Calculating Percentage Yield
Percentage yield measures a reaction's practical efficiency by comparing the mass of product actually created with the maximum mass theoretically possible. This is crucial for evaluating experimental methods and understanding why reactions rarely achieve 100% conversion.
Part of the ESAT Chemistry syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The 'predicted yield' is the maximum possible mass of a product, calculated from the balanced chemical equation assuming perfect conditions and complete reaction.
- The 'actual yield' is the real, measured mass of the product obtained after performing the experiment and isolating it.
- Percentage yield is almost always less than 100% due to factors like reversible reactions not going to completion, loss of product during separation, or competing side reactions.
- Calculating percentage yield is a two-step process: first, determine the predicted yield using moles and stoichiometry, then apply the percentage yield formula.
- The masses in the formula must both refer to the product, not the reactants.
Formulae
Percentage Yield = (Actual Yield (g) / Predicted Yield (g)) × 100 To find the efficiency of a chemical reaction by comparing the measured product mass to the stoichiometrically calculated maximum possible mass.
Definitions
- Predicted Yield
- Also known as theoretical yield. It is the maximum mass of product that can be formed from a given mass of reactants, calculated using stoichiometry. This assumes the reaction is 100% efficient.
- Actual Yield
- The mass of the desired product that is physically measured and collected at the end of a chemical reaction.
- Limiting Reactant
- The reactant that is completely consumed in a reaction and therefore determines the maximum amount of product that can be formed.
Worked example
In the Haber process, nitrogen reacts with hydrogen to form ammonia: N₂(g) + 3H₂(g) → 2NH₃(g). If 5.6 g of nitrogen reacts with excess hydrogen to produce an actual yield of 3.4 g of ammonia, what is the percentage yield? (Ar values: N=14, H=1)
- 1
Step 1:
Find the molar masses.
Mr(N₂) = 2 × 14 = 28Mr(NH₃) = 14 + (3 × 1) = 17 - 2
Step 2:
Calculate the moles of the limiting reactant, nitrogen.
Moles = mass / Mr = 5.6 g / 28 g/mol = 0.2 mol - 3
Step 3:
Use the balanced equation's molar ratio to find the theoretical moles of ammonia.
The ratio of N₂ to NH₃ is 1:2.
So, 0.2 mol of N₂ will produce 0.2 × 2 = 0.4 mol of NH₃.
- 4
Step 4:
Calculate the predicted (theoretical) mass of ammonia.
Mass = moles × Mr = 0.4 mol × 17 g/mol = 6.8 g - 5
Step 5:
Use the percentage yield formula.
Percentage Yield = (Actual Yield / Predicted Yield) × 100 = (3.4 g / 6.8 g) × 100.
- 6
Step 6:
Simplify the fraction and calculate.
3.4 / 6.8 is exactly 1/2.
So, (1/2) × 100 = 50%.
Answer: 50%
Common mistakes
- ×Incorrectly inverting the formula, placing predicted yield over actual yield.
- ×Forgetting to use the molar ratios from the balanced equation to find the moles of product.
- ×Using the initial mass of a reactant as the 'predicted yield' in the denominator, instead of calculating the predicted mass of the product.
- ×Mixing up the molar masses, e.g., using the Mr of a reactant to calculate the mass of the product.
No-calculator tips
- ✓Always simplify the fraction of (Actual Yield / Predicted Yield) before multiplying by 100. Look for common factors or simple relationships, like one number being double the other.
- ✓ESAT questions often use masses that are simple multiples or fractions of the molar mass (e.g., 5.6 g of N₂ with Mr=28 is 56/280 = 1/5 = 0.2 mol), so you can calculate moles quickly.
- ✓If the numbers are difficult, estimate the fraction. For example, 6.1g / 8.2g is roughly 6/8 or 3/4, so the yield will be around 75%. This can help you eliminate incorrect multiple-choice options.