Percent Yield Worked Examples
Below are nine fully worked percent yield problems, each carried from given data through to a rated result: straightforward gram calculations, moles-only cases, mixed-unit conversions, limiting-reagent and multi-step syntheses, plus an above-100-percent result and a very low yield. Every example follows the same formula and rating scale used throughout this site.
How These Worked Examples Are Set Up
Each example below starts from given data, masses, moles, or both, and ends with a percent yield rated against the same quality bands used by the percent yield calculator on the home page: excellent for 90% and above, very good for 80% to 89%, good for 70% to 79%, fair for 50% to 69%, and poor for anything under 50%. Every calculation uses the same percent yield formula, actual yield divided by theoretical yield times 100, so once you can follow one example you can follow all of them.
If you need a refresher on how actual yield is measured in the lab or how theoretical yield is calculated from a balanced equation, those pages cover the method in more depth. This page focuses on complete problems, given data through to a rated answer, including a few awkward cases that trip students up on homework and lab reports.
Quick-Reference Table of All Nine Examples
| Example | Scenario | Actual yield | Theoretical yield | Percent yield | Rating |
|---|---|---|---|---|---|
| 1 | Grams only | 8.4 g | 10.0 g | 84.0% | Very good |
| 2 | Limiting reagent mass to theoretical yield | 12.9 g | 16.7 g | 77.1% | Good |
| 3 | Moles only | 0.184 mol | 0.220 mol | 83.6% | Very good |
| 4 | Mixed units, grams and moles | 4.98 g | 5.40 g | 92.2% | Excellent |
| 5 | Two reactants, limiting reagent | 17.5 g | 21.3 g | 82.2% | Very good |
| 6 | Two-step synthesis | 41.2 g | 63.0 g | 65.4% | Fair |
| 7 | Above 100 percent, undried sample | 24.9 g | 22.4 g | 111% | Above 100, recheck drying |
| 8 | Very low yield | 1.4 g | 5.41 g | 25.9% | Poor |
| 9 | Milligram scale | 210 mg | 250 mg | 84.0% | Very good |
Example 1: Percent Yield from Given Masses Only
A student runs a synthesis and recovers 8.4 g of dried product. The reaction's balanced equation predicts a theoretical yield of 10.0 g for the scale used. Both numbers are already in grams, so no conversion is needed before applying the percent yield formula.
- Percent yield = (actual yield / theoretical yield) x 100
- Percent yield = (8.4 g / 10.0 g) x 100
- Percent yield = 84.0%
A result of 84.0% falls in the very good band, 80% to 89%, the kind of number a well-run small-scale synthesis typically produces once minor losses from transfers and filtration are accounted for.
Example 2: Computing Theoretical Yield from a Limiting Reagent Mass
Benzoic acid is heated with excess methanol and a trace of acid catalyst to form methyl benzoate by Fischer esterification: C6H5COOH + CH3OH -> C6H5COOCH3 + H2O. The reaction starts with 15.0 g of benzoic acid as the limiting reagent, and the student isolates 12.9 g of methyl benzoate after workup. Here the theoretical yield has to be calculated before percent yield can be found.
- Moles of benzoic acid = 15.0 g / 122.12 g/mol = 0.1228 mol
- The mole ratio of benzoic acid to methyl benzoate is 1:1, so moles of product = 0.1228 mol
- Theoretical mass of methyl benzoate = 0.1228 mol x 136.15 g/mol = 16.7 g
- Percent yield = (12.9 g / 16.7 g) x 100 = 77.1%
77.1% sits in the good band, 70% to 79%, a reasonable result for a single esterification with one purification step.
Example 3: Percent Yield Using Moles Only
Mixing aqueous lead(II) nitrate with sodium chloride precipitates lead(II) chloride: Pb(NO3)2 + 2 NaCl -> PbCl2 + 2 NaNO3. A lab worksheet reports the results directly in moles rather than mass: stoichiometry predicts 0.220 mol of PbCl2, and 0.184 mol is actually recovered after filtration and drying. Because both figures already share a unit, the mole version of the formula applies without any mass conversion.
- Percent yield = (actual moles / theoretical moles) x 100
- Percent yield = (0.184 mol / 0.220 mol) x 100
- Percent yield = 83.6%
83.6% rates as very good. Working in moles instead of grams changes nothing about the calculation; the ratio only cares that both numbers use the same unit.
