Theoretical Yield Calculator

What theoretical yield means

Theoretical yield is the mass of product a balanced equation allows when the limiting reactant is used up and nothing is lost along the way. OpenStax Chemistry 2e defines it as the amount of product that may be produced under specified conditions, calculated from the stoichiometry of a balanced equation (OpenStax Chemistry 2e, section 4.4, 2019 edition, current as of 2026). It is a ceiling, not a forecast of what you will scrape out of the flask.

Three numbers describe one reaction. The theoretical yield is the ceiling. The actual yield is what the balance reads after the product is filtered and dried. Percent yield puts them in one figure: percent yield = (actual / theoretical) x 100. If the equation allows 0.5072 grams of copper and you collect 0.392 grams, the percent yield is (0.392 / 0.5072) x 100 = 77.3 percent.

The IUPAC Gold Book keeps the vocabulary tight: chemical yield is the fraction of the amount of a compound obtained after a specified reaction or separation. A fraction of 0.773 is the same statement as 77.3 percent, and the decimal form is what lands in a lab notebook.

Theoretical yield earns its place when you plan rather than report. It tells you how much product a batch can make, how much reagent to buy for a target mass, and whether a disappointing result came from the chemistry or from the workup.

Five steps, and the reagent that runs out first

Five steps carry any problem, and the fourth is the one people skip.

1. Balance the equation so every coefficient is known.
2. Convert each reactant mass to moles: moles = grams / molar mass.
3. Convert each reactant's moles into moles of the product: moles of product = moles of reactant x (product coefficient / reactant coefficient).
4. Compare the results. The smaller product amount is the theoretical yield in moles, and its reactant is the limiting reagent.
5. Multiply by the product's molar mass to get grams.

The comparison in step 4 is the step people skip, usually by assuming the smaller pile limits. Mass does not decide the race, because a light molecule packs more moles into the same grams. Two grams of hydrogen holds about 0.99 moles, while two grams of zinc holds about 0.03 moles, a difference of more than 30 times. Moles and coefficients decide, never the size of the pile.

A shortcut gives the same answer: divide each reactant's moles by its own coefficient and compare the quotients. The smallest quotient belongs to the limiting reagent, because it runs out first relative to what the equation asks for. Both routes compare the same ratio, which is why they never disagree.

One edge case: if a reactant appears on both sides of the equation, count only the amount consumed as a reactant. The calculator above treats each entry as a one way input, which covers most textbook problems and most bench recipes.

Worked example: copper sulfate and zinc

Zinc metal dropped into copper sulfate solution displaces the copper: CuSO4 + Zn gives Cu + ZnSO4. Every coefficient is 1, so one mole of copper sulfate makes one mole of copper.

Dissolve 1.274 grams of copper sulfate and add 2.00 grams of zinc. Copper sulfate has a molar mass of 159.62 grams per mole, so the solution holds 1.274 / 159.62 = 0.007981 moles. Zinc at 65.38 grams per mole is 2.00 / 65.38 = 0.03059 moles.

Run both paths. The copper sulfate can make 0.007981 x (1/1) = 0.007981 moles of copper, which is 0.007981 x 63.55 = 0.5072 grams. The zinc can make 0.03059 moles, which is 1.944 grams. Copper sulfate gives the smaller product mass, so it is the limiting reagent and 0.5072 grams is the theoretical yield.

The student filters, dries and weighs 0.392 grams of copper. Percent yield is (0.392 / 0.5072) x 100 = 77.3 percent, the same figure OpenStax works through in section 4.4. Zinc is the excess reagent: 0.007981 moles of it reacted, which is 0.522 grams, leaving 1.478 grams in the beaker.

Note what the 2.00 grams of zinc did not do. It did not raise the yield, and adding more would not raise it either, because the copper sulfate was already spent.

Worked example: ammonia, where the smaller pile wins

Nitrogen and hydrogen combine as N2 + 3 H2 gives 2 NH3. The coefficients are 1, 3 and 2, so the mole ratios to ammonia are 2/1 from nitrogen and 2/3 from hydrogen.

