Theoretical Yield Calculator

Find the theoretical product mass a reaction can give from its limiting reagent.

Any of the eight boxes can be solved, across 13 mass, 4 molar-mass and 5 mole units.

Updated October 3, 2026

CreatorFrank Zhao

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Introduction

Every reaction has a hard ceiling. Weigh out the reagents, let them go to completion with nothing lost and nothing unwanted, and the balanced equation fixes the largest mass of product you could possibly isolate. That number is the theoretical yield, and knowing it before you start tells you whether the charge you planned is sensible or whether you are about to waste the expensive half of it.

This page works that ceiling out in either direction. Start from what you have weighed out and it tells you what you will get; start from the mass you need and it tells you what to weigh out. The work is split into two groups — the limiting reagent you have, and the product you want — and five of the eight boxes accept your numbers. The rest fill themselves, so you never have to decide in advance which equation to apply or which box to leave blank.

Eight boxes

Five of them accept your numbers

Three relationships

Each solved in either direction

Thirteen mass units

Micrograms through to long tons

One prerequisite, though. The yield comes out of the coefficients in a balanced equation, so if you have not balanced yours yet, settle that first with the chemical equation balancer. An unbalanced equation does not fail here — it returns a confident, tidy and completely wrong answer.

A ceiling, never a prediction

The calculation assumes 100% efficiency, which means every molecule reacted the way you wanted, nothing diverted into a side product, and nothing left stuck to the glassware. Real reactions fall short of that every time, which is exactly why a yield calculated here is a benchmark rather than a forecast.

Nothing on this page tries to estimate the shortfall. The moment you have weighed your product, the percent yield is what tells you how far short you landed, and that is a separate calculation.

How to Use

Five steps from a balanced equation to a number on the screen. You supply five values; the calculator works out the other three boxes.

Get the coefficients and the molar masses ready

From your balanced equation you need two coefficients: the one in front of the reagent that runs out, and the one in front of your product. You also need a molar mass for each. If you have the formulas but not the numbers, the molar mass calculator will give you both.

Decide which reagent is limiting

Convert each reagent to moles, then divide by its own coefficient. The smallest result is the limiting reagent. This is the step people get wrong most often, and the Calculation Method section below shows why moles on their own are not enough.

Fill the Limiting reagent group

Enter the mass you weighed out and its molar mass. Leave Moles blank and it will be worked out for you, or type it in directly if you already know it. Then enter the coefficient for that reagent from the equation.

Fill the Desired product group

Enter the coefficient for your product and its molar mass. The Theoretical yield box is the one you normally leave empty — it is the answer this page exists to produce.

Worked example: acetone cyanohydrin from sodium cyanide

For acetone plus cyanide you weigh out 2.00 g of sodium cyanide, whose molar mass is 49.01 g/mol, and you want acetone cyanohydrin at 85.11 g/mol. Both coefficients are 1, so the moles carry straight across and the mass follows:

n_{\mathrm{lim}} = \frac{2.00}{49.01}

=

0.040808\ \mathrm{mol}

m_{p} = 0.040808 \times 85.11

=

3.47317\ \mathrm{g}

The 5 g of acetone you also weighed out is not competing. That is 0.0860882 mol against the 0.040808 mol of cyanide available, so cyanide is what runs out and acetone sits there in excess.

Which boxes are yours

Whichever box the calculator fills in for you is highlighted, so you can always see at a glance which numbers you supplied and which ones are results. Edit any filled-in box and the solver moves its output somewhere else instead of fighting you for it.

When to Use It

This is a stoichiometry tool, and it is at its best when a balanced equation, one clear limiting reagent and a real reagent charge are all in hand.

Reach for it when

Four jobs it does well

Before you start. Deciding whether a planned charge makes sense, and which reagent to buy in excess deliberately.
Scaling up. Finding out whether doubling a reaction doubles the output or whether it hits a new bottleneck.
Working backwards. Sizing a reagent charge from a target product mass.
Checking a suspicion. Confirming that a cheap reagent really is the one limiting you, rather than the precious one sitting untouched in the flask.

Keep away when

Four jobs it cannot do

Equilibrium conversions. Those are set by the equilibrium constant, not by stoichiometry, and the two can differ enormously.
Gas volumes away from standard conditions. A gas-to-gas ratio is a mole ratio, so use this to get moles first and convert afterwards.
Reactions fed gradually. If a reagent is added in portions, the limiting reagent can change partway through, and a single calculation cannot describe that.
Percent recovery. If you have already isolated and weighed the product, the percent yield calculator turns your real mass into a percentage of the figure you would get here.

