Percentage Yield and Atom Economy (Part 1/2) Revision Notes | A Level AQA Chemistry

1. Why Reactions Never Give You Everything You Expect

Picture this: you run a reaction in the lab, collect your product, and weigh it — only to find you have far less than your calculations predicted. This happens in every real reaction, and chemists have two precise tools for measuring how well a reaction actually performed: percentage yield and atom economy.

These are not the same thing, and confusing them is one of the most common errors at A Level. Percentage yield tells you how much of the theoretical product you actually collected. Atom economy tells you how efficiently the atoms in your reactants end up in the product you want — even before you run the reaction.

Both quantities matter enormously in industrial chemistry, where wasted atoms mean wasted money and increased environmental impact. By the end of this lesson, you will be able to calculate both with confidence and explain what each one reveals about a reaction.

2. Percentage Yield — Measuring Real-World Efficiency

The theoretical yield is the mass of product you would obtain if every mole of limiting reactant converted perfectly into product, with zero losses. In practice, reactions rarely go to completion, products are lost during purification, and side reactions consume some of your starting material.

To find the theoretical yield, you must work through a moles calculation using the balanced equation. The stoichiometric ratio from the equation tells you exactly how many moles of product form per mole of limiting reactant. Multiply by the molar mass of the product to convert moles into grams.

Worked example: Calculating Percentage Yield: Magnesium to Magnesium Oxide

0.486 g of magnesium is burned completely in excess oxygen. The magnesium oxide collected has a mass of 0.712 g. Calculate the percentage yield. ($M_r$: Mg = 24.3, O = 16.0)

1. $2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$ — The mole ratio of Mg to MgO is 2:2, which simplifies to 1:1. Every mole of Mg produces one mole of MgO.
2. $n(\text{Mg}) = \frac{0.486}{24.3} = 0.0200 \text{ mol}$ — Moles = mass ÷ molar mass. Mg is the limiting reactant here because O2 is in excess.
3. $n(\text{MgO}) = 0.0200 \text{ mol}$ (1:1 ratio) — Since the ratio is 1:1, the theoretical moles of MgO equals the moles of Mg.
4. $M_r(\text{MgO}) = 24.3 + 16.0 = 40.3$ $\text{theoretical yield} = 0.0200 \times 40.3 = 0.806 \text{ g}$ — Multiply moles by molar mass to get the maximum possible mass of product.
5. $\text{percentage yield} = \frac{0.712}{0.806} \times 100 = 88.3\%$ — Actual yield divided by theoretical yield, multiplied by 100. The answer is given to 3 significant figures, matching the data.

Answer: Percentage yield = 88.3%

Notice that the percentage yield can never exceed 100% — if your calculated answer is above 100%, check whether you have used the correct theoretical yield or whether your actual yield value has been misread. A yield above 100% is physically impossible and always signals an arithmetic error.

3. Atom Economy — Efficiency Built Into the Equation

Percentage yield measures what happened in your specific experiment. Atom economy is different: it is a property of the reaction itself, calculated entirely from the balanced equation, before any experiment takes place.

The key idea is this: in any reaction, atoms from the reactants end up either in the desired product or in by-products. Atom economy asks what fraction of the total reactant mass ends up where you actually want it — in the useful product.

A reaction with 100% atom economy produces only the desired product — every atom in the reactants ends up in the product. Addition reactions, where two molecules combine to form a single product, often have high atom economy. Substitution and elimination reactions, which produce by-products, tend to have lower atom economy.

Worked example: Calculating Percentage Atom Economy

Calcium carbonate decomposes on heating: $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$. The desired product is calcium oxide (CaO). Calculate the percentage atom economy. ($A_r$: Ca = 40.1, C = 12.0, O = 16.0)

1. $M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 40.1 + 12.0 + 48.0 = 100.1$ — There is one mole of CaCO3 in the equation, so the sum of Mr of all reactants = 100.1.
2. $M_r(\text{CaO}) = 40.1 + 16.0 = 56.1$ — CaO is the desired product. Its stoichiometric coefficient is 1, so we use 56.1 directly.
3. $\text{atom economy} = \frac{56.1}{100.1} \times 100 = 56.0\%$ — 56.0% of the reactant mass ends up in the desired product CaO. The remaining 44.0% leaves as CO2 by-product.

