Calculate percent yield, actual yield, or theoretical yield for a reaction, including mixed gram and mole comparisons with molar mass, step-by-step working.
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Interpretation
Percent yield compares what you actually isolated with the reaction’s theoretical maximum. If one value is a mass and the other is an amount in moles, the molar mass converts both yields before the ratio is calculated.
Result
Moderate recovery. This is common when purification or transfer losses matter.
Shortfall to theoretical: 7.5 g
Step-by-step working
Science — Chemistry
A percent yield calculator compares the product you actually isolated with the reaction’s theoretical maximum. It can solve percent yield directly or work backwards to recover the actual yield or theoretical yield when the other two values are already known, and it can use molar mass when one yield is entered in grams and the other is entered in moles.
Percent yield measures how much product you recovered relative to the amount predicted by stoichiometry. It is one of the standard ways chemistry students and lab workers judge reaction efficiency and recovery.
Low values often reflect incomplete reaction, side reactions, or transfer and purification losses. Values above 100% usually indicate residual solvent, impurities, or an incorrect theoretical-yield assumption.
The standard expression compares actual yield with theoretical yield and converts the ratio into a percentage. In practice, the calculator applies this percent yield calculator relationship to the user inputs, keeps the units and assumptions consistent, and then surfaces the supporting context needed to interpret the output responsibly.
Percent yield = (actual yield / theoretical yield) × 100
Compares the isolated product against the theoretical maximum from stoichiometry.
Actual yield = (percent yield / 100) × theoretical yield
Back-solves the isolated product amount when percent yield and theoretical yield are known.
Actual yield and theoretical yield must be compared on the same basis. If both values are in grams, compare grams. If both values are in moles, compare moles. If one value is a mass and the other is an amount of substance, convert through molar mass first.
For example, an actual yield of 3.50 g of glucose and a theoretical yield of 0.025 mol cannot be divided directly. Using glucose's molar mass of 180.16 g/mol, the theoretical yield is 0.025 × 180.16 = 4.504 g, so the percent yield is 3.50 / 4.504 × 100 = 77.7%.
mass = moles × molar mass
Converts a mole-based theoretical yield into grams before the percent-yield ratio is applied.
moles = mass / molar mass
Converts a mass-based actual yield into moles when the theoretical value is given as an amount of substance.
Percent loss is the unused part of the theoretical yield when the actual yield is lower than expected. It is often easier to interpret alongside percent yield because 85% yield corresponds to a 15% shortfall on the same comparison basis.
A result above 100% is not treated as a successful reaction by itself. It usually means the measured product includes something besides pure dry product, such as solvent, water, salts, filter-paper fibres, or another impurity. It can also point to an incorrect limiting-reactant or theoretical-yield setup.
Percent loss = 100% − percent yield
Shows the recovery gap when actual yield is below theoretical yield.
A low percent yield is a signal to review the chemistry workflow, not just the arithmetic. Common causes include incomplete conversion, side reactions, poor crystallisation, product left in glassware, losses during filtration or transfer, and weighing before the product is fully dry.
When the reaction is part of a homework or lab-report chain, check the stoichiometry step first. The theoretical yield depends on a balanced equation and the limiting reactant. If that setup is wrong, the percent-yield calculator will faithfully report a misleading percentage.
Suppose the theoretical yield is 50 g and the isolated product mass is 42.5 g. The percent yield is (42.5 / 50) × 100 = 85%.
That result indicates good recovery for a teaching-lab or bench-scale workflow, while still leaving room for expected purification and handling losses.
Frequently asked questions
Because the isolated material may contain solvent, water, impurities, filter-paper residue, or other non-product mass. A percent yield above 100% usually means the product was not fully dry or the theoretical-yield setup needs review.
Yes. Percent yield is a ratio, so the actual and theoretical values must refer to the same measurement basis. This page assumes you are comparing like with like.
No. High percent yield only shows the amount recovered relative to the theoretical maximum. Purity still requires separate analytical confirmation such as melting point, chromatography, spectroscopy, or another validated test.
Actual yield is the amount you actually isolated from the reaction. Theoretical yield is the maximum amount the balanced equation predicts you could make if the reaction went perfectly and nothing was lost.
Low yield can happen because the reaction did not go to completion, side products formed, or product was lost during transfer, filtration, or purification. The calculator measures recovery, but it cannot tell you which step caused the loss.
You only need molar mass when one yield is entered as a mass and the other is entered as moles. If actual yield and theoretical yield already use the same unit basis, the units cancel in the percent-yield ratio.
Yes. Actual and theoretical yields may be compared as moles, grams, or another matching quantity. The key rule is that both sides must describe the same product on the same basis before the ratio is taken.
No. The limiting reactant is the reactant that runs out first. The theoretical yield is the maximum product amount predicted from that limiting reactant and the balanced equation.
Percent loss is the shortfall from the theoretical maximum when actual yield is lower than theoretical yield. It is usually 100% minus percent yield, so an 85% yield corresponds to about 15% loss on the same comparison basis.
It depends on the reaction, scale, purification method, and product stability. A simple teaching-lab reaction may be expected to recover much more than a difficult multistep synthesis, so interpret the number against the procedure rather than a universal cutoff.
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