Example

This lesson works through examples of theoretical yield and percent yield, including finding the limiting reactant first and working backwards from a known percent yield.

The theoretical yield is the maximum mass of product the limiting reactant can give; the actual yield is what you really collect; the percent yield compares the two. This lesson applies these ideas to four worked problems: burning magnesium, a reaction where the limiting reactant must be identified, a pharmaceutical synthesis of aspirin, and a calculation that works backwards from a percent yield to the mass of reactant needed. It also explains why real yields fall short of 100% and what a yield above 100% tells you.

Theoretical Yield and Percent Yield: Worked Examples

The two key relationships are:

• Theoretical yield: the maximum mass of product, calculated from the limiting reactant using the balanced equation.
• Percent yield = (actual yield ÷ theoretical yield) × 100

Rearranged, actual yield = (percent yield ÷ 100) × theoretical yield, and theoretical yield = actual yield ÷ (percent yield ÷ 100).

Relative atomic masses used: H = 1.0, C = 12.0, O = 16.0, Mg = 24.3, Ca = 40.1.

Worked problem 1: burning magnesium A student heats 4.86 g of magnesium in a crucible and collects 7.50 g of magnesium oxide. Equation: 2Mg + O₂ → 2MgO

1. n(Mg) = 4.86 ÷ 24.3 = 0.200 mol
2. Mg : MgO = 2 : 2 = 1 : 1, so n(MgO) = 0.200 mol
3. Theoretical yield = 0.200 × 40.3 = 8.06 g
4. Percent yield = (7.50 ÷ 8.06) × 100 = 93.1%

Likely losses: some white magnesium oxide smoke escapes when the lid is lifted, and some magnesium reacts with nitrogen in the air instead of oxygen.

Worked problem 2: find the limiting reactant first 4.0 g of hydrogen gas and 16.0 g of oxygen gas react, and 15.3 g of water is collected. Equation: 2H₂ + O₂ → 2H₂O

1. n(H₂) = 4.0 ÷ 2.0 = 2.0 mol and n(O₂) = 16.0 ÷ 32.0 = 0.500 mol
2. 2.0 mol of H₂ would need 1.0 mol of O₂, but only 0.500 mol is available, so oxygen is the limiting reactant and hydrogen is in excess.
3. O₂ : H₂O = 1 : 2, so n(H₂O) = 1.00 mol and the theoretical yield = 1.00 × 18.0 = 18.0 g
4. Percent yield = (15.3 ÷ 18.0) × 100 = 85.0%

Basing the calculation on hydrogen instead would give a theoretical yield of 36.0 g, which is impossible with so little oxygen.

Worked problem 3: making aspirin Salicylic acid (C₇H₆O₃) reacts with ethanoic anhydride to make aspirin (C₉H₈O₄) in a 1 : 1 mole ratio. A student uses 2.00 g of salicylic acid with the anhydride in excess and obtains 2.15 g of dry aspirin.

1. M(C₇H₆O₃) = 84.0 + 6.0 + 48.0 = 138.0 g mol⁻¹, so n = 2.00 ÷ 138.0 = 0.01449 mol
2. 1 : 1 ratio, so n(aspirin) = 0.01449 mol
3. M(C₉H₈O₄) = 108.0 + 8.0 + 64.0 = 180.0 g mol⁻¹, so the theoretical yield = 0.01449 × 180.0 = 2.61 g
4. Percent yield = (2.15 ÷ 2.61) × 100 = 82.4%

Some aspirin stays dissolved in the liquid when the crystals are filtered, and more is lost when the product is purified by recrystallisation.

Worked problem 4: working backwards Heating calcium carbonate gives calcium oxide with an 80.0% yield. What mass of calcium carbonate is needed to obtain 28.0 g of calcium oxide? Equation: CaCO₃ → CaO + CO₂

1. Theoretical yield needed = 28.0 ÷ 0.800 = 35.0 g of CaO
2. n(CaO) = 35.0 ÷ 56.1 = 0.6239 mol
3. CaCO₃ : CaO = 1 : 1, so n(CaCO₃) = 0.6239 mol
4. m(CaCO₃) = 0.6239 × 100.1 = 62.5 g

Why real yields are lower than 100%

• The reaction does not go to completion, for example because it is reversible.
• Side reactions turn some reactant into other products.
• Product is lost during transfer, filtration, crystallisation or purification.
• The reactants are not pure.

A percent yield above 100% does not mean extra product was made. It means the product is impure, most often still wet with water or solvent, so it weighs more than it should.