Reacting masses and volumes
Section: 2 Atoms, Molecules and Stoichiometry | Syllabus: Cambridge AS Level Chemistry (9701)
Reacting Masses
Stoichiometric coefficients in a balanced equation give the molar ratios in which substances react and are produced. These ratios are used to calculate reacting masses.
- Step 1: Write the balanced equation.
- Step 2: Convert given mass to moles: n = m ÷ M
- Step 3: Use the molar ratio from the equation to find moles of the required substance.
- Step 4: Convert back to mass: m = n × M
CaCO₃(s) → CaO(s) + CO₂(g)
M(CaCO₃) = 100 g mol⁻¹ → n(CaCO₃) = 10 ÷ 100 = 0.10 mol
Ratio 1:1 → n(CO₂) = 0.10 mol
m(CO₂) = 0.10 × 44 = 4.4 g
Percentage Yield
Percentage Yield The percentage of the theoretical maximum yield that is actually obtained in a reaction.% yield = (actual yield ÷ theoretical yield) × 100
- Theoretical yield is the maximum mass of product calculated from the stoichiometry, assuming complete reaction.
- Actual yield is the mass of product collected in the experiment.
- Yield is always less than 100% due to: reversible reactions, side reactions, losses during purification, incomplete reaction.
Theoretical yield of CaO from 10 g CaCO₃ = 5.6 g. Actual yield obtained = 4.8 g.
% yield = (4.8 ÷ 5.6) × 100 = 85.7%
Volumes of Gases
Molar Gas Volume The volume occupied by one mole of any ideal gas under specified conditions. At s.t.p. (0 °C, 1 atm) = 22.4 dm³ mol⁻¹. At r.t.p. (25 °C, 1 atm) = 24.0 dm³ mol⁻¹.- All gases occupy the same molar volume under the same conditions of temperature and pressure (ideal gas assumption).
- Volume of gas: V = n × molar gas volume
- Moles of gas: n = V ÷ molar gas volume
- 1 dm³ = 1 litre = 1000 cm³
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
1 mol CH₄ produces 1 mol CO₂
V(CO₂) = 1 × 24.0 = 24.0 dm³
Moles from Volume of GasWhat amount (in mol) is 6.0 dm³ of O₂ at r.t.p.?
n = 6.0 ÷ 24.0 = 0.25 mol
Figure 2.3: Molar Gas Volume at r.t.p.
(Visual showing a cube of 24.0 dm³ volume. Inside, label examples: 1 mol H₂ (2 g), 1 mol O₂ (32 g), 1 mol CO₂ (44 g) - all occupy the same volume at r.t.p. Emphasise the different masses but same volume.)
Volumes and Concentrations of Solutions
Concentration The amount of solute dissolved per unit volume of solution. Can be expressed in mol dm⁻³ or g dm⁻³.c = n ÷ V where c is in mol dm⁻³, n in mol, V in dm³
- 1 dm³ = 1000 cm³, so V(dm³) = V(cm³) ÷ 1000
- To convert between mol dm⁻³ and g dm⁻³: multiply by molar mass M
0.050 mol of NaOH dissolved in 250 cm³ of solution.
V = 250 ÷ 1000 = 0.250 dm³
c = 0.050 ÷ 0.250 = 0.20 mol dm⁻³
Moles from Concentration and Volumen = c × V = 0.20 × 0.250 = 0.050 mol
Figure 2.4: Concentration, Moles and Volume Triangle
(Triangle: top = n (moles), bottom-left = c (concentration, mol dm⁻³), bottom-right = V (volume, dm³). Arrows: n = c × V, c = n ÷ V, V = n ÷ c.)
Limiting Reagent and Excess Reagent
Limiting Reagent The reactant that is completely consumed first in a reaction, determining the maximum amount of product that can be formed. Excess Reagent A reactant present in greater quantity than needed to react with the limiting reagent. Some of it remains unreacted at the end.- To identify the limiting reagent: calculate the moles of each reactant available, then compare with the stoichiometric ratio required.
- The product yield is calculated from the moles of the limiting reagent only.
N₂(g) + 3H₂(g) → 2NH₃(g)
Available: 2.0 mol N₂ and 4.0 mol H₂
Stoichiometry requires 1 mol N₂ : 3 mol H₂
For 2.0 mol N₂, need: 2.0 × 3 = 6.0 mol H₂ - but only 4.0 mol available.
H₂ is the limiting reagent. N₂ is in excess.
Moles of NH₃ produced = 4.0 ÷ 3 × 2 = 2.67 mol
Exam TipAlways state which reagent is limiting and which is in excess. Show your working clearly - divide moles available by the stoichiometric coefficient to find the "mole ratio value"; the reactant with the smallest value is the limiting reagent.
Deducing Stoichiometric Relationships
Experimental data from reacting masses, gas volumes, or solution concentrations can be used to determine the stoichiometric ratio of a reaction - and hence the balanced equation.
- Calculate moles of each substance from the experimental data.
- Find the simplest whole-number ratio of moles.
- Use this to write the balanced equation.
25.0 cm³ of 0.100 mol dm⁻³ NaOH reacts exactly with 12.5 cm³ of 0.200 mol dm⁻³ H₂SO₄.
n(NaOH) = 0.100 × 0.0250 = 0.00250 mol
n(H₂SO₄) = 0.200 × 0.0125 = 0.00250 mol
Ratio NaOH : H₂SO₄ = 0.00250 : 0.00250 = ... wait - the actual ratio for H₂SO₄ + 2NaOH:
Ratio = 2 : 1 → Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Significant FiguresAnswers should reflect the number of significant figures in the data given. Do not round prematurely during multi-step calculations - carry extra figures through and round only at the final answer.
Interactive revision notes, videos and practice questions load below.