Example 4: Mixed Units, Grams and Moles Together
This is the case that catches most students off guard: one yield is given in grams and the other in moles. Zinc metal displaces copper from copper(II) sulfate solution: Zn + CuSO4 -> ZnSO4 + Cu. Stoichiometry predicts 0.0850 mol of copper metal, but the copper collected on the filter paper is weighed on a balance and reported as 4.98 g. The two numbers cannot be compared until they share a unit.
- Convert the theoretical yield to grams using the molar mass of copper, 63.55 g/mol: 0.0850 mol x 63.55 g/mol = 5.40 g
- Percent yield = (4.98 g / 5.40 g) x 100
- Percent yield = 92.2%
92.2% is excellent, 90% and above. The lesson here matters more than the answer: always check that actual and theoretical yield are expressed in the same unit before dividing, whether that means converting moles to grams or grams to moles.
Example 5: Two Reactants, Finding the Limiting Reagent First
Sodium hydroxide and sulfuric acid react to form sodium sulfate: 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O. A student starts with 12.0 g of NaOH and 15.0 g of H2SO4 and needs to find which one is the limiting reagent before theoretical yield can be found.
- Moles of NaOH = 12.0 g / 40.00 g/mol = 0.300 mol
- Moles of H2SO4 = 15.0 g / 98.09 g/mol = 0.153 mol
- The equation needs 2 mol NaOH per 1 mol H2SO4, so 0.153 mol H2SO4 would require 0.306 mol NaOH; only 0.300 mol is available, so NaOH is the limiting reagent
- Moles of Na2SO4 from 0.300 mol NaOH = 0.300 / 2 = 0.150 mol
- Theoretical mass of Na2SO4 = 0.150 mol x 142.04 g/mol = 21.3 g
- Percent yield = (17.5 g / 21.3 g) x 100 = 82.2%
82.2% is very good. Skipping the limiting reagent check is the most common way to get this style of problem wrong, since basing theoretical yield on the wrong reactant shifts every number that follows.
Example 6: A Two-Step Synthesis Route
Multi-step syntheses carry a theoretical yield across more than one reaction. Ethanol is first oxidized to acetic acid, then the acetic acid is esterified with 1-butanol to form butyl acetate: step one, C2H5OH + 2[O] -> CH3COOH + H2O; step two, CH3COOH + C4H9OH -> C4H9OOCCH3 + H2O. The synthesis starts with 25.0 g of ethanol, and 41.2 g of butyl acetate is isolated at the end of the second step.
- Moles of ethanol = 25.0 g / 46.07 g/mol = 0.5427 mol
- Step one is 1:1, so 0.5427 mol of acetic acid is theoretically available for step two
- Step two is also 1:1, so the overall theoretical moles of butyl acetate = 0.5427 mol
- Theoretical mass of butyl acetate = 0.5427 mol x 116.16 g/mol = 63.0 g
- Percent yield = (41.2 g / 63.0 g) x 100 = 65.4%
65.4% lands in the fair band. Multi-step routes like this one tend to report lower overall percent yield than a single reaction, because losses in step one carry forward and compound with losses in step two rather than resetting between steps.
Example 7: A Result Above 100 Percent
Barium chloride solution is combined with sodium sulfate solution to precipitate barium sulfate: BaCl2 + Na2SO4 -> BaSO4 + 2 NaCl. Starting from 20.0 g of BaCl2 as the limiting reagent, the theoretical yield of BaSO4 is calculated first.
- Moles of BaCl2 = 20.0 g / 208.23 g/mol = 0.0961 mol
- Theoretical mass of BaSO4 = 0.0961 mol x 233.39 g/mol = 22.4 g
- The precipitate is filtered and weighed straight off the filter paper at 24.9 g, without a full drying step
- Percent yield = (24.9 g / 22.4 g) x 100 = 111%
A percent yield over 100% is not a valid final result on its own; it is a sign that the measured mass includes something besides the pure product, almost always residual water, filter paper fiber, or unreacted starting material trapped in the crystals. The fix is to dry the sample fully in an oven or desiccator and re-weigh before recalculating. If a dried, re-weighed sample still reads oddly high or low, the common lab errors page covers the usual causes in more depth.
Example 8: A Very Low Percent Yield
1-bromobutane is treated with aqueous sodium hydroxide to substitute the bromine for a hydroxyl group: C4H9Br + NaOH -> C4H9OH + NaBr. Starting from 10.0 g of 1-bromobutane, only 1.4 g of butan-1-ol is isolated after workup.
