How do I find the theoretical yield?

Three steps. Convert the limiting reagent's mass into moles by dividing by its molar mass. Multiply by the mole ratio — how many moles of product the balanced equation gives per mole of that reagent. Multiply by the product's molar mass to get grams. Starting from 28.014 grams of nitrogen, which is one mole, the ammonia synthesis gives two moles of ammonia at 17.031 grams per mole, so 34.062 grams.

Does this page balance the chemical equation for me?

No, and it is worth being clear about that. It reads chemical formulas — Fe2O3 is parsed into two iron atoms and three oxygens — but an equation with an arrow in it is a different object, and balancing one means solving a system of equations. So the mole ratio is a field you fill in. The reference table lists seven reactions with their ratios so you can see the shape of the answer: one to one, two to one, or two thirds.

How do I know which reagent is limiting?

Divide each reactant's moles by its coefficient in the balanced equation; the smallest result is the limiting reagent. That needs the amount of every reactant, and this page only collects one, so the identification is yours to make and to enter in the formula field. Getting it wrong does not produce an error — it produces a theoretical yield based on the wrong reagent, which is usually too high, and a percent yield that then looks correspondingly low.

Can I enter moles instead of grams?

Yes. The limiting reagent has a mass field and a moles field and either one is enough. Fill in both and the page converts the mass to moles and compares: if they differ by more than one percent it reports a contradiction rather than choosing. When both are given and they agree, the mass field is the one the answer is built on. Entering both is the habit that catches a molar mass copied wrong.

Why is the answer smaller than the mass I started with?

Because the reaction gives off something. Burning 100.086 grams of calcium carbonate leaves 56.077 grams of lime, and the missing 44.009 grams is carbon dioxide that went up the stack. The atoms are conserved but the product you are weighing is only part of what went in. When the product is heavier than the reagent, the reaction has added something from the air or from the other reactant, as the thermite reaction does with oxygen.

Theoretical Yield Calculator

Result

34.062 g

Theoretical yield

Moles of product: 2.000000 mol
Moles of limiting reagent: 1.000000 mol
Molar mass of limiting reagent: 28.014 g/mol
Molar mass of product: 17.031 g/mol

A theoretical yield is worked out on paper before anything is weighed, and this calculator does the three-step arithmetic: turn the limiting reagent into moles, multiply by how many moles of product each mole of it gives, and turn that into grams of the product. Type the two chemical formulas and the molar masses come from them. Type the limiting reagent's mass, or its moles, or both — giving both makes the page check one against the other, which catches a mistyped molar mass. The mole ratio is yours to enter, because this page cannot balance an equation for you.

Seven reactions, and the ratio that decides the yield

Reaction