Flashcards · H432 · Topic 2

Foundations in chemistry flashcards

42 cards on Topic 2, subtopics 2.1.1 to 2.2.2. Pick a subtopic or study them all.

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2.1.1 Atomic structure and isotopes

Isotopes: Atoms of the same element with the same proton number but different numbers of neutrons, so different masses. They have the same chemistry because the electron arrangement is the same.
Relative isotopic mass: The mass of an atom of an isotope compared with 1/12th the mass of an atom of carbon-12.
Relative atomic mass: The weighted mean mass of an atom of the element, weighted by the abundance of each isotope as it occurs, compared with 1/12th the mass of an atom of carbon-12.
Protons, neutrons and electrons in ⁵⁶Fe³⁺ (atomic number 26): 26 protons, 30 neutrons, 23 electrons.
Peaks at m/z 46, 47 and 48 with abundances 12.0%, 28.0% and 60.0%. Calculate Ar.: (46 × 12.0 + 47 × 28.0 + 48 × 60.0) ÷ 100 = 4748 ÷ 100 = 47.5 (3 s.f.)
Relative molecular mass vs relative formula mass: Relative molecular mass, Mr: simple molecules (e.g. SO₂). Relative formula mass: giant structures, including ionic compounds (e.g. Ca(NO₃)₂).

2.1.2 Compounds, formulae and equations

Formulae of the nitrate, carbonate, sulfate, hydroxide and ammonium ions: NO₃⁻, CO₃²⁻, SO₄²⁻, OH⁻, NH₄⁺.
Charges on simple ions of Groups 1, 2, 13, 15, 16 and 17: 1+, 2+, 3+, 3−, 2−, 1−.
Formulae of magnesium nitride and ammonium sulfate: Mg₃N₂ (three Mg²⁺ balance two N³⁻). (NH₄)₂SO₄ (two NH₄⁺ balance one SO₄²⁻).
Ionic equation for aqueous barium chloride + aqueous sodium sulfate: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Na⁺ and Cl⁻ are spectator ions.
How do you balance an equation without changing a substance?: Change only the coefficients in front of formulae, never the subscripts: a subscript changes the substance.

2.1.3 Amount of substance

Amount of substance relationships (mass, solution, gas at RTP, particles): n = m ÷ M n = c × V (V in dm³) n = V ÷ 24.0 (V in dm³ at RTP) n = N ÷ Nᴀ (Nᴀ = 6.02 × 10²³ mol⁻¹)
Ideal gas equation and its units: pV = nRT, with p in Pa, V in m³, T in K and R = 8.314 J mol⁻¹ K⁻¹.
A gas occupies 250 cm³ at 100 kPa and 298 K and has mass 0.44 g. Find Mr.: n = (1.00 × 10⁵ × 2.50 × 10⁻⁴) ÷ (8.314 × 298) = 1.01 × 10⁻² mol; Mr = 0.44 ÷ 0.0101 ≈ 44.
Percentage yield vs atom economy: Yield = actual moles ÷ theoretical moles × 100 (what was obtained). Atom economy = Mr of desired product ÷ sum of Mr of all reactants × 100 (what share of reactant mass ends up in the product).
How do you find the limiting reagent?: Convert both reactants to moles and divide each by its coefficient in the equation; the smaller value is the limiting reagent.
Finding x in a hydrated salt, salt·xH₂O: Find the moles of anhydrous salt and the moles of water lost on heating, then take the simplest whole-number ratio.

2.1.4 Acids

Strong acid vs weak acid: A strong acid is almost completely dissociated in solution (HCl → H⁺ + Cl⁻). A weak acid is only partially dissociated (CH₃COOH ⇌ CH₃COO⁻ + H⁺).
Base vs alkali: A base neutralises an acid. An alkali is a base that dissolves in water, releasing OH⁻ ions. All alkalis are bases; CuO and CaCO₃ are bases but not alkalis.
Ionic equation for neutralisation of an acid by an alkali: H⁺(aq) + OH⁻(aq) → H₂O(l)
Why does ammonia + an acid give no water?: NH₃ accepts H⁺ to form NH₄⁺, giving an ammonium salt only, e.g. NH₃ + HCl → NH₄Cl.
Mass of Na₂CO₃ (M = 106.0) for 250.0 cm³ of a 0.0500 mol dm⁻³ standard solution: 0.2500 × 0.0500 = 0.01250 mol; 0.01250 × 106.0 = 1.325 g
Back titration: when and how?: For an insoluble or slow-reacting base. Add a known excess of acid, titrate the leftover acid with standard alkali; moles reacting with the sample = moles added − moles left.

2.1.5 Redox

Oxidation and reduction in terms of electrons and oxidation number: Oxidation: loss of electrons; oxidation number goes up. Reduction: gain of electrons; oxidation number goes down.
Oxidation number of H in KH, and of O in Li₂O₂: H in a metal hydride: −1. O in a peroxide: −1.
Oxidation number of P in P₄O₆: 4x + 6(−2) = 0, so x = +3 per P atom.
Formulae of the chlorate(I) and chlorate(V) ions, and sulfate(IV): Chlorate(I): ClO⁻. Chlorate(V): ClO₃⁻. Sulfate(IV): SO₃²⁻.
Balanced equation: aluminium with ethanoic acid: 2Al(s) + 6CH₃COOH(aq) → 2(CH₃COO)₃Al(aq) + 3H₂(g). Al is oxidised (0 → +3); H is reduced (+1 → 0).
Iron with a dilute acid: iron(II) or iron(III) salt?: Iron(II), e.g. Fe + 2HCl → FeCl₂ + H₂.

2.2.1 Electron structure

Atomic orbital: A region around the nucleus that can hold up to two electrons, with opposite spins.
Shapes of s- and p-orbitals: s: a sphere centred on the nucleus. p: a dumbbell with two lobes either side of the nucleus; three p-orbitals at 90° to each other.
Maximum number of electrons in shells n = 1 to 4: 2, 8, 18 and 32 (2n² electrons in n² orbitals).
Order of filling sub-shells up to 4p: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. Electrons occupy orbitals singly before pairing.
Isoelectronic: Having the same electron configuration (for example Mg²⁺, Ne and F⁻).
Electron configuration of Ge²⁺ (Z = 32): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s². The two 4p electrons are removed; 3d is never removed from a p-block ion.

2.2.2 Bonding and structure

Covalent bond: The strong electrostatic attraction between a shared pair of electrons and the nuclei of the bonded atoms.
Dative covalent (coordinate) bond: A covalent bond in which the shared pair is supplied entirely by one of the two atoms. Example: the third pair in CO comes from oxygen.
Shapes and angles: NH₃ and H₂O: NH₃: pyramidal, 107° (3 bonding pairs, 1 lone pair). H₂O: non-linear, 104.5° (2 bonding pairs, 2 lone pairs). Each lone pair reduces the angle by about 2.5°.
Why is CO₂ non-polar although its bonds are polar?: It is linear, so the two bond dipoles are equal and opposite and cancel.
Van der Waals’ forces vs London forces: Van der Waals’ forces cover both permanent dipole–dipole and induced dipole–dipole interactions. London (dispersion) forces are induced dipole–dipole interactions only.
Why does water have a higher boiling point than predicted from the other Group 16 hydrides?: Hydrogen bonds between H₂O molecules must be broken as well as London forces, which needs more energy.
Properties of a simple molecular substance: Low melting and boiling points (only intermolecular forces are overcome); no conductivity in any state (no mobile charged particles); solubility follows "like dissolves like".