Percentage yield: actual versus theoretical
Calculate percentage yield from actual and theoretical amounts, distinguish it from purity, and explain low or apparently excessive experimental yields.
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Percentage yield asks: how much of the maximum possible product did we actually obtain? Start with the balanced equation and the limiting reactant, then compare the collected product with that maximum.
Actual versus theoretical yield
The theoretical yield is the maximum product amount calculated from the limiting reactant, assuming complete conversion to that product with no competing reactions. The actual yield is the amount of that product obtained experimentally.
Percentage yield = (actual yield)/(theoretical yield) × 100%
Compare the same product in the same form and unit. For a mass calculation, use the mass of dry, pure product; water or impurities in the weighed sample would inflate the apparent yield. A hydrate and its anhydrous salt have different molar masses, so they cannot be treated as the same product. Gas volumes must be measured at the same temperature and pressure if you compare them directly.
How purity differs
Percentage purity compares the mass of a desired substance with the total mass of a sample. That sample could be a reactant or a collected product. Purity tells you what is in the sample; yield tells you how much product you obtained relative to the calculated maximum.
The next lesson, percentage purity, develops the pure-part calculation from gas and titration data. If a reactant sample is impure, use only its pure reacting mass to calculate the theoretical yield.
Why yield can be below 100%
A low yield can result from different stages of the experiment:
- Less product forms: the reaction is incomplete, reaches equilibrium, or competing reactions consume some reactant.
- Some product is not collected: it is lost during transfer or separation, escapes as a gas, or remains dissolved in the mother liquor after crystallisation.
Choose an explanation supported by the experiment. For example, crystals remaining in solution reduce the collected yield even if the chemical reaction has already finished.
What does an apparent yield above 100% mean?
For the stated reaction and a correctly calculated theoretical yield, you cannot obtain more pure product than the maximum. An apparent result above 100% suggests a problem with the measured mass, product identity or calculation. Possible causes include wet crystals, impurities, an incorrect molar mass, or incorrect units. The percentage alone does not identify one unique cause.
Convert the limiting reactant to moles, use the equation ratio, and convert the product to the same unit as the actual yield. Then divide actual by theoretical. Keep the unrounded theoretical value for the final calculation.
From a direct comparison to a reacting-mass calculation
Modelled example 1
Percentage Yield (Direct)
Problem
The theoretical yield of iron is 50.0 g and the actual mass of dry, pure iron collected is 42.5 g. Calculate the percentage yield.
Study the worked solution
Identify actual and theoretical quantities
Method
Use 42.5 g as actual yield and 50.0 g as theoretical yield.
Reason
Yield compares collected product with the calculated maximum for the same substance and unit.
Working
Actual = 42.5 g; theoretical = 50.0 g.Form the yield percentage
Method
Divide actual by theoretical and multiply by 100.Reason
The actual amount is the achieved fraction of the maximum.
Working
Percentage yield = (42.5/50.0) × 100 = 85.0%.
Guided practice 2
Percentage Yield (Theoretical Yield From Stoichiometry)
Problem
Find the maximum before comparing
Hints
Hint 1: calculate theoretical product
n(Mg) = 4.80/24 and the equation gives a 1:1 Mg:MgO amount ratio.
Hint 2: compare like masses
M(MgO) = 40 g mol⁻¹; convert theoretical moles to grams before using actual/theoretical.
View solution step by step
Calculate magnesium amount
Method
Divide magnesium mass by molar mass.Reason
The balanced equation relates reactants and products in moles.
Working
n(Mg) = 4.80/24 = 0.200 mol.Find theoretical magnesium oxide
Method
Use the 2:2 amount ratio and convert product amount to mass.
Reason
Excess oxygen means magnesium determines the maximum product.
Working
n(MgO) = 0.200 mol; m_theoretical = 0.200(40) = 8.00 g.
Calculate percentage yield
Method
Divide actual 7.20 g by theoretical 8.00 g.Reason
Both values now refer to the same product in the same unit.
Working
Percentage yield = (7.20/8.00) × 100 = 90.0%.
Check the meaning of your percentage
Common misconception 3
Yield Is Not Purity
Learner response
A 10.0 g impure calcium carbonate sample contains 8.0 g pure CaCO₃. A learner calculates (8.0/10.0) × 100 = 80% and calls it percentage yield. Explain why the label is wrong and state the correct comparison.
Name the numerator and denominator
View solution step by step
Identify the quantities
Method
Recognise 8.0 g as pure substance and 10.0 g as total impure sample.
Reason
Neither value is an actual or theoretical product yield.
Working
Pure part / total sample = 8.0/10.0.Correct the interpretation
Method
Name the calculation percentage purity.Reason
Purity measures the desired substance within a sample; yield measures actual product against a theoretical maximum.
Working
Percentage purity = (8.0/10.0) × 100 = 80%.
The calcium carbonate calculation, zinc calculation and combined purity and yield problem continue in the separate purity lesson.
Try independently
Mind stretcher 1: Work backwards from percentage yieldExtension
A reaction has a theoretical yield of 12.5 g of a product and a percentage yield of 76.0%. Find the actual mass of dry, pure product collected.
Show answer
Actual yield is 12.5 × 76.0/100 = 9.50 g. Multiplying by the yield fraction gives the collected amount; dividing by it would give an amount greater than the maximum.
Mind stretcher 2: Investigate an apparent yield above 100%Extension
A student reports a yield of 112% for CuSO₄ * 5H₂O crystals. The calculation uses the correct theoretical yield for this hydrate. The crystals are weighed immediately after filtering, with visible liquid between them. Explain a plausible cause, state a suitable action, and explain why strong heating would be unsuitable.
Show answer
Liquid adhering to the crystals adds mass that is not copper(II) sulfate pentahydrate, so using the whole wet mass inflates the apparent yield. Blot off surface liquid with filter paper, allow the crystals to dry appropriately, then reweigh. Strong heating can remove water of crystallisation and change the product from the stated hydrate, so it would make the comparison invalid. The observation supports wet crystals as a cause; the percentage alone would not prove it.
Use the chemical calculations topic check to practise choosing the correct comparison. Continue to percentage purity to work back from experimental data to the pure substance in a sample.
Syllabus and review details
- SEC G3 Pure Chemistry 2027 · 2027
Content structure and subject content, PDF pages 9–24
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