How to Distinguish Actual, Theoretical and Percentage Yield

A student heats 6.35 g of copper in excess oxygen and recovers 7.20 g of dry copper(II) oxide. The product weighs more than the copper they started with. Has the reaction somehow produced more than 100% yield?

No. Oxygen has combined with the copper, adding mass to the product. The important question is not whether the product weighs more than the starting metal. It is how much product was predicted to form compared with how much was actually recovered.

Before calculating, consider three possible quantities:

• 6.35 g: the mass of copper used.
• 7.20 g: the mass of copper(II) oxide actually recovered.
• A third value: the maximum mass of copper(II) oxide predicted by the balanced equation.

The distinction is straightforward once you know what each number represents:

• Theoretical yield: the maximum amount of product that could form, calculated using stoichiometry.
• Actual yield: the amount of product actually obtained in an experiment, determined by measurement.
• Percentage yield: the actual yield expressed as a percentage of the theoretical yield.

These three quantities describe different stages of the same reaction. Let’s work through how to identify and calculate each one.

01What is the difference between actual, theoretical and percentage yield?

Imagine preparing a product in a chemistry laboratory.

Before the reaction, you know how much reactant you have. You can use its mass, molar mass, and the balanced chemical equation to predict how much product should form.

After the reaction, you collect the product and measure it.

You now have two different quantities: a calculated prediction and an experimental measurement. Percentage yield compares them.

Theoretical yield: what should form?

The theoretical yield is the maximum amount of a specified product that can form from the available reactants.

It is calculated using the balanced chemical equation.

The calculation assumes:

• The limiting reagent reacts completely.
• The reactants follow the specified reaction.
• No reactant is diverted into competing reactions.
• The desired product is formed according to the equation.

It is a prediction, not an experimental measurement.

For example, if stoichiometry predicts that a reaction can produce 8.00 g of a substance, its theoretical yield is 8.00 g.

You might collect less than this amount in practice.

Actual yield: what did you actually obtain?

The actual yield is the amount of desired product obtained experimentally.

For an isolated solid, this generally means the mass of the pure, dry product recovered after separation and purification.

For a gas, it might be the measured volume or calculated number of moles of gas collected.

There is an important detail here: the total mass of material collected is not automatically the actual yield.

If your solid contains water, unreacted reactant, or other impurities, you must determine how much of the sample is actually the desired product.

For example, collecting 5.00 g of a solid that contains only 4.50 g of the desired product does not give an actual yield of 5.00 g. The actual yield is 4.50 g.

Percentage yield: how successful was the reaction?

Percentage yield compares the actual yield with the theoretical yield.

The formula is:

\[ \%\text{ yield}= \frac{\text{actual yield}} {\text{theoretical yield}} \times100 \]

You must compare the same product in the same units.

For example, if the theoretical yield is 8.00 g and the actual yield is 6.40 g:

\[ \begin{aligned} \%\text{ yield} &=\frac{6.40}{8.00}\times100\\ &=80.0\% \end{aligned} \]

This means that 80.0% of the theoretical amount was recovered as the desired product.

It does not mean that 80.0% of the reactant necessarily reacted. Incomplete reaction, competing reactions, and product loss can all affect the result.

02How to calculate theoretical yield

The most important calculation skill is understanding what the coefficients in a balanced chemical equation represent.

They give you mole ratios, not mass ratios.

Consider the oxidation of copper:

\[ \ce{2Cu(s) + O2(g) -> 2CuO(s)} \]

The coefficients tell us:

\[ 2\text{ mol Cu}:1\text{ mol O}_2: 2\text{ mol CuO} \]

This means two moles of copper react with one mole of oxygen to produce two moles of copper(II) oxide.

It does not mean 2 g of copper reacts with 1 g of oxygen to produce 2 g of copper oxide.

You must convert masses into moles before using the equation.

The mass-to-moles-to-moles-to-mass method

When a question gives you the mass of a reactant and asks for the theoretical mass of a product, follow this sequence:

Mass of reactant → Moles of reactant → Moles of product → Mass of product

The two essential formulas are:

\[ n=\frac{m}{M} \]

and

\[ m=nM \]

where:

• \(n\) = amount of substance, in mol.
• \(m\) = mass, in g.
• \(M\) = molar mass, in \(\mathrm{g\,mol^{-1}}\).

The steps are:

1. Write and balance the chemical equation.
2. Convert the known reactant mass into moles.
3. Use the coefficient ratio to calculate moles of product.
4. Convert the product amount into mass.

