Theoretical Yield Calculator
Theoretical yield is the maximum mass of product a reaction can form, calculated by converting the limiting reagent's mass to moles, applying the mole ratio from the balanced equation, then converting moles of product back to mass. Enter the reagent mass, both molar masses, and the two coefficients above to get the theoretical yield in grams.
What Theoretical Yield Tells You
Theoretical yield is the largest possible mass of product a reaction can produce, assuming the limiting reagent converts completely and nothing is lost to side reactions, spills, or incomplete recovery. It is a calculated ceiling, not a measured result: you never weigh out theoretical yield on a balance, you compute it from the balanced equation before or after running the experiment.
The mass you actually collect after the reaction and any purification is the actual yield, and it will always be equal to or less than theoretical yield under normal lab conditions. Dividing actual by theoretical and multiplying by 100 gives percent yield, which the percent yield formula on the homepage covers in full.
How the Calculator Converts Mass to Mass
The theoretical yield calculator above runs the same three-step conversion every stoichiometry problem uses. First, the mass of the limiting reagent (entered as mg, g, or kg) is divided by its molar mass to get moles of reagent, exactly as described on the molar mass page. Second, moles of reagent are converted to moles of product using the mole ratio: multiply by the product's coefficient and divide by the reagent's coefficient, the same ratio math explained on the mole ratio page. Third, moles of product are multiplied by the product's molar mass to get the theoretical yield in grams.
This mass-to-mole-to-mass sequence is the general method behind every mass-to-mass stoichiometry problem, laid out step by step on the stoichiometry page. If more than one reactant is listed in your reaction, you must first work out which one is the limiting reagent, since only its mass belongs in the calculator; the excess reactant's mass is irrelevant to theoretical yield.
Worked Example: Aspirin from Salicylic Acid
Aspirin (acetylsalicylic acid) forms from salicylic acid and acetic anhydride in a 1:1 reaction: salicylic acid + acetic anhydride -> acetylsalicylic acid + acetic acid. Salicylic acid has a molar mass of 138.12 g/mol and acetylsalicylic acid has a molar mass of 180.16 g/mol.
Starting from 5.00 g of salicylic acid as the limiting reagent: 5.00 g / 138.12 g/mol = 0.03620 mol salicylic acid. Because the coefficients are 1 and 1, moles of aspirin also equal 0.03620 mol. Converting to mass: 0.03620 mol x 180.16 g/mol = 6.52 g. The theoretical yield of aspirin is 6.52 g.
Worked Example: Water from Hydrogen and Oxygen
The synthesis 2 H2 + O2 -> 2 H2O has a 2:2 mole ratio between hydrogen gas and water, which simplifies to 1:1. Hydrogen gas has a molar mass of 2.016 g/mol and water has a molar mass of 18.015 g/mol.
Starting from 4.00 g of H2 as the limiting reagent: 4.00 g / 2.016 g/mol = 1.984 mol H2. Applying the 2:2 ratio gives 1.984 mol H2O. Converting to mass: 1.984 mol x 18.015 g/mol = 35.7 g. The theoretical yield of water is 35.7 g, even though the starting hydrogen mass was only 4.00 g, because each mole of the much lighter H2 pairs with oxygen to form a heavier mole of H2O.
Worked Example: Carbon Dioxide from Calcium Carbonate
The reaction CaCO3 + 2 HCl -> CaCl2 + H2O + CO2 has a 1:1 mole ratio between calcium carbonate and carbon dioxide. Calcium carbonate has a molar mass of 100.09 g/mol and carbon dioxide has a molar mass of 44.01 g/mol.
Starting from 10.0 g of CaCO3 as the limiting reagent: 10.0 g / 100.09 g/mol = 0.0999 mol CaCO3. With a 1:1 ratio, moles of CO2 also equal 0.0999 mol. Converting to mass: 0.0999 mol x 44.01 g/mol = 4.40 g. The theoretical yield of carbon dioxide is 4.40 g, even though HCl is written in the equation with a coefficient of 2, because HCl here is the reactant present in excess, not the one used to size the product.
