Percent Yield Calculator
Percent yield tells you how much of a reaction’s theoretical product you actually managed to isolate, as a single percentage. Enter the theoretical yield from your stoichiometry and the actual yield you weighed, and this calculator returns the percent yield plus a plain-English efficiency reading, so a number like 84% immediately tells you whether the run was strong, ordinary or worth troubleshooting.
Yield loss (theoretical − actual): 8.00 in your unit
The efficiency reading is a guide, not a verdict on your technique: some reactions simply cannot approach 100% because of equilibrium or unavoidable side products.
Quick Answer
Percent yield = (actual yield ÷ theoretical yield) × 100. With a theoretical yield of 50 g and an actual yield of 42 g, the percent yield is 84.00% and the yield loss is 8.00 g. Type your own two numbers into the calculator to get the percentage, the loss and an efficiency reading.
Key takeaways
- Percent yield compares what you isolated against the stoichiometric maximum: actual ÷ theoretical × 100. The worked example on this page, 42 g from 50 g, gives 84.00%.
- Both yields must use the same unit, grams with grams or moles with moles, because the units cancel in the division.
- A result over 100% is a red flag, not a triumph: the product is usually impure, incompletely dried, or weighed with an error.
- The calculator also reports the yield loss (theoretical − actual), which tells you the physical amount that never made it into the flask.
- Quick Answer
- How to use the percent yield calculator
- The percent yield formula, and what each term really is
- Worked example: 42 g recovered from a 50 g ceiling
- Where the theoretical yield comes from
- Why real yields fall short, and what over 100% means
- What competing calculators miss
- Frequently asked questions
- Add This Calculator to Your Website
- References
How to use the percent yield calculator
Two numbers in, one percentage out, plus a reading of what the percentage means. The results stay hidden until you calculate.
- Enter the theoretical yield: the stoichiometric maximum from your balanced equation and limiting reactant, pre-filled as 50. It must be greater than zero; anything else gets a clear error message instead of a nonsense answer.
- Enter the actual yield: the pure, dry product you actually weighed, pre-filled as 42. Use the same unit as the theoretical yield, and it cannot be negative.
- Calculate and read three things: the percent yield as a large headline number, the result value for copying, and the yield loss in your unit. Underneath, the badge rates the run from Poor (under 50%) to Excellent (90% or above).
- Copy or share: the buttons under the results copy a one-line summary for a lab report or hand it to the device share sheet.
Blank or non-numeric entries stop with the message “Please enter valid numbers for both theoretical and actual yield.” The tool accepts any consistent unit, grams, moles, kilograms, because the ratio is dimensionless. It does not adjust for product purity: weigh the product dry and as pure as your procedure allows, since an impure weighing is the most common route to a yield above 100%.
The percent yield formula, and what each term really is
Percent yield = (actual yield ÷ theoretical yield) × 100. The formula is a ratio, which is why the units cancel, and it is the definition used by the OpenStax Chemistry textbook and by LibreTexts introductory chemistry.
Yield loss = Theoretical yield − Actual yield | Loss % = 100% − Percent yield
The theoretical yield is the ceiling: the product mass (or moles) the balanced equation promises if the limiting reactant converts completely and every molecule is recovered. The actual yield is the floor you stand on: what survived filtration, transfer, recrystallisation and drying. The gap between them is where chemistry happens, and the yield loss figure the calculator prints (theoretical − actual) puts a physical number on that gap. In the worked example the loss is 8.00 g, which is 16.00% of the theoretical 50 g.
Worked example: 42 g recovered from a 50 g ceiling
The calculator is pre-filled with this example. Step one: divide the actual yield by the theoretical yield, 42 ÷ 50 = 0.84. Step two: multiply by 100, giving 84.00%. The yield loss is 50 − 42 = 8.00 g, and the badge reads Very good, the 80–89% band for an efficient run with only modest losses.
| Reaction | Theoretical | Actual | Percent yield |
|---|---|---|---|
| Worked example (this page) | 50 g | 42 g | 84.00% |
| Aspirin synthesis (salicylic acid route) | 17.4 g | 13.9 g | 79.89% |
| Water from hydrogen combustion | 322.1 g | 290 g | 90.03% |
| Copper from CuSO4 + Zn (OpenStax) | 0.5072 g | 0.392 g | 77.29% |
| Generic check (50 from 60) | 60 g | 50 g | 83.33% |
Every row is a real run of this calculator: type the two numbers in and you get the same percentage back. The copper row is the textbook example from OpenStax (0.392 g Cu from a 0.5072 g ceiling = 77.29%), and the water row reproduces the hydrogen-combustion arithmetic used by stoichiometry references. Notice how the badge bands read them: 90.03% is Excellent, 77.29% and 79.89% are Good, 84.00% is Very good.
