This page offers ten self-testing percent yield practice cards, each with a hidden step-by-step solution you reveal only after attempting the problem yourself. Problems range from a one-line percent yield calculation to limiting-reagent and multi-step syntheses. For calculators use the homepage tools; for taught examples or printable worksheets, separate pages cover those instead.
Every percent yield problem rewards a few seconds of reading before any arithmetic starts. Work through these checks in order, every time, and the numbers tend to fall into place on the first try.
Reading a problem in this order turns a wall of text into a short checklist, and it is the same order used to work through the ten cards below.
Most wrong answers on a percent yield problem do not come from a misunderstood concept. They come from one of the same five slips, repeated across countless homework sets and lab reports. The table below lists each one next to the fix that prevents it.
| Common mistake | One-line fix |
|---|---|
| Treating whichever reactant is listed first as the limiting reagent | Convert every reactant to moles and compare mole-to-coefficient ratios before deciding which one limits the reaction |
| Plugging a mass straight into a mole ratio without converting | Divide mass by molar mass to get moles before applying any stoichiometric ratio |
| Swapping actual and theoretical yield in the formula | Actual yield, the amount measured in the lab, always goes on top; theoretical yield, the calculated maximum, always goes on the bottom |
| Dropping or mixing units partway through a solution | Carry grams, moles, and grams-per-mole labels on every line so a mismatched unit is caught immediately |
| Answering the wrong quantity because the question was skimmed | Reread the last sentence of the problem and confirm the units of the requested answer before writing anything down |
None of these mistakes require more chemistry knowledge to avoid. They require slowing down at the exact point where the slip happens, which is why each of the ten cards below is built to force that same slow-down before the answer is revealed.
Working a problem cold and then checking the answer teaches more than reading a solved example line by line. Each card below states a problem and nothing else; the full worked solution sits behind a reveal so you can commit to an answer before seeing how it was reached. That is a different exercise from a written walkthrough, and a different one again from a printed sheet with an answer key at the bottom.
For a fully narrated explanation of the reasoning behind each step, including why a particular conversion or ratio was chosen, the worked examples page covers eight complete problems in that format. For working on paper with a separate answer key, the printable worksheets page has fifteen tiered problems built for that purpose. This page sits between the two: short enough to work through in one sitting, interactive enough to check yourself one problem at a time.
The ten cards below move from a single plug-into-the-formula calculation to a two-reaction sequence with two separate yield losses. Working them in order mirrors how a stoichiometry unit is usually taught: percent yield in isolation first, then rearranging the formula for a missing yield or theoretical value, then a real reaction with molar masses, then a second reactant that might be limiting, then a result over 100 percent that has to be explained rather than just calculated, and finally a sequence of two reactions chained together. If any single card feels harder than it should, the strategy checklist above is the fastest way to find the missed step.
These ten cards assume the percent yield formula is already known and the goal is practicing it under slightly different conditions each time. If the formula itself, its derivation, or its rearrangements for actual or theoretical yield need a refresher first, that material is on the formula page. For a plain-language walkthrough of what actual yield, theoretical yield, and limiting reagents mean, and how they connect to the rest of a stoichiometry unit, see the chemistry guide. And if a live calculator would be faster than working the arithmetic by hand for a real assignment, the percent yield calculator on the homepage handles all three versions of the formula automatically.
Once the ten problems below start feeling routine rather than tricky, that is usually a sign it is time to move to problems with more reactants, longer reaction sequences, or lab-based sources of yield loss, several of which show up in the harder cards near the end of this set.
Problem 1
A student runs a synthesis and recovers 3.5 g of product. The theoretical yield for the reaction, calculated from the limiting reagent, is 5.0 g. What is the percent yield?
Percent yield equals actual yield divided by theoretical yield, multiplied by 100.
Percent yield = (3.5 g / 5.0 g) x 100 = 70.0%
The reaction produced 70.0% of the maximum amount of product that stoichiometry allows, which is a normal result for a student-run synthesis.
Problem 2
A reaction has a theoretical yield of 12.0 g. The percent yield for this run was 80.0%. How many grams of product were actually recovered?
Rearrange the percent yield formula to solve for actual yield: actual yield = (percent yield / 100) x theoretical yield.
Actual yield = 0.800 x 12.0 g = 9.60 g
The student recovered 9.60 g of product from this run.
Problem 3
A synthesis recovers 6.30 g of product at a percent yield of 90.0%. What was the theoretical yield for this reaction?
Rearrange the formula to solve for theoretical yield: theoretical yield = actual yield / (percent yield / 100).
