Audited —·Last updated 29 Jul 2026·6 citations·Tier 1·0 uses

Percent Yield Calculator

Free percent yield calculator: enter actual and theoretical yield, or solve for either one. Explains yields above 100% instead of rejecting them.

Percent Yield Calculator

Solve for
The mass or amount of purified product you actually weighed. Use ANY unit you like — grams, milligrams, moles — as long as the theoretical yield below uses the SAME unit. Percent yield is a dimensionless ratio, so the unit cancels.
The maximum the balanced equation allows from your limiting reactant, in the SAME unit as the actual yield. If you do not have this number yet, compute it first with the theoretical yield calculator or the limiting reagent calculator.
Only used when you are solving for actual or theoretical yield. Enter it as a percentage (77.29), not a fraction (0.7729).
%
Percent yield
77.29
Actual yield divided by theoretical yield, times 100. Dimensionless — the unit you entered cancels. Values above 100% are reported, not rejected, because they diagnose an impure or incompletely dried product.
Actual yield
0.392
Theoretical yield
0.5072
Amount lost
0.1152
Loss
22.71
Reading
A good preparative yield. Losses in this band are normally mechanical — transfer losses, material left in the mother liquor after recrystallisation, and losses on the filter.

Background.

Percent yield is the single number that tells you how much of the product your reaction was theoretically capable of making actually ended up in the flask. It is defined as the actual yield divided by the theoretical yield, multiplied by 100. Enter the mass you weighed and the mass the stoichiometry allowed, and this calculator returns the percentage, the amount lost, and a plain-language reading of what a result in that range usually means at the bench.

Three things about this page are worth knowing before you trust the number. First, percent yield is dimensionless: the unit cancels. You can enter grams and grams, milligrams and milligrams, or moles and moles, and you will get the same answer — but the two figures must be in the same unit. This tool never converts units and never assumes grams. Second, this page takes the theoretical yield as an input; it does not derive it from a balanced equation. If you do not have that number yet, compute it first with the theoretical yield calculator, which converts a mass of limiting reactant into a maximum mass of product, or with the limiting reagent calculator when two reactants are competing and you do not yet know which one caps the reaction. Third, results above 100 percent are reported rather than rejected. That is a deliberate design decision and it is explained in detail below.

Why does a real reaction almost never give 100 percent? Four causes dominate, and they are worth separating because they call for different fixes. Incomplete conversion means the reaction simply stopped short of consuming all the limiting reactant — a reversible reaction that reached equilibrium, or a run that was quenched too early. Competing side reactions divert some of the limiting reactant into products you did not want. Mechanical losses are the boring ones and often the biggest: material left on the walls of the flask, product still dissolved in the mother liquor after recrystallisation, a few milligrams lost on the sinter or the column. And purification losses are the price of a clean product — every recrystallisation, every chromatographic separation, every wash trades yield for purity. A synthesis that reports 95 percent crude and 62 percent after column has not made a mistake; it has made a choice.

A yield above 100 percent is the case most calculators get wrong by refusing to display it. Conservation of mass makes a yield above 100 percent impossible for pure, dry, isolated product, so if your calculation gives 108 percent the arithmetic is not the problem — the sample is. Something other than your product is contributing to the mass on the balance. In descending order of likelihood: residual solvent or water that has not been driven off, co-isolated unreacted starting material, inorganic salts carried through from the work-up, or a theoretical yield that was computed from the wrong reactant. The fix is diagnostic rather than mathematical: dry the sample to constant mass, re-weigh, and check purity by melting point, NMR, or chromatography. Because that diagnosis is the useful output, this calculator computes and displays the true figure and tells you what it means, rather than clamping it at 100 or throwing an error.

On reporting conventions: the ACS Journal of Organic Chemistry author guidelines require that a reported yield be the weighed amount of ISOLATED, PURIFIED product, given as both a weight and a percentage — not a yield inferred from an NMR integral or a chromatogram — and they flag yields above 95 percent for extra editorial scrutiny of the accompanying purity data. If you are writing up a result, that is the standard your number will be read against.

On rounding: this calculator performs no intermediate rounding. The ratio is carried at arbitrary precision and rounded once, at the end, to ten decimal places. That matters more than it sounds. If you round the theoretical yield to four significant figures before dividing, as most textbooks print it, the worked example below gives 77.29 percent; carrying the theoretical yield at full precision gives 77.28 percent. Both round to 77.3 percent at three significant figures, which is the honest precision for a bench balance reading to the milligram — and significant figures, not decimal places, are what you should report.

