Theoretical Actual and Percent Yield Worksheet: How to Solve It
The percent yield formula is: percent yield = (actual yield ÷ theoretical yield) × 100%. To solve a theoretical, actual and percent yield worksheet, calculate the theoretical yield from the balanced equation and limiting reactant, take the actual yield from the problem statement (or lab data), then divide actual by theoretical and multiply by 100. If the question gives percent yield instead, rearrange: theoretical = actual ÷ (percent yield ÷ 100).
Worksheets on this topic build on a few core ideas. Theoretical yield is the maximum product predicted by stoichiometry; actual yield is what is really obtained; percent yield expresses how close the two are.
The Three Yield Concepts
Theoretical yield is the amount of product predicted by the balanced chemical equation when the limiting reactant is completely consumed. It assumes perfect conditions: no side reactions, no losses during transfer or purification, and 100% conversion. You calculate it, you never measure it.
Actual yield is the amount of product actually isolated in the lab or stated in the worksheet. It is always a measured or given value, and in real experiments it is almost always less than the theoretical yield.
Percent yield is the ratio of actual to theoretical, expressed as a percentage. It measures the efficiency of a reaction or a student’s technique.
In a worksheet problem, the three are connected in a fixed order. You are usually given masses or moles of reactants and an actual yield. You convert reactants to moles, find the limiting reactant, convert to moles of product, convert to grams — that is theoretical yield. Then you apply the percent yield formula to the given actual yield. Some problems flip the direction: they give the theoretical yield and the percent yield, and ask you to work backward to the actual yield.
A percent yield above 100% signals an error rather than a super-efficient reaction. Common causes are an impure or wet product, incomplete drying, or weighing the product while it still contains solvent.
Solving Theoretical Yield
Theoretical yield comes from stoichiometry, and the steps are always the same. Get the balanced equation first — mole ratios depend entirely on the coefficients.
- Balance the equation. This establishes the mole ratio between reactants and products.
- Convert each reactant to moles. Use moles = mass ÷ molar mass.
- Identify the limiting reactant. Divide each reactant’s moles by its coefficient in the balanced equation; the smallest result is limiting.
- Use the mole ratio. Multiply the limiting reactant’s moles by the ratio of product coefficient to reactant coefficient.
- Convert to grams. Multiply the product’s moles by its molar mass.
Worked example
Take the reaction of hydrogen with oxygen to form water: 2H₂ + O₂ → 2H₂O. Suppose 4.0 g of H₂ reacts with 32.0 g of O₂ (molar masses: H₂ = 2.02 g/mol, O₂ = 32.00 g/mol, H₂O = 18.02 g/mol).
Moles of H₂ = 4.0 ÷ 2.02 ≈ 1.98 mol. Divided by its coefficient of 2, that gives ≈ 0.99. Moles of O₂ = 32.0 ÷ 32.00 = 1.00 mol. Divided by its coefficient of 1, that gives 1.00. Since 0.99 < 1.00, hydrogen is the limiting reactant.
From the ratio, 2 mol H₂ produce 2 mol H₂O, so 1.98 mol H₂ produces 1.98 mol H₂O. Converting: 1.98 × 18.02 ≈ 35.7 g. That is the theoretical yield.
The general method is the same for any reaction, which is why Khan Academy treats limiting reactant problems as the foundation for yield calculations. The Royal Society of Chemistry notes that this predicted amount assumes complete conversion of the limiting reagent, which is rarely achieved in practice.
Calculating Percent Yield
Percent yield compares what you got with what you expected: percent yield = (actual yield ÷ theoretical yield) × 100%.
Both values must be in the same unit. Grams and grams works, moles and moles works, but grams over moles will give a meaningless answer.
Plugging in values is straightforward once you identify the right numbers. If a worksheet says a reaction has a theoretical yield of 35.7 g and the student obtained 30.0 g, the percent yield is (30.0 ÷ 35.7) × 100% ≈ 84.0%.
Two rules catch students out. First, actual yield is the smaller number in most problems, so the result should normally sit below 100%. If it comes out higher, recheck the subtraction, the units, or whether the product was fully dry. Second, if a problem asks for the actual yield and gives percent yield, rearrange the formula: actual yield = theoretical yield × (percent yield ÷ 100). A theoretical yield of 50.0 g at 80% yield gives 40.0 g.
OpenStax’s chemistry text, available at OpenStax, emphasizes that percent yield reflects all the losses in a process — from side reactions to product left behind on glassware — so a result near 100% is a sign of careful technique.
Quick reference
| What you are given | What you solve | Formula to use |
|---|---|---|
| Reactant masses | Theoretical yield | Stoichiometry via limiting reactant |
| Theoretical and actual yield | Percent yield | (actual ÷ theoretical) × 100% |
| Theoretical and percent yield | Actual yield | theoretical × (percent ÷ 100) |
| Actual and percent yield | Theoretical yield | actual ÷ (percent ÷ 100) |
Every worksheet problem is a variation on one of these four rows.
Worksheet Practice Walkthrough
These two problems model the typical progression from a basic percent yield question to a full stoichiometry setup.
Problem 1 — Simple percent yield
A reaction has a theoretical yield of 12.5 g and produces 10.8 g. Find the percent yield.
Solution: (10.8 ÷ 12.5) × 100% = 86.4%.
Problem 2 — Full stoichiometry
Calcium carbonate decomposes: CaCO₃ → CaO + CO₂. If 50.0 g of CaCO₃ (molar mass 100.09 g/mol) decomposes and 25.0 g of CaO is collected, what is the percent yield? (CaO molar mass = 56.08 g/mol.)
Step 1 — moles of CaCO₃ = 50.0 ÷ 100.09 ≈ 0.500 mol.
Step 2 — the ratio is 1:1, so theoretical moles of CaO = 0.500 mol.
Step 3 — theoretical mass of CaO = 0.500 × 56.08 ≈ 28.0 g.
Step 4 — percent yield = (25.0 ÷ 28.0) × 100% ≈ 89.3%.
Work through the limiting reactant step before touching the percent yield formula; most lost marks come from skipping straight to the division. The American Chemical Society notes that stoichiometry problems of this type are standard in introductory courses precisely because they combine several skills at once. Checking units at every conversion is the fastest way to catch mistakes before they propagate into the final answer.
Key Takeaways
- Percent yield = (actual yield ÷ theoretical yield) × 100%, and it answers how efficient a reaction was.
- Theoretical yield is always calculated from the balanced equation and the limiting reactant; actual yield is always measured or given.
- Find the limiting reactant by dividing each reactant’s moles by its coefficient and taking the smallest value.
- Keep actual and theoretical yields in the same unit before dividing.
- Rearrange the formula when the problem gives percent yield: theoretical = actual ÷ (percent ÷ 100), or actual = theoretical × (percent ÷ 100).
- A percent yield above 100% points to an error such as an impure or wet product, not to better-than-perfect chemistry.