If you want the whole equation, not one reaction

This page deliberately follows a single limiting reagent. If you would rather hand over a balanced equation plus one quantity and get masses for every species at once, the molar ratio calculator does that, and includes the same limiting-reagent ranking.

Calculation Method

Three short relationships do all the work. They are the ones shown in the calculator's equation panel, in the same order.

1 · Moles of the limiting reagent

The definition of the mole: mass divided by molar mass. This is how you find how many moles you actually hold, and it is why a reagent in solution needs its moles from concentration rather than from mass.

n_{\mathrm{lim}} = \frac{m_{\mathrm{lim}}}{M_{\mathrm{lim}}}

2 · Moles of product from the coefficients

The coefficient ratio from the balanced equation. Note that the limiting reagent's coefficient goes on the bottom: three moles of hydrogen make two moles of ammonia, not three. When both coefficients are 1 this reduces to the two amounts being equal.

n_{p} = n_{\mathrm{lim}} \times \frac{c_{p}}{c_{\mathrm{lim}}}

3 · Theoretical yield

The mole definition again, now for the product. Combine all three and the whole chain collapses into one step, which is convenient when you are working backwards from a target mass.

m_{p} = M_{p} \times n_{p}

m_{p} = m_{\mathrm{lim}} \times \frac{M_{p} \times c_{p}}{M_{\mathrm{lim}} \times c_{\mathrm{lim}}}

What each symbol means

m_{\mathrm{lim}}

Mass of the limiting reagent you have

M_{\mathrm{lim}}

Molar mass of the limiting reagent

n_{\mathrm{lim}}

Moles of the limiting reagent

c_{\mathrm{lim}}

Coefficient of the limiting reagent in the balanced equation

c_{p}

Coefficient of the product in the balanced equation

n_{p}

Moles of product the reaction can form

M_{p}

Molar mass of the product

m_{p}

Theoretical yield — the mass of product

How the limiting reagent is chosen

Each reagent is ranked by its own amount divided by its own coefficient. That number is how many times the balanced equation can be run as far as that reagent allows, and the smallest one is the limiting reagent. Working an ammonia synthesis through it makes the reason obvious:

\mathrm{N_2}:\ \frac{0.30\ \mathrm{mol}}{1} = 0.30

\qquad

\mathrm{H_2}:\ \frac{1.50\ \mathrm{mol}}{3} = 0.50

Hydrogen has five times as many moles and weighs less than the nitrogen (3.024 g against 8.4042 g), yet nitrogen is limiting. Ranking by moles, or by mass, or by anything other than amount over coefficient gets this wrong — and it is the single most common error in yield calculations.

Which way you can run it

Because all three relationships solve both ways, any five of the eight boxes will do. Type a mass and a molar mass to get moles; type moles and a molar mass to get mass; type a product yield and the two coefficients to recover the reagent charge you would need. The calculator decides which box to solve for based on what you left empty.

Units and how much they matter

Masses cover thirteen units from micrograms to imperial tons, molar masses four and amounts five, each with its own menu. Two of the molar-mass units — kilograms per kilomole and pounds per pound-mole — read the same number as grams per mole, because the definitions cancel exactly rather than approximately: a kilomole is a thousand moles against a kilogram being a thousand grams, and one pound-mole is 453.59237 moles against one pound weighing 453.59237 grams.

Every conversion factor is an exact definition rather than a measurement — the avoirdupois pound is exactly 453.59237 g — so changing a unit changes how the number reads and nothing else. The arithmetic also runs at full precision internally no matter how the value is displayed, so rounding a figure on screen never quietly rounds the answer underneath.

Real-World Examples

Two situations that need a different route through the same eight boxes. The numbers below were read straight off the calculator.

Sizing a reagent charge backwards

\mathrm{(CH_3)_2CO + CN^- + H_2O \rightarrow (CH_3)_2C(OH)CN + OH^-}

You need 25.0 g of acetone cyanohydrin and the reaction is already known to be 1:1. Rather than starting from whatever happens to be on the shelf, work out what you actually need on the bench.

