Difficulty: Introductory | Prerequisites: Molar mass conversions, stoichiometry basics (mole ratios), limiting reagent identification.
Percent yield sits at the junction of stoichiometry and real laboratory work. Every reaction you have balanced so far assumed that all reactants convert perfectly into products, but in practice that never happens. Side reactions, incomplete mixing, losses during transfer and purification all eat into your output. This topic gives you the tools to quantify that gap. If you are comfortable converting grams to moles and identifying a limiting reagent, you are ready for this material.
Percent yield measures how much product you actually get from a reaction compared to what stoichiometry predicts you should get. You calculate a theoretical yield from the limiting reagent, then divide the actual (experimental) yield by that number and multiply by 100. A value under 100% is normal; a value over 100% signals an error.
Theoretical yield
The maximum mass of product that can form from a given amount of limiting reagent, calculated by stoichiometry. Think of it as the best-case scenario, assuming perfect conversion and zero losses.
Actual yield
The mass of product you physically collect in the lab after the reaction and any purification steps. In simple terms, this is what you weigh on the balance at the end.
Percent yield
The ratio of actual yield to theoretical yield, expressed as a percentage. It tells you how efficient your reaction was.
Limiting reagent (limiting reactant)
The reactant that runs out first and therefore caps the amount of product that can form. In simple terms, this is the ingredient you do not have enough of. All theoretical yield calculations start from this reactant.
Excess reagent (excess reactant)
The reactant present in more than enough quantity; some of it remains unreacted when the limiting reagent is consumed.
The standard workflow has four steps:
Convert each reactant mass to moles using its molar mass.
Identify the limiting reagent by comparing the mole ratio each reactant can supply against the balanced equation.
Use the limiting reagent's moles and the stoichiometric ratio to calculate the theoretical yield in moles, then convert to grams.
Apply the percent yield formula: (actual yield / theoretical yield) x 100.
Example (aspirin synthesis): 3.03 g salicylic acid (C7H6O3, M = 138.1) and 4.01 g acetic anhydride (C4H6O3, M = 102.1) react in a 1 : 1 ratio. Salicylic acid gives 0.02194 mol; acetic anhydride gives 0.03928 mol. Salicylic acid is limiting. Theoretical yield of aspirin = 0.02194 mol x 180.2 g/mol = 3.953 g. Actual yield = 2.04 g. Percent yield = (2.04 / 3.953) x 100 = 51.6%.
Sometimes you know the percent yield and a desired actual yield, and need to find how much reactant is required.
Rearrange the formula: theoretical yield = actual yield / (percent yield / 100).
Convert the theoretical yield in grams to moles of product.
Use the stoichiometric ratio to find moles of reactant needed.
Convert to grams.
Example: 24.8% yield, want 12.5 g NO from 4 NH3 + 5 O2 → 4 NO + 6 H2O. Theoretical yield of NO needed = 12.5 / 0.248 = 50.40 g. Moles NO = 50.40 / 30.01 = 1.680 mol. Mole ratio NH3 : NO = 4 : 4 = 1 : 1, so 1.680 mol NH3 needed. Mass = 1.680 x 17.03 = 28.6 g.
When two reactant masses are given alongside a percent yield, you must find the limiting reagent first, calculate the theoretical yield from it, and then apply the percent yield.
Example: 15.0 g N2 and 2.00 g H2, 9.82% yield, N2 + 3 H2 → 2 NH3. Moles N2 = 0.5353; moles H2 = 0.9921. H2 needs 3 mol per mol N2; available ratio = 0.9921 / 0.5353 = 1.853, which is less than 3, so H2 is limiting. From 0.9921 mol H2, theoretical NH3 = (2/3)(0.9921)(17.03) = 11.26 g. Actual = 11.26 x 0.0982 = 1.11 g.
