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This lesson covers percentage yield and atom economy as required by the AQA GCSE Chemistry specification (4.3.2). This is Higher Tier only content. You need to understand why actual yield is often less than theoretical yield, be able to calculate percentage yield and atom economy, and explain why these measures are important in industrial chemistry.
The theoretical yield is the maximum mass of product that could be formed in a chemical reaction, calculated from the balanced equation and the mass of the limiting reactant. The actual yield is the mass of product that is actually obtained when the reaction is carried out in practice.
| Term | Definition |
|---|---|
| Theoretical yield | The maximum mass of product calculated from the balanced equation |
| Actual yield | The mass of product actually obtained from the experiment |
| Percentage yield | The actual yield expressed as a percentage of the theoretical yield |
The actual yield is always less than or equal to the theoretical yield. It can never exceed it.
Exam Tip: The theoretical yield is what you calculate on paper. The actual yield is what you measure in the lab. The percentage yield tells you how close you got to the theoretical maximum.
There are several reasons why the actual yield of a reaction is usually less than the theoretical yield:
| Reason | Explanation |
|---|---|
| Incomplete reaction | The reaction may not go to completion — not all the reactants are converted to products |
| Side reactions | Unwanted reactions may occur, producing different products and using up some of the reactants |
| Loss during transfer | Product may be lost when transferring between containers (e.g. residue left on filter paper or in the beaker) |
| Loss during purification | Steps such as filtration, evaporation, or crystallisation may result in some product being lost |
| Reversible reactions | If the reaction is reversible, an equilibrium is established and not all reactants are converted to products |
Percentage yield = (actual yield / theoretical yield) x 100
A student calculated that 10 g of copper sulfate should be produced (theoretical yield). The student actually obtained 7 g. Calculate the percentage yield.
56 g of iron reacts with excess sulfur. The theoretical yield of iron sulfide (FeS) is 88 g. The student obtained 66 g. Calculate the percentage yield.
12 g of magnesium reacts with excess hydrochloric acid. 38 g of magnesium chloride was obtained. Calculate the percentage yield.
Balanced equation: Mg + 2HCl --> MgCl2 + H2
Step 1: Calculate the theoretical yield.
Step 2: Calculate percentage yield.
Exam Tip: If a percentage yield question does not give you the theoretical yield directly, you will need to calculate it first using the reacting masses method. Always show this calculation clearly.
graph TD
A["Balanced equation"] --> B["Calculate theoretical yield"]
B --> C["Note actual yield from question"]
C --> D["Percentage yield = actual / theoretical x 100"]
D --> E["Express as a percentage"]
F["Reasons for low yield"] --> G["Incomplete reaction"]
F --> H["Side reactions"]
F --> I["Losses during transfer"]
F --> J["Losses during purification"]
F --> K["Reversible reactions"]
style A fill:#3498db,color:#fff
style D fill:#27ae60,color:#fff
style F fill:#e74c3c,color:#fff
Atom economy is a measure of how much of the total mass of the reactants ends up as the desired product. It tells you how efficient a reaction is in terms of atoms — how many of the atoms in the reactants end up in the useful product, rather than in waste by-products.
Atom economy = (Mr of desired product / sum of Mr of all products) x 100
Note: some textbooks and mark schemes use:
Atom economy = (total Mr of useful products / total Mr of all reactants) x 100
Both give the same result for a balanced equation because total Mr of reactants = total Mr of products (conservation of mass).
Exam Tip: The AQA mark scheme accepts either formula. The key is to identify the "desired product" and the total Mr of all products. If there are no by-products, the atom economy is 100%.
Calculate the atom economy for the production of calcium oxide from calcium carbonate.
CaCO3 --> CaO + CO2
Atom economy = (56 / 100) x 100 = 56%
This means 56% of the mass of the reactants ends up as the useful product (CaO), and 44% is "wasted" as CO2.
Calculate the atom economy for the production of hydrogen from the reaction of zinc with hydrochloric acid.
Zn + 2HCl --> ZnCl2 + H2
Atom economy = (2 / 138) x 100 = 1.4%
This is a very low atom economy — most of the mass ends up as zinc chloride, not hydrogen.
Calculate the atom economy for the reaction: N2 + 3H2 --> 2NH3
Atom economy = (34 / 34) x 100 = 100%
Exam Tip: Addition reactions (where two reactants combine to form one product with no by-products) always have an atom economy of 100%. This is why they are considered the most efficient type of reaction.
| Feature | Percentage Yield | Atom Economy |
|---|---|---|
| What it measures | How much desired product you actually obtained compared to the maximum | How much of the reactant atoms end up in the desired product |
| Depends on | Practical factors (losses, incomplete reactions) | The balanced equation (theoretical) |
| Can be improved by | Better technique, more complete reaction | Choosing a different reaction pathway with fewer by-products |
| Maximum value | 100% (all theoretical yield obtained) | 100% (all atoms end up in the desired product) |
| Affected by practical losses? | Yes | No |
In industry, both percentage yield and atom economy are important for economic and environmental reasons:
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