How to Calculate Theoretical Yield

You start with a mass of reactants from a lab balance and need to know the maximum grams of product possible, which is how to calculate theoretical yield. Theoretical yield is a ceiling, not a target. Real reactions almost never hit it, but it tells you what perfect conditions would produce. Calculate theoretical yield by hand using only a balanced equation, molar masses from the CIAAW standard atomic weights (IUPAC), and a five-step process.

Theoretical Yield (g) = Moles of Limiting Reactant × (Product Coefficient ÷ Reactant Coefficient) × Molar Mass of Product

Moles cancel to leave grams. The formula works because the balanced equation’s coefficients define the mole ratio between any two substances. Once you know how many moles of the limiting reactant you have, the ratio scales that number to moles of product, and the molar mass turns moles into a weighable mass. The same formula applies whether you are working with solids, liquids, or gases, the units are consistent throughout.

OpenStax Chemistry 2e (sections 4.3-4.4) emphasises that a balanced equation is non-negotiable. Without it, the stoichiometric ratios are wrong. Count atoms of each element on both sides and adjust coefficients until they match. For the examples here, water from H₂ and O₂, and aluminium chloride from Al and Cl₂, the equations are already balanced: 2H₂ + O₂ → 2H₂O and 2Al + 3Cl₂ → 2AlCl₃. If you are given an unbalanced equation, balance it before proceeding to the next step.

Use the formula: Moles = Mass (g) ÷ Molar Mass (g/mol). Molar masses come from the CIAAW standard atomic weights, updated biennially. For hydrogen gas (H₂), the molar mass is 2.016 g/mol (2 × 1.008). For oxygen gas (O₂), it is 31.998 g/mol (2 × 15.999). For chlorine gas (Cl₂), it is 70.90 g/mol (2 × 35.45). For aluminium, 26.98 g/mol. For aluminium chloride (AlCl₃), 133.34 g/mol (26.98 + 3 × 35.45). Convert every given reactant mass to moles before you compare them.

The limiting reactant is the one that produces the least product when the stoichiometric ratio is applied. To find it, divide the moles of each reactant by its coefficient from the balanced equation. The smallest result is the limiting reactant. This method works because it normalises each reactant to the same reference point, moles per coefficient. Do not compare absolute moles directly; a small amount of a reactant with a large coefficient can be limiting even if its absolute moles are higher.

Multiply the moles of the limiting reactant by the stoichiometric ratio of product to limiting reactant. The ratio is (Coefficient of Product ÷ Coefficient of Limiting Reactant) from the balanced equation. For the water example, with H₂ as limiting reactant: 2 mol H₂O ÷ 2 mol H₂ = 1. So moles of H₂O = moles of H₂. For the aluminium chloride example, with Cl₂ as limiting reactant: 2 mol AlCl₃ ÷ 3 mol Cl₂ = 0.6667. So moles of AlCl₃ = 0.6667 × moles of Cl₂.

Multiply the moles of product (from Step 4) by its molar mass. This gives the theoretical yield in grams. For water, molar mass is 18.015 g/mol (2 × 1.008 + 15.999). For aluminium chloride, molar mass is 133.34 g/mol. The result is the maximum mass of product that could form if the reaction goes to completion, all product is recovered, and no side reactions occur. In practice, the actual yield, the mass you measure on the balance, will be lower.

Reaction: 2H₂ + O₂ → 2H₂O. Given 4.0 g H₂ and 32.0 g O₂.

Step 1: Balanced Equation

The equation is balanced as written.

Step 2: Grams To Moles

H₂: 4.0 g ÷ 2.016 g/mol = 1.984 mol. O₂: 32.0 g ÷ 31.998 g/mol = 1.000 mol.

Step 3: Limiting Reactant

For H₂: 1.984 mol ÷ 2 = 0.992. For O₂: 1.000 mol ÷ 1 = 1.000. Smaller value is H₂, so H₂ is limiting.

Step 4: Moles Of Product

1.984 mol H₂ × (2 mol H₂O ÷ 2 mol H₂) = 1.984 mol H₂O.

Step 5: Grams Of Product

1.984 mol × 18.015 g/mol = 35.74 g H₂O. Theoretical yield: 35.7 g (rounded).

Reaction: 2Al + 3Cl₂ → 2AlCl₃. Given 5.0 g Al and 10.0 g Cl₂.

Step 1: Balanced Equation

The equation is balanced as written.

Step 2: Grams To Moles

Al: 5.0 g ÷ 26.98 g/mol = 0.1853 mol. Cl₂: 10.0 g ÷ 70.90 g/mol = 0.1410 mol.

Step 3: Limiting Reactant

Al: 0.1853 mol ÷ 2 = 0.09265. Cl₂: 0.1410 mol ÷ 3 = 0.0470. Smaller value is Cl₂, so Cl₂ is limiting.

Step 4: Moles Of Product

0.1410 mol Cl₂ × (2 mol AlCl₃ ÷ 3 mol Cl₂) = 0.0940 mol AlCl₃.

Step 5: Grams Of Product

0.0940 mol × 133.34 g/mol = 12.53 g AlCl₃. Theoretical yield: 12.5 g (rounded).

• Unbalanced equation: Skipping this step invalidates all ratios. Balance first, then proceed.
• Wrong molar mass: Use CIAAW standard atomic weights, not an outdated periodic table. For example, oxygen is 15.999, not 16.00.
• Using the excess reactant: Theoretical yield is always based on the limiting reactant. Using the excess reactant gives a larger, incorrect value.
• Ignoring the stoichiometric ratio: Do not multiply by a 1:1 ratio if the coefficients are different. The ratio comes from the balanced equation.
• Rounding too early: Carry extra digits through intermediate steps. Round only the final answer to the correct significant figures.

The five-step calculation can be written as a single chain of conversion factors:

Mass of Reactant (g) ÷ Molar Mass of Reactant (g/mol) = Moles of Reactant × (Coefficient of Product ÷ Coefficient of Reactant) = Moles of Product × Molar Mass of Product (g/mol) = Theoretical Yield (g)

For the water example: 4.0 g H₂ ÷ 2.016 g/mol = 1.984 mol H₂ × (2 ÷ 2) = 1.984 mol H₂O × 18.015 g/mol = 35.74 g H₂O. Every unit cancels except grams of product, which is the result you want.