Limiting Reactant and Percent Yield Explained

Why moles, not grams, decide which reactant runs out

A balanced equation counts particles. 2 H₂ + O₂ → 2 H₂O means two molecules of hydrogen react with one molecule of oxygen, or 2 mol with 1 mol. It says nothing directly about grams, because different substances have different molar masses. Comparing the masses of two reactants therefore cannot tell you which one runs out first.

The mole ratio from the coefficients works like an exchange rate between substances. To use it, convert each given mass or volume to moles first, apply the ratio, and convert back to grams only if the question asks for a mass.

A kitchen analogy shows the idea. If each sandwich needs 2 slices of bread and 1 slice of cheese, then 10 slices of bread and 7 slices of cheese make only 5 sandwiches. Bread limits the output even though there are more slices of it, because the recipe uses bread twice as fast.

Two reliable tests for the limiting reactant

Compare product amounts. For each reactant, calculate how much product it could form if everything else were unlimited. The reactant that gives the smallest amount of product is limiting, and that smallest amount is the theoretical yield. This test answers the most common follow-up question at the same time.

Compare required with available. Pick one reactant, calculate how much of a second reactant it would need, and compare with what is present. If less is present than needed, the second reactant is limiting; otherwise the first is. An equivalent shortcut divides each reactant's amount in moles by its coefficient; the smallest quotient marks the limiting reactant.

Both tests need a balanced equation and amounts in moles. With three or more reactants, the product-amount test or the divide-by-coefficient shortcut is less error-prone than comparing reactants in pairs.

Worked example: copper wire in silver nitrate solution

Copper displaces silver from silver nitrate: Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2 Ag(s). A 3.00 g piece of copper wire is placed in a solution containing 8.50 g of AgNO₃. Which reactant is limiting, what mass of silver can form, and how much copper remains? Molar masses use IUPAC standard atomic weights rounded to two decimals (Cu 63.55, Ag 107.87, N 14.01, O 16.00), so AgNO₃ is 169.88 g/mol.

Convert to moles, carrying extra digits: 3.00 g Cu ÷ 63.55 g/mol = 0.047207 mol Cu, and 8.50 g AgNO₃ ÷ 169.88 g/mol = 0.050035 mol AgNO₃. The copper would need 2 × 0.047207 = 0.094414 mol AgNO₃, but only 0.050035 mol is present, so AgNO₃ is limiting, even though its mass is almost three times that of the copper.

Theoretical yield of silver: 0.050035 mol AgNO₃ × (2 mol Ag / 2 mol AgNO₃) × 107.87 g/mol = 5.397 g, reported as 5.40 g. Basing the calculation on copper instead would predict 10.18 g of silver, which cannot form because the silver nitrate runs out first.

Excess copper: the AgNO₃ consumes 0.050035 mol ÷ 2 = 0.025018 mol Cu, or 0.025018 mol × 63.55 g/mol = 1.590 g. Copper remaining = 3.00 g − 1.590 g = 1.41 g, reported to the hundredths place because 3.00 g is known to the hundredths place. Mass is conserved: 5.397 g Ag + 4.693 g Cu(NO₃)₂ (187.57 g/mol) + 1.410 g Cu = 11.50 g, equal to the 3.00 g + 8.50 g at the start.

Theoretical, actual and percent yield

The theoretical yield is the maximum amount of product the limiting reactant can form, assuming the reaction goes to completion as written with no side reactions and no losses. The actual yield is the amount you isolate and measure. Percent yield = (actual yield ÷ theoretical yield) × 100%.

Yields below 100% are normal. A reaction may reach equilibrium before the limiting reactant is used up, side reactions may consume reactants, and product is lost during filtration, transfer, washing or recrystallization. A yield above 100% for a supposedly pure product usually means it still contains solvent or impurities, or that a mass was recorded incorrectly. Treat it as a signal to dry and reweigh, not as a result.

Both yields must describe the same substance in the same units. If the theoretical yield is in moles, convert the actual yield to moles as well, or express both in grams.

Worked example: percent yield of an ester synthesis

Isopentyl acetate, an ester with a banana-like odor, forms from acetic acid and isopentyl alcohol (3-methyl-1-butanol): C₂H₄O₂ + C₅H₁₂O → C₇H₁₄O₂ + H₂O. A mixture of 6.00 g of acetic acid and 7.50 g of isopentyl alcohol gives 8.12 g of purified ester. What is the percent yield?

Molar masses (C 12.01, H 1.01, O 16.00): acetic acid 60.06 g/mol, isopentyl alcohol 88.17 g/mol, ester 130.21 g/mol. Amounts: 6.00 g ÷ 60.06 g/mol = 0.09990 mol acid and 7.50 g ÷ 88.17 g/mol = 0.08506 mol alcohol. The ratio is 1:1, so the alcohol is limiting and 0.01484 mol of acid is in excess. Using one reactant in excess is a common way to push an equilibrium-limited reaction like this one toward product.

Theoretical yield = 0.08506 mol × 130.21 g/mol = 11.076 g. Percent yield = (8.12 g ÷ 11.076 g) × 100% = 73.3%, three significant figures from 8.12 g and 7.50 g. Rounding the theoretical yield to 11.1 g before dividing would give 73.2%, so carry the unrounded value forward.

Percent yield also works backward. To collect 5.00 g of ester at the same 73.3% yield, you need a theoretical yield of 5.00 g ÷ 0.733 = 6.82 g, which is 0.0524 mol of ester. That requires at least 0.0524 mol, or 4.62 g, of isopentyl alcohol, with the acid in excess.

Yield and excess-reactant errors to check for

Most errors in these problems are setup errors rather than arithmetic, so each has a check that does not require redoing the whole calculation.

After finishing, confirm three things: the limiting reactant was chosen from moles, the theoretical yield came from that reactant, and the percent yield is 100% or less.

• Choosing the limiting reactant by comparing masses. In the copper example, the reactant with the larger mass was limiting.
• Comparing amounts in moles without the coefficients. Divide each amount by its coefficient, or compute the product from each reactant.
• Calculating the theoretical yield from the excess reactant. The theoretical yield is always the smaller of the possible product amounts.
• Finding leftover excess by subtracting the mass of one substance from the mass of another. Compute how much of the excess reactant is consumed, in its own units, and subtract that from the amount of the same substance present.
• Swapping actual and theoretical yield in the formula. A percent yield above 100% for a dry, pure product means something is wrong.
• Rounding intermediate values. Rework the last step with unrounded numbers and see whether the reported digits change.