- Moles of 1-bromobutane = 10.0 g / 137.02 g/mol = 0.0730 mol
- Theoretical mass of butan-1-ol = 0.0730 mol x 74.12 g/mol = 5.41 g
- Percent yield = (1.4 g / 5.41 g) x 100 = 25.9%
25.9% is poor by any reading of the scale. This particular substitution competes with an elimination side reaction that turns some of the starting material into an alkene instead of the alcohol, one plausible explanation for a result this low, but it is also worth checking the more ordinary causes covered on the lab errors page: an incomplete reaction, product lost during a transfer, or a filtration step that let product through the paper.
Example 9: A Milligram-Scale Example
Not every synthesis happens on the gram scale. A small-scale organic synthesis run in a research lab predicts a theoretical yield of 250 mg of purified product; 210 mg is recovered after column purification.
- Percent yield = (actual yield / theoretical yield) x 100
- Percent yield = (210 mg / 250 mg) x 100
- Percent yield = 84.0%
84.0% is very good, the same rating as example 1's gram-scale result. The unit, milligrams, grams, or moles, never changes the method described on the percent yield formula page; it only changes how many decimal places show up along the way.
Common Mistakes Across These Problems
Three mistakes account for most wrong answers in problems like these. The first is dividing theoretical yield by actual yield instead of the other way around, which produces a number that looks plausible but is inverted. The second is comparing a mass to a mole count without converting one of them first, the mistake built into example 4 on purpose. The third is skipping the limiting reagent check when two reactants are given, the point of example 5.
For more practice applying the same formula to fresh numbers, the practice problems page has additional questions to work through.
Frequently asked questions
Where can I find worked percent yield examples?
This page collects nine complete worked examples, each carried from given data through to a rated percent yield result. They cover grams-only, moles-only, mixed-unit, limiting-reagent, multi-step synthesis, above-100-percent, and low-yield scenarios, so most homework and lab-report situations have a matching case to compare against on this page.
What does a percent yield example with mixed units look like?
A mixed-units example gives actual yield in one unit, usually grams, and theoretical yield in another, usually moles, or vice versa. Example 4 on this page walks through converting 0.0850 mol of copper to 5.40 g using its molar mass before dividing, since percent yield only works once both numbers share the same unit.
How do you solve a percent yield problem with two reactants given?
Convert both reactant masses to moles, compare each to the mole ratio in the balanced equation, and identify whichever reactant runs out first as the limiting reagent. Example 5 does this with sodium hydroxide and sulfuric acid, finding sodium hydroxide is limiting before calculating theoretical yield of sodium sulfate and comparing it to the actual yield given.
What does a multi-step synthesis percent yield calculation look like?
It carries the limiting reagent's moles through each reaction step using the mole ratios in every balanced equation, then applies the final product's molar mass only at the last step. Example 6 tracks ethanol through an oxidation and an esterification to reach an overall theoretical yield of butyl acetate before comparing it to the actual mass isolated.
Why would a percent yield example come out above 100 percent?
A result above 100% almost always means the measured actual yield includes more than the pure product, commonly residual water, solvent, or unreacted starting material trapped in the sample. Example 7 shows a barium sulfate precipitate weighed before fully drying, giving 111%. The fix is drying the sample completely and re-weighing before recalculating percent yield.
What counts as a good result in these examples?
The same rating scale applies across every example on this page and on the home page calculator: 90% and above is excellent, 80% to 89% is very good, 70% to 79% is good, 50% to 69% is fair, and below 50% is poor. Most single-step lab syntheses in these examples land in the good to very good range.
How do you handle a percent yield problem given only in moles?
When both actual and theoretical yield are already in moles, as in example 3's lead(II) chloride precipitation, no conversion is needed. Divide actual moles by theoretical moles and multiply by 100, exactly as the formula works with grams. The unit only matters when actual and theoretical yield are given in different units from each other.
What is the most common mistake in these types of problems?
The most common mistake is dividing theoretical yield by actual yield instead of actual by theoretical, which flips the answer. Close behind is comparing a mass to a mole count without converting first, and skipping the limiting reagent check when two starting materials are given, the two issues built into examples 4 and 5 on this page.
Can these examples be solved without a calculator?
Most of them, yes, with rounded molar masses and a bit of long division, though a calculator makes the mole conversions in examples 2, 4, 5, 6, 7, and 8 faster and less error-prone. The simple grams-only and moles-only cases, examples 1, 3, and 9, can be worked entirely by hand with basic arithmetic.
Where can I practice more problems on my own?
The practice problems page (/problems/) has additional percent yield questions with different numbers to work through, and the worksheets page (/worksheets/) has printable problem sets for repeated practice. Both use the same formula and rating scale demonstrated in the worked examples on this page.