If multiple reactants are supplied, you must identify the limiting reagent first.

The limiting reagent is the reactant that runs out first according to the balanced equation. It determines the maximum possible amount of product.

03What if there are two reactants?

So far, we have considered a reaction where oxygen is supplied in excess.

Many HSC questions give you the amounts of two reactants. You must then determine which reactant limits the theoretical yield.

A common mistake is choosing the reactant with the smaller starting mass.

That doesn’t work because different substances have different molar masses, and the balanced equation may require different mole ratios.

Instead, convert both reactants into moles and account for their coefficients.

A useful method is to divide each reactant’s mole amount by its coefficient.

The smallest result identifies the limiting reagent.

04A quick decision method for HSC yield questions

Before starting a calculation, identify what the question is asking you to find.

You can also rearrange the percentage yield equation.

To calculate actual yield:

\[ \text{Actual yield} = \frac{\%\text{ yield}}{100} \times\text{Theoretical yield} \]

To calculate theoretical yield:

\[ \text{Theoretical yield} = \frac{\text{Actual yield}} {\%\text{ yield}/100} \]

For example, if a reaction has an actual yield of 6.00 g and a percentage yield of 75.0%:

\[ \begin{aligned} \text{Theoretical yield} &=\frac{6.00}{0.750}\\ &=8.00\text{ g} \end{aligned} \]

You do not always need to calculate theoretical yield from the original reactant masses. Sometimes the question gives you enough information to work backwards.

Three mistakes worth avoiding

Mistake 1: Comparing the product mass with the starting reactant mass.

The product can contain atoms from multiple reactants. Use the theoretical product mass as the denominator when calculating percentage yield.

Mistake 2: Applying coefficients directly to masses.

Coefficients describe mole ratios. Convert masses into moles before applying them.

Mistake 3: Treating the entire recovered sample as actual yield.

If the product contains impurities, excess solvent, or unreacted material, the total sample mass may not equal the actual yield of the desired product.

If a question gives a purity percentage, you may need to calculate the mass of pure product first. This uses the same reasoning as percentage composition by mass for HSC Chemistry.

05Questions and solutions

The following questions progress from straightforward yield calculations to problems that combine stoichiometry, experimental measurements, and product composition.

Question 1

Hydrogen reacts with oxygen to produce water:

\[ \ce{2H2(g) + O2(g) -> 2H2O(l)} \]

A student reacts 0.0400 mol of hydrogen with excess oxygen. The experiment produces 0.612 g of pure water.

Calculate:

(a) The theoretical yield of water in grams.

(b) The percentage yield.

Use \(M(\ce{H2O})=18.0\ \mathrm{g\,mol^{-1}}\).

Solution 1

The theoretical yield is 0.720 g of water, and the percentage yield is 85.0%.

Step 1: Calculate the theoretical moles of water.

The balanced equation gives a 2:2 ratio between hydrogen and water.

Therefore:

\[ n(\ce{H2O})=0.0400\text{ mol} \]

Step 2: Convert the theoretical moles into mass.

\[ \begin{aligned} m(\ce{H2O}) &=nM\\ &=0.0400\times18.0\\ &=0.720\text{ g} \end{aligned} \]

The theoretical yield is 0.720 g.

Step 3: Identify the actual yield.

The experiment produced 0.612 g of water.

Therefore:

\[ \text{Actual yield}=0.612\text{ g} \]

Step 4: Calculate the percentage yield.

\[ \begin{aligned} \%\text{ yield} &=\frac{0.612}{0.720}\times100\\ &=85.0\% \end{aligned} \]

The student recovered 85.0% of the theoretical amount of water.

Question 2

Magnesium chloride reacts with sodium hydroxide to produce magnesium hydroxide:

\[ \ce{MgCl2(aq) + 2NaOH(aq) -> Mg(OH)2(s) + 2NaCl(aq)} \]

A student mixes:

• 40.0 mL of 0.100 mol L\(^{-1}\) magnesium chloride.
• 60.0 mL of 0.120 mol L\(^{-1}\) sodium hydroxide.

After filtration and drying, 0.176 g of pure magnesium hydroxide is recovered.

Calculate:

(a) The limiting reagent.

(b) The theoretical yield of magnesium hydroxide in grams.

(c) The percentage yield.

Use \(M(\ce{Mg(OH)2})=58.3\ \mathrm{g\,mol^{-1}}\).

