Theoretical Yield Calculator

The best case a reaction can do

Theoretical yield answers a hypothetical: if every molecule of the limiting reagent reacted, none went down a side path, and not a milligram was lost in the flask or the filter, how much product would you have? It’s not a prediction of what you’ll get — that’s always less — but the benchmark that makes the actual result meaningful. A 4 g product is impressive if theory allowed 4.2 g and disappointing if it allowed 12 g.

The calculation is a short chain of unit conversions:

The first and last arrows use molar masses; the middle one uses the mole ratio from the balanced equation.

The theoretical yield formula

is the limiting reagent’s mass, and are the molar masses, and is the mole ratio (product coefficient ÷ limiting-reagent coefficient) from the balanced equation.

Worked example

The calculator’s defaults are 10 g of a limiting reagent with molar mass 40 g/mol, a product molar mass of 100 g/mol, and a 1:1 mole ratio.

Moles of reactant:

Moles of product (ratio 1):

Mass of product:

How the mole ratio changes the result

Same 10 g reactant (40 g/mol) and 100 g/mol product, different balanced-equation ratios:

Getting this ratio wrong is the most common error in the whole calculation.

Where theoretical yield is used

• Before a reaction, to estimate how much product to expect and whether it’s worth the effort.
• After a reaction, as the denominator in percent yield.
• Scale-up planning, to work out how much limiting reagent is needed for a target amount of product.
• Comparing routes — two syntheses to the same target, judged partly on their theoretical (and then actual) yields.

Common mistakes

• Using an excess reagent instead of the limiting one — always identify the limiting reagent first.
• Inverting the mole ratio (reactant:product instead of product:reactant).
• Swapping the two molar masses.
• Forgetting to balance the equation before reading coefficients off it.

Feed the result into the percent yield calculator once you have an actual mass. For the full multi-reactant path, see the stoichiometry calculator, and to compute a molar mass, the molar mass calculator.

Finding the limiting reagent: a worked case

For ammonia synthesis, , start with 28.0 g of N₂ and 9.0 g of H₂.

• Moles of N₂: mol.
• Moles of H₂: mol.
• Reacting all the N₂ would need mol of H₂. There are 4.46 mol, so H₂ is in excess and N₂ is limiting.

Now run the calculator’s chain on the limiting reagent, with a mole ratio of 2 (two NH₃ per N₂):

As a check, running the same chain on H₂ gives g. The smaller answer always wins, because the reaction stops when the limiting reagent runs out.

Theoretical yield in moles

Stop one step early. In the example above, the theoretical yield is mol of NH₃. Moles are the better unit when the product goes straight into another reaction or when you compare yields across products of different molar mass.