Theoretical Yield Calculator

Background.

The theoretical yield is the most product a reaction could possibly make from what you put in it — the ceiling that a balanced equation and a limiting reactant impose, before any real-world loss is taken into account. This calculator does the classic mass–mole–mass conversion: enter the formula, balanced-equation coefficient and mass of your limiting reactant, plus the formula and coefficient of the product you want, and it returns the maximum mass of that product together with every intermediate quantity so you can check each step.

The chain has exactly three steps and it is worth being able to do it by hand. First, convert grams to moles by dividing by the molar mass. Second, apply the mole ratio taken from the balanced equation — the product's coefficient divided by the limiting reactant's — to get moles of product. Third, convert moles back to grams by multiplying by the product's molar mass. That middle step is the only chemistry in the calculation; the first and third are unit conversions. This page prints all three intermediates, so if the answer looks wrong you can see immediately whether the problem is a molar mass, the ratio, or the input mass.

WHERE THIS PAGE SITS IN THE CHAIN, so you use the right tool. If two reactants are competing and you do not yet know which one runs out first, start at the limiting reagent calculator, which compares them and also tells you how much of the excess is left over. Bring the limiting one here to get the maximum product mass. Then take that maximum and the mass you actually isolated to the percent yield calculator. This page deliberately assumes the reactant you enter IS the limiting one — it takes a single reactant and does not check for a competitor.

UNITS AND CONVENTIONS, stated where the answer is rather than buried in an FAQ. Masses are in grams, molar masses in g/mol, amounts in moles; the two ratio outputs are dimensionless. There is no temperature or pressure basis anywhere on this page, and that is deliberate: a mole balance holds at any T and p, and gases are entered as masses precisely so that no STP-versus-SATP reference state has to be chosen. Molar masses come from the CIAAW Abridged Standard Atomic Weights 2024 table — the IUPAC Atomic Weights 2021 recommendation with the 2024 revisions to gadolinium, lutetium and zirconium. Elements with no IUPAC standard atomic weight (technetium, promethium, polonium, astatine, radon, francium, radium, actinium and everything heavier than uranium) are rejected with an explicit error rather than given a guessed value, because a natural-abundance molar mass does not exist for them.

WHAT THE NUMBER DOES AND DOES NOT MEAN. The theoretical yield assumes three things that are almost never all true: that conversion is complete, that the reaction proceeds by one channel with no side products, and that you recover every gram of what forms. Real reactions stop at equilibrium, run competing reactions, and lose material in the flask, on the filter and in the mother liquor. So the theoretical yield is an upper bound you measure against, not a forecast of what you will weigh. A run that returns 77 percent of theory has not failed; it has performed normally.

THE MOST COMMON ERROR, by a wide margin, is dropping the product coefficient. For 2 H₂ + O₂ → 2 H₂O, one mole of extent produces TWO moles of water, so entering 1 for the product coefficient halves the answer. Enter 2.000 g of hydrogen with coefficients 2 and 2 and the calculator returns 17.8720 g of water; enter the product coefficient as 1 and you get 8.9360 g, which is wrong and looks entirely plausible. The second most common error is the equation not being balanced at all, which this tool cannot detect — it takes your coefficients on trust. Balance on paper first.

ON ROUNDING AND PRECISION. Nothing is rounded in the middle of the calculation. Molar masses, moles and the mole ratio are all carried at arbitrary decimal precision and each returned value is rounded exactly once, at the end, to ten decimal places. Report your answer to the significant figures your least precise input justifies — normally the balance, so three or four figures. One consequence of rounding to ten decimal places rather than to significant figures is worth knowing: masses below about a microgram lose resolution, because 1 × 10⁻¹⁰ g is the smallest increment the outputs can express. For anything at milligram scale or above this never matters; for genuinely trace work, enter your masses in milligrams and read the results as milligrams.

What is theoretical yield calculator?

The theoretical yield of a reaction is the amount of product that may be produced under the specified conditions, as calculated from the stoichiometry of an appropriate balanced chemical equation and from the amount of limiting reactant available. It is a calculated quantity, never a measured one — no balance ever reads a theoretical yield. Formally it follows from the IUPAC extent of reaction ξ: for a balanced equation aA → pP, running the reaction to completion on the limiting reactant gives ξ = n(A)/a, and the product formed is n(P) = ξ × p = n(A) × (p/a). Converting back to mass, m(P) = n(A) × (p/a) × M(P). The ratio p/a is the mole ratio, and it is the only place the chemistry enters; everything else is unit conversion through molar mass. Three assumptions are baked in and should be stated whenever the number is quoted. Conversion is complete — every mole of limiting reactant reacts. Selectivity is perfect — all of it goes to the product you named, with no side reactions. And recovery is complete — nothing is lost to the walls of the flask, the filter, or the mother liquor. Because none of these is exactly true in practice, the theoretical yield is the denominator against which real performance is measured, which is precisely what the percent yield does: percent yield = actual / theoretical × 100.