Answer: Percentage atom economy = 56.0%

4. Worked Examples Across AQA Contexts

AQA exam questions apply these calculations to a range of real chemical contexts. You need to be comfortable working through the moles logic in each case, not just substituting into a formula. The following examples cover the key contexts specified in the AQA course.

Worked example: Finding the Mr of MHCO3 by Back-Calculation from Yield Data

A student reacts 0.500 g of a Group 1 metal hydrogen carbonate MHCO3 with excess hydrochloric acid. The reaction produces 0.220 g of CO2. Use this data to calculate the Mr of MHCO3. ($M_r$: CO2 = 44.0)

1. $\text{MHCO}_3(s) + \text{HCl}(aq) \rightarrow \text{MCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$ — The mole ratio of MHCO3 to CO2 is 1:1.
2. $n(\text{CO}_2) = \frac{0.220}{44.0} = 0.00500 \text{ mol}$ — Moles = mass ÷ Mr. CO2 is the product we can measure directly.
3. $n(\text{MHCO}_3) = 0.00500 \text{ mol}$ (1:1 ratio) — One mole of MHCO3 produces exactly one mole of CO2, so the moles are equal.
4. $M_r(\text{MHCO}_3) = \frac{\text{mass}}{\text{moles}} = \frac{0.500}{0.00500} = 100$ — Mr = mass ÷ moles. This gives the relative formula mass of the unknown hydrogen carbonate.
5. $M_r(\text{MHCO}_3) = M_r(\text{M}) + 1.0 + 12.0 + (3 \times 16.0) = M_r(\text{M}) + 61.0$ $M_r(\text{M}) = 100 - 61 = 39$ — Subtracting the known atomic masses of H, C, and 3O from 100 gives the relative atomic mass of M = 39, which corresponds to potassium (K).

Answer: $M_r$ of MHCO3 = 100; the metal M is potassium (K), so the compound is KHCO3.

The same back-calculation logic applies when finding the $M_r$ of succinic acid or the $M_r$ of a hydrated salt. In each case, you measure a quantity you can determine experimentally — moles of product, or mass lost on heating — and use the mole ratio to work backwards to the unknown $M_r$.

Worked example: Finding the Mr of a Hydrated Salt by Heating to Constant Mass

A student heats a sample of hydrated magnesium sulfate, $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$, to constant mass. The initial mass is 2.46 g and the final mass of anhydrous MgSO4 is 1.20 g. Find the value of $x$. ($M_r$: MgSO4 = 120.4, H2O = 18.0)

1. $\text{mass of H}_2\text{O} = 2.46 - 1.20 = 1.26 \text{ g}$ — Heating to constant mass drives off all the water of crystallisation. The mass difference equals the mass of water.
2. $n(\text{MgSO}_4) = \frac{1.20}{120.4} = 0.00997 \text{ mol}$ — Use the anhydrous mass and the Mr of anhydrous MgSO4.
3. $n(\text{H}_2\text{O}) = \frac{1.26}{18.0} = 0.0700 \text{ mol}$ — Use the mass of water lost and the Mr of water.
4. $x = \frac{n(\text{H}_2\text{O})}{n(\text{MgSO}_4)} = \frac{0.0700}{0.00997} \approx 7.02 \approx 7$ — The ratio gives the number of water molecules per formula unit. Round to the nearest integer — x must be a whole number.

Answer: $x = 7$, so the hydrated salt is $\text{MgSO}_4 \cdot 7\text{H}_2\text{O}$ (magnesium sulfate heptahydrate).

Heating to constant mass is the key experimental technique here. The phrase means you heat, cool, and weigh repeatedly until two consecutive readings agree — confirming all the water of crystallisation has been removed. A single heating is not sufficient, because some water may remain.

