{8.31\times300}=1.00\times10^{-2}\text{ mol}\).
  2. \(M=\dfrac{0.320}{1.00\times10^{-2}}=\mathbf{32.0\text{ g mol}^{-1}}\), so it is probably oxygen, O₂.

Exam tip: Most pV = nRT errors are units: convert kPa to Pa and dm³ to m³ before substituting.

Practise Structure 1 · Models of the particulate nature of matter with exam-style questions →

Structure 2 · Models of bonding and structure →

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