Audited 29 Jul 2026·Last updated 28 Oct 2026·6 citations·Tier 1·0 uses

Theoretical Yield Calculator

Maximum grams of product from a mass of limiting reactant and the balanced-equation coefficients, with moles, mole ratio, mass ratio and both molar masses.

Theoretical Yield Calculator

Hill/IUPAC notation, case sensitive: Co is cobalt, CO is carbon monoxide. Parentheses work, e.g. Al2(SO4)3. Write hydrates in expanded form — CuSO4·5H2O becomes CuSO4(H2O)5.
Its coefficient in the BALANCED equation. For 3 Si + 2 N2 -> Si3N4 with silicon as the limiting reactant, enter 3. The calculator cannot check that your equation is balanced.
Grams of the reactant that runs out first. If you do not yet know which of your reactants is limiting, find out with the limiting reagent calculator and bring that one here.
g
The product whose maximum mass you want. If the reaction makes several, enter the one you are isolating.
Its coefficient in the BALANCED equation. For 2 H2 + O2 -> 2 H2O this is 2, not 1 — forgetting it halves the answer, and that is the single most common error in this calculation.
Theoretical yield
0.5072
The largest mass of product, in grams, that the balanced equation allows from the limiting reactant you entered; real recovered yields fall below it.
Theoretical yield (amount)
0.008 mol
Moles of limiting reactant
0.008 mol
Mole ratio (product : reactant)
1
Mass ratio (g product per g reactant)
0.3982
Molar mass of reactant
159.602 g/mol
Molar mass of product
63.546 g/mol

Background.

Drop zinc into a blue solution of copper(II) sulfate and the colour fades as a reddish-brown powder of copper metal settles out. Start with 1.274 g of copper sulfate and add zinc in excess, and the most copper that can possibly form is 0.5072 g. That ceiling is the theoretical yield, and it is fixed before you touch a beaker: by how much of the limiting reactant you used, by the coefficients of the balanced equation, and by the molar masses of the two substances.

The calculation is a chain of three conversions. Grams of reactant become moles by dividing by its molar mass. Moles of reactant become moles of product through the mole ratio, the product's coefficient divided by the reactant's. Moles of product become grams by multiplying by the product's molar mass. Only the middle step involves the chemistry of the reaction; the other two are bookkeeping between mass and amount of substance. Molar masses are worked out from the formulas you type, using the current IUPAC standard atomic weights, and every intermediate value is shown so that a wrong formula or a forgotten coefficient is visible rather than buried in the answer.

What you recover at the bench will be less. Reactions stop short of completion, side reactions divert some reactant elsewhere, and product is lost on the filter paper, in the washings and on the walls of the flask. In OpenStax's version of the copper experiment, 0.392 g was isolated against the 0.5072 g ceiling, a percent yield of about 77.3%, which is an ordinary result rather than a failure.

What is theoretical yield calculator?

Theoretical yield is the amount of product a reaction could produce from the amount of limiting reactant present, calculated from the stoichiometry of the balanced chemical equation. It is a calculated quantity, never a measured one. Formally it rests on the extent of reaction, ξ: for a balanced equation a A → p P, using up all of the limiting reactant A sets ξ = n(A) ⁄ a, and the product formed is n(P) = p × ξ = n(A) × p ⁄ a. Three assumptions are built in: every mole of the limiting reactant reacts, all of it goes to the product you named rather than to side products, and none of the product is lost in isolation. The measured mass divided by the theoretical yield, times 100, is the percent yield.

How to use this calculator.

  1. Balance the equation on paper and decide which reactant runs out first. If two reactants could be limiting, settle it with the limiting reagent calculator; a reactant added in deliberate excess is never the limiting one.
  2. Type the limiting reactant's formula in ordinary notation, minding capitals: Co is cobalt and CO is carbon monoxide. Parentheses are allowed, as in Al2(SO4)3, but dots, charges and state labels are not, so write a hydrate such as CuSO4·5H2O as CuSO4(H2O)5.
  3. Enter that reactant's coefficient from the balanced equation and its mass in grams. For amounts below about a milligram, enter milligrams in the mass field and read every mass result as milligrams; the calculation is proportional, so the answer scales exactly.
  4. Type the product's formula and its own coefficient. For 2 H2 + O2 → 2 H2O the water coefficient is 2.
  5. Before trusting the theoretical yield, glance at the two molar masses and the mole ratio. A molar mass that disagrees with your own working points to a mistyped formula, and a mole ratio that is not p ⁄ a points to a coefficient error.
  6. Report the theoretical yield to the significant figures of your balance reading, then take it with the mass you actually isolated to the percent yield calculator.

The formula.

m(P) = [ m(A) ÷ M(A) ] × (p ÷ a) × M(P)

For a balanced equation a A → p P with m(A) grams of limiting reactant, the three steps are n(A) = m(A) ⁄ M(A), then n(P) = n(A) × p ⁄ a, then m(P) = n(P) × M(P). The units run g ÷ g/mol = mol, then mol × a pure number = mol, then mol × g/mol = g. Collapsed into one line, m(P) = m(A) × (p ⁄ a) × M(P) ⁄ M(A).

