For percent yield practice problems, first identify two amounts of the same product: the amount actually collected and the amount the balanced reaction predicts. Divide actual yield by theoretical yield, then multiply by 100%. The arithmetic is usually short. The harder part is deciding which numbers belong in that fraction.
These eight original exercises move from straightforward substitution to problems with mixed units, missing information, and misleading measurements. They are fictional classroom examples, not laboratory instructions. Try each one before reading its answer, and keep the labels beside every number. A correct number without the right product or unit can hide a mistaken setup.
The three quantities you need to separate
Theoretical yield is the predicted product amount under the reaction’s stated assumptions. Actual yield is the amount obtained. Percent yield compares them: actual yield / theoretical yield x 100%. Both amounts must describe the same product in compatible units. OpenStax’s reaction-yields section explains these definitions and why incomplete reactions, competing reactions, or collection losses can reduce yield.
Make a three-line box on your scratch paper: actual, theoretical, percent. Fill in what the question gives, then circle the blank you need. This small step prevents treating the largest mass as theoretical yield simply because it is largest. A reactant’s starting mass is not automatically the theoretical mass of product.
If the question gives amounts of multiple reactants, you may need stoichiometry before calculating the percentage. Use the limiting-reactant practice problems for that prerequisite. The exercises below emphasize what happens after you have identified the product and its predicted amount.
Eight percent yield problems with worked answers
1. Choose the numerator and denominator
Problem: A reaction should produce 18.0 g of a solid. A student collects 13.5 g of that same dry, pure solid. What is the percent yield?
Answer: The actual yield is 13.5 g and the theoretical yield is 18.0 g. Calculate (13.5 g / 18.0 g) x 100% = 75.0%. The grams cancel because the two quantities use the same unit and refer to the same product.
Check: Three quarters of 18.0 is 13.5, so a result of 75.0% makes sense. Reversing the fraction produces about 133%, which does not match the stated collection of less product than predicted. Write the words actual and theoretical before entering numbers into your calculator.
2. Convert units before dividing
Problem: The theoretical yield is 2.40 g. The measured product is 1800 mg, recorded to three significant figures. Find the percent yield.
Answer: Convert 1800 mg to 1.80 g. Then (1.80 / 2.40) x 100% = 75.0%. Alternatively, convert the theoretical yield to 2400 mg and divide the two milligram values.
Check: Dividing 1800 by 2.40 without conversion gives an enormous percentage because it compares different scales. Unit conversion does not change how much product exists. If your result differs by a factor of one thousand from a reasonable estimate, inspect the units before changing the chemistry.
3. Find actual yield from a percentage
Problem: A worksheet predicts 16.0 g of product. It asks you to estimate the amount collected if the percent yield is 82.5%. What actual yield corresponds to that assumption?
Answer: Convert 82.5% to the fraction 0.825. Actual yield = 0.825 x 16.0 g = 13.2 g. This is a prediction conditional on the supplied yield, not a measurement of a real experiment.
Check: Multiplying by 82.5 instead of 0.825 gives 1320 g. That answer confuses a percentage with a proportion. You can check the correct result by putting 13.2 and 16.0 into the original fraction: the calculation returns 82.5%.
4. Work backward to theoretical yield
Problem: A student obtains 7.20 g of product at a reported yield of 60.0%. What theoretical product amount was used to calculate that percentage?
Answer: Start with 0.600 = 7.20 g / theoretical yield. Rearranging gives theoretical yield = 7.20 g / 0.600 = 12.0 g.
Check: The predicted amount must be larger than the actual amount when the yield is 60.0%. Multiplying 7.20 by 0.600 would give 4.32 g, which fails that check. Before rearranging an equation, write a sentence about whether your answer should increase or decrease. That sentence can catch an algebra mistake immediately.
5. Calculate theoretical yield from a balanced equation
Problem: For 2Mg + O2 → 2MgO, assume oxygen is in excess and 2.43 g of magnesium reacts. Use molar masses of 24.3 g/mol for Mg and 40.3 g/mol for MgO. The actual MgO collected is 3.22 g. Find the percent yield.
Answer: Magnesium amount = 2.43 / 24.3 = 0.100 mol. The equation has a 2:2 ratio, so the predicted MgO amount is also 0.100 mol. Theoretical MgO mass = 0.100 x 40.3 = 4.03 g. Percent yield = (3.22 / 4.03) x 100% = 79.9%, rounded to three significant figures.
