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🧪 Stoichiometry

Theoretical Yield Calculator

From the limiting reactant's mass and molar mass, the mole ratio and the product's molar mass, find the maximum (theoretical) product yield in grams.

Theoretical yield
Limiting reactant moles
Mole ratio
Product moles
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Theoretical yield — Quick answer

Turn the limiting reactant into moles, apply the mole ratio to get moles of product, then convert back to grams.

nreact = mass / molar mass · nprod = nreact × ratio
theoretical yield = nprod × product molar mass

Worked example: 25 g reactant (M = 100) → 0.25 mol. Ratio 1:1, product M = 120 → 0.25 × 1 × 120 = 30 g.

Yield vs limiting-reactant mass (M=100, 1:1, prod M=120)

Reactant massMolesYield
25 g0.25 mol30 g
50 g0.50 mol60 g
100 g1.00 mol120 g

Used for: reaction planning, lab prep, scale-up, percent-yield checks.

🧪 Theoretical Yield Calculator

Enter the limiting reactant's mass and molar mass, the mole ratio (product ÷ reactant), and the product's molar mass.

Theoretical yield
Reactant moles
Product moles
Mole ratio used

⚠️ Use the limiting reactant (the one that runs out first). The mole ratio is product coefficient ÷ reactant coefficient from the balanced equation. Actual yield will be lower — divide actual by this for percent yield.

The theoretical yield is the most product a reaction could possibly make — the ceiling, assuming nothing is wasted. It is fixed by the limiting reactant, the ingredient that runs out first, because once it is gone the reaction stops. The recipe is always the same three steps: turn the limiting reactant's mass into moles, scale by the mole ratio from the balanced equation to get moles of product, then convert back to grams with the product's molar mass. Compare your real result against this ceiling and you have the percent yield.

Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: stoichiometric mole-ratio yield calculation.

The theoretical-yield equations

Limiting reactant to moles
nreactant = mass / molar mass
Moles of product
nproduct = nreactant × (product coeff / reactant coeff)
Theoretical yield (mass)
yield = nproduct × product molar mass

Balanced equations relate substances by moles, so you must work in moles, not grams — the coefficients are the mole ratio. The product-to-reactant coefficient ratio scales the limiting reactant's moles into product moles: 1 if they react one-for-one, 2 if two products form per reactant, 0.5 if two reactants make one product. Multiplying by the product's molar mass returns a mass you can weigh, the theoretical maximum for that amount of limiting reactant.

Worked example — a one-step synthesis

Scenario: 25 g of a limiting reactant (molar mass 100 g/mol) reacts 1:1 to give a product of molar mass 120 g/mol.

Moles of reactant & product
n = 25 / 100 = 0.25 mol · nproduct = 0.25 × 1 = 0.25 mol
Theoretical yield
yield = 0.25 × 120 = 30 g

The reaction could make at most 30 g of product. Double the limiting reactant to 50 g and the yield scales straight to 60 g, since yield is proportional to the limiting reactant. If the balanced equation instead gave two product molecules per reactant (ratio 2), the same 0.25 mol would make 0.5 mol of product — 60 g. And if you actually isolated only 25.5 g, your percent yield would be 25.5 / 30 × 100 = 85%.

Frequently Asked Questions

How do you calculate theoretical yield?

moles of limiting reactant × mole ratio × product molar mass. 25 g/100 = 0.25 mol → ×1 ×120 = 30 g.

What is the limiting reactant?

The one that runs out first. Convert each reactant to moles ÷ its coefficient; the smallest is limiting.

What is the mole ratio?

Product coefficient ÷ reactant coefficient from the balanced equation. 2H₂+O₂→2H₂O: H₂O:H₂ = 1, H₂O:O₂ = 2.

Theoretical vs actual yield?

Theoretical = max from stoichiometry; actual = what you isolate (lower). actual ÷ theoretical × 100 = percent yield.

Why convert to moles?

Balanced equations relate moles, not masses. Mole is the common currency: mass → moles → ratio → moles → mass.

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