Solution 2

Sodium hydroxide is the limiting reagent. The theoretical yield is 0.210 g of magnesium hydroxide, and the percentage yield is 83.9%.

Step 1: Convert the solution volumes into litres.

For magnesium chloride:

\[ 40.0\text{ mL}=0.0400\text{ L} \]

For sodium hydroxide:

\[ 60.0\text{ mL}=0.0600\text{ L} \]

Step 2: Calculate the moles of each reactant.

Use:

\[ n=cV \]

where \(c\) is concentration in mol L\(^{-1}\), and \(V\) is volume in litres.

For magnesium chloride:

\[ \begin{aligned} n(\ce{MgCl2}) &=0.100\times0.0400\\ &=0.00400\text{ mol} \end{aligned} \]

For sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=0.120\times0.0600\\ &=0.00720\text{ mol} \end{aligned} \]

Step 3: Identify the limiting reagent.

The equation requires two moles of sodium hydroxide for every mole of magnesium chloride.

Compare the available moles with their coefficients:

\[ \begin{aligned} \ce{MgCl2}:&\quad 0.00400/1=0.00400\\ \ce{NaOH}:&\quad 0.00720/2=0.00360 \end{aligned} \]

Sodium hydroxide has the smaller result.

Therefore, sodium hydroxide is the limiting reagent.

Step 4: Calculate the theoretical moles of magnesium hydroxide.

The equation gives a 2:1 ratio between sodium hydroxide and magnesium hydroxide.

\[ \begin{aligned} n(\ce{Mg(OH)2}) &=0.00720\times\frac{1}{2}\\ &=0.00360\text{ mol} \end{aligned} \]

Step 5: Convert the theoretical amount into mass.

\[ \begin{aligned} m(\ce{Mg(OH)2}) &=0.00360\times58.3\\ &=0.20988\text{ g}\\ &\approx0.210\text{ g} \end{aligned} \]

The theoretical yield is 0.210 g.

Step 6: Calculate the percentage yield.

\[ \begin{aligned} \%\text{ yield} &=\frac{0.176}{0.20988}\times100\\ &=83.9\% \end{aligned} \]

The experiment recovered 83.9% of the theoretical magnesium hydroxide.

Notice that sodium hydroxide was not the reactant with the smaller mole amount. It was limiting because the equation requires twice as many moles of sodium hydroxide as magnesium chloride.

Question 3

Calcium oxide reacts with water to form calcium hydroxide:

\[ \ce{CaO(s) + H2O(l) -> Ca(OH)2(s)} \]

Water is supplied in excess.

A student recovers 8.40 g of pure, dry calcium hydroxide. The percentage yield is calculated to be 70.0%.

Calculate:

(a) The theoretical yield of calcium hydroxide.

(b) The mass of calcium oxide initially required.

Use:

• \(M(\ce{Ca(OH)2})=74.1\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{CaO})=56.1\ \mathrm{g\,mol^{-1}}\)

Solution 3

The theoretical yield is 12.0 g of calcium hydroxide, and the initial mass of calcium oxide was 9.08 g.

Step 1: Identify the actual yield and percentage yield.

The actual yield is the experimentally recovered mass:

\[ \text{Actual yield}=8.40\text{ g} \]

The percentage yield is 70.0%.

Step 2: Rearrange the percentage yield equation to calculate theoretical yield.

\[ \begin{aligned} \text{Theoretical yield} &=\frac{\text{Actual yield}} {\%\text{ yield}/100}\\ &=\frac{8.40}{0.700}\\ &=12.0\text{ g} \end{aligned} \]

The theoretical yield is 12.0 g.

Step 3: Convert the theoretical product mass into moles.

\[ \begin{aligned} n(\ce{Ca(OH)2}) &=\frac{12.0}{74.1}\\ &=0.1619\text{ mol} \end{aligned} \]

Step 4: Use the mole ratio to determine the required moles of calcium oxide.

The balanced equation gives a 1:1 ratio.

Therefore:

\[ n(\ce{CaO})=0.1619\text{ mol} \]

Step 5: Calculate the initial mass of calcium oxide.

\[ \begin{aligned} m(\ce{CaO}) &=nM\\ &=0.1619\times56.1\\ &=9.08\text{ g} \end{aligned} \]

The initial mass of calcium oxide was 9.08 g.

This question works backwards from the percentage yield. Remember that the actual yield is not the mass of starting reactant. It is the mass of product recovered.

