Percentage Yield

When a chemist makes a product, they rarely obtain every last gram the equation says is possible. Some product is always lost, or the reaction does not go fully to completion. Percentage yield measures how successful the preparation was — how much product was actually obtained compared with the maximum the equation predicts. It matters in industry because a low yield means wasted raw materials and higher costs. This lesson, part of Topic C5 of OCR Gateway Science A, defines yield, theoretical yield and percentage yield, shows how to calculate them, and explains why the yield is almost always below 100%.

By the end of this lesson you should be able to define yield and theoretical yield, calculate a percentage yield, work out a theoretical yield from a balanced equation, and explain why yields are less than 100%.

Percentage yield is one of the ways a chemist monitors how well a reaction has gone. A high yield tells you the reaction was efficient and little was wasted; a low yield is a signal to look for losses or for a reaction that did not go to completion. Being able to calculate it, and to explain the result, is therefore a core skill in this topic.

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Yield and Theoretical Yield

The yield of a reaction is the amount of product actually obtained. The theoretical yield is the maximum amount of product that could be made from the given amount of reactant, calculated from the balanced equation — assuming the reaction goes perfectly and nothing is lost.

Working out a theoretical yield is just a reacting-mass calculation, exactly as in Topic C3: convert the reactant mass to moles, use the mole ratio, and convert back to the mass of product. (Recall $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​.) In practice the actual yield is almost always less than the theoretical yield.

Exam Tip: Keep the two terms separate: the theoretical yield is the maximum from the equation (a calculation); the actual yield is what you really collected (a measurement). Percentage yield compares the two.

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Calculating Percentage Yield

The percentage yield is found from:

$\% \text{ yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$% yield=theoretical yieldactual yield​×100

Both yields must be in the same units (both in grams, or both in moles). A high percentage yield means the preparation was efficient; a low one means a lot of product was lost or not formed. A yield of $100\%$100% would mean every gram the equation predicts was actually obtained — which, as we will see, almost never happens in practice.

Worked Example 1 — A straightforward percentage yield

A reaction has a theoretical yield of $8.0 \text{ g}$8.0 g, but only $6.0 \text{ g}$6.0 g of product is actually obtained. Calculate the percentage yield.

Step 1 — write the equation: $\% \text{ yield} = \dfrac{\text{actual}}{\text{theoretical}} \times 100$% yield=theoreticalactual​×100.

Step 2 — substitute: $\% \text{ yield} = \dfrac{6.0}{8.0} \times 100$% yield=8.06.0​×100.

Step 3 — calculate: $\% \text{ yield} = 75\%$% yield=75%.

Answer: the percentage yield is $75\%$75%.

Worked Example 2 — Finding the actual yield

A preparation has a theoretical yield of $20 \text{ g}$20 g and a percentage yield of $85\%$85%. What mass of product is actually obtained?

Step 1 — rearrange: $\text{actual yield} = \dfrac{\% \text{ yield}}{100} \times \text{theoretical yield}$actual yield=100% yield​×theoretical yield.

Step 2 — substitute: $\text{actual yield} = \dfrac{85}{100} \times 20$actual yield=10085​×20.

Step 3 — calculate: $\text{actual yield} = 17 \text{ g}$actual yield=17 g.

Answer: $17 \text{ g}$17 g of product.

Exam Tip: Make sure both yields are in the same unit before you divide. If the question gives a theoretical yield in grams and an actual yield in grams, you are ready to divide; never mix grams with moles.

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Calculating the Theoretical Yield First

Often the theoretical yield is not given — you have to calculate it from a balanced equation before you can find the percentage yield.

Worked Example 3 — Theoretical yield, then percentage yield

Calcium carbonate is decomposed by heating: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$CaCO3​→CaO+CO2​. A student heats $50 \text{ g}$50 g of calcium carbonate and obtains $21 \text{ g}$21 g of calcium oxide. Calculate the percentage yield. ($M_r$Mr​: $\text{CaCO}_3 = 100$CaCO3​=100, $\text{CaO} = 56$CaO=56.)

Step 1 — moles of $\text{CaCO}_3$CaCO3​: $\dfrac{\text{mass}}{M_r} = \dfrac{50}{100} = 0.5 \text{ mol}$Mr​mass​=10050​=0.5 mol.

Step 2 — mole ratio $\text{CaCO}_3 : \text{CaO} = 1 : 1$CaCO3​:CaO=1:1, so moles of CaO $= 0.5 \text{ mol}$=0.5 mol.

Step 3 — theoretical yield of CaO: $\text{mass} = \text{moles} \times M_r = 0.5 \times 56 = 28 \text{ g}$mass=moles×Mr​=0.5×56=28 g.

Step 4 — percentage yield: $\dfrac{\text{actual}}{\text{theoretical}} \times 100 = \dfrac{21}{28} \times 100 = 75\%$theoreticalactual​×100=2821​×100=75%.

Answer: the percentage yield is $75\%$75%.

Worked Example 4 — A reaction with a mole ratio

Magnesium burns in oxygen: $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO. Burning $6.0 \text{ g}$6.0 g of magnesium gives $8.0 \text{ g}$8.0 g of magnesium oxide. Calculate the percentage yield. ($A_r$Ar​: Mg = 24; $M_r$Mr​ of MgO $= 40$=40.)

Step 1 — moles of Mg: $\dfrac{6.0}{24} = 0.25 \text{ mol}$246.0​=0.25 mol.

Step 2 — mole ratio $\text{Mg} : \text{MgO} = 2 : 2 = 1 : 1$Mg:MgO=2:2=1:1, so moles of MgO $= 0.25 \text{ mol}$=0.25 mol.

Step 3 — theoretical yield of MgO: $0.25 \times 40 = 10 \text{ g}$0.25×40=10 g.

Step 4 — percentage yield: $\dfrac{8.0}{10} \times 100 = 80\%$108.0​×100=80%.

Answer: the percentage yield is $80\%$80%.

Exam Tip: For "calculate the percentage yield" questions where the theoretical yield is not given, always calculate the theoretical yield first by the reacting-mass route (moles → ratio → mass), then divide the actual by the theoretical.

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Why Yields Are Less Than 100%

In practice the actual yield is nearly always below the theoretical yield. The main reasons are:

Reason	What happens
Incomplete reaction	Not all the reactant reacts — some is left over when the reaction is stopped
Reversible reaction	The reaction goes both ways and does not fully convert reactants to products
Side reactions	The reactants also react in other ways, forming unwanted by-products instead of the desired product
Loss in handling	Product is lost during filtering, transferring between containers, or left stuck to apparatus
Impurities / loss in purification	Some product is lost when it is washed and purified

So a percentage yield below 100% does not mean the equation is wrong or that atoms have been destroyed — it simply reflects these practical losses and the limits of the reaction.