5. Further AQA Calculation Contexts

AQA specifies several real-world contexts in which you must apply percentage yield and moles calculations. These include finding the concentration of ethanoic acid in vinegar, the mass of calcium carbonate in an indigestion tablet, the $M_r$ of succinic acid, and the mass of aspirin in a tablet. Each follows the same underlying logic: balanced equation, mole ratio, and conversion between mass and moles.

Worked example: Finding the Mass of Calcium Carbonate in an Indigestion Tablet

An indigestion tablet is dissolved in 25.0 cm³ of 1.00 mol dm⁻³ HCl. The excess acid is then titrated against 0.500 mol dm⁻³ NaOH, requiring 14.0 cm³ to reach the endpoint. Calculate the mass of CaCO3 in the tablet. ($M_r$: CaCO3 = 100.1. Equations: $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$; $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$)

1. $n(\text{HCl})_{\text{initial}} = 1.00 \times \frac{25.0}{1000} = 0.0250 \text{ mol}$ — Moles = concentration × volume (in dm³). Convert cm³ to dm³ by dividing by 1000.
2. $n(\text{NaOH}) = 0.500 \times \frac{14.0}{1000} = 0.00700 \text{ mol}$ — The NaOH neutralises only the excess HCl that did not react with CaCO3.
3. $n(\text{HCl})_{\text{excess}} = 0.00700 \text{ mol}$ (1:1 ratio with NaOH) — NaOH and HCl react in a 1:1 ratio, so moles of excess HCl = moles of NaOH.
4. $n(\text{HCl})_{\text{reacted}} = 0.0250 - 0.00700 = 0.0180 \text{ mol}$ — Subtract excess HCl from initial HCl to find how much reacted with the tablet.
5. $n(\text{CaCO}_3) = \frac{0.0180}{2} = 0.00900 \text{ mol}$ — From the equation, CaCO3: HCl = 1:2, so divide moles of HCl by 2.
6. $\text{mass} = 0.00900 \times 100.1 = 0.901 \text{ g}$ — Mass = moles × Mr. This is the mass of calcium carbonate in the tablet.

Answer: Mass of CaCO3 in the tablet = 0.901 g

This back-titration method is used whenever the analyte (here, CaCO3) does not dissolve readily or react quickly enough for a direct titration. You add a known excess of acid, let it react fully, then titrate the leftover acid. The same approach applies to finding the mass of aspirin in a tablet, where aspirin reacts with a known volume of NaOH and the excess is back-titrated.

6. Atom Economy in Green Chemistry

Atom economy was introduced by Barry Trost in 1991 as a way to evaluate reaction efficiency from an environmental perspective. A reaction with low atom economy produces large quantities of by-products, which must be disposed of — often at financial and environmental cost.

Consider two routes to the same product. Route A has 95% atom economy but 60% yield. Route B has 40% atom economy but 90% yield. Route B gives more of the product per experiment, but Route A wastes far fewer atoms overall. Industrial chemists must weigh both factors when selecting a synthesis route.

A high atom economy is desirable because it means fewer by-products to separate, treat, or dispose of. This reduces raw material costs, lowers waste disposal costs, and decreases the environmental footprint of the process — all central principles of green chemistry.

7. Exam Technique and Common Errors

Percentage yield and atom economy questions appear regularly on AQA papers, often carrying 3–5 marks. The marks are distributed across the method steps, so showing clear working — even if your final answer is wrong — can still earn method marks. Always write the formula, substitute with units, and state your final answer with appropriate significant figures.

When an exam question asks you to 'evaluate' a reaction route, you must address both atom economy and percentage yield, consider the implications of any by-products, and reach a reasoned conclusion. A one-sided answer — discussing only atom economy or only yield — will not access the top mark band.

Percentage yield and atom economy together give a complete picture of reaction efficiency: yield tells you how well the experiment was conducted; atom economy tells you how well the reaction was designed. Mastering both calculations, and knowing when to apply each, is central to the AQA Amount of Substance topic.

Worked example: How to Avoid Mixing Up Yield and Atom Economy

1. Use the balanced equation and moles to find theoretical yield.
2. Compare actual yield with theoretical yield to get percentage yield.
3. Separately, use Mr values from the balanced equation to calculate atom economy before considering experimental losses.