Theoretical yield = 6.30 g / 0.900 = 7.00 g
The maximum amount of product this reaction could have produced, based on the limiting reagent, was 7.00 g.
Problem 4
Compound X (molar mass 150.0 g/mol) is converted to Compound Y (molar mass 198.0 g/mol) in a reaction with a 1:1 mole ratio. A chemist starts with 6.00 g of Compound X and recovers 6.34 g of Compound Y. Find the theoretical yield and the percent yield.
Step 1: Convert the starting mass of X to moles.
Moles X = 6.00 g / 150.0 g/mol = 0.0400 mol
Step 2: Use the 1:1 mole ratio to find moles of Y, then convert to grams for the theoretical yield.
Moles Y = 0.0400 mol (same as X, since the ratio is 1:1)
Theoretical yield of Y = 0.0400 mol x 198.0 g/mol = 7.92 g
Step 3: Apply the percent yield formula.
Percent yield = (6.34 g / 7.92 g) x 100 = 80.1%
Problem 5
A reaction has a theoretical yield of 0.250 mol of product. The chemist actually isolates 0.205 mol of that same product. What is the percent yield? No mass or molar mass conversion is needed here.
Because both the actual and theoretical amounts are already given in moles of the same substance, the mole values can go directly into the percent yield formula without converting to mass first.
Percent yield = (0.205 mol / 0.250 mol) x 100 = 82.0%
Converting both values to grams first would give the identical answer, since the molar mass would cancel out of the ratio, so it can be skipped entirely.
Problem 6
The theoretical yield of a reaction is 0.150 mol of a product with a molar mass of 106.0 g/mol. In the lab, the actual yield is measured on a balance as 12.7 g. What is the percent yield?
Step 1: The theoretical yield is in moles and the actual yield is in grams, so one of them has to be converted before they can be compared. Convert the theoretical yield to grams.
Theoretical yield = 0.150 mol x 106.0 g/mol = 15.9 g
Step 2: Now both values are in grams, so apply the percent yield formula.
Percent yield = (12.7 g / 15.9 g) x 100 = 79.9%
Problem 7
Aluminum reacts with chlorine gas: 2 Al + 3 Cl2 -> 2 AlCl3. A reaction combines 5.40 g of Al (molar mass 27.0 g/mol) with 14.20 g of Cl2 (molar mass 71.0 g/mol). The chemist recovers 15.0 g of AlCl3 (molar mass 133.5 g/mol). Identify the limiting reagent and find the percent yield.
Step 1: Convert both reactants to moles.
Moles Al = 5.40 g / 27.0 g/mol = 0.200 mol
Moles Cl2 = 14.20 g / 71.0 g/mol = 0.200 mol
Step 2: Compare each to its coefficient in the balanced equation (2 Al : 3 Cl2). Divide each mole amount by its coefficient.
Al: 0.200 / 2 = 0.100
Cl2: 0.200 / 3 = 0.0667
Cl2 gives the smaller value, so Cl2 is the limiting reagent, even though both reactants started at the same number of moles.
Step 3: Use the limiting reagent to find the theoretical yield of AlCl3.
Moles AlCl3 = 0.200 mol Cl2 x (2 mol AlCl3 / 3 mol Cl2) = 0.1333 mol
Theoretical yield = 0.1333 mol x 133.5 g/mol = 17.80 g
Step 4: Apply the percent yield formula.
Percent yield = (15.0 g / 17.80 g) x 100 = 84.3%
Problem 8
A student calculates a theoretical yield of 4.50 g for a reaction. After filtering, drying, and weighing the product, the balance reads 4.92 g of actual yield. Calculate the percent yield and explain what a result like this means.
Percent yield = (4.92 g / 4.50 g) x 100 = 109.3%
A percent yield above 100% is not a valid chemical result on its own, since a reaction cannot produce more product than the limiting reagent allows. It signals that the measured mass includes something other than pure, dry product, most often leftover solvent, residual water, or an unreacted starting material that did not fully separate out.
The fix is not to report 109.3% as the final answer. The product should be dried further and reweighed, and the calculation should be checked for an error in the theoretical yield itself, such as a missed limiting-reagent check or a molar mass mistake. Once the product is properly dried, the recalculated percent yield should fall back below 100%.
Problem 9
Compound P (molar mass 100.0 g/mol) is converted to Compound Q (molar mass 114.0 g/mol) in a 1:1 reaction with a 75.0% yield. The Q produced is then carried into a second 1:1 reaction that converts it to Compound R (molar mass 130.0 g/mol) at an 80.0% yield. Starting from 10.0 g of Compound P, find the overall percent yield for the two-step sequence and the final mass of Compound R obtained.
