What theoretical yield means, and what it assumes
Theoretical yield is the mass of product you would isolate if every molecule of the limiting reactant were converted to the product you want, and none of it were lost anywhere. It is a ceiling, not a prediction. No real preparation reaches it.
It matters because it is the benchmark. Weighing 4.2 g of product tells you nothing on its own; 4.2 g against a theoretical 5.0 g is an 84% yield, which for a multi-step organic synthesis is a good result and for a simple precipitation is a poor one. The percent yield calculator makes that comparison, and this page supplies the denominator.
Three assumptions are built in. The reaction goes to completion, which equilibria do not. Only the reaction you wrote occurs, which side reactions violate. And nothing is lost in filtration, transfer, recrystallisation or drying, which is never true. Every one of those pushes the real yield below this number, which is why a measured yield above 100% always means an error — usually a product that is still wet with solvent.
The three conversions, and the one that trips people
Mass to moles. Divide the mass of limiting reactant by its molar mass. If the reagent is not pure, multiply by the purity first — 16.043 g of a solid assayed at 90% supplies only 14.439 g of actual reactant.
Moles to moles. Multiply by the coefficient ratio, product over reactant. This is the only step that involves the reaction at all, and it is the one people invert. The rule that resolves it: the ratio is written the way you want the units to cancel, so if the equation says one nitrogen makes two ammonias, you multiply moles of nitrogen by 2/1 to get moles of ammonia. Multiplying by 1/2 would tell you that a mole of nitrogen makes half a mole of ammonia, which is out by a factor of four.
Moles to mass. Multiply by the product's molar mass. Nothing subtle here beyond getting the product formula right, hydrates included.
Two warnings about inputs. The equation must be balanced first — the balancer does that — because unbalanced coefficients give a wrong ratio and therefore a wrong ceiling. And the reactant you enter must genuinely be the limiting one; if you are not certain, the limiting reagent calculator settles it in one step.
Worked example: carbon dioxide from burning methane
Burn 16.043 g of methane in excess oxygen: CH₄ + 2 O₂ → CO₂ + 2 H₂O. How much carbon dioxide can form?
- Molar mass of methane. 12.011 + 4 × 1.008 = 16.043 g/mol.
- Moles of methane. 16.043 ÷ 16.043 = 1.000 mol.
- Mole ratio. The equation gives 1 CO₂ per 1 CH₄, so the ratio is 1/1 and moles of carbon dioxide = 1.000 mol.
- Molar mass of carbon dioxide. 12.011 + 2 × 15.999 = 44.009 g/mol.
- Theoretical yield. 1.000 × 44.009 = 44.009 g.
Notice that 16 g of fuel produces 44 g of carbon dioxide — 2.74 g per gram burned. That is not a violation of mass conservation: the extra mass is oxygen taken from the air, 64.0 g of it, of which 32.0 g leaves as the two waters. This ratio is why a vehicle's carbon dioxide output vastly exceeds the mass of fuel in its tank.
Now add purity. If the methane stream were only 90% methane, the effective mass would be 14.439 g, giving 0.9000 mol and 39.61 g of carbon dioxide. Purity scales the answer linearly, which is why an uncorrected assay quietly inflates every theoretical yield derived from it.
Judging the number and what to do next
Sanity-check the ratio of product mass to reactant mass. If the product is larger than the reactant, atoms from another reactant have been incorporated — combustion, oxidation and hydration all do this and the result is normal. If the product is smaller, something has been eliminated, which is what a condensation or a decomposition does. A product ten times the mass of the reactant, on the other hand, usually means the coefficient ratio is inverted.
Then compare against what you actually isolate. Rough rules of thumb by reaction class: a clean inorganic precipitation reaches the low nineties, a straightforward organic transformation the seventies to nineties, and a difficult coupling or a multi-step sequence considerably less. Treat those as expectations rather than standards — the honest benchmark is the yield reported in the procedure you are following. The compounding is the important part — five steps at 80% each leave 0.80⁵ = 32.8% overall.
Finally, remember that the theoretical yield is per product. A reaction producing both carbon dioxide and water has a theoretical yield for each, computed from the same limiting reactant with different coefficients and molar masses. Run this calculator once per product you care about.
Theoretical yields from one mole of limiting reactant
| Balanced equation | Limiting reactant | Mole ratio | Product | Yield from 1.000 mol |
|---|---|---|---|---|
| CH₄ + 2 O₂ → CO₂ + 2 H₂O | CH₄ (16.043 g) | 1 : 1 | CO₂ | 44.009 g |
| CH₄ + 2 O₂ → CO₂ + 2 H₂O | CH₄ (16.043 g) | 2 : 1 | H₂O | 36.030 g |
| N₂ + 3 H₂ → 2 NH₃ | N₂ (28.014 g) | 2 : 1 | NH₃ | 34.062 g |
| 2 H₂ + O₂ → 2 H₂O | H₂ (2.016 g) | 2 : 2 | H₂O | 18.015 g |
| CaCO₃ → CaO + CO₂ | CaCO₃ (100.086 g) | 1 : 1 | CaO | 56.077 g |
| 4 Fe + 3 O₂ → 2 Fe₂O₃ | Fe (55.845 g) | 2 : 4 | Fe₂O₃ | 79.844 g |
| C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O | C₃H₈ (44.097 g) | 3 : 1 | CO₂ | 132.027 g |
| 2 Al + 3 Cl₂ → 2 AlCl₃ | Al (26.982 g) | 2 : 2 | AlCl₃ | 133.332 g |
Row five is the lime kiln reaction: calcining a tonne of limestone yields 560 kg of quicklime and releases 440 kg of carbon dioxide.
What makes a theoretical yield wrong
- An unbalanced equation. The coefficient ratio is the whole calculation. Balance first, every time.
- Inverting the mole ratio. Product coefficient over reactant coefficient, not the other way round. Check by asking whether the answer moves in the direction the equation implies.
- Using the wrong reactant. Theoretical yield is set by the limiting reactant only. Basing it on a reactant present in excess gives a ceiling the reaction can never approach.
- Ignoring purity and water content. A hydrated or impure reagent supplies fewer moles than its mass suggests, and the error is linear in the yield.
- Forgetting that this is a ceiling. Comparing an isolated mass to a theoretical yield and getting more than 100% means the product is wet, impure, or weighed wrongly.
- Applying it to an equilibrium as if it were a prediction. A reaction with a small equilibrium constant may stop far short of consuming the limiting reactant.
Atom economy, conversion and selectivity
Theoretical yield answers "how much product can I get from this reactant". Three related measures answer questions it cannot.
Atom economy is the molar mass of the desired product divided by the sum of the molar masses of all reactants, expressed as a percentage. It measures how much of what you put in ends up in what you want, and unlike yield it cannot be improved by better technique — it is fixed by the reaction you chose. A rearrangement has 100% atom economy; a substitution that discards a large leaving group may have well under 50% even at quantitative yield.
Conversion is the fraction of the limiting reactant that reacted at all. Selectivity is the fraction of what reacted that became the product you wanted. Yield is the product of the two, which is why a process can have excellent selectivity and still deliver little product if conversion is low. Industrial plants routinely run at deliberately low conversion with a recycle, because that maximises selectivity.
Once you have both the theoretical yield and a weighed product, the percent yield calculator completes the picture. If you are still working out which reactant limits, start at the limiting reagent calculator; if you need the mass of a different species entirely, use the mole-ratio calculator.