Theoretical Yield Calculator – Find the Limiting Reagent, Then the Real Maximum

See the limiting reagent at a glance

Each bar is the amount of a reactant divided by its coefficient — the number of times its share of the reaction can run. The shortest bar runs out first, and it sets the whole yield.

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How it works, in 3 steps

Worked example, step by step

Deep dive

Reading the result table, with the default water example

The result on this page answers two questions in order: which reactant runs out first, and how much product that reactant can make at most. The default example is the water reaction, 2H₂ + O₂ → 2H₂O, starting from 10 g of hydrogen and 100 g of oxygen, and the page shows a theoretical yield of 89.36 g of water with hydrogen as the limiting reagent. The table underneath the headline is the proof, and it is worth reading row by row rather than trusting the headline alone. Hydrogen converts to 4.9603 mol, and dividing by its coefficient of 2 gives 2.4802 reaction runs. Oxygen converts to 3.1251 mol, and dividing by its coefficient of 1 gives 3.1251 runs. The smaller number of runs wins, in the sense that it runs out first, so hydrogen limits. The yield then follows from the limiting runs: 2.4802 runs times 2 mol of water per run gives 4.9603 mol of water, and 4.9603 mol times the molar mass of water, 18.015 g/mol, gives 89.36 g. Every step is shown on the page, and every step can be checked with a hand calculator from the molar masses the page states.

The bars display is the same comparison drawn visually. Each bar is the moles of a reactant divided by its coefficient, which is the number of times that reactant can cover its share of the reaction. The shortest bar runs out first and sets the whole yield. That is why the page insists on division rather than on comparing masses directly: grams mislead, because atoms have different weights. Ten grams of hydrogen is about 5 mol, while ten grams of oxygen is only about 0.31 mol, yet the water equation needs twice as much hydrogen as oxygen. Only after converting to moles and dividing by the coefficient can the two reactants be compared fairly. A student who picks the limiting reagent by smallest mass will be wrong whenever the molar masses differ enough to reverse the order, which is most of the time. The division step is the whole method, and everything else is arithmetic built on it.

The leftover row in the table is just as important as the yield, because it stops a common misunderstanding: the excess reactant is not used up. In the default example, only 79.36 g of the 100 g of oxygen actually reacts with the 10 g of hydrogen, leaving 20.64 g unused. That leftover is real material. In a lab it is recovered, not thrown away by the arithmetic, and in a costing exercise it is material you already paid for that did not end up in the product. A yield calculation that ignores the leftover understates what you started with and overstates how efficient the reaction was. Read the leftover row every time, especially when you are comparing two ways of running the same reaction, because the option with the higher yield is not automatically the option with the least wasted starting material.

Four probed cases, and what each one teaches

Every figure in this table was produced by this page’s own calculator and is quoted exactly as the page displays it. The cases are chosen so that each one changes one thing at a time.

The second case is the most instructive. Doubling only the hydrogen, from 10 g to 20 g, does not double the yield. It rises from 89.36 g to 112.60 g, and the limiting reagent flips from hydrogen to oxygen. The page shows why: with 20 g of hydrogen, hydrogen covers 4.9603 runs, but oxygen still covers only 3.1252 runs, so oxygen now runs out first, and 3.1252 runs times 2 mol of water per run, scaled by the molar mass, gives 112.60 g. The leftover flips too: hydrogen now has 7.40 g left over, where oxygen previously had 20.64 g left over. The lesson generalises to every limiting-reagent problem: adding more of the limiting reactant helps only until a different reactant becomes limiting, and after that point extra material of the first reactant is simply leftover. If you are costing a reaction, the cheapest improvement is usually to add more of whichever reactant currently limits, and the point where that stops helping is exactly where the limiting reagent changes.

The limestone case shows the single-reactant situation. Heating limestone follows CaCO₃ → CaO + CO₂, and with only one reactant there is nothing else that can run out, so limestone is the limiting reagent by definition. From 100 g of CaCO₃ the page shows 0.9991 mol, and the product coefficient ratio of 1 gives 0.9991 mol of CaO, which at its molar mass is 56.03 g. The page’s own worked text also gives the companion figure for carbon dioxide, 43.97 g, from the same 0.9991 mol. The percent-yield case closes the loop between theory and lab: collecting 70 g of water against a theoretical 89.36 g gives a percent yield of 78.33%, shown by the page as 70.00 g divided by 89.36 g. A percent yield below 100% is normal, because real reactions are rarely complete, side reactions consume reactants, and product is lost in filtering, washing, and transfer. A percent yield above 100%, by contrast, means a measurement error or an impure product, not a reaction that beat its own ceiling.