Molar Masses Used in These Examples
Each worked example above depends on molar mass values pulled from the periodic table or computed from a molecular formula, the same skill covered on the molar mass calculator page. The table below collects the values used so you can check your own setup against them.
| Substance | Formula | Molar Mass (g/mol) |
|---|---|---|
| Salicylic acid | C7H6O3 | 138.12 |
| Acetylsalicylic acid (aspirin) | C9H8O4 | 180.16 |
| Hydrogen gas | H2 | 2.016 |
| Oxygen gas | O2 | 32.00 |
| Water | H2O | 18.015 |
| Calcium carbonate | CaCO3 | 100.09 |
| Hydrochloric acid | HCl | 36.46 |
| Carbon dioxide | CO2 | 44.01 |
Reading the Formula and Fixing Common Setup Errors
The theoretical yield calculation above is one arrangement of the general percent yield relationship; the full formula and its three algebraic rearrangements (solving for theoretical yield, actual yield, or percent yield) are laid out on the percent yield formula page. The setup errors that most often produce a wrong theoretical yield are using the excess reagent's mass instead of the limiting reagent's, swapping the two coefficients, or entering the product's molar mass in the limiting reagent field.
Because coefficients come directly from the balanced equation, an unbalanced equation will always produce an incorrect mole ratio and therefore an incorrect theoretical yield, regardless of how carefully the rest of the arithmetic is done.
Frequently asked questions
How do you calculate theoretical yield?
Theoretical yield is calculated by converting the mass of the limiting reagent to moles using its molar mass, then applying the mole ratio from the balanced equation to find moles of product, then converting that mole amount back to mass using the product's molar mass. This mass-to-mole-to-mass sequence is exactly what the calculator above performs when you enter the reagent mass, both molar masses, and the two equation coefficients.
Can theoretical yield be less than actual yield?
No, theoretical yield cannot be lower than a correctly measured actual yield under normal reaction conditions, because theoretical yield is the fixed maximum set by stoichiometry. If your actual yield calculation comes out higher than the theoretical yield, the product is likely wet, impure, or was weighed with contaminants included, not that the reaction outperformed its stoichiometric limit. See the actual yield page for how that figure is measured.
Do you use the limiting or excess reagent to calculate theoretical yield?
Always use the limiting reagent, the reactant that runs out first and therefore caps how much product can form. Using the reagent present in excess will overstate theoretical yield because it ignores the reagent that actually restricts the reaction. If you have not identified which reactant is limiting yet, use the limiting reactant calculator before entering values here.
Does theoretical yield have to be entered or reported in grams?
No, grams are just the default display unit; the calculator lets you enter the limiting reagent's mass in milligrams, grams, or kilograms and reports the theoretical yield in grams regardless. Internally the math runs through moles, so any consistent mass unit works as long as you convert the final answer to whatever unit your lab report requires.
How is theoretical yield different from actual yield?
Theoretical yield is a calculated number derived entirely from stoichiometry before you ever run the reaction, while actual yield is the mass you physically weigh out of the product after the experiment and any purification. Percent yield compares the two by dividing actual by theoretical and multiplying by 100. Full details on measuring and reporting actual yield are on /actual/.
What if my chemical equation isn't balanced yet?
Balance it first; the coefficients you enter into the calculator only give correct mole ratios once the equation is balanced, and an unbalanced equation will produce a theoretical yield that does not match reality. Count atoms of each element on both sides and adjust coefficients until they match, then read the reagent and product coefficients straight from the balanced equation. See the stoichiometry page for the full mass-to-mass method.
Can theoretical yield be zero?
Yes, if the limiting reagent's mass or mole count entered is zero, or if the reagent does not participate in the reaction shown, the calculated theoretical yield is zero. In practice this usually signals a data-entry mistake, such as leaving the mass field blank or pairing the wrong molar mass with the wrong coefficient.
Why is my theoretical yield larger than the mass of my starting reagent?
This happens when the product's molar mass, multiplied by its coefficient, is larger than the reagent's molar mass multiplied by its coefficient, so each mole of reagent converts into a heavier mole of product. For example, 5.00 g of salicylic acid (138.12 g/mol) yields 6.52 g of aspirin (180.16 g/mol) at a 1:1 ratio, since 180.16 exceeds 138.12 per mole even though mole count stays the same.
How many decimal places or significant figures should I use for theoretical yield?
Match the significant figures of your least precise input, which is usually the measured mass of the limiting reagent. If that mass was weighed to three significant figures, such as 5.00 g, report the theoretical yield to three significant figures as well, typically two decimal places for masses in the 1-99 gram range.
What happens if I have two reagents that both look limiting?
Compare how many moles of each reactant you actually have to how many moles the balanced equation requires, using the mole ratio between them; whichever reactant falls short first is the true limiting reagent. Two reagents can look limiting by mass alone but not by moles, since molar masses differ. Use the limiting reactant calculator to run this comparison directly, then the mole ratio calculator to check the required ratio.