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This cheat sheet covers the core calculation formulas used in GCSE Combined Science Chemistry. Students need these formulas to answer quantitative exam questions clearly and accurately. It helps connect chemical equations, masses, solutions, gases, and reaction efficiency.

The sheet is designed as a quick reference for revision, homework, and exam practice.

The most important ideas are conservation of mass, using the mole as a counting unit, and linking quantities through balanced equations. Key formulas include n=mMrn = \frac{m}{M_r}, c=nVc = \frac{n}{V}, and percentage yield=actual yieldtheoretical yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100. Students should always check units before substituting values.

Balanced symbol equations are often needed before mole ratios can be used correctly.

Key Facts

  • Relative formula mass is found by adding the relative atomic masses in a formula: Mr=ArM_r = \sum A_r.
  • The number of moles is calculated using n=mMrn = \frac{m}{M_r}, where nn is in mol, mm is in g, and MrM_r is in g mol1^{-1}.
  • Mass can be found from moles using m=nMrm = nM_r.
  • Concentration in mol dm3^{-3} is calculated using c=nVc = \frac{n}{V}, where VV must be in dm3^3.
  • Mass concentration is calculated using concentration=mass of solutevolume of solution\text{concentration} = \frac{\text{mass of solute}}{\text{volume of solution}}.
  • At room temperature and pressure, gas volume can be estimated using V=24nV = 24n, where VV is in dm3^3 and nn is in mol.
  • Percentage yield is calculated using percentage yield=actual yieldtheoretical yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100.
  • Atom economy is calculated using atom economy=Mr of desired productMr of all products×100\text{atom economy} = \frac{M_r\text{ of desired product}}{M_r\text{ of all products}} \times 100.

Vocabulary

Relative atomic mass
Relative atomic mass, ArA_r, is the average mass of atoms of an element compared with 112\frac{1}{12} of the mass of a carbon-12 atom.
Relative formula mass
Relative formula mass, MrM_r, is the total of the relative atomic masses of all atoms in a compound formula.
Mole
A mole is an amount of substance containing 6.02×10236.02 \times 10^{23} particles.
Concentration
Concentration is the amount of solute dissolved in a given volume of solution, often measured in mol dm3^{-3} or g dm3^{-3}.
Theoretical yield
The theoretical yield is the maximum mass of product predicted from a balanced chemical equation.
Atom economy
Atom economy is the percentage of reactant atoms that become part of the desired product.

Common Mistakes to Avoid

  • Using volume in cm3^3 directly in c=nVc = \frac{n}{V} is wrong because this formula usually needs VV in dm3^3. Convert using 1000 cm3=1 dm31000\text{ cm}^3 = 1\text{ dm}^3.
  • Forgetting to balance the chemical equation before using mole ratios gives the wrong reacting amounts. The coefficients in the balanced equation give the mole ratio.
  • Confusing ArA_r and MrM_r leads to incorrect masses. Use ArA_r for single elements and add all atoms in the formula to find MrM_r.
  • Putting actual yield and theoretical yield the wrong way round can give a percentage yield above 100%100\%. The correct formula is actual yieldtheoretical yield×100\frac{\text{actual yield}}{\text{theoretical yield}} \times 100.
  • Rounding too early changes the final answer. Keep extra digits during working and round only at the end to the required number of significant figures.

Practice Questions

  1. 1 Calculate the number of moles in 9.0 g9.0\text{ g} of water, H2O\text{H}_2\text{O}, given Mr=18M_r = 18.
  2. 2 A solution contains 0.25 mol0.25\text{ mol} of sodium chloride in 0.50 dm30.50\text{ dm}^3 of solution. Calculate the concentration using c=nVc = \frac{n}{V}.
  3. 3 A reaction has a theoretical yield of 12.0 g12.0\text{ g} but produces 9.6 g9.6\text{ g} of product. Calculate the percentage yield.
  4. 4 Explain why a process with high atom economy is usually better for industry and the environment than a process with low atom economy.