Question 4

Zinc reacts with hydrochloric acid to produce hydrogen gas:

\[ \ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)} \]

A student reacts 0.100 mol of zinc with excess hydrochloric acid.

The hydrogen is collected over water at 25.0°C. The following measurements are recorded:

• Total gas pressure: 101.3 kPa.
• Water vapour pressure: 3.17 kPa.
• Collected gas volume: 2.20 L.
• Temperature: 298 K.

Use the ideal gas equation:

\[ PV=nRT \]

where:

• \(P\) = pressure in kPa.
• \(V\) = volume in L.
• \(n\) = amount of gas in mol.
• \(R=8.31\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\).
• \(T\) = temperature in K.

Calculate:

(a) The actual amount of hydrogen collected.

(b) The theoretical amount of hydrogen.

(c) The percentage yield.

Assume ideal gas behaviour.

Solution 4

The actual amount of hydrogen collected is 0.0872 mol. The theoretical amount is 0.100 mol, giving a percentage yield of 87.2%.

Step 1: Determine the partial pressure of hydrogen.

The collected gas contains hydrogen and water vapour.

The total pressure is the sum of their partial pressures:

\[ P_{\text{total}}=P_{\ce{H2}}+P_{\ce{H2O}} \]

Rearrange:

\[ \begin{aligned} P_{\ce{H2}} &=101.3-3.17\\ &=98.13\text{ kPa} \end{aligned} \]

We must use the hydrogen partial pressure, not the total pressure.

Step 2: Calculate the actual moles of hydrogen.

Rearrange the ideal gas equation:

\[ n=\frac{PV}{RT} \]

Substitute the hydrogen pressure:

\[ \begin{aligned} n(\ce{H2}) &=\frac{98.13\times2.20}{8.31\times298}\\ &=0.0872\text{ mol} \end{aligned} \]

This is the actual yield in moles.

Step 3: Calculate the theoretical amount of hydrogen.

The balanced equation gives a 1:1 ratio between zinc and hydrogen.

The student started with 0.100 mol of zinc, and hydrochloric acid is in excess.

Therefore:

\[ n_{\text{theoretical}}(\ce{H2}) =0.100\text{ mol} \]

Step 4: Calculate the percentage yield.

\[ \begin{aligned} \%\text{ yield} &=\frac{0.0872}{0.100}\times100\\ &=87.2\% \end{aligned} \]

The percentage yield is 87.2%.

A tempting mistake is to use 101.3 kPa in the ideal gas equation. This would give approximately 0.0900 mol of gas and an apparent yield of 90.0%.

The problem is that the collected gas is not entirely hydrogen. Some of the measured pressure comes from water vapour, which is not the desired product.

This is why actual yield must represent the amount of the target substance, rather than an uncorrected measurement of the entire sample.

Question 5

Anhydrous copper(II) sulfate can form blue crystals of copper(II) sulfate pentahydrate when crystallised from water.

The hydration process is represented by:

\[ \mathrm{CuSO_4(s)+5H_2O(l) \rightarrow CuSO_4\cdot5H_2O(s)} \]

A student starts with 4.00 g of pure, anhydrous copper(II) sulfate and uses excess water.

After crystallisation, the collected crystals have a mass of 4.85 g.

Further analysis establishes that:

• 0.15 g of the collected sample is free surface water.
• The remaining material is pure copper(II) sulfate pentahydrate.
• A separate analysis of the remaining solution finds that 1.00 g of anhydrous copper(II) sulfate remains dissolved.

Use:

• \(M(\ce{CuSO4})=159.6\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{CuSO4.5H2O})=249.6\ \mathrm{g\,mol^{-1}}\)

Assume all the copper(II) sulfate is either in the recovered crystals or remains dissolved in the solution. No decomposition occurs.

Calculate:

(a) The theoretical yield of copper(II) sulfate pentahydrate.

(b) The actual yield and percentage yield.

(c) The mass of anhydrous copper(II) sulfate represented by the recovered crystals. Check whether the remaining amount is consistent with the analysis of the solution.

(d) Explain why calculating \(4.85/4.00\times100\) does not give a valid percentage yield.

Solution 5

The theoretical yield is 6.26 g of copper(II) sulfate pentahydrate, the actual yield is 4.70 g, and the percentage yield is 75.1%. The remaining 0.995 g of anhydrous copper(II) sulfate is consistent with the 1.00 g found in the solution.