Target yield

25 g

Product molar mass

85.11 g/mol

Product moles needed

0.293738 mol

Sodium cyanide required

14.3961 g

You entered the yield and the product molar mass and left the reagent mass blank. The product and reagent moles filled in together, and the mass came out of the definition of the mole:

n_{p} = \frac{25.0}{85.11} = 0.293738\ \mathrm{mol}

m_{\mathrm{lim}} = 0.293738 \times 49.01

=

14.3961\ \mathrm{g}

To keep cyanide as the limiting reagent you also need at least 0.293738 mol of acetone, which is about 17.06 g. That is the check the calculator cannot do for you, because it only ever holds one reagent.

Sodium cyanide to weigh out14.3961 g

When the coefficient changes the answer

\mathrm{N_2 + 3\,H_2 \rightarrow 2\,NH_3}

For ammonia, nitrogen and hydrogen react as one nitrogen to three hydrogen giving two ammonia. Suppose you have 0.30 mol of nitrogen and 1.50 mol of hydrogen on hand.

Nitrogen available

0.30 mol (8.4042 g)

Hydrogen available

1.50 mol (3.024 g)

Ranks (moles ÷ coefficient)

0.30 vs 0.50

Limiting reagent

Nitrogen

Name nitrogen as the limiting reagent with a coefficient of 1 and ammonia at 2, and the moles of product follow the coefficient ratio:

n_{p} = 0.30 \times \frac{2}{1} = 0.6\ \mathrm{mol}

m_{p} = 0.6 \times 17.031

=

10.2186\ \mathrm{g}

Now check it the other way. Name hydrogen as limiting instead and give it exactly 0.90 mol — 1.8144 g — which is what a 1:3 ratio against 0.30 mol of nitrogen requires. You get the same 0.6 mol and the same 10.2186 g, from a charge that is a quarter the size. The coefficients, not the label you picked, decide the answer.

Theoretical yield of ammonia10.2186 g

Tips & Best Practices

Six habits that keep the arithmetic honest. None of them need a calculator to check — they are all things to do before you type.

Rank by amount over coefficient

Moles alone mislead you, and so does mass. In the ammonia case hydrogen has more moles and less mass and is still not the limiting reagent.

Ignore spectator ions

Sodium cyanide puts one cyanide ion in solution per formula unit, not two usable reagents. If spectators make your equation hard to read, the net ionic equation view strips them out and leaves the reaction you are actually doing.

Get moles from concentration when the reagent is in solution

There is no mass to divide by a molar mass when your reagent is a volume of a known solution. The molarity calculator converts concentration and volume into moles, which then drops straight into the Moles box.

Copy the coefficients across carefully

Writing 2 in the wrong box does not produce an obviously wrong answer — it produces a perfectly plausible one. Read both coefficients straight off your balanced equation each time.

Use standard molar masses

A slip in a product molar mass passes straight through to the yield. If you are unsure of one, look it up rather than reconstructing it from memory.

Treat the warning as a real signal

If the yield comes out more than 2,000 times the mass of the limiting reagent that produced it, something is wrong — no balanced reaction comes close. The usual culprits are a molar mass entered in kilograms per mole where grams per mole was meant, and a misplaced decimal point.

Limitations

The model is deliberately narrow: one balanced equation, one limiting reagent, complete conversion. Those assumptions are what make the answer trustworthy, and they are also where it stops.

100% is the ceiling, by definition

Nothing here models incomplete conversion, side reactions, solubility limits, reaction rate or losses during isolation. Use this figure as the benchmark you measure against, not as the mass you expect to weigh.

Every value must be greater than zero

A coefficient of zero, a molar mass of zero or a missing reagent is rejected rather than guessed at, and a result that would need to divide by one of them stays blank instead of showing a placeholder.

Your equation must already be balanced

The coefficients are taken on trust. Nothing here checks that they are correct, so an unbalanced equation produces a confident answer with no warning at all.

One limiting reagent at a time

Reagents added in portions, or reactions that run in competing steps, are outside what a single calculation can describe.

Report to the precision you actually have

Standard molar masses usually carry four significant figures, so a yield quoted to ten decimal places is arithmetic rather than chemistry. Quote the answer to the precision your masses deserve.

Before you run anything

This page tells you what the stoichiometry allows. It does not tell you whether your procedure, solvent, temperature or workup will get you anywhere near it. For anything you intend to put on the bench, follow a real written procedure — and if that procedure carries a safety case, that reasoning is yours to check, not this page's.

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