What is percent yield calculator?

Percent yield measures the extent to which a reaction's theoretical yield is actually achieved. Three terms have to be kept apart. The THEORETICAL YIELD is the amount of product that may be produced under the specified conditions, calculated from the stoichiometry of a balanced chemical equation and the limiting reactant — it is a calculated ceiling, not a measurement. The ACTUAL YIELD is the amount of purified product you obtained and weighed; it is a measurement and it is almost always less than the theoretical yield. The PERCENT YIELD is the ratio of the two, expressed as a percentage: percent yield = (actual yield / theoretical yield) x 100 percent. Because both quantities appear in the same unit, the ratio is dimensionless — a percent yield has no units, and it does not matter whether you work in grams, milligrams, or moles as long as you are consistent. Percent yield is not the same thing as CONVERSION (the fraction of the limiting reactant consumed, whatever it turned into) or SELECTIVITY (the fraction of the consumed reactant that became the desired product). A reaction can have 100 percent conversion and a 40 percent yield if most of the substrate went to a side product. Nor is percent yield an efficiency in the green-chemistry sense: a reaction can give a 99 percent yield and still be extremely wasteful if the stoichiometry throws most of the reactants' atoms away as by-products, which is what atom economy measures instead.

How to use this calculator.

  1. Leave 'Solve for' on 'Percent yield' if you already know both the actual and theoretical amounts. Switch it to 'Actual yield' or 'Theoretical yield' to back-calculate one from a known percentage instead.
  2. Enter the actual yield: the mass of purified, dried product you weighed. Use isolated product, not a crude weight, if you are going to report the number.
  3. Enter the theoretical yield in the SAME unit. Grams with grams, moles with moles. The tool does not convert, and mixing units is the most common way to get a nonsense answer.
  4. If you do not have a theoretical yield yet, compute it first: use the theoretical yield calculator for a single limiting reactant, or the limiting reagent calculator when two reactants are competing.
  5. Read the percent yield at the top and the 'Amount lost' figure below it — the second tells you how many grams went missing, which is often more actionable than the percentage.
  6. Read the 'Reading' line. It names the band your result falls into and the loss mechanisms typical of that band.
  7. If the result is above 100 percent, do not adjust the arithmetic. Dry the sample to constant mass, re-weigh, and re-check which reactant is actually limiting.
  8. Report the answer to the significant figures your balance justifies — usually three. The calculator shows two decimal places so you can see where rounding would bite, not because that precision is real.

The formula.

% yield = (m_actual / m_theoretical) × 100
m_actual · isolated, purified product (any unit)m_theoretical · stoichiometric maximum (same unit)% yield · dimensionless, expressed in percent

The relation has three quantities and this calculator solves it for any one of them.

percent yield = (actual yield / theoretical yield) x 100 actual yield = theoretical yield x (percent yield / 100) theoretical yield = actual yield / (percent yield / 100)

UNITS AND DIMENSIONS. Both yields carry the same dimension — mass (g, mg, kg) or amount of substance (mol, mmol) — and it cancels in the ratio, so percent yield is dimensionless and is quoted in percent. The calculator therefore imposes no unit on the inputs; it only requires that they match. Mixing a mass with an amount, or grams with milligrams, produces a number that is arithmetically valid and chemically meaningless, and no software can detect it for you.

ROUNDING STAGE. There is no intermediate rounding anywhere in this module. Both inputs are converted to arbitrary-precision decimals, the division and the multiplication by 100 are performed at full precision, and the result is rounded exactly once, at the return boundary, to ten decimal places. The interpretation band is selected from the UNROUNDED ratio, so the band and the number can never disagree at a boundary. This is not a cosmetic detail. In the worked example below, a theoretical yield of 0.5072 g (four significant figures, as OpenStax prints it) gives 77.28706624 percent, while the same calculation carried at full precision from 1.274 g of copper(II) sulfate gives 0.50724680141852859 g and 77.27993531 percent. The two answers differ in the second decimal place purely because of where the rounding happened. Both round to 77.3 percent at three significant figures.

SIGNIFICANT FIGURES. Report the percent yield to the number of significant figures justified by the less precise of your two inputs — in practice, the balance reading. A three-place balance weighing 0.392 g justifies three significant figures, so 77.3 percent is the honest report and 77.28706624 percent is false precision. The calculator displays two decimal places deliberately, so that you can see the digit at which rounding starts to matter, not because that precision is physically meaningful.

