To calculate theoretical yield, use the limiting reactant, mole ratios, and molar masses to predict the maximum amount of product.
The phrase how to calculate theoretical yield shows up in almost every set of stoichiometry notes, yet many students still feel unsure when numbers hit the page. The good news is that theoretical yield is just a tidy name for the biggest amount of product you could make if everything in the reaction worked perfectly.
In this guide you will see what theoretical yield means, how it links to limiting reactants and percent yield, and how to move from a word problem to a clean numerical answer. You will also see a full worked problem and a compact checklist you can use in lab or during exams.
What Theoretical Yield Means In A Reaction
In chemistry, theoretical yield is the amount of product predicted by the balanced equation when the limiting reactant reacts completely. Textbooks describe it as the maximum possible mass or moles of product for a given set of starting amounts, assuming the reaction goes to completion and nothing is lost on glassware or during transfers.
Many open textbooks describe theoretical yield in this way and contrast it with actual yield, the mass you measure in the lab, and percent yield, which compares the two values as a percentage. The LibreTexts section on theoretical yield gives the same core message: equations set the ideal target, experiments deliver something smaller.
| Term | Simple Meaning | Role In Yield Problems |
|---|---|---|
| Theoretical yield | Predicted amount of product from the equation | Target value you calculate from the limiting reactant |
| Actual yield | Measured amount of product from the experiment | Used with theoretical yield to find percent yield |
| Percent yield | Actual yield divided by theoretical yield times one hundred | Shows how efficient the reaction was in practice |
| Balanced equation | Chemical equation with equal atoms on both sides | Supplies the mole ratios between reactants and products |
| Molar mass | Mass of one mole of a substance | Converts between grams and moles |
| Limiting reactant | Reactant that runs out first | Sets the maximum amount of product that can form |
| Excess reactant | Reactant that is left over | Does not control the size of theoretical yield |
| Stoichiometric ratio | Mole ratio from the balanced equation | Links moles of limiting reactant to moles of product |
How To Calculate Theoretical Yield Step By Step
When a question asks how to calculate theoretical yield, the safest plan is to move through the same short sequence every time. The numbers may change, but the route stays the same: balance, convert, compare, predict, convert back.
Step 1: Write And Balance The Equation
Every yield calculation starts with a balanced equation. Coefficients in that equation show the mole ratios that tie reactants to products. Without those coefficients you cannot reliably match up moles of starting materials with moles of product.
Step 2: Convert Each Given Amount To Moles
Word problems usually give masses, volumes, or sometimes moles. Turn every reactant amount into moles before you compare anything. Use molar mass for solids and liquids, and for gases use either molar mass or, when conditions match, the molar volume from the gas law chapter.
Step 3: Find The Limiting Reactant
Use the balanced equation and your mole values to decide which reactant limits the reaction. One method promoted in many tutorials is this: using each reactant in turn, compute the moles of a chosen product, then pick the smallest predicted amount. The reactant that yields the smaller product amount is the limiting reactant. Open educational resources on limiting reagents, such as the Open Oregon section on reaction yields, outline the same logic.
Step 4: Use The Mole Ratio To Get Product Moles
Once you know the limiting reactant, ignore the others for yield purposes. Starting from the moles of limiting reactant, apply the mole ratio that links that reactant to the desired product in the balanced equation. This gives the theoretical moles of product.
Step 5: Convert Product Moles To Grams
Most questions ask for theoretical yield in grams. Multiply the moles of product by the molar mass of that product. Label units carefully at each step so that you can follow the trail from grams of reactant to moles of reactant, then to moles of product, then to grams of product.
Worked Problem Using Masses Of Reactants
To see the steps in action, walk through a complete problem. Say hydrogen gas reacts with oxygen gas to form liquid water.
Step 1: Balanced Equation For Water Formation
The balanced equation is:
2 H2(g) + O2(g) → 2 H2O(l)
Assume you start with 4.00 g of H2 and 32.0 g of O2. You want the theoretical yield of water in grams.
