Theoretical yield is the ceiling, not the prediction

Theoretical yield is the answer to a counterfactual: if every molecule of the limiting reagent reacted exactly once, with no losses, no side products, and perfect recovery, how much product would you get? You will never hit it in real life. A pristine textbook reaction might come in at 95%; a messy multistep synthesis might land at 30%. The point of calculating it is not to predict your bench result — it is to give you the denominator you need to evaluate your bench result.

Two things go wrong here for almost everyone: skipping the limiting reagent check and using mass ratios when the equation only gives you mole ratios. Get those right and the rest is unit cancellation.

The four-step pattern

Every theoretical-yield problem reduces to the same workflow:

Balance the equation. Coefficients are mole ratios — they are wrong if the equation is unbalanced.
Convert each given mass to moles by dividing by molar mass.
Compare available moles against the stoichiometric ratio to find the limiting reagent.
Multiply limiting-reagent moles by (mole product / mole limiting reagent), then by the product’s molar mass.

Percent yield is the easy part: actual / theoretical × 100.

Worked example with one reagent in excess

Calculate the theoretical yield of water from 10.0 g H₂ reacting with excess O₂.

2 H₂ + O₂ → 2 H₂O

When one reactant is “in excess,” skip the limiting-reagent check — H₂ is automatically limiting.

Moles of H₂ = 10.0 / 2.016 = 4.960 mol Mole ratio: 2 mol H₂ → 2 mol H₂O, so 4.960 mol H₂ → 4.960 mol H₂O Mass of H₂O = 4.960 × 18.015 = 89.4 g

The 1:1 ratio here makes the middle step look trivial, but the mole ratio is doing real work whenever it is not 1:1 — see the next example.

Worked example with a real limiting-reagent check

25.0 g Fe₂O₃ reacts with 10.0 g Al. Find the theoretical yield of Fe.

2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe

Convert both reactants to moles:

Al: 10.0 / 26.98 = 0.3707 mol
Fe₂O₃: 25.0 / 159.69 = 0.1565 mol

Check stoichiometry against the 2:1 Al:Fe₂O₃ ratio. There are two equivalent ways to do this — pick one and stick with it:

Method A — how much Al would you need? 0.1565 mol Fe₂O₃ × (2 mol Al / 1 mol Fe₂O₃) = 0.3130 mol Al required. You have 0.3707 mol. Al is in excess; Fe₂O₃ is limiting.

Method B — divide each by its coefficient. Al: 0.3707/2 = 0.1854. Fe₂O₃: 0.1565/1 = 0.1565. Smaller value is the limiting reagent. Fe₂O₃ wins.

Now run the limiting reagent through the rest of the equation: Moles of Fe = 0.1565 × (2 mol Fe / 1 mol Fe₂O₃) = 0.3130 mol Mass of Fe = 0.3130 × 55.85 = 17.5 g

Side check: how much Al was consumed? 0.3130 mol, leaving 0.0577 mol (about 1.56 g) of unreacted aluminum in the crucible. Tracking the leftover excess is a habit worth building — it lets you sanity-check the limiting-reagent assignment.

Percent yield from the bench

Same reaction. The student weighs the iron after reduction and gets 15.8 g.

Percent yield = (15.8 / 17.5) × 100 = 90.3%

For thermite-style reactions in a controlled setup, 90% is excellent. For a multistep organic synthesis, the same number would suggest you got lucky or made a measurement error.

Working backward from percent yield

A reaction has a theoretical yield of 50.0 g and a typical percent yield of 75.0%. How much product will you actually pull from the round-bottom?

Actual = 50.0 × 0.750 = 37.5 g

This is the planning calculation — useful for deciding whether your starting material gets you enough product to move on to the next step.

The traps that bite hardest

Skipping the limiting-reagent check. When two reactant masses are given, never assume the larger mass is in excess. Hydrogen is light; an excess of H₂ might still be the limiting reagent on a mole basis.
Mass ratios instead of mole ratios. Coefficients in a balanced equation are moles, not grams. Convert to moles, apply the ratio, convert back.
Forgetting the coefficient when computing moles of product. 0.1565 mol Fe₂O₃ gives 0.3130 mol Fe, not 0.1565 mol Fe. The factor of 2 is in the equation for a reason.
Confusing theoretical with actual. Theoretical comes from arithmetic. Actual comes from a balance. Percent yield compares the two — never the other way around.
Rounding intermediate steps. Carry four significant figures through the calculation; round the final answer.

Practice problems

Try these, then check with the Stoichiometry Calculator:

Theoretical yield of CO₂ from 25.0 g propane combustion: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O.
5.00 g Zn + excess HCl → ZnCl₂ + H₂. Theoretical yield of H₂?
20.0 g NaOH + 20.0 g HCl → NaCl + H₂O. Limiting reagent and theoretical yield of NaCl?
A student isolates 3.50 g aspirin (C₉H₈O₄); theoretical yield is 4.20 g. Percent yield?
A reaction averages 80.0% yield. To collect 100.0 g of product, what theoretical yield do you need?

How to Calculate Theoretical Yield