What is theoretical yield?

Theoretical yield is the maximum mass of product that could form if every molecule of the limiting reagent reacted completely and no product was lost. It is calculated from stoichiometry using the balanced chemical equation, the moles of each reactant, and the molar mass of the product.

How is theoretical yield calculated?

First find the limiting reagent by dividing each reactant's available moles by its stoichiometric coefficient — the smallest result is the limiting reagent. Then use the mole ratio to find moles of product: (moles of limiting reagent / its coefficient) × product coefficient. Finally multiply by the product molar mass: theoretical yield (g) = moles of product × M (g/mol).

What is percent yield and why is it never 100%?

Percent yield = (actual yield / theoretical yield) × 100. In practice yields fall short of 100% because of side reactions that consume reactants, reversible equilibria that do not go to completion, product lost during filtration, transfer or purification, and measurement uncertainty. A percent yield above 100% usually means the product contains impurities or was not fully dried.

What is the limiting reagent and how do I find it?

The limiting reagent is the reactant that runs out first and therefore sets the ceiling on how much product can form. To find it, divide the moles of each reactant by its stoichiometric coefficient. The reactant with the lowest quotient is the limiting reagent. All other reactants are in excess.

What units does the calculator use?

Moles (mol) for amounts of substance, grams per mole (g/mol) for molar mass, and grams (g) for yield. Make sure to use consistent units throughout — if you have masses instead of moles, divide each mass by the molar mass first to convert.

Can I use this for reactions with more than two reactants?

Yes. Click "Add reactant" to include up to five reactants. The limiting-reagent logic applies to all of them simultaneously — the one with the smallest (moles / coefficient) ratio limits the reaction regardless of how many others are present.

Theoretical Yield Calculator

Theoretical yield is the cornerstone quantity in every quantitative chemistry calculation: it tells you the most product your reaction can possibly produce, given the amounts of starting materials and the balanced stoichiometric equation. Whether you are a student confirming a lab result, a researcher optimising a synthesis, or an industrial chemist sizing reactor feed ratios, knowing the theoretical yield lets you judge how efficiently your reaction actually ran.

How it works

The calculation follows three well-defined steps rooted in the mole concept.

Step 1 — Find the limiting reagent. For each reactant, divide its available moles by its stoichiometric coefficient from the balanced equation. The reactant that gives the smallest quotient is the limiting reagent because it is the first to be consumed. Every other reactant is present in excess — extra moles of excess reactants do not increase product.

Step 2 — Convert to moles of product. Using the mole ratio from the balanced equation:

moles of product = (moles of limiting reagent / coefficient of limiting reagent) × coefficient of product

Step 3 — Convert to grams. Multiply the moles of product by its molar mass M (in g/mol):

theoretical yield (g) = moles of product × M

Percent yield is then defined as:

percent yield (%) = (actual yield / theoretical yield) × 100

The calculator accepts up to five reactants and can solve in three directions: find theoretical yield from moles, find percent yield from an actual yield, or back-calculate the actual yield corresponding to a given percent yield.

Worked example

Reaction: combustion of ethanol — C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

Suppose you start with 2.00 mol ethanol (M = 46.07 g/mol) and 5.00 mol O₂ (M = 32.00 g/mol), and you want the theoretical yield of CO₂ (M = 44.01 g/mol, coefficient = 2).

Reactant	Moles available	Coefficient	Moles / coeff
C₂H₅OH	2.00	1	2.00
O₂	5.00	3	1.667

Limiting reagent: O₂ (smallest ratio = 1.667)

Moles of CO₂ = (5.00 / 3) × 2 = 3.333 mol

Theoretical yield of CO₂ = 3.333 × 44.01 = 146.7 g

If the experiment actually produced 114 g of CO₂:

Percent yield = (114 / 146.7) × 100 = 77.7%

Formula note

The core stoichiometric identity used here is:

n(product) = n(limiting) × (coeff_product / coeff_limiting)

where n is amount in moles. Combined with m = n × M (mass = moles × molar mass), this gives the theoretical yield in grams. No approximations are involved — the only source of difference from experiment is real-world loss (incomplete reaction, product retained on glassware, etc.).

Constants used: Avogadro’s number N_A = 6.022 × 10²³ mol⁻¹ is implicit in the molar mass values (atomic masses on the IUPAC 2021 scale). The calculation itself requires only the mole ratio and M, not N_A explicitly.