Step 2: Convert Masses To Moles
First convert hydrogen to moles:
moles H2 = 4.00 g ÷ 2.02 g mol-1 ≈ 1.98 mol
Then convert oxygen to moles:
moles O2 = 32.0 g ÷ 32.00 g mol-1 = 1.00 mol
Step 3: Decide Which Reactant Limits The Reaction
The equation shows a 2:1 mole ratio between hydrogen and oxygen. That means the reaction needs two moles of H2 for every one mole of O2. You currently have about 1.98 mol of H2 and 1.00 mol of O2. To match the stoichiometric ratio you would need 2.00 mol of H2, so hydrogen is slightly short. Hydrogen is the limiting reactant, while oxygen is present in slight excess.
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Step 4: Use The Limiting Reactant To Find Product Moles
The balanced equation shows a 2:2 ratio between H2 and H2O, so the mole ratio is one to one. That means the theoretical moles of water equal the moles of hydrogen that react.
moles H2O (theoretical) ≈ 1.98 mol
Step 5: Convert Product Moles To Grams Of Water
Now convert moles of water to grams using the molar mass of H2O, about 18.02 g mol-1:
mass H2O (theoretical) ≈ 1.98 mol × 18.02 g mol-1 ≈ 35.7 g
This value, 35.7 g of water, is the theoretical yield for the stated starting masses. Any real experiment with the same setup would give an actual yield that is equal to or smaller than this prediction.
How To Calculate Theoretical Yield From Moles Directly
Sometimes a problem hands you moles instead of grams. In that case you can shorten the plan. After you find the limiting reactant using mole ratios, jump straight from moles of limiting reactant to moles of product, then stop unless the question asks for grams.
Sample Mole Based Scenario
Suppose 0.50 mol of A reacts with 0.30 mol of B to produce compound C according to the reaction
2 A + B → 3 C
A quick scan of the stoichiometric ratio shows that the reaction needs twice as many moles of A as B. The required ratio is 2:1, yet the starting ratio is 0.50:0.30, or about 1.67:1, so A is limiting. Use that mole value with the 2:3 ratio to find moles of C, then multiply by the molar mass of C if you need grams.
Common Mistakes When You Calculate Theoretical Yield
Yield questions share a group of frequent pitfalls. Running through them before you start a calculation can save marks and reduce rework at the lab bench.
Skipping The Balancing Step
Many errors start when the equation is not balanced. If coefficients are wrong, every mole ratio that follows will be off by the same factor. Take a moment to adjust coefficients so atoms match on each side before you touch a calculator.
Comparing Grams Instead Of Moles
Another regular source of confusion comes from trying to spot the limiting reactant by comparing masses directly. Different substances have different molar masses, so grams alone do not show how many particles you have. Always work in moles when you compare reactant amounts.
Forgetting To Label Units
Even strong problem solvers lose track of steps when units vanish from the page. Label every number with its unit and write each conversion as a fraction. That habit lets you see grams cancel to leave moles, then moles cancel to leave grams again, which gives a quick check on your setup.
Rounding Too Early In The Calculation
Rounding at every stage can nudge the final theoretical yield away from the value the instructor expects. Carry an extra digit or two during your working steps, then round once at the end to the requested number of digits.
| Reaction | Given Data | Theoretical Yield |
|---|---|---|
| 2 H2 + O2 → 2 H2O | 4.00 g H2, excess O2 | About 35.6 g H2O |
| N2 + 3 H2 → 2 NH3 | 5.00 mol N2, 12.0 mol H2 | 10.0 mol NH3 |
| 2 Al + 3 Cl2 → 2 AlCl3 | 10.0 g Al, excess Cl2 | About 49.4 g AlCl3 |
| CaCO3 → CaO + CO2 | 15.0 g CaCO3 | About 8.41 g CaO |
Quick Checklist For Yield Questions
Before you step into lab or open an online homework set, scan this short list tied to how to calculate theoretical yield:
- Write the balanced equation and confirm atom counts on both sides.
- Turn every given reactant amount into moles.
- Use mole ratios to find which reactant limits the reaction.
- From the limiting reactant, compute theoretical moles of the desired product.
- Convert those moles to grams if the question asks for a mass.
- If you have an experimental mass, combine it with theoretical yield to find percent yield.
- Check that your answer has the correct unit and a sensible number of digits.