Topic 1
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Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, follow the link back to the full page.
Constants and conversions
| Quantity | Value |
|---|---|
| Avogadro constant, L | 6.02 × 10²³ mol⁻¹ |
| molar volume at RTP | 24.0 dm³ mol⁻¹ = 24 000 cm³ mol⁻¹ |
| gas constant, R | 8.31 J mol⁻¹ K⁻¹ |
| temperature | K = °C + 273 |
| pressure | Pa = kPa × 10³ |
| volume | dm³ = cm³ ÷ 1000; m³ = dm³ ÷ 1000 = cm³ × 10⁻⁶ |
| ppm and % | ppm = % × 10⁴ |
The equations
| To find | Use | Page |
|---|---|---|
| amount from mass | n = m ÷ M | Relative mass and the mole |
| number of particles | N = n × L | Relative mass and the mole |
| amount in solution | n = c × V (V in dm³) | Solutions and ppm |
| mass concentration | g dm⁻³ = mol dm⁻³ × M | Solutions and ppm |
| parts per million | part ÷ whole × 10⁶ | Solutions and ppm |
| amount of gas at RTP | n = V ÷ 24.0 (V in dm³) | Gas volumes |
| amount of gas, any conditions | pV = nRT (Pa, m³, K) | Gas volumes |
| molar mass of a vapour | M = mRT ÷ pV | Gas volumes |
| percentage yield | actual ÷ theoretical × 100 | Yield and atom economy |
| atom economy | M(desired) ÷ ΣM(all products) × 100 | Yield and atom economy |
Definitions to learn word for word
- Relative atomic mass: the weighted mean mass of an atom of an element, relative to one twelfth of the mass of an atom of carbon-12.
- Mole: the amount of substance containing 6.02 × 10²³ particles, the same number as atoms in exactly 12 g of carbon-12.
- Molar mass: the mass per mole of a substance, in g mol⁻¹.
- Empirical formula: the simplest whole-number ratio of atoms of each element in a compound.
- Molecular formula: the actual number of atoms of each element in one molecule.
- Atom economy: the percentage, by mass, of the products that is the desired product.
Methods
- Reacting masses: mass → moles → ratio → moles → mass.
- Limiting reactant: convert both to moles, allow for the ratio, and the one that runs out first decides the product.
- Empirical formula: mass of each element → ÷ → ÷ smallest → multiply to whole numbers.
- Molecular formula: ÷ mass of one empirical unit = n.
- Ionic equation: full equation → split aqueous ionic substances → cancel spectators → check atoms and charge.
- Confirming an equation: predict the result of each candidate, compare with the measurement in moles.
The ten traps
- Changing a subscript instead of a coefficient to balance an equation.
- Writing H₂O(aq), or (aq) for a precipitate.
- Splitting a solid (CaCO₃, CuO, AgCl) into ions in an ionic equation.
- Using M = 16.0 for oxygen gas instead of 32.0.
- Multiplying M by the coefficient in a reacting-mass calculation.
- Rounding 1.5 or 1.33 to a whole number in an empirical formula.
- Leaving volumes in cm³ in c = n ÷ V or pV = nRT.
- Using kPa or °C in pV = nRT.
- Forgetting the coefficients in atom economy.
- Naming a product ("hydrogen forms") when asked for an observation ("fizzing").
Mixed test
Try these with no notes, then check the answers.
- Write the ionic equation, with state symbols, for zinc reacting with dilute hydrochloric acid.
- What mass of calcium carbonate is needed to make 2.40 dm³ of CO₂ at RTP, with excess acid?
- 0.200 g of a volatile liquid gives 61.0 cm³ of vapour at 373 K and 101 kPa. Calculate its molar mass.
- A compound is 52.2% carbon, 13.0% hydrogen and 34.8% oxygen by mass. Find its empirical formula.
- 25.0 cm³ of a solution contains 0.00500 mol of sodium carbonate. Give its concentration in mol dm⁻³ and in g dm⁻³.
- Calculate the atom economy for making ammonia: N₂ + 3H₂ → 2NH₃. Explain your answer.
- Barium chloride solution is added to a solution of magnesium sulfate. Give the observation and the ionic equation.
- Seawater contains about 0.40 g of potassium in every 1.0 kg. Express this in ppm.
Answers cover these until you have tried the questions
- n(CO₂) = 2.40 ÷ 24.0 = 0.100 mol = n(CaCO₃). m = 0.100 × 100.1 = 10.0 g
- n = (101 000 × 61.0 × 10⁻⁶) ÷ (8.31 × 373) = 1.99 × 10⁻³ mol. M = 0.200 ÷ 1.99 × 10⁻³ = 101 g mol⁻¹
- C: 52.2 ÷ 12.0 = 4.35. H: 13.0 ÷ 1.0 = 13.0. O: 34.8 ÷ 16.0 = 2.175. ÷ 2.175: 2 : 6 : 1, so C₂H₆O.
- c = 0.00500 ÷ 0.0250 = 0.200 mol dm⁻³. M(Na₂CO₃) = 106.0 g mol⁻¹, so 0.200 × 106.0 = 21.2 g dm⁻³
- 100%. Ammonia is the only product, so all the mass of the reactants ends up in the desired product.
- A white precipitate.
- 0.40 ÷ 1000 × 10⁶ = 400 ppm (4.0 × 10² ppm)