The important distinction is that water of crystallisation is part of the product, while free surface water is an impurity.

Step 1: Calculate the initial moles of anhydrous copper(II) sulfate.

The starting mass is 4.00 g.

\[ \begin{aligned} n(\ce{CuSO4}) &=\frac{4.00}{159.6}\\ &=0.0251\text{ mol} \end{aligned} \]

Step 2: Calculate the theoretical moles of copper(II) sulfate pentahydrate.

The equation gives a 1:1 ratio between anhydrous copper(II) sulfate and the pentahydrate.

Therefore:

\[ n(\ce{CuSO4.5H2O}) =0.0251\text{ mol} \]

Step 3: Calculate the theoretical mass of the hydrate.

Use the molar mass of the complete hydrated product, not the anhydrous salt.

\[ \begin{aligned} m(\ce{CuSO4.5H2O}) &=0.0251\times249.6\\ &=6.26\text{ g} \end{aligned} \]

The theoretical yield is 6.26 g.

Notice that this is greater than the starting mass of anhydrous copper(II) sulfate.

That is expected because five water molecules become part of each formula unit of the crystalline product.

Step 4: Calculate the actual yield.

The collected crystals have a total mass of 4.85 g.

However, 0.15 g is free surface water, not part of the desired crystalline product.

Therefore:

\[ \begin{aligned} \text{Actual yield} &=4.85-0.15\\ &=4.70\text{ g} \end{aligned} \]

The actual yield is 4.70 g of pure copper(II) sulfate pentahydrate.

Do not subtract the five waters of crystallisation. They are part of the correct product formula.

Step 5: Calculate the percentage yield.

\[ \begin{aligned} \%\text{ yield} &=\frac{4.70}{6.26}\times100\\ &=75.1\% \end{aligned} \]

The percentage yield is 75.1%.

Step 6: Check the mass of copper(II) sulfate represented by the recovered crystals.

This requires another mole calculation.

First, calculate the moles of recovered pentahydrate:

\[ \begin{aligned} n(\ce{CuSO4.5H2O}) &=\frac{4.70}{249.6}\\ &=0.01883\text{ mol} \end{aligned} \]

The equation gives a 1:1 ratio between the hydrate and anhydrous copper(II) sulfate.

Therefore, the crystals contain 0.01883 mol of copper(II) sulfate units.

Convert this amount into the equivalent mass of anhydrous copper(II) sulfate:

\[ \begin{aligned} m(\ce{CuSO4}) &=0.01883\times159.6\\ &=3.01\text{ g} \end{aligned} \]

The crystals therefore contain the equivalent of 3.01 g of anhydrous copper(II) sulfate.

Step 7: Check the amount remaining in solution.

The initial mass was 4.00 g.

The amount represented by the recovered crystals was 3.01 g.

Therefore:

\[ \begin{aligned} m_{\text{remaining}} &=4.00-3.01\\ &=0.99\text{ g} \end{aligned} \]

Using unrounded intermediate values gives approximately 0.995 g.

This agrees with the independently measured 1.00 g remaining in solution, to the precision of the supplied data.

The mass balance is consistent.

Step 8: Explain why the 121% calculation is incorrect.

A student might calculate:

\[ \frac{4.85}{4.00}\times100=121\% \]

This is not a valid percentage yield because the numerator and denominator represent different substances.

The numerator is the mass of wet copper(II) sulfate pentahydrate crystals.

The denominator is the mass of anhydrous copper(II) sulfate originally used.

They are not directly comparable.

There are two separate issues:

• The five waters of crystallisation are part of the intended product. They must be included in both the theoretical and actual product masses.
• The 0.15 g of free surface water is not part of the product. It must be excluded from the actual yield.

The 3.01 g calculated in Step 6 is also not the actual yield. It is simply the mass of anhydrous copper(II) sulfate represented within the 4.70 g of recovered hydrate.

This is why identifying the exact chemical formula of the product matters before choosing a molar mass.

06Where to go next

Once you can distinguish actual yield, theoretical yield, and percentage yield, the next challenge is explaining why the actual yield falls below the theoretical maximum.

A percentage yield of 75% tells you how much product was recovered, but it cannot independently tell you whether the reaction was incomplete, a competing reaction occurred, or product was lost during filtration and purification.

To investigate those possibilities, you need additional experimental evidence and a proper mass balance.

For that next step, see Theoretical vs Actual Yield: Calculate Yield and Explain Losses.