Where the theoretical yield comes from
The theoretical yield is not a guess; it is the product of a three-step stoichiometry calculation. First, identify the limiting reactant, the reactant that runs out first. Second, convert its mass to moles and apply the mole ratio from the balanced equation. Third, convert product moles back to mass using the product’s molar mass. For that last step you need a reliable molar mass, which is exactly what a compound mass calculator provides, and if you want to confirm the product’s formula against its mass, an exact mass calculator derives the monoisotopic mass atom by atom.
The most common student error is computing the ceiling from the wrong reactant, usually the one present in the largest mass rather than the limiting one. The ChemPRIME percent yield page walks through the Haber ammonia synthesis, where only about a quarter of the reactants convert to product even under industrial conditions, as a reminder that the ceiling itself depends on the reaction’s nature. Once you have product in solution rather than as a weighed solid, the follow-on arithmetic, turning moles into concentrations, belongs to a dilution molarity calculator.
Why real yields fall short, and what over 100% means
Yields below 100% are the norm, and the shortfall has familiar causes. OpenStax lists the main ones: side reactions that consume reactants into unwanted products, reactions that are inherently incomplete (equilibrium never reaches the finish line), and products that are simply difficult to collect without loss, so less-than-perfect recovery drags the actual yield down. On the bench, add transfer losses between vessels, product left on filter paper, and incomplete drying of the final weighing, which cuts the other way.
An over-100% result deserves suspicion, not celebration. As the LibreTexts yield page puts it, proper percent yields sit between 0% and 100%, and a figure above 100% means an error has been made: the usual culprits are a product that is still wet, impurities weighed along with the product, or a miscalculated theoretical yield. The calculator’s badge says this plainly when it happens. One more subtlety: percent yield is not the same as atom economy, which compares product mass against the mass of all reactants and measures how efficient the process itself is, a distinction the LibreTexts green chemistry page explains well.
What competing calculators miss
Percent yield tools fall into a few camps, and each leaves something out:
- Three-way solvers: the strongest tools let you solve for any one of the three quantities, actual, theoretical or percent, given the other two. This page is honest about its scope: it solves the classic direction only (yields in, percent out). A reverse-solve mode is on the future list, not silently faked.
- Purity-aware tools: a few calculators adjust for product purity, which matters when the weighed solid is not pure. Ours assumes you weighed dry, pure product and says so, rather than hiding a purity guess inside the maths.
- Basic two-box tools: most competitors give you the percentage and stop. They print no yield loss, no efficiency reading, and their error handling is a dead number or a blank field. Here the validation messages match the classic tool’s exactly (“Theoretical yield must be greater than zero”, “Actual yield cannot be negative”), while the results add the loss figure and a verdict badge.
- The gap this page fills: the familiar two-input calculation, kept exactly as chemists expect it, with the yield loss shown beside the percentage, a live efficiency reading wired to the same consensus bands textbooks use, worked checks that reproduce to the second decimal, and an explicit over-100% flag instead of a silently impossible number.
Frequently asked questions
Percent yield is the actual yield of a reaction expressed as a percentage of the theoretical yield: (actual yield ÷ theoretical yield) × 100. It measures how efficiently a reaction was carried out, combining the chemistry itself and the experimenter’s technique into one number.
As a rough guide, 90% or above is excellent, 80–89% is very good, 70–79% is good, and 50–69% is fair. Under 50% is poor and worth investigating. These bands depend on the reaction: some syntheses, like the industrial Haber process for ammonia, convert only about a quarter of their reactants even when run well.
Proper percent yields fall between 0% and 100%, so a figure above 100% means something went wrong. The usual causes are a product that was not fully dried, impurities weighed along with the product, or an error in the theoretical yield calculation. Re-dry and re-weigh the product, then re-check the stoichiometry.
Because real reactions lose product at every stage. Side reactions consume reactants into unwanted products, some reactions are inherently incomplete, and products are difficult to collect without loss: material stays on filter paper, in transfer vessels, and in purification steps. A yield in the 80s or 90s is usually a genuinely strong result.
Yes. Both yields can be expressed as masses or as molar amounts, and the units cancel when the percent yield is calculated. Grams with grams, moles with moles: mixing them gives a wrong answer.
Identify the limiting reactant, convert its amount to moles, multiply by the mole ratio from the balanced equation, and convert the product moles to mass using the product’s molar mass. The theoretical yield is that mass: the maximum the reaction could produce.
Yes. The formula is a ratio, so any consistent unit works: grams, moles, kilograms. The result is the same percentage either way, as long as both inputs share the unit.
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References
- OpenStax. Chemistry 2e, Section 4.4: Reaction Yields. Accessed October 23, 2026.
- Chan, B. Beginning Chemistry, Section 7.06: Yields, LibreTexts. Accessed October 23, 2026.
- University of Kentucky. General Chemistry, Section 4.4: Reaction Yields, LibreTexts. Accessed October 23, 2026.
- University of Toronto Scarborough. First-Year Chemistry, Section 10.04: Reaction Yields (Green Chemistry and Atom Economy), LibreTexts. Accessed October 23, 2026.
- Moore, J. W. et al. ChemPRIME, Section 3.04: Percent Yield, LibreTexts. Accessed October 23, 2026.