Step 1: Convert the starting mass of P to moles.
Moles P = 10.0 g / 100.0 g/mol = 0.100 mol
Step 2: Find the actual moles and mass of Q produced in step one, using its 75.0% yield.
Theoretical Q = 0.100 mol (1:1 ratio) x 114.0 g/mol = 11.4 g
Actual Q = 0.750 x 11.4 g = 8.55 g, which is 8.55 g / 114.0 g/mol = 0.0750 mol
Step 3: Carry the actual moles of Q (0.0750 mol) into the second reaction and find the actual mass of R, using its 80.0% yield.
Theoretical R from this step = 0.0750 mol (1:1 ratio) x 130.0 g/mol = 9.75 g
Actual R = 0.800 x 9.75 g = 7.80 g
Step 4: Find the overall percent yield for the whole sequence. Since the reactions are 1:1:1 all the way through, the theoretical mass of R starting from all 0.100 mol of P with no losses at all would be 0.100 mol x 130.0 g/mol = 13.0 g.
Overall percent yield = (7.80 g / 13.0 g) x 100 = 60.0%
This matches multiplying the two individual step yields directly: 75.0% x 80.0% = 60.0%. The final mass of Compound R obtained is 7.80 g.
Problem 10
A reaction has a theoretical yield of 8.40 g. The chemist recovers 6.30 g of actual product. Find the percent yield, the mass of product lost, and the percent lost.
Percent yield = (6.30 g / 8.40 g) x 100 = 75.0%
Mass lost = theoretical yield minus actual yield = 8.40 g - 6.30 g = 2.10 g
Percent lost = (2.10 g / 8.40 g) x 100 = 25.0%, which also equals 100% minus 75.0%, since the yield and the loss have to add up to the full theoretical amount.
The fastest way to improve is deliberate practice with immediate feedback: attempt a problem fully before checking the answer, then compare your steps line by line against the solution rather than just the final number. Working through the strategy checklist on this page for every problem, even easy ones, builds the habit that carries over to harder multi-step and limiting-reagent questions.
Read the question twice before calculating anything: once to identify every given number and its unit, and once to identify exactly what the question asks for. Then check, in order, whether the equation needs balancing, whether a limiting reagent needs identifying, and whether any mass needs converting to moles. The formula itself is the last and easiest step.
The most common mistake is assuming the reactant listed first, or the one with the larger starting mass, is automatically the limiting reagent. Limiting reagent status depends on moles compared against the balanced equation's coefficients, not on which reactant sounds bigger. Converting both reactants to moles and comparing mole-to-coefficient ratios is the only reliable way to check.
Balance the equation first, always. Percent yield depends on a correct theoretical yield, and the theoretical yield depends on the mole ratio taken directly from the balanced equation's coefficients. An unbalanced equation gives an incorrect ratio, which produces an incorrect theoretical yield and an incorrect percent yield even if every later calculation step is done correctly.
A reagent is limiting if it runs out first and stops the reaction before the other reactant is used up. To find it, convert every reactant's mass to moles, divide each by its coefficient in the balanced equation, and compare the results. The reactant with the smallest divided value is the limiting reagent, regardless of its starting mass.
A percent yield above 100% means the measured product mass included something other than pure product, since a reaction cannot exceed its theoretical maximum. Check first for a calculation error, such as a missed limiting-reagent check, then consider lab causes like residual solvent, moisture, or unreacted starting material still clinging to the sample. Drying and reweighing the product usually resolves it.
Yes, when both actual and theoretical yield are already given in the same unit, percent yield is a single division: actual divided by theoretical, multiplied by 100. The shortcut disappears once a problem involves a mole ratio, a limiting reagent, or a mass-to-mole conversion, since those steps have to happen before the percent yield formula can be applied at all.
There is no fixed number that works for everyone, but working through problems until each difficulty level (simple percent yield, rearranged formula, limiting reagent, multi-step) can be completed without checking the solution first is a reasonable bar. The ten cards on this page cover that full range in one sitting, and the <a href="/worksheets/">printable worksheets</a> add fifteen more for extra repetition.
The <a href="/worksheets/">worksheets page</a> has fifteen tiered percent yield problems formatted for printing, with a plain answer key listed separately at the bottom of the page instead of a per-problem reveal. It is built for working on paper, while this page is built for checking yourself one problem at a time on screen.
The <a href="/examples/">worked examples page</a> covers eight complete percent yield problems with full prose explanations of the reasoning behind each step, not just the arithmetic. It is the better starting point if a concept still feels unclear, while this page's practice cards are built for testing whether that reasoning can already be applied independently.