The mole ratio p ⁄ a follows from the IUPAC definition of the extent of reaction. Every species changes by its stoichiometric number times the same extent ξ, negative for reactants and positive for products, so when A is used up, ξ = n(A) ⁄ a, and the product formed is p times that. The ratio is exact because both coefficients are pure numbers, and since the 2019 SI redefinition fixed the Avogadro constant at exactly 6.02214076 × 10²³ per mole, the mole itself is an exactly defined count.

The factor (p ⁄ a) × M(P) ⁄ M(A) is given separately as the mass ratio: grams of product per gram of reactant. For copper from copper(II) sulfate it is 0.3982, so every gram of the salt can give at most 0.3982 g of copper. Below 1 means the product is lighter than the reactant it came from, above 1 heavier, and once you know it, any starting mass times the ratio is the theoretical yield.

Molar masses are summed from the CIAAW abridged standard atomic weights, 2024 revision, so Cu is 63.546, S 32.06, O 15.999 and H 1.008. Elements without a standard atomic weight, such as technetium, radon and the elements beyond uranium, have no natural-abundance molar mass and are rejected. Coefficients need not be whole numbers, which allows equations written per mole of fuel.

Two things are taken on trust. The calculator does not know the rest of your equation, so it cannot check that it balances, and it does not know whether another reactant would run out sooner. Temperature and pressure play no part, because a mole balance holds under any conditions; that is why gases are entered as masses here.

A worked example.

Example

The reaction is CuSO4 + Zn → Cu + ZnSO4, with 1.274 g of anhydrous copper(II) sulfate as the limiting reactant and zinc in excess. Both coefficients are 1. The molar mass of CuSO4 is 63.546 + 32.06 + 4 × 15.999 = 63.546 + 32.06 + 63.996 = 159.602 g/mol, and copper's is 63.546 g/mol. Step one: 1.274 ÷ 159.602 = 0.007982 mol of copper sulfate, shown as 0.008 mol. Step two: the mole ratio is 1 ÷ 1 = 1, so 0.007982 mol of copper can form. Step three: 0.007982 × 63.546 = 0.5072 g of copper, the theoretical yield, matching the figure OpenStax gives for this reaction. The mass ratio, 63.546 ÷ 159.602 = 0.3982, gives the same answer in one step: 1.274 × 0.3982 = 0.5072 g. If the 1.274 g had been weighed from the familiar blue crystals instead, the reactant would be the pentahydrate, CuSO4(H2O)5, with a molar mass of 249.677 g/mol. The same mass then holds only 0.005103 mol of copper sulfate, and the ceiling drops to 0.3242 g of copper. A student who recovered 0.392 g from that sample and divided by 0.5072 g would see a plausible 77% and miss that the result is impossible, since it exceeds the true ceiling.

reactant Coefficient1
product FormulaCu
reactant FormulaCuSO4
product Coefficient1
reactant Mass1.274

Frequently asked questions.

