What is the difference between theoretical yield and actual yield?

Theoretical yield is the calculated maximum mass of product from the limiting reactant. Actual yield is the mass of product recovered from the experiment, measured on a balance. Actual yield is always less than or equal to theoretical yield.

How do I identify the limiting reactant?

Convert the mass of each reactant to moles. Then divide each mole value by its coefficient from the balanced equation. The reactant with the smallest result is the limiting reactant. This determines the stoichiometric proportions from limiting reactant to product.

What is the mole ratio and why does it matter?

The mole ratio is the ratio of coefficients of two substances in a balanced equation. It scales the amount of limiting reactant to the amount of product. Without the correct mole ratio, the theoretical yield calculation will be wrong.

Can the theoretical yield be greater than the actual yield?

Yes, always. Theoretical yield is a ceiling; the actual yield is what you recover. If your actual yield appears greater, you have a measurement error, wet product, impure compound, or a calculation mistake.

How do I handle reagent purity in the calculation?

Multiply the mass of the reagent by its purity fraction. For a 95% pure reagent, the usable mass is 0.95 × the total mass. This affects the moles of limiting reactant and therefore the theoretical yield.

What is the theoretical yield of a gas product?

At IUPAC STP (273.15 K, 100 kPa), one mole of an ideal gas occupies 22.7 L. Calculate moles of gas from the limiting reactant using the mole ratio, then multiply by 22.7 L/mol to get volume. Remember this is a theoretical maximum.

What is a good percent yield?

In organic chemistry, above 90% is excellent, 70-90% is good, 50-70% is fair, and below 50% is poor. Context matters: a 50% yield in a difficult multistep synthesis can be acceptable.

Stoichiometric Proportions From Limiting Reactant to Product

The Wrong Assumption About Theoretical Yield

The most common mistake is thinking theoretical yield is the amount of product you will weigh on the balance. It is not. Theoretical yield is the maximum mass of product the balanced equation predicts from the limiting reactant, assuming perfect conditions. The real yield is always lower. The calculation uses stoichiometric proportions from limiting reactant to product, which directly scales the moles of the limiting reactant into the moles of product using the balanced coefficients. No other reactant limits this step; only the limiting reactant matters.

The balanced coefficients define the relevant mole ratio. For the reaction 2H₂ + O₂ → 2H₂O, the mole ratio of H₂ to H₂O is 2:2, which simplifies to 1:1. If you have 1.984 mol of H₂ (the limiting reactant), you get 1.984 mol of H₂O. Then multiply by the molar mass of water (18.015 g/mol) to get 35.74 g of theoretical yield. That is the ceiling. Real reactions almost never hit it due to side reactions, transfer losses, or incomplete conversion.

Stoichiometric Proportions From Limiting Reactant to Product Explained

The phrase means exactly what it says: the stoichiometric coefficients from the balanced equation scale the available amount of the limiting reactant into the maximum product. The formula is n_product = n_limiting_reactant × (coefficient_product / coefficient_limiting_reactant), then m_product = n_product × M_product. The coefficients are dimensionless; they scale moles, not grams. You must convert the mass of the limiting reactant to moles first using its molar mass.

For the hydrogen combustion example, start with 4.000 g of H₂. Molar mass of H₂ is 2.016 g/mol (from CIAAW standard atomic weights). n_H₂ = 4.000 g / 2.016 g/mol = 1.984 mol. The mole ratio H₂:H₂O is 2:2 = 1:1, so n_H₂O = 1.984 mol. Molar mass of H₂O is 18.015 g/mol. m_H₂O = 1.984 mol × 18.015 g/mol = 35.74 g. That is the theoretical yield. The mole ratio scales the available H₂ directly to the product amount. No other reactant matters at this step.

Failure Case: Wrong Limiting Reactant

If you misidentify the limiting reactant, you apply the mole ratio to the wrong substance. Suppose you have 4.000 g H₂ and 32.00 g O₂. Moles of O₂ = 32.00 g / 32.00 g/mol = 1.000 mol. From the balanced equation, 2 mol H₂ reacts with 1 mol O₂. For 1.000 mol O₂, you need 2.000 mol H₂. You have 1.984 mol H₂, so H₂ is limiting. If you mistakenly use O₂ as limiting, n_O₂ = 1.000 mol, then n_H₂O = 1.000 mol × (2/1) = 2.000 mol, giving 36.03 g H₂O. That is 0.29 g too high. The error comes from ignoring the mole ratio requirement of the balanced equation.

Calculation Steps

To calculate the theoretical yield, follow these steps. First, write the balanced chemical equation. Second, convert the mass of each reactant to moles using its molar mass. Third, identify the limiting reactant by comparing the mole ratios. Fourth, use the mole ratio from the limiting reactant to the product. Fifth, convert moles of product to mass using the product's molar mass.

For a liquid reagent, you must convert volume to mass using density. The CRC Handbook of Chemistry and Physics is the standard source for density data. For example, if a reaction uses 10.0 mL of ethanol (C₂H₅OH, density 0.789 g/mL at 20 °C), the mass is 10.0 mL × 0.789 g/mL = 7.89 g. Then convert to moles: molar mass of ethanol is 46.07 g/mol, n = 7.89 g / 46.07 g/mol = 0.171 mol. At this point, use the site's proportion calculator to apply the mole ratio from the balanced equation.

Unit Mismatch Failure

If you use grams for one reactant and milligrams for another without conversion, the mole calculation will be off by a factor of 1000. Always convert all masses to the same unit (grams) before calculating moles. For a 500 mg sample of NaCl, that is 0.500 g. Molar mass is 58.44 g/mol, so n = 0.500 g / 58.44 g/mol = 0.00856 mol. Missing the conversion gives 8.56 mol, which is impossible.