Chemical Formulae, Equations and Calculations

Reacting masses

• A balanced equation tells you the mole ratio in which substances react
• Using that ratio, you can work out the mass of any product (or reactant) from the mass of any other species in the reaction

The three-step recipe:

1. Find the moles of the substance you are given (mass ÷ M_r)
2. Use the equation's mole ratio to convert to moles of the substance you want
3. Convert moles back to mass (moles × M_r)

Example. What mass of magnesium oxide forms when 2.4 g of magnesium burns completely in oxygen?

2Mg + O₂ → 2MgO

1. Moles of Mg = 2.4 / 24 = 0.10 mol
2. Mole ratio Mg : MgO = 2 : 2 = 1 : 1, so 0.10 mol of MgO is formed
3. M_r(MgO) = 24 + 16 = 40, so mass of MgO = 0.10 × 40 = 4.0 g

Percentage yield

• The actual yield is the mass of product you actually recover in the lab
• The theoretical yield is the mass that would form if the reaction went perfectly with no loss — calculated from the balanced equation and reacting masses
• The percentage yield compares the two:

$\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

• Yields are never 100 % in practice. Common reasons:
  • Some of the reactant is left unreacted
  • The reaction is a reversible reaction so the products partially turn back into reactants
  • Product is lost during filtration, washing, drying or transfer between vessels
  • Side reactions form other products you didn't want

Example. A student decomposes 10.0 g of calcium carbonate by heating:

CaCO₃ (s) → CaO (s) + CO₂ (g)

They recover 4.5 g of calcium oxide. What is the percentage yield?

• M_r(CaCO₃) = 100, M_r(CaO) = 56
• Theoretical: moles of CaCO₃ = 10.0 / 100 = 0.10 mol → 0.10 mol of CaO → 0.10 × 56 = 5.6 g theoretical
• Percentage yield = (4.5 / 5.6) × 100 = 80.4 %