INVALID DOMAIN AND SINGULARITIES. A theoretical yield of zero is a genuine singularity: the ratio is undefined, and the calculator raises a field error rather than returning infinity. Solving backwards for a theoretical yield from a percent yield of zero is the same singularity in the other direction and is refused for the same reason. Negative masses are refused — you cannot isolate a negative amount. A zero actual yield IS accepted and returns 0 percent, because a failed reaction is a real and reportable result.

VALUES ABOVE 100 PERCENT. The calculator does not clamp them. Mass conservation forbids a yield above 100 percent for pure, dry, isolated product, so any value above 100 percent is information about the sample rather than about the arithmetic: residual solvent or water, co-isolated starting material or salts, or a theoretical yield computed from the wrong limiting reactant. In that band the 'Amount lost' output goes negative, which is the correct sign — the excess is the mass of whatever is in the solid that is not your product.

A worked example.

Example

Worked example — displacement of copper by zinc, CuSO4(aq) + Zn(s) -> Cu(s) + ZnSO4(aq). You start with 1.274 g of copper(II) sulfate. Its molar mass, from the IUPAC CIAAW abridged atomic weights, is 63.546 + 32.06 + 4 x 15.999 = 159.602 g/mol, so 1.274 g is 1.274 / 159.602 = 0.00798236 mol. The equation is 1:1 in copper, so the maximum possible amount of copper metal is also 0.00798236 mol, which at 63.546 g/mol is 0.5072 g. That is the theoretical yield. You run the reaction, filter, dry the copper and weigh 0.392 g. Entering actual = 0.392 and theoretical = 0.5072 gives a percent yield of 0.392 / 0.5072 x 100 = 77.29 percent, an amount lost of 0.5072 - 0.392 = 0.1152 g, and a loss of 22.71 percentage points. Reported to the three significant figures a milligram balance justifies, that is 77.3 percent — the figure OpenStax prints for this same reaction. The reading places it in the 'good preparative yield' band, where the usual culprits are mechanical: copper left clinging to the zinc strip, fines lost through the filter, and a little product still in the wash water. Now switch 'Solve for' to 'Theoretical yield', leave the actual yield at 0.392 g and enter 77.29 percent: the calculator returns 0.50718 g, recovering the input to within the rounding of the percentage you typed, which is the round-trip check that the three modes are consistent. Finally, a contrast worth trying: set the actual yield to 0.560 g against the same 0.5072 g theoretical. The result is 110.41 percent with an amount lost of -0.0528 g. Nothing in the arithmetic is wrong; the copper is wet. Dry it to constant mass and weigh again.

actual Yield0.392
theoretical Yield0.507
percent Yield Input77.29
solve ForpercentYield

Frequently asked questions.