Generic mode: where the method is yours to get right

The built-in reactions are safe by construction: water, limestone, and ammonia use fixed, balanced equations and IUPAC CIAAW 2021 conventional molar masses, so you only type the masses you start with. Generic mode is different, and the page says so openly: you type moles and coefficients yourself, and the equation you describe must already be balanced, or the answer will be wrong. The calculator cannot check your balancing for you, because it only sees the numbers you typed. If you enter coefficients from an unbalanced equation, the division step will still run, the limiting reagent will still be named, and a yield will still be shown, but the yield will describe a reaction that does not exist. The discipline is therefore to balance first, on paper or with the balancing tool linked from this page, and only then to enter the coefficients. Treat any generic-mode answer as provisional until you have re-read each coefficient against the balanced equation.

There is a second generic-mode trap, subtler than balancing: the units of what you type. In generic mode the reactant fields are moles, not grams. If you type a mass in grams into a moles field, the division step will treat grams as moles and the answer will be wrong by the ratio of the molar masses, which can be a factor of ten or more. The built-in modes hide this trap by doing the grams-to-moles conversion for you, but generic mode hands it to you. The safe workflow is to convert each mass to moles first, using the molar mass of that reactant, write the moles down, and only then enter them. The same care applies to the product: the product coefficient and the product molar mass must belong to the same product, in the same balanced equation, or the final scaling step multiplies by the wrong mass. A yield that is wrong by a clean factor, such as exactly the ratio of two molar masses, is almost always a units or product-identity error rather than a chemistry error, and it is worth checking those two things before doubting the method.

Percent yield deserves the same care. The actual yield field is optional, and the percent yield stays blank until you enter it, which is the correct behaviour: a percent yield without an actual result is meaningless, and the page does not invent one. When you do enter an actual yield, make sure it is the mass of the same product, dried and purified to the same standard as the theoretical figure assumes. Weighing a wet product, or a product still mixed with unreacted starting material, inflates the actual yield and can push it above 100%. That does not mean the reaction outperformed theory; it means the weighing does not measure what the calculation assumes. The same applies in reverse: product lost in transfer or left on filter paper lowers the actual yield without saying anything about how completely the reaction itself ran. Percent yield measures the whole process, from reaction to weighing, not the reaction alone, and a low figure is a prompt to look at every step, not just the chemistry.

What this page does not cover

Three omissions are stated here so that they are not mistaken for errors. First, this page covers two-reactant generic reactions and the single-reactant limestone case; it does not provide a three-reactant generic mode. A reaction with three reactants needs the same division step applied to all three, with the smallest moles-per-coefficient value limiting, but you cannot enter three reactants on this page, and you should not force a three-reactant problem into two fields by combining reactants. Second, this page works from grams and moles, not from solutions: it does not take concentrations and volumes. If your reactants are in solution, convert concentration times volume to moles first, or use the molarity tool linked from this page, and then enter moles in generic mode. Third, the theoretical yield is a ceiling for a reaction that goes to completion with no losses. It is not a prediction of what you will collect, and it says nothing about reaction rate, equilibrium position, or safety. A reaction with a high theoretical yield can still be slow, reversible, or hazardous, and none of those properties appears in this calculation.

A worked routine for every yield problem

Use the same routine whether you are in a classroom or a lab, because it catches the errors that a single headline figure hides. First, write the balanced equation and circle the coefficients; do not start typing until the equation balances. Second, convert every starting mass to moles and write the moles down with the reactant name attached, so that a mass is never mistaken for moles later. Third, divide each mole amount by its coefficient and compare the runs, not the masses or the moles; the smallest runs value is the limiting reagent. Fourth, scale from the limiting runs to the product: multiply by the product coefficient ratio, then by the product molar mass, and check the units at each step, moles to moles to grams. Fifth, read the leftover row and ask whether the excess makes sense: if the leftover is larger than what you started with, or negative, something was mistyped. Sixth, only when you have an actual collected mass, enter it and read the percent yield, and if that figure is above 100%, treat it as a measurement problem to investigate, not as a result to quote. Run through those six steps on the default example once, and the method becomes a habit rather than a formula to remember.

Scope note, stated plainly: molar masses on this page use IUPAC CIAAW 2021 conventional values, as the page itself states, and the built-in reactions are the standard balanced equations shown. Nothing on this page is a laboratory instruction, and no safety guidance is given or implied: handling real reactants, heating limestone, or collecting gases needs proper supervision, equipment, and safety assessment outside this page. The figures in the table above come from this page’s own calculator, and the discussion of generic mode, leftovers, and percent yield is an explanation of that calculator’s method, not a claim about any particular laboratory result.

Reading a yield backwards: from product wanted to reactant needed

Most yield problems run forwards, from reactants to product, but real planning often runs backwards: you know how much product you want, and you need to know how much reactant to start with. The default water example shows how the same arithmetic answers the backwards question, because every step in it is reversible. The page shows that 10 g of hydrogen, at 4.9603 mol, gives 4.9603 mol of water, which is 89.36 g. Run that chain in reverse: to make 89.36 g of water you need 4.9603 mol of water, which needs 4.9603 mol of hydrogen by the one-to-one ratio in the balanced equation, which is 10 g of hydrogen. The oxygen side follows the same logic: 4.9603 mol of water needs half as many moles of oxygen, because the equation uses one oxygen for every two waters, and the page’s own default shows that only 79.36 g of the 100 g of oxygen is actually consumed, with 20.64 g left over. Working backwards therefore tells you not just the limiting reactant you need, but also how much of the excess reactant will be left, which matters when you are ordering material or costing a batch.

