Percentage Yield and Atom Economy

(Separate Chemistry only) The whole of section 4.3.3, yield and atom economy, is assessed in AQA GCSE Chemistry but not in Combined Science.

When you carry out a reaction, you almost never collect the full mass of product the equation predicts. The amount you do collect is the actual yield. The maximum the reactants could possibly give is the theoretical yield. Three reasons explain the gap between them.

• The reaction may be reversible and never go fully to completion, so some reactants remain unreacted.
• Product is lost during separation and transfer, for example when filtering, or when solid is left stuck to glassware.
• Reactants react in unexpected ways and form by-products instead of the product you wanted.

None of these means atoms have been destroyed. The law of conservation of mass still holds; the missing product is simply present as unreacted starting material, waste by-products, or losses to the apparatus.

Percentage yield compares what you actually made against the theoretical maximum.

$\text{percentage yield}=\frac{\text{mass of product actually made}}{\text{maximum theoretical mass of product}}\times 100$

Worked example. A student heats a sample and the equation predicts a maximum of 8.0 g of product. They collect 6.0 g. Find the percentage yield.

$\text{percentage yield}=\frac{6.0}{8.0}\times 100=75\%$

Both masses must be in the same unit before dividing. Because the theoretical yield is the largest mass possible, a correctly calculated percentage yield is always between 0% and 100%.

A yield above 100% is a signal that something is wrong: the product is still wet, still impure, or the arithmetic has slipped. Check before writing it down.

(Higher Tier only) Calculating theoretical mass from a given mass of reactant and the balanced equation is assessed at Higher tier.

Percentage-yield questions often give you the reactant mass rather than the theoretical yield, so you must work the theoretical mass out yourself. Convert reactant mass to moles, use the equation ratio, then convert product moles back to mass.

Worked example. Calcium carbonate decomposes on heating:

${\text{CaCO}}_3\to \text{CaO}+{\text{CO}}_2$

Calculate the theoretical mass of calcium oxide (CaO) from 50 g of calcium carbonate. Use $A_r$: Ca = 40, C = 12, O = 16.

Relative formula masses:

• $M_r({\text{CaCO}}_3)=40+12+(3\times 16)=100$
• $M_r(\text{CaO})=40+16=56$

Step 1 — moles of CaCO₃:

$n=\frac{\text{mass}}{M_r}=\frac{50}{100}=0.50\text{ mol}$

Step 2 — the equation ratio is 1 : 1, so moles of CaO = 0.50 mol.

Step 3 — mass of CaO:

$\text{mass}=n\times M_r=0.50\times 56=28\text{ g}$

The theoretical yield is 28 g. If the student actually collected 21 g, the percentage yield would be $(21÷28)\times 100=75\%$.

Percentage yield tells you how efficiently a reaction was carried out. Atom economy tells you something different: what proportion of the atoms in the reactants end up in the product you actually want, rather than in waste.

$\text{atom economy}=\frac{M_r\text{ of desired product from the equation}}{\text{sum of }{M}_r\text{ of all reactants from the equation}}\times 100$

Use the amounts shown in the balanced equation, so a reactant written as $3\text{CO}$ contributes $3\times M_r$. A high atom economy matters for sustainability and economics: it means less raw material is wasted, less waste needs disposal, and the process is cheaper to run.

A reaction that makes only one product has an atom economy of 100%, because every reactant atom ends up in that product.

Worked example. Iron is extracted in the blast furnace by reducing iron(III) oxide with carbon monoxide:

${\text{Fe}}_2{\text{O}}_3+3\text{CO}\to 2\text{Fe}+3{\text{CO}}_2$

Calculate the atom economy for iron. Use $A_r$: Fe = 56, C = 12, O = 16.

Step 1 — $M_r$ of the desired product. The desired product is iron; the equation makes $2\text{Fe}$:

$2\times 56=112$

Step 2 — sum of $M_r$ of all reactants:

• $M_r({\text{Fe}}_2{\text{O}}_3)=(2\times 56)+(3\times 16)=112+48=160$
• $3\times M_r(\text{CO})=3\times (12+16)=3\times 28=84$
• Total = $160+84=244$

Step 3 — divide and multiply by 100:

$\text{atom economy}=\frac{112}{244}\times 100=45.9\%$

Just under half the mass of the reactants ends up as iron; the rest leaves as carbon dioxide.

(Higher Tier only) Explaining why a particular reaction pathway is chosen is assessed at Higher tier.

Industry often has more than one route to the same product, and atom economy is only one factor in the choice. A pathway is judged on atom economy, percentage yield, rate of reaction, equilibrium position, and how useful any by-products are.

Consider two hypothetical routes to the same product:

Factor	Route A	Route B
Atom economy	90%	55%
Percentage yield	40%	95%
By-products	none useful	sold on

Route A wastes fewer atoms per reaction, but its low yield means much reactant is wasted anyway. Route B wastes atoms into by-products, but those by-products are sold, and its high yield and faster rate may make it cheaper overall. The best pathway balances all these factors, so the answer depends on the data given, not on atom economy alone.

Read the data in the question. Do not simply pick the highest atom economy; a route with a saleable by-product or far higher yield can be the better industrial choice.

1. Confusing percentage yield with atom economy

Yield depends on how the reaction is carried out and the mass of reactant used. Atom economy depends only on the balanced equation. A reaction can have 100% atom economy and still give a poor yield, or a high yield and a low atom economy.

2. Forgetting the balancing numbers in atom economy

A reactant written as $3\text{CO}$ contributes $3\times M_r$ to the reactant total, and a product written as $2\text{Fe}$ has $M_r$ of $2\times 56$. Leaving these multipliers out gives the wrong ratio.

3. Dividing the wrong way round

Percentage yield is actual over theoretical, and atom economy is desired product over total reactants. Putting the larger number on the bottom gives an answer below what it should be; putting it on top gives an impossible figure over 100%.

4. Quoting a yield above 100%

The theoretical yield is the maximum possible, so the actual yield cannot exceed it. A result over 100% means the sample is wet or impure, or a calculation error has crept in.

5. Mixing up desired product and total products

Atom economy uses the $M_r$ of the desired product only, not the sum of all products. The denominator is the sum of the reactants, which equals the sum of all products by conservation of mass, but the numerator is just the one product you want.