Flashcards · 9620 · AS Topic 2

Amount of substance (AS Level) flashcards

22 cards on AS Topic 2, subtopics 3.1.2.1 to 3.1.2.5. Pick a subtopic or study them all.

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Relative atomic mass, Ar: The weighted mean mass of an atom of an element (allowing for its isotopes) divided by one twelfth of the mass of an atom of carbon-12.
Relative molecular mass, Mr: The mean mass of a molecule divided by one twelfth of the mass of an atom of carbon-12.
Why is “relative formula mass” used for ionic compounds?: Ionic compounds exist as lattices, not as separate molecules.
Why do Ar and Mr have no units?: Each is a ratio of masses (compared with one twelfth of the mass of a carbon-12 atom).

The Avogadro constant, L: The number of entities (atoms, molecules, ions, electrons…) in one mole of a substance: 6.02 × 10²³ mol⁻¹.
Equations linking amount (n) with mass, number of particles and concentration: n = m ÷ Mr N = n × L c = n ÷ V (c in mol dm⁻³, V in dm³)
0.0500 mol of Na₂CO₃ is dissolved to make 250 cm³ of solution. Concentration?: c = 0.0500 ÷ 0.250 = 0.200 mol dm⁻³
How many moles of ions are in 0.0200 mol of Al₂(SO₄)₃?: 0.0400 mol Al³⁺ + 0.0600 mol SO₄²⁻ = 0.100 mol of ions.

The ideal gas equation, with SI units: pV = nRT p in Pa, V in m³, n in mol, R = 8.31 J K⁻¹ mol⁻¹, T in K.
Unit conversions for pV = nRT: kPa × 1000 = Pa cm³ × 10⁻⁶ = m³ dm³ × 10⁻³ = m³ K = °C + 273
Find p for 0.0500 mol of gas at 400 K in a 2.00 dm³ vessel (R = 8.31 J K⁻¹ mol⁻¹): p = nRT ÷ V = (0.0500 × 8.31 × 400) ÷ (2.00 × 10⁻³) = 83 100 Pa = 83.1 kPa
How do you find Mr of a gas from pV = nRT?: Find n = pV ÷ RT (all in SI units), then Mr = mass ÷ n.

Empirical formula and molecular formula: Empirical: the smallest whole-number ratio of atoms of each element in a compound. Molecular: the actual number of atoms of each element in one molecule (a whole-number multiple of the empirical formula).
Method for an empirical formula from masses or percentages: Divide each mass (or % as grams in 100 g) by Ar to get moles, divide all by the smallest, then multiply to whole numbers if needed (×2 for .5, ×3 for .33 or .67).
An oxide is 69.9% Fe and 30.1% O. Empirical formula?: Fe: 69.9 ÷ 55.8 = 1.25; O: 30.1 ÷ 16.0 = 1.88. Ratio 1 : 1.50, multiply by 2 → Fe₂O₃. Do not round 1.50 down.
Empirical formula CH₂O, Mr = 60.0. Molecular formula?: CH₂O has mass 30.0. 60.0 ÷ 30.0 = 2, so C₂H₄O₂.

Percentage yield: (actual mass of product ÷ theoretical mass) × 100
Percentage atom economy: (molecular mass of desired product ÷ sum of molecular masses of all reactants) × 100
Yield vs atom economy: Percentage yield depends on what happened in one experiment. Atom economy is fixed by the balanced equation.
Advantages of a high atom economy: Economic: less raw material ends up as unwanted product, and waste disposal and separation costs are lower. Environmental/ethical: fewer natural resources used and less waste produced.
Ionic equation and spectator ions: An ionic equation shows only the species that change, for example Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s). Spectator ions are present unchanged on both sides and are left out.
Required practical 1: making up a volumetric (standard) solution: Weigh the solid by difference, dissolve in a beaker, transfer with rinsings to a volumetric flask, make up to the graduation line, then invert to mix.