\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\text{Theoretical Yield (g)} = \text{mol of limiting reagent} \times \frac{\text{mol product}}{\text{mol reagent}} \times M_{\text{product}}Rearranged for reverse problems:
\text{Theoretical Yield} = \frac{\text{Actual Yield}}{\text{Percent Yield} / 100}Pharmaceutical manufacturing tracks percent yield at every synthetic step because even a small improvement, from 48% to 52% say, saves millions of dollars worth of raw material per year. In industrial ammonia production (the Haber process), engineers optimise temperature, pressure and catalyst to push the yield as high as thermodynamics allows.
Students often use the excess reagent instead of the limiting reagent to calculate theoretical yield. The excess reagent will not all react, so it gives a theoretical yield that is too high.
Students sometimes forget to convert grams to moles before applying the stoichiometric ratio. Grams do not cancel with mole ratios.
A percent yield above 100% does not mean you made extra product. It means the product is impure (it contains solvent, unreacted reagent, or by-products) or there is an experimental error.
When a problem gives a percent yield and asks for the mass of reactant needed, students sometimes multiply by the yield instead of dividing. If the yield is less than 100%, you need more reactant than stoichiometry alone suggests, not less.
⚠️ Expect at least one problem that requires you to identify the limiting reagent before calculating percent yield. Both masses of reactants will be given.
⚠️ Reverse problems (given percent yield and desired actual yield, find mass of reactant) appear frequently. Know how to rearrange the formula.
⚠️ Units matter. Converting grams → moles → moles (via ratio) → grams is the bread-and-butter pathway. Skipping a step is the most common source of wrong answers.
True or false: the theoretical yield is always greater than or equal to the actual yield. (True.)
Fill in the blank: percent yield = (______ / theoretical yield) x 100. (Actual yield.)
True or false: if one reactant is in excess, you use its mass to find the theoretical yield. (False; use the limiting reagent.)
Fill in the blank: to find how much reactant you need when percent yield is known, you ______ the desired actual yield by the decimal form of the percent yield. (Divide.)
True or false: a percent yield of 110% is a perfectly normal experimental result. (False; it signals impurity or error.)
Q: If 2.04 g of aspirin (M = 180.2) is produced from 3.03 g salicylic acid (M = 138.1) and 4.01 g acetic anhydride (M = 102.1) in a 1 : 1 reaction, what is the percent yield?
A: Salicylic acid is limiting (0.02194 mol vs 0.03928 mol). Theoretical yield = 0.02194 x 180.2 = 3.953 g. Percent yield = (2.04 / 3.953) x 100 = 51.6%.
Q: The reaction 4 NH3 + 5 O2 → 4 NO + 6 H2O has a 24.8% yield. What mass of NH3 must react to yield 12.5 g NO?
A: Theoretical NO needed = 12.5 / 0.248 = 50.40 g = 1.680 mol. Mole ratio is 1 : 1, so 1.680 mol NH3 = 1.680 x 17.03 = 28.6 g.
Q: 15.0 g N2 reacts with 2.00 g H2 at 9.82% yield via N2 + 3 H2 → 2 NH3. What mass of NH3 forms?
A: H2 is limiting (0.9921 mol H2 needs 0.3307 mol N2, but provides only a 1.85 : 1 ratio vs the required 3 : 1). Theoretical NH3 = (2/3)(0.9921)(17.03) = 11.26 g. Actual = 11.26 x 0.0982 = 1.11 g.
Q: S8 + 24 F2 → 8 SF6. If 18.3 g SF6 is isolated from 10.0 g S8 and 30.0 g F2, what is the percent yield?
A: Moles S8 = 10.0 / 256.5 = 0.03899; moles F2 = 30.0 / 38.00 = 0.7895. F2 is limiting (needs 24 x 0.03899 = 0.9357 mol, only 0.7895 available). Theoretical SF6 = (8/24)(0.7895)(146.1) = 38.47 g. Percent yield = (18.3 / 38.47) x 100 = 47.6%.
This connects to stoichiometry (Unit 3) because every percent yield problem begins with a balanced equation and mole ratios. It also ties into limiting reagent problems directly: without correctly identifying the limiting reagent, the theoretical yield is wrong and the percent yield follows.
In later units, percent yield reappears in solution stoichiometry and titration calculations, where concentrations replace masses but the logic is identical.
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