Why is my answer exactly half, double or a third of the textbook's?
Almost always a coefficient. For 2 H2 + O2 → 2 H2O, 2.000 g of hydrogen gives 17.8720 g of water with the product coefficient set to 2, but 8.9360 g if it is left at 1. For 3 Si + 2 N2 → Si3N4, 2.00 g of silicon gives 3.3300 g of silicon nitride with the reactant coefficient at 3, and three times that, 9.9899 g, at 1. The mole ratio output shows which coefficient is wrong.
Do I have to know which reactant is limiting before I start?
Yes. The calculator assumes the reactant you enter is the one that runs out. Base the calculation on a reactant that is actually in excess and the theoretical yield comes out too high, so every percent yield measured against it looks too low. When two reactants are present in amounts that are not obviously lopsided, compare moles divided by coefficient for each, or use the limiting reagent calculator, and bring the smaller one here.
My copper sulfate is blue crystals. Does that change the calculation?
Yes, considerably. Blue copper sulfate is the pentahydrate, which carries five waters of crystallisation; the anhydrous salt is a greyish-white powder. Enter it as CuSO4(H2O)5 and its molar mass becomes 249.677 g/mol instead of 159.602, so 1.274 g yields at most 0.3242 g of copper rather than 0.5072 g. The same applies to any hydrate: the formula you type must match the substance you weighed.
How can my actual yield come out higher than the theoretical yield?
It cannot really, so a result above 100% means something else is in the product or the ceiling was calculated wrongly. The usual causes are a product still wet with solvent or water, contamination with another solid such as unreacted excess reagent, a theoretical yield based on the wrong limiting reactant, or a hydrate entered as the anhydrous formula. Drying to constant mass fixes the first; the others need the inputs checked.
Does this work for gases, and do I need STP?
It works for any phase, and no standard conditions are needed, because everything is entered and returned as mass or moles, which do not depend on temperature or pressure. If you measured a gas by volume, turn it into moles with the ideal gas law at your actual conditions, multiply by the molar mass for grams, and enter that. To find the volume a gaseous product would occupy, take the theoretical yield in moles back to the ideal gas law.
The reaction makes two products. How do I get the yield of both?
Run the calculation once for each product, changing only the product formula and coefficient. Burning 10 g of ethane in C2H6 + 3.5 O2 → 2 CO2 + 3 H2O gives at most 29.27 g of carbon dioxide and 17.97 g of water. Their total, 47.24 g, equals the 10 g of ethane plus the 37.24 g of oxygen it consumes, which is a quick check that the equation balances.
Can I enter a coefficient like 3.5?
Yes. Equations are often written per mole of one substance, which leaves fractional coefficients, as in C2H6 + 3.5 O2 → 2 CO2 + 3 H2O. Only the ratio of the two coefficients you enter matters, so 3.5 and 2 give the same answer as the doubled equation's 7 and 4. Keep both coefficients from the same version of the equation; mixing a coefficient from the halved form with one from the doubled form is a factor-of-two error.
Why do the molar masses differ slightly from the ones in my textbook?
Because atomic weights are revised as isotope measurements improve, and older tables keep older values. The calculator uses the CIAAW abridged standard atomic weights in their 2024 revision. Lithium, for example, is 6.94 there, while many older books print 6.941. Differences of this kind change answers only in the third or fourth significant figure, well inside the precision of an ordinary lab balance, but they explain why a hand calculation can disagree in the last digit.
How is theoretical yield different from atom economy?
Theoretical yield measures how much product the limiting reactant can give; atom economy measures how much of all the reactants' mass ends up in the product you want. In the copper example a perfect 0.5072 g of copper is still accompanied by 1.2886 g of zinc sulfate, so only 63.546 ÷ (159.602 + 65.38), about 28.2%, of the reactant mass becomes copper. A reaction can achieve 100% yield and still generate mostly by-product.
How many digits of the theoretical yield belong in a lab report?
As many as your least precise measurement supports, usually the balance reading. A mass of 1.274 g has four significant figures, so the theoretical yield is reported as 0.5072 g, and the atomic weights, known far more precisely, do not limit it. For a series of runs with the same reaction, keep the mass ratio from the first calculation and multiply each new starting mass by it.

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, CC BY 4.0. PRIMARY SOURCE: defines theoretical yield as 'the amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation', and works Example 4.13 (1.274 g CuSO4 -> 0.5072 g Cu), which is this page's worked example, plus Example 4.12 (2.00 g Si in 3 Si + 2 N2 -> Si3N4), used as an independent check. Retrieved 2026-07-29. Open access.
  2. [2]IUPAC. Compendium of Chemical Terminology (the Gold Book), 'stoichiometric number, ν' (S06025). Sourced there to Pure Appl. Chem. 1996, 68, 149 ('A glossary of terms used in chemical kinetics, including reaction dynamics', IUPAC Recommendations 1996) at p. 187 and to the Green Book 2nd ed. p. 42: stoichiometric numbers are positive for products and negative for reactants, with dξ = dn_B / ν_B. It agrees with the textbook mass–mole–mass rule and shows the mole ratio p/a to be a consequence of a single extent of reaction rather than an independent postulate. Retrieved 2026-07-29 via the legacy IUPAC host (the current goldbook.iupac.org returns HTTP 403 to automated retrieval). Open access.
  3. [3]IUPAC (2007). Quantities, Units and Symbols in Physical Chemistry ('the Green Book'), 3rd edition (2nd printing 2008), section 2.12 'Chemical reactions' — formal definitions of stoichiometric number, extent of reaction and amount of substance underpinning the mole-ratio step. International Union of Pure and Applied Chemistry / RSC Publishing. Retrieved 2026-07-29. Bibliographic reference: print and PDF edition rather than a live web page.
  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' (named revision: 2024, incorporating the Gd, Lu and Zr revisions on the Atomic Weights 2021 base). The named, versioned source of every molar mass this calculator computes — Cu 63.546, S 32.06, O 15.999, Si 28.085, N 14.007, H 1.0080. Retrieved 2026-07-29. Open access.
  5. [5]BIPM (2019). The International System of Units (SI), 9th edition — the 2019 redefinition fixing the Avogadro constant at exactly 6.02214076 × 10^23 mol⁻¹, which is what makes 'amount of substance' an exactly defined base quantity and the mole ratio an exact rather than measured conversion. Bureau International des Poids et Mesures. Retrieved 2026-07-29. Open access.
  6. [6]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 records measured yields across twenty-eight industrial reaction classes (71–96 %), the evidence for this page's statement that a theoretical yield is a ceiling routinely missed by 10–30 % even in optimised manufacturing. Retrieved 2026-07-29. Publisher page paywalled; an open copy was consulted.

How this page was produced

Published by
Quanta Calculator
Primary sources
6 cited below
Method
m(P) = [ m(A) ÷ M(A) ] × (p ÷ a) × M(P)
Published
Last verified

Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.

In this category

Embed

Quanta Pro

Paid features are coming later.

  • All 1560 calculators remain free
  • No billing is enabled
Coming soon