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1.3.2Physical Chemistry

Reacting Masses — AQA AS Chemistry Revision Notes & Topical Questions

5 min read 6 parts Lesson 15 of 102

📚 AQA AS Chemistry Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 6The Method

Every reacting mass calculation follows the same pattern.

1
Write the balanced equation for the reaction.
2
Calculate the moles of the substance whose mass is known, using n = m ÷ M.
3
Use the coefficients to convert to the moles of the substance you want.
4
Convert these moles to mass using m = n × M.
5
Give the answer to the correct number of significant figures with units.

Part 2 of 6Reading the Equation

Mole ratio
  • Mole ratio The coefficients show the ratio in which moles of reactants and products react, also called the stoichiometry.
Not mass ratio
  • Not mass ratio The coefficients give mole ratios and not mass ratios, so masses must always be converted to moles first.
Formula units
  • Formula units Moles of a compound refer to moles of the whole formula; CaF₂ gives 1 mol Ca²⁺ and 2 mol F⁻.
Scaling
  • Scaling Doubling the moles of one substance doubles the moles of every other substance in the equation.

Part 3 of 6Limiting and Excess Reagents

Limiting and Excess Reagents
Limiting reagent
  • Limiting reagent The reactant that is used up first, so it determines how much product forms.
Excess reagent
  • Excess reagent The reactant left over when the reaction stops.
Finding it
  • Finding it Divide the moles of each reactant by its coefficient;
    • the smallest value shows the limiting reagent.
Use
  • Use All product calculations are based on the limiting reagent, never the excess reagent.

Part 4 of 6Practical Context

Economics
  • Economics Chemists calculate reacting masses to avoid wasting expensive reagents.
Safety
  • Safety Correct masses prevent adding too much of a hazardous reactant to a reaction.
Purity
  • Purity Impure reactants mean the real moles are lower than calculated from the mass weighed.
Scaling up
  • Scaling up The same equation lets a laboratory method be scaled up to kilogram or tonne quantities.

Part 5 of 6Worked Example

Question Calculate the mass of carbon dioxide produced when 25.0 g of calcium carbonate reacts completely with hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ (Mr: CaCO₃ = 100.1, CO₂ = 44.0).

  1. n(CaCO₃) = 25.0 ÷ 100.1 = 0.2498 mol.
  2. The ratio CaCO₃CO₂ is 1 : 1, so n(CO₂) = 0.2498 mol.
  3. Mass of CO₂ = 0.2498 × 44.0 = 10.99 g.
  4. Round to 3 significant figures.

Answer Mass of CO₂ = 11.0 g

Part 6 of 6Exam Checklist & Tips

1.3.2 – Key Points for the Exam

  • Stoichiometry The mole ratio between substances in a balanced equation.
  • Limiting reagent The reactant that is completely used up first.
  • Excess reagent A reactant present in more than the amount needed.
  • Molar mass The mass of one mole of a substance in g mol⁻¹.

Common Mistakes

  • Comparing masses directly instead of converting to moles and using the mole ratio.
  • Using the coefficient as a mass multiplier rather than a mole multiplier.
  • Basing the answer on the excess reactant instead of the limiting reagent.
  • Rounding the moles too early, which gives a final answer with the wrong third significant figure.

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