The backwards routine also exposes a planning error that the forwards routine hides. Suppose you plan a batch from the product you want and you buy exactly the limiting reactant you calculated, with no margin. Any loss in handling, any incomplete reaction, and any side reaction then leaves you short of the product you wanted, because the theoretical yield is a ceiling, not a promised output. The page’s percent-yield discussion makes the same point from the other side: collecting 70 g against a theoretical 89.36 g is a percent yield of 78.33%, which is normal for a real process. Planning backwards from a target therefore needs a margin, and the size of that margin is a practical judgement about your process, not something this calculator computes. The calculator tells you the ceiling and the limiting step; how far below the ceiling you plan to land is a decision for the person running the reaction, informed by experience with that reaction’s losses.

Why a ceiling is more useful than a prediction

It is tempting to wish the calculator predicted what you will actually collect, but a ceiling is the more honest and more useful figure. A prediction would have to assume a reaction rate, an equilibrium position, side reactions, and handling losses, none of which can be known from masses and coefficients alone. A ceiling, by contrast, is exact given the inputs: no arrangement of the same starting materials, under the same balanced equation, can make more product than the limiting reagent allows. That exactness is what makes the ceiling a diagnostic tool. If you collect far below it, something in the process deserves investigation, whether that is incomplete reaction, a side reaction, or product lost in handling. If you collect above it, the measurement deserves investigation, because the ceiling cannot be beaten by a better technique, only by a weighing that does not measure what it claims to measure. Used that way, the theoretical yield is not a disappointing overestimate; it is the reference point that makes actual results interpretable at all.

One final check ties the whole routine together. Before you trust any yield, re-read the balanced equation, re-read the moles you entered, and re-read the runs comparison, because those three readings catch nearly every error this page can produce: an unbalanced equation, a mass entered as moles, and a limiting reagent chosen by mass instead of by runs. If all three readings agree with the page, the ceiling and the leftover are as sound as the inputs, and any gap between the ceiling and what you collect is information about the process, not about the arithmetic.

Keep the same discipline with the leftover figures. The leftover is exact in the same sense as the ceiling: it is what the arithmetic says remains when the limiting reagent is used up, assuming the reaction goes to completion. In a real run the leftover will differ, because the reaction may not go to completion and side reactions may consume the excess. The value of the calculated leftover is as a planning figure and a check: it tells you how much excess you ordered, it tells you what recovery is worth attempting, and it tells you when a measured leftover is so different from the calculated one that the reaction did not behave as assumed. Treat both the ceiling and the leftover as references to compare against reality, not as promises about it, and the calculator becomes a tool for learning from each run rather than a machine for producing a single number.

Limiting reagent first, yield second: the two-step routine

Yield is a ceiling set by the scarcest reactant. Find the ceiling before measuring anything against it.

Consider water from hydrogen and oxygen: 2H₂ + O₂ → 2H₂O. Start with 10 g of hydrogen (5 mol) and 32 g of oxygen (1 mol). The equation needs two moles of hydrogen per mole of oxygen, so 5 mol of hydrogen would need 2.5 mol of oxygen — but only 1 mol is present, so oxygen is the limiting reagent. One mole of oxygen makes two moles of water, 36 g: that is the theoretical yield. Hydrogen is in excess, and about 6 g of it (3 mol) remains unreacted. If 30 g of water is actually collected, the percent yield is 30 ÷ 36 × 100 = 83.3%.

The routine generalises to every stoichiometry problem. Balance the equation first — an unbalanced equation gives wrong mole ratios and therefore a wrong ceiling. Convert every given mass to moles. Divide each reactant’s moles by its coefficient and compare: the smallest result is the limiting reagent. Use only the limiting reagent to work out the product moles, convert back to mass, and only then compare the actual yield with that theoretical figure. Skipping the limiting-reagent step and assuming the first reactant listed is limiting is the commonest error in yield problems.

Percent recovery vs percent yield: same arithmetic, different question

Yield compares what you made with what the reaction could make. Recovery compares what you kept with what you started with.

Percent yield asks how completely a reaction ran: actual product divided by theoretical product. Percent recovery asks how completely a purification kept what was already there: mass recovered divided by mass started with, times 100. A recrystallisation that starts with 10 g of impure solid and returns 7 g of pure crystals has a 70% recovery, even though no reaction happened at all. The two figures share the same division, but they answer different questions, and a low recovery does not mean a reaction failed — it usually means product stayed in the solvent, on the filter, or in the mother liquor.

Keeping them apart prevents a common misreading. A reaction can have a 95% yield and a 60% recovery if the product is hard to isolate, and the combined result — 57% of the theoretical maximum in hand — is the product of the two. Reporting both figures separately, with the losses named (transfer, solubility, side reactions), says far more than either figure alone, and it points at the step worth improving next time.

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