Flashcards · 7405 · AS Topic 2

Amount of substance (AS Level) flashcards

19 cards on AS Topic 2, subtopics 3.1.2.1 to 3.1.2.4. 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, accounting for isotopic abundance, on a scale where an atom of carbon-12 has a mass of exactly 12.
Relative molecular mass (Mr): The weighted mean mass of a molecule on the carbon-12 scale; the sum of the Ar values of its atoms. Used for covalently bonded molecules.
Relative formula mass is used for…: Ionic compounds (e.g. NaCl) — use this term, not Mr.
Why do Ar and Mr have no units?: They are ratios: masses relative to one-twelfth of the mass of a carbon-12 atom, not actual masses.

The three mole equations: moles = mass ÷ Mr moles = concentration (mol dm⁻³) × volume (dm³) number of particles = moles × Avogadro constant
The Avogadro constant: The number of particles in one mole of a substance, about 6.02 × 10²³ mol⁻¹. You do not need to recall its value; it is given when needed.
How many molecules are in 8.8 g of CO₂ (Mr 44)? Take L = 6.02 × 10²³ mol⁻¹.: 8.8 ÷ 44 = 0.20 mol; 0.20 × 6.02 × 10²³ = 1.2 × 10²³ molecules (2 s.f.).
How many ions are in 250 cm³ of 0.20 mol dm⁻³ NaCl? Take L = 6.02 × 10²³ mol⁻¹.: 0.20 × 0.250 = 0.050 mol NaCl → 3.0 × 10²² formula units. Each gives 2 ions, so 6.0 × 10²² ions.
To how many significant figures should a calculated answer be given?: To the number of significant figures in the least precise (least accurate) piece of data used.

The ideal gas equation and its SI units: pV = nRT p in Pa, V in m³, n in mol, T in K, R in J K⁻¹ mol⁻¹ (8.31, given in the question).
Convert to SI: 600 cm³, 2.5 dm³, 101 kPa, 25 °C: 600 cm³ = 6.00 × 10⁻⁴ m³ 2.5 dm³ = 2.5 × 10⁻³ m³ 101 kPa = 1.01 × 10⁵ Pa 25 °C = 298 K
0.0250 mol of argon is in 600 cm³ at 300 K. Find the pressure in kPa (R = 8.31 J K⁻¹ mol⁻¹).: p = nRT ÷ V = (0.0250 × 8.31 × 300) ÷ (6.00 × 10⁻⁴) = 1.04 × 10⁵ Pa = 104 kPa
How do you find the Mr of a volatile liquid with a heated gas syringe?: Inject a weighed mass of liquid into a syringe above its boiling point and read the vapour volume. Then M = mRT ÷ pV, with T the syringe temperature and p the room pressure.
Classic unit trap when converting cm³ for pV = nRT: cm³ to m³ is × 10⁻⁶. Dividing by 1000 only gives dm³.

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 the formula.
Method: empirical formula from percentage composition: Treat percentages as masses in 100 g. Divide each by Ar, divide all by the smallest, then clear fractions (× 2 for .5, × 3 for .33 or .67, × 4 for .25 or .75).
A carbide is 75.0% Al and 25.0% C by mass (Al 27.0, C 12.0). Find its empirical formula.: Al 75.0 ÷ 27.0 = 2.778; C 25.0 ÷ 12.0 = 2.083. Ratio 1.333 : 1, × 3 gives 4 : 3. Al₄C₃.
How do you get the molecular formula from the empirical formula?: Multiplier = Mr ÷ empirical formula mass. Multiply every subscript by it. Composition alone cannot do this: CF₂ and C₂F₄ have the same percentages.
Finding an oxide’s formula: how do you know the reaction is complete?: Heat to constant mass. Then mass of oxygen = mass of oxide − mass of metal.