Flashcards · 9701 · AS Topic 2

Atoms, molecules and stoichiometry (AS Level) flashcards

23 cards on AS Topic 2, subtopics 2.1 to 2.4. Pick a subtopic or study them all.

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Unified atomic mass unit: One twelfth of the mass of an atom of carbon-12.
Relative atomic mass, Ar: The weighted mean mass of an atom of an element (taking account of its isotopes), compared with one twelfth of the mass of a carbon-12 atom.
Relative isotopic mass: The mass of an atom of a particular isotope compared with one twelfth of the mass of a carbon-12 atom.
Relative molecular mass and relative formula mass: Relative molecular mass, Mr: the mass of a molecule compared with one twelfth of the mass of a carbon-12 atom. Relative formula mass: the same idea for one formula unit of an ionic compound.
Boron is 20% ¹⁰B and 80% ¹¹B (illustrative values). Calculate Ar.: Ar = (10 × 20 + 11 × 80) ÷ 100 = 1080 ÷ 100 = 10.8

The mole: The amount of substance containing 6.02 × 10²³ particles (the Avogadro constant, L) of that substance.
Link between moles, mass and molar mass: n = m ÷ M n in mol, m in g, M in g mol⁻¹.
Number of particles from moles: N = n × 6.02 × 10²³
Molar volume of a gas at rtp, and when it applies: 24 dm³ mol⁻¹, so n = V ÷ 24 (V in dm³). Gases at rtp only, never solids or liquids.
Moles of solute in 25.0 cm³ of 0.200 mol dm⁻³ solution: V = 0.0250 dm³; n = cV = 0.200 × 0.0250 = 5.00 × 10⁻³ mol

Empirical formula and molecular formula: Empirical: the simplest whole-number ratio of atoms of each element in a compound. Molecular: the actual number of atoms of each element in a molecule.
Method: empirical formula from percentage composition: 1 Take the mass or % of each element. 2 Divide by Ar to get moles. 3 Divide by the smallest to get the ratio. 4 Multiply to whole numbers if needed.
How do you get the molecular formula from the empirical formula?: Molecular formula = empirical formula × n, where n = Mr(molecular) ÷ Mr(empirical).
Formulas and charges of the polyatomic ions 9701 expects you to recall: NO₃⁻, CO₃²⁻, SO₄²⁻, OH⁻, NH₄⁺, HCO₃⁻, PO₄³⁻ (plus the monatomic ions Zn²⁺ and Ag⁺).
Ionic equation: An equation showing only the species that change, omitting spectator ions. Precipitates, gases and covalent molecules stay as full formulae. e.g. Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
Anhydrous, hydrated and water of crystallisation: Hydrated: a crystalline compound containing water molecules in its structure (water of crystallisation), e.g. CuSO₄·5H₂O. Anhydrous: the compound without its water of crystallisation.
Write the formula of iron(III) oxide from its ions: Fe³⁺ and O²⁻. The lowest common multiple of 3 and 2 is 6, so 2 Fe³⁺ and 3 O²⁻: Fe₂O₃.

The four-step method for reacting-quantity calculations: 1 Balance the equation. 2 Convert the known quantity to moles. 3 Apply the mole ratio. 4 Convert back to the quantity asked for.
How do you find the limiting reagent?: Convert both reactants to moles, divide each by its coefficient in the balanced equation; the smaller value is limiting. Base all product calculations on it.
Percentage yield: Percentage yield = (actual yield ÷ theoretical yield) × 100. Theoretical yield comes from the limiting reagent. A value above 100% means an impure or wet product, or an error.
4.00 g H₂ reacts with 4.00 g O₂ (2H₂ + O₂ → 2H₂O). Which is limiting, and what mass of water forms?: H₂: 2.00 mol ÷ 2 = 1.00. O₂: 0.125 mol ÷ 1 = 0.125. O₂ is limiting. Water = 2 × 0.125 = 0.250 mol = 4.50 g.
Concentration, moles and volume of a solution: n = cV, with c in mol dm⁻³ and V in dm³ (cm³ ÷ 1000).
Significant figures in 9701 calculations: Give answers to the number of significant figures given or asked for in the question, and round only at the end, not part-way through.