How to use this calculator.

1. 01Balance your chemical equation on paper first, and identify the limiting reactant. If two reactants compete, use the limiting reagent calculator to find out which one it is before you come here.
2. 02Enter the limiting reactant's formula in Hill/IUPAC notation. Capitalisation matters: Co is cobalt, CO is carbon monoxide.
3. 03Enter its coefficient from the balanced equation, and the mass in grams you are starting with.
4. 04Enter the product's formula and — this is the step people skip — its own coefficient from the balanced equation. For 2 H2 + O2 -> 2 H2O the product coefficient is 2.
5. 05Read the theoretical yield in grams at the top. The moles version below it is more convenient if you are going to compare several products.
6. 06Check the two molar masses in the breakdown against a periodic table. If the answer is surprising, a mistyped formula is the usual reason.
7. 07Use the 'mass ratio' output as a shortcut for repeat runs: multiply any starting mass of that reactant by it to get the theoretical yield directly, with no mole arithmetic.
8. 08Take the theoretical yield and the mass you actually isolated to the percent yield calculator to get the number you will report.
9. 09Round your answer to the significant figures your balance justifies — normally three or four, not the ten decimal places on screen.

The formula.

For a balanced equation a A → p P, with m(A) grams of limiting reactant:

STEP 1 — grams to moles. n(A) = m(A) / M(A). Dimensionally, g ÷ (g/mol) = mol. The molar mass M(A) is the sum over the parsed formula of each element's standard atomic weight times its count, using the CIAAW Abridged Standard Atomic Weights 2024 table.

STEP 2 — the mole ratio. n(P) = n(A) × (p / a). This is the only step that contains chemistry. It is a direct consequence of the IUPAC extent of reaction: since Δn(X) = ν(X) × ξ with ν negative for reactants and positive for products, driving the limiting reactant to zero fixes ξ = n(A)/a, and the product formed is ξ × p. Both p and a are pure numbers, so the ratio is dimensionless and n(P) is in mol.

STEP 3 — moles to grams. m(P) = n(P) × M(P), giving mol × g/mol = g.

Collapsed into one line, m(P) = m(A) × (p/a) × M(P)/M(A). The factor (p/a) × M(P)/M(A) is reported separately as the MASS RATIO, in grams of product per gram of reactant: multiply any starting mass by it and you have the theoretical yield with no intermediate arithmetic. For the worked example below that factor is 0.3981529054, so every gram of copper(II) sulfate can yield at most 0.398 g of copper metal — a useful sanity check, since a mass ratio below 1 means you must lose mass and one above 1 means the product is heavier per mole than the reactant.

ROUNDING STAGE. No intermediate rounding of any kind. Molar masses, moles, the mole ratio, the mass ratio and the yield are all computed at arbitrary decimal precision, and each returned value is rounded exactly once, at the return boundary, to ten decimal places. One documented consequence: because the rounding is to ten DECIMAL PLACES rather than to significant figures, a mass output smaller than about 1 × 10⁻⁸ g keeps only three or four significant figures. At milligram scale and above this is invisible; for trace-scale work, enter masses in milligrams and read the outputs as milligrams.

SIGNIFICANT FIGURES. Report to the significant figures of your least precise input. In the worked example 1.274 g is a four-figure measurement, so 0.5072 g is the honest theoretical yield; the further digits on screen exist so you can see where rounding starts to bite, not to be transcribed into a lab report.

INVALID DOMAIN. A reactant coefficient of zero is a true singularity — the mole ratio divides by it — and is refused, as are negative coefficients and a zero or negative mass. Unknown element symbols, unbalanced parentheses, stray characters and elements with no IUPAC standard atomic weight raise a field error naming the fault instead of returning a plausible-looking wrong molar mass. Non-integer coefficients ARE accepted, because half-equations such as C₂H₆ + 3.5 O₂ → 2 CO₂ + 3 H₂O are legitimate ways to write a balanced combustion.

A worked example.

Frequently asked questions.

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 the independent check in the test suite. Retrieved 2026-07-29. Open access.
2. [2]IUPAC. Compendium of Chemical Terminology (the Gold Book), 'stoichiometric number, ν' (S06025). SECOND, INDEPENDENT AUTHORITY consulted for this page. 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.