Flashcards · IB Diploma Programme Chemistry · Topic 1

Structure 1 – Models of the particulate nature of matter flashcards

34 cards on Topic 1, subtopics 1.1 to 1.5. Pick a subtopic or study them all.

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1.1 Introduction to the particulate nature of matter

Element, compound and mixture: Element: one type of atom only. Compound: atoms of different elements chemically bonded in a fixed ratio; separated only by a chemical reaction. Mixture: not chemically bonded, no fixed ratio; separable by physical methods.
Homogeneous vs heterogeneous mixture: Homogeneous: uniform composition, one phase (vinegar, bronze, filtered air). Heterogeneous: non-uniform, two or more phases (muddy water).
Recrystallization: method: Dissolve the impure solid in the minimum of hot solvent; filter hot to remove insoluble impurities; cool so the product crystallizes (soluble impurities stay in solution); filter, wash with a little cold solvent, dry.
Solvent front 9.0 cm, spot 2.7 cm from the base line. Rf?: Rf = 2.7 ÷ 9.0 = 0.30. Rf always lies between 0 and 1; both distances are measured from the base line.
Evaporation vs boiling: Evaporation: at the surface only, below the boiling point; the most energetic particles escape, so the liquid cools. Boiling: throughout the liquid, at a fixed temperature for a given external pressure.
What does kelvin temperature measure? Convert −18 °C to kelvin.: The average kinetic energy of the particles (double the kelvin temperature, double the average kinetic energy). −18 + 273.15 = 255.15 K.
Why is a heating curve flat during melting and boiling?: The energy supplied overcomes the attractions between particles, so the average kinetic energy and temperature stay constant while two states coexist.

1.2 The nuclear atom

Relative charge, relative mass and location of the three subatomic particles: Proton: +1, 1, nucleus. Neutron: 0, 1, nucleus. Electron: −1, negligible (about 1/1836), outside the nucleus.
Protons, neutrons and electrons in ¹¹⁹₅₀Sn²⁺: 50 protons, 119 − 50 = 69 neutrons, 50 − 2 = 48 electrons.
Why do isotopes have the same chemical properties but different physical properties?: Same number of protons, so the same number and arrangement of electrons, which decide reactions. Different numbers of neutrons, so different masses: density, melting and boiling points and diffusion rate differ.
Rhenium is 37.4% ¹⁸⁵Re (184.95) and 62.6% ¹⁸⁷Re (186.96). Ar?: (184.95 × 37.4 + 186.96 × 62.6) ÷ 100 = 186.21
Isotopes of masses 64.00 and 66.00 give Ar = 64.70. Percentage abundances?: 64.00x + 66.00(1 − x) = 64.70, so 66.00 − 2.00x = 64.70 and x = 0.650. 65.0% of mass 64.00 and 35.0% of mass 66.00.
Reading a mass spectrum of an element: Each peak is one isotope (m/z = isotope mass for 1+ ions); peak height is relative abundance. Ar is the weighted mean of the m/z values. (HL only)

1.3 Electron configurations

Hydrogen emission spectrum: which transitions give ultraviolet, visible and infrared lines?: Falls to n = 1: ultraviolet. To n = 2: visible. To n = 3: infrared. In each series the lines converge at higher frequency, because the levels get closer together as n increases.
Orbitals per sublevel, and the maximum electrons in main level n: s 1, p 3, d 5, f 7 orbitals. Main level n holds n² orbitals and up to 2n² electrons. Each orbital holds two electrons of opposite spin.
Aufbau principle, Pauli exclusion principle, Hund’s rule: Aufbau: fill the lowest-energy orbital first. Pauli: at most two electrons per orbital, with opposite spins. Hund: occupy equal-energy orbitals singly with parallel spins before pairing.
Configurations of chromium and copper: Cr: [Ar] 3d⁵ 4s¹ (half-filled 3d). Cu: [Ar] 3d¹⁰ 4s¹ (filled 3d). Each gives a lower overall energy than the expected form.
Filling order vs loss order: 4s fills before 3d but empties before 3d.
Configuration of Ni²⁺ (Z = 28), and its unpaired electrons: Ni: [Ar] 3d⁸ 4s². Ni²⁺: [Ar] 3d⁸ (4s emptied first), with 2 unpaired electrons.
Lithium’s convergence limit is at 1.30 × 10¹⁵ Hz (h = 6.63 × 10⁻³⁴ J s). First ionization energy?: E = hf = 6.63 × 10⁻³⁴ × 1.30 × 10¹⁵ = 8.62 × 10⁻¹⁹ J. × 6.02 × 10²³ = 5.19 × 10⁵ J mol⁻¹ = 519 kJ mol⁻¹. (HL only)
Successive ionization energies of boron: 801, 2427, 3660, 25 026 kJ mol⁻¹. Group?: The biggest increase comes after the third electron, so boron has three outer electrons: group 13. Look for the biggest increase, not the biggest value. (HL only)

1.4 Counting particles by mass: The mole

The mole and the Avogadro constant: The mole is the SI unit for amount of substance; one mole contains NA elementary entities. NA = 6.02 × 10²³ mol⁻¹.
Ar and Mr versus molar mass M: The same number, but Ar and Mr are ratios with no units, and M is the mass of one mole in g mol⁻¹. n = m ÷ M.
How many sodium ions are in 0.0200 mol of Na₂CO₃?: 2 × 0.0200 = 0.0400 mol Na⁺; 0.0400 × 6.02 × 10²³ = 2.41 × 10²² ions.
5.55 g of CaCl₂ (M = 110.98 g mol⁻¹) is made up to 500.0 cm³. [CaCl₂] and [Cl⁻]?: n = 5.55 ÷ 110.98 = 0.0500 mol. [CaCl₂] = 0.0500 ÷ 0.5000 = 0.100 mol dm⁻³; [Cl⁻] = 0.200 mol dm⁻³.
Empirical formula: what to do with a mole ratio ending in .5, .33 or .25: .5: multiply by 2. .33 or .67: multiply by 3. .25 or .75: multiply by 4. Molecular formula multiple = M ÷ empirical formula mass.
Avogadro's law: At equal temperature and pressure, equal volumes of gases contain equal numbers of molecules, so gas volume ratios equal mole ratios.
Dilution: The amount of solute stays the same, so C₁V₁ = C₂V₂. Concentration uses the final volume of solution.

1.5 Ideal gases

Assumptions of the ideal gas model: Particles move constantly and randomly; their own volume is negligible; there are no intermolecular forces; collisions are elastic.
When and why do real gases deviate from ideal behaviour?: High pressure: the particles’ own volume is no longer negligible. Low temperature: intermolecular attractions are no longer negligible. Most ideal at high T, low P, with small non-polar particles.
Molar volume of an ideal gas at STP: 22.7 dm³ mol⁻¹ at 273.15 K and 100 kPa. Under other conditions Vm = RT ÷ P.
PV = nRT: the units to use: P in Pa, V in m³, T in K, R = 8.31 J K⁻¹ mol⁻¹. kPa × 10³ = Pa; dm³ × 10⁻³ = m³; cm³ × 10⁻⁶ = m³.
A gas at 100 kPa and 20.0 °C is compressed to half its volume; the pressure becomes 230 kPa. New temperature?: T₂ = T₁ × P₂V₂ ÷ (P₁V₁) = 293.15 × (230 × 0.5) ÷ (100 × 1) = 337 K (64.0 °C).
Volume against temperature in °C for a fixed mass of gas at constant pressure: A straight line that does not pass through the origin; extrapolated to V = 0 it meets the temperature axis near −273 °C (absolute zero). V is proportional to T only in kelvin.