What is the formula for percent yield?
Percent yield = (actual yield / theoretical yield) x 100 percent. The actual yield is the mass of purified product you weighed; the theoretical yield is the maximum the balanced equation allows from the limiting reactant. Both must be in the same unit — grams with grams, or moles with moles — because the ratio is dimensionless and the unit cancels. For the copper displacement in the worked example, 0.392 g isolated against a 0.5072 g theoretical maximum gives 0.392 / 0.5072 x 100 = 77.29 percent, reported as 77.3 percent at three significant figures. If you are working in moles rather than grams the formula is identical and the answer is identical, provided you do not mix the two.
Can percent yield be more than 100 percent?
Arithmetically yes, physically no — and that gap is exactly what makes the number useful. Conservation of mass means a pure, dry, isolated product cannot weigh more than the stoichiometry allows, so a calculated yield above 100 percent tells you something is in your sample that is not your product. The most common cause by a wide margin is residual solvent or water that was never driven off; drying to constant mass and re-weighing usually resolves it. Other causes are co-isolated unreacted starting material, inorganic salts carried through from the aqueous work-up, or a theoretical yield that was computed from the wrong reactant — if you assumed reactant A was limiting when in fact reactant B was, your theoretical yield is too low and every yield you compute against it will be inflated. This calculator deliberately displays values above 100 percent instead of rejecting them, and the 'Amount lost' output goes negative to show how much excess mass is unaccounted for.
What is the difference between actual yield, theoretical yield, and percent yield?
The theoretical yield is a calculation, not a measurement: it is the maximum amount of product the balanced equation permits from the amount of limiting reactant you started with, assuming complete conversion and no losses. The actual yield is a measurement: the mass of purified product you weighed at the end. The percent yield is the ratio of the two as a percentage. In the worked example, 1.274 g of CuSO4 has a theoretical yield of 0.5072 g of copper (a calculation), 0.392 g was actually isolated (a measurement), and 77.29 percent is the ratio. Confusing the first two is the most common student error: the theoretical yield never comes from a balance, and the actual yield never comes from a periodic table.
Why is my percent yield so low?
Four mechanisms account for nearly all low yields, and they need different fixes. INCOMPLETE CONVERSION means the reaction stopped short — an equilibrium that never went to completion, or a run quenched too early; the fix is time, temperature, catalyst, or driving the equilibrium. SIDE REACTIONS divert the limiting reactant into something else; the fix is selectivity — milder conditions, a protecting group, a different reagent. MECHANICAL LOSSES are material left in the flask, on the filter, or dissolved in the mother liquor; the fix is careful transfer and cold washes. PURIFICATION LOSSES are the deliberate price of purity, and every recrystallisation or column costs yield. If your yield is under about 40 percent, start by checking that you identified the limiting reactant correctly — an error there changes the denominator and can make a perfectly good reaction look like a failure.
Is percent yield the same as percent conversion or selectivity?
No, and process chemists keep them strictly apart. CONVERSION is the fraction of the limiting reactant that was consumed, regardless of what it became. SELECTIVITY is the fraction of the consumed reactant that became the desired product. YIELD is the product of the two: yield = conversion x selectivity. A reaction with 100 percent conversion and 45 percent selectivity has a 45 percent yield — every molecule of starting material reacted, but most went somewhere you did not want. A reaction with 50 percent conversion and 100 percent selectivity also has a 50 percent yield, but the diagnosis and the fix are completely different: the first needs a better catalyst or milder conditions, the second needs more time. Percent yield alone cannot distinguish them, which is why kinetics work reports all three.
Should I calculate percent yield in grams or in moles?
Either — the answer is identical, because percent yield is a dimensionless ratio and the unit cancels. What you must not do is mix them. Comparing 0.392 g of isolated product against a theoretical yield expressed as 0.00798 mol gives a numerically enormous and completely meaningless answer. In practice grams are more convenient because that is what a balance reads directly, and moles are more convenient when you are comparing several products of different molar masses on a common footing. If your reaction makes more than one mole of product per mole of limiting reactant, be careful: both yields must refer to the same species and the same stoichiometric basis.
What counts as a good percent yield?
It depends entirely on the chemistry, and any universal threshold is marketing rather than science. Simple acid-base salt formations and clean displacements routinely run above 90 percent. Multi-step syntheses of complex molecules are often celebrated at 40 to 60 percent per step, because compounding matters: ten steps at 80 percent each give an overall yield of 0.8 to the tenth power, about 11 percent. In the published fine-chemicals data of Constable, Curzons and Cunningham, individual reaction classes averaged yields in the high 70s to low 90s. For grading a teaching lab, 70 to 90 percent is usually treated as a competent result. Note the other end of the scale too: the ACS Journal of Organic Chemistry author guidelines flag reported yields above 95 percent for additional editorial scrutiny of the purity data, so an unusually high yield is a claim you should be prepared to defend with a melting point or a spectrum.
How do I find the theoretical yield to put into this calculator?
Convert the mass of your limiting reactant to moles by dividing by its molar mass, multiply by the mole ratio of product to limiting reactant taken from the BALANCED equation, then multiply by the product's molar mass. For the worked example: 1.274 g CuSO4 / 159.602 g/mol = 0.00798236 mol; the equation is 1:1 so 0.00798236 mol of copper; times 63.546 g/mol = 0.5072 g. The theoretical yield calculator does this chain for you, and the limiting reagent calculator does it when you have two reactants and need to find out which one runs out first. The single most common error in this step is using the unbalanced equation — if the coefficient on your product is not 1, the mole ratio is not 1.
Does the calculator round, and where?
It performs no intermediate rounding at all. Both inputs are converted to arbitrary-precision decimals, the division and the multiplication by 100 happen at full precision, and the answer is rounded exactly once at the end, to ten decimal places. The interpretation band is chosen from the unrounded ratio so the label can never contradict the number at a boundary. This matters at the third significant figure: the textbook value of 0.5072 g for the theoretical yield gives 77.28706624 percent, while carrying that same theoretical yield at full precision from 1.274 g gives 0.50724680141852859 g and 77.27993531 percent. Both correctly report as 77.3 percent. Report to the significant figures your balance justifies, not to the digits the screen shows.
My yield is exactly 100 percent. Is that suspicious?
Not automatically, but it deserves a second look. Genuinely quantitative yields do occur, particularly for simple salt metatheses, precipitations of very insoluble products, and reactions where the product is isolated by filtration with no recrystallisation step. What makes an exact 100 percent worth checking is that it sits precisely on the boundary that mass conservation forbids exceeding, so a small amount of retained solvent would push it over. Confirm purity before reporting it — melting point, NMR, or elemental analysis — and confirm that the theoretical yield was computed from the correct limiting reactant. The ACS/JOC guidance that yields above 95 percent attract extra scrutiny applies here in full.