Feed 10.0 grams of nitrogen (28.02 grams per mole) and 2.00 grams of hydrogen (2.016 grams per mole). Nitrogen is 10.0 / 28.02 = 0.3569 moles and can make 0.3569 x 2 = 0.7138 moles of ammonia, which is 12.16 grams. Hydrogen is 2.00 / 2.016 = 0.9921 moles and can make 0.9921 x (2/3) = 0.6614 moles, which is 11.26 grams.

Hydrogen limits, even though it is the lighter substance and the smaller pile. The theoretical yield is 11.26 grams of ammonia. A run that collects 9.50 grams reports (9.50 / 11.26) x 100 = 84.3 percent. The nitrogen that reacted is 0.6614 / 2 = 0.3307 moles, which is 9.27 grams, so 0.73 grams of nitrogen is left over.

This example is the reason the calculator asks for both reactants. With one reactant entered it can only divide by a coefficient, and the question of which reagent limits never comes up.

Leftover reagent: the cost you can see

Excess reagent is not free, and the arithmetic is short. Convert the theoretical product moles back through the ratio: moles of excess consumed = product moles x (excess coefficient / product coefficient). Subtract that from the starting moles, multiply by the molar mass, and the leftover grams appear.

In the copper example, 0.007981 moles of copper means 0.007981 moles of zinc reacted, which is 0.522 grams. Starting from 2.00 grams, 1.478 grams of zinc remains. Scale the batch to 100 grams of copper sulfate and the ratios hold: the multiplier is 100 / 1.274 = 78.5, so the zinc charge becomes 157 grams and about 116 grams of it comes back out.

At hobby prices that leftover is a few dollars. At production scale it is the difference between a process that runs and one that does not, which is why excess is usually set deliberately at 5 to 20 percent rather than by pouring from a scoop. The calculator reports the leftover so the next batch can be charged closer to the stoichiometric amount.

Where the molar masses come from

Molar mass is the sum of standard atomic weights, in grams per mole. Copper sulfate is copper 63.546, sulfur 32.06 and four oxygens at 15.999 each, which adds to 159.60. The OpenStax example uses 159.62, a difference in the last digit that comes from the atomic weight table behind it, and either value moves the theoretical yield in the fourth figure only.

Use one table for the whole calculation. Mixing a periodic table value of 63.55 for copper with 63.546 from a data sheet creates differences that look like experimental error. The NIST Chemistry WebBook is a dependable source for individual compounds, and the table printed inside a textbook is fine as long as it is the only one you use.

Hydrates are the common trap. Copper sulfate pentahydrate, CuSO4 with five waters attached, has a molar mass of about 249.7 grams per mole, not 159.62, because the five waters at 18.015 each add 90.08. Weighing the blue crystals and dividing by 159.62 overstates the moles by 56 percent and inflates the theoretical yield to match.

Round at the end, not in the middle. Carry at least four significant figures through the mole step, because the percent yield divides two numbers of similar size and small rounding in the denominator shows up in the answer.

Checks before you report the number

Six checks catch most wrong answers before they reach a report.

• Same substance on both sides of the division. Actual grams of copper over theoretical grams of copper is a yield. Actual copper over theoretical copper oxide is a mistake.
• Units matched. Grams over grams, or moles over moles, never one of each.
• Coefficients read from the balanced equation, not from the subscripts in a formula.
• Limiting reagent chosen by product moles, not by the smaller pile.
• Hydrate water counted in the molar mass.
• A result above 100 percent explained before it is reported, since wet or impure product is the usual cause.

What a percent yield usually means, before anything is blamed on technique:

A yield in the 70 to 90 percent range is normal for a teaching lab, where transfers, filtration and solubility each take a cut. Below 50 percent usually points at an incomplete reaction, a loss during workup, or product that stayed dissolved in the mother liquor. Above 100 percent points at the balance, not the chemistry.

Run the numbers once with the calculator above, then check the limiting reagent by hand with the quotient shortcut from the second section. When both routes agree, the theoretical yield is as solid as the molar masses that went into it.