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1.2.5Physical Chemistry

Balanced Equations and Associated Calculations

5 min read 6 parts Lesson 11 of 63

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Part 1 of 6Using a Balanced Equation

1.2.5 Balanced Equations and Associated Calculations
Coefficients
  • The numbers in front of formulae give the ratio of moles of each substance that react or form.
State symbols
  • (s), (l), (g) and (aq) show whether each substance is a solid, liquid, gas or aqueous solution.
Reacting masses
  • Convert the known mass to moles, use the mole ratio, then convert to mass of the required substance.
Limiting reagent
  • The reactant that is used up first controls how much product forms.

Part 2 of 6Calculation Route

1
Write a balanced equation for the reaction, including state symbols if needed.
2
Convert the quantity given for one substance into moles.
3
Use the coefficients to find the moles of the substance wanted.
4
Convert moles into mass, gas volume or concentration as the question requires.
5
Check the answer is sensible and give units and significant figures.

Part 3 of 6Yield and Atom Economy

Yield and Atom Economy
Percentage yield
  • Actual mass obtained divided by theoretical mass, multiplied by 100.
Why yield is below 100%
  • Reactions may be reversible, incomplete, have side reactions, or lose product during separation and purification.
Atom economy
  • Mr of desired product divided by the sum of Mr of all products, multiplied by 100.
Importance
  • High atom economy means less waste, which makes a process more sustainable.

Part 4 of 6Gas Volumes

Molar volume
  • At room temperature and pressure, one mole of any gas occupies about 24.0 dm³.
Avogadro's law
  • Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
Volume ratios
  • For gases at the same conditions, the volume ratio in a reaction equals the mole ratio.
General case
  • For other conditions use pV = nRT after finding the moles.

Part 5 of 6Worked Example

Question

  • 5.00 g of calcium carbonate reacts with excess hydrochloric acid: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g).
  • 4.20 g of calcium chloride is isolated after crystallisation.
  • Calculate the percentage yield (Mr CaCO₃ = 100.1, Mr CaCl₂ = 111.1).

  1. n(CaCO₃) = 5.00 / 100.1 = 0.0500 mol.
  2. The ratio CaCO₃CaCl₂ is 1 : 1, so the theoretical n(CaCl₂) = 0.0500 mol.
  3. Theoretical mass = 0.0500 × 111.1 = 5.55 g.
  4. Percentage yield = (4.20 / 5.55) × 100 = 75.6%.

Answer Percentage yield = 75.6%.

Part 6 of 6Exam Checklist & Tips

1.2.5 – Key Points for the Exam

  • Limiting reagent The reactant that is completely used up and so limits the amount of product.
  • Theoretical yield The maximum mass of product calculated from the balanced equation.
  • Percentage yield (Actual yield / theoretical yield) × 100.
  • Atom economy The proportion of the total mass of reactants that ends up in the desired product.

Common Mistakes

  • Calculating from the unbalanced equation, so the mole ratio is wrong.
  • Using the mass of the excess reagent to calculate the theoretical yield.
  • Forgetting that percentage yield uses masses or moles of the same substance.
  • Calculating atom economy from the reactants rather than all the products.
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