References& sources.

  1. [1]Flowers, P., Theopold, K., Langley, R. & Robinson, W. R. (2019). Chemistry 2e, Section 4.4 'Reaction Yields'. OpenStax, Rice University. Peer-reviewed, openly licensed (CC BY 4.0). PRIMARY DEFINITION for this page: 'percent yield = (actual yield / theoretical yield) x 100 %', plus the definitions of limiting reactant, theoretical yield and actual yield, and Example 4.13 (CuSO4 + Zn -> Cu + ZnSO4; 1.274 g CuSO4, theoretical yield 0.5072 g Cu, 0.392 g isolated, 77.3 %) which is reproduced as this page's worked example. Retrieved 2026-07-29. Open access.
  2. [2]American Chemical Society, The Journal of Organic Chemistry — Author Guidelines, section on reporting yields and purity of compounds. SECOND, INDEPENDENT AUTHORITY consulted for this page: requires that reported yields be weighed amounts of isolated and purified product, given as both weight and percentage rather than inferred from spectroscopic or chromatographic integration, and identifies yields above 95 % as warranting additional editorial scrutiny of purity documentation. Agrees with the OpenStax definition and supplies the reporting convention used by this page's interpretation bands. Retrieved 2026-07-29. Open access.
  3. [3]Constable, D. J. C., Curzons, A. D. & Cunningham, V. L. (2002). 'Metrics to "green" chemistry — which are the best?' Green Chemistry 4(6), 521–527, doi:10.1039/B206169B. Royal Society of Chemistry, peer-reviewed. Table 1 reports measured yields and atom economies by reaction class across GlaxoSmithKline processes (for example: acid salt formation 83 % yield, hydrogenation 89 %, Grignard 71 %, resolution 36 %), which is the source for this page's statement about what yields are normal in industrial fine-chemicals practice. Retrieved 2026-07-29. Publisher page is paywalled; an author-accessible copy was consulted.
  4. [4]Meija, J. et al. (2021). 'Atomic weights of the elements 2021 (IUPAC Technical Report)'. Pure and Applied Chemistry 93(5), 573–600, doi:10.1515/pac-2019-0603, as maintained in the CIAAW table 'Abridged Standard Atomic Weights 2024' (revision: 2024, based on the Atomic Weights 2021 report). Source of the molar masses used in the worked example: Cu 63.546, S 32.06, O 15.999, giving M(CuSO4) = 159.602 g/mol. Retrieved 2026-07-29. Open access.
  5. [5]IUPAC (2007). Quantities, Units and Symbols in Physical Chemistry ('the Green Book'), 3rd edition, section 2.12 on chemical reactions — the formal basis for stoichiometric numbers and the extent of reaction that underlie any theoretical-yield calculation feeding this page. International Union of Pure and Applied Chemistry / RSC Publishing. Retrieved 2026-07-29. Bibliographic reference; the full text is a print/PDF edition rather than a live web page.
  6. [6]Sheldon, R. A. (2018). 'Metrics of Green Chemistry and Sustainability: Past, Present, and Future'. ACS Sustainable Chemistry & Engineering 6(1), 32–48, doi:10.1021/acssuschemeng.7b03505. Peer-reviewed. Used here for the distinction this page draws between chemical yield and mass-efficiency metrics: Sheldon notes explicitly that chemical yield 'needed to be replaced or, at least, supplemented' by metrics that assign value to eliminating waste, which is why a high percent yield does not by itself mean an efficient reaction. Retrieved 2026-07-29. Publisher version paywalled; the author's institutional repository copy at TU Delft is open.
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