Check: Dividing 3.22 by the starting 2.43 g would compare product with reactant, not actual product with theoretical product. MgO contains oxygen as well as magnesium, so its mass can exceed the magnesium starting mass without violating conservation of mass. Keep the chemical formula attached to each quantity.
6. Interpret an apparent yield above 100%
Problem: A calculation predicts 5.00 g of dry product. A wet sample weighs 5.60 g. Calculate the apparent yield and explain whether this proves that the reaction made more pure product than predicted.
Answer: The apparent yield is (5.60 / 5.00) x 100% = 112%. It does not prove an excess of pure product. The numerator includes a wet sample, while the denominator describes dry product. They are not comparable measurements of the intended substance.
Check: Report the calculation and the measurement mismatch separately. Do not silently replace 112% with 100%, invent a corrected dry mass, or decide how much solvent is present from these numbers alone. In an actual course lab, follow the supplied measurement procedure and discuss the discrepancy using recorded observations.
7. Recognize when actual yield is missing
Problem: A question provides a balanced equation and enough reactant information to calculate a theoretical yield of 9.00 g. It gives no measured product amount and no yield percentage. Can you calculate percent yield?
Answer: No. You know the denominator but not the numerator. A theoretical yield tells you what the model predicts, not what was collected. You need actual yield or another statement that determines it.
Check: Writing 100% assumes complete conversion and perfect recovery without evidence. A useful answer says exactly what is missing: the actual amount of the specified product. Identifying insufficient information is part of solving the problem, not a failure to remember another formula.
8. Combine yields for two sequential steps
Problem: In a simplified two-step sequence, each step has a 1:1 mole ratio. Step one converts starting material to an intermediate with 80.0% yield. All recovered intermediate goes into step two, whose yield is 75.0%. Relative to the ideal final amount from the original starting material, what is the overall yield?
Answer: Use fractions: 0.800 x 0.750 = 0.600, so the overall yield is 60.0%. For an ideal path starting with 1.00 mol, step one supplies 0.800 mol of intermediate and step two supplies 0.600 mol of final product.
Check: Averaging the percentages gives 77.5%, which describes neither step nor the complete sequence. The second percentage applies only to material that reaches step two. This example assumes the stated mole relationships and use of all recovered intermediate; extra aliquots or different bookkeeping require a revised setup.
Turn incorrect answers into useful notes
After the first pass, label each mistake as a quantity error, unit error, algebra error, or interpretation error. A quantity error means you used a reactant mass or wet sample where pure product was required. A unit error means comparable amounts were written on different scales. An algebra error means you rearranged the fraction incorrectly. An interpretation error means your conclusion went beyond the information.
Then write a correction in one sentence. For problem five, you might write, “I must calculate grams of MgO from moles of Mg before comparing the two product amounts.” That sentence is more useful than copying 79.9% into a notebook without explaining where the denominator came from.
For a second practice pass, cover the worked answers and change one number in each problem. Check your new solution by substituting it back into the formula. Keep one ordinary question, one reverse calculation, and one missing-information question together so you must decide what to do before calculating.
Frequently asked questions
Is percent yield the same as percent error?
No. Yield compares obtained product with predicted product. Percent error compares a measurement with a reference value using the error convention assigned in your course. Choose the formula from the question’s meaning, not from the presence of a percent sign.
Does a low yield identify the cause?
No. A percentage alone cannot distinguish incomplete reaction, competing products, or losses during collection. Use observations and the experimental procedure when explaining possible causes. Keep a plausible explanation separate from a demonstrated cause.
When should I round?
Keep extra calculator digits during intermediate steps and round the final answer according to the supplied data and your course’s rules. In problem five, retaining the unrounded ratio before reporting 79.9% avoids unnecessary intermediate rounding.
Build a chemistry practice page you can share
Use the chemistry document page to find related material, or browse student documents for your course. Check each resource against your instructor’s notation and assumptions.
Students should upload their own notes to help other students. A useful original yield sheet includes the question, labeled quantities, calculation, units, and one explanation of the likely mistake. Share work you have permission to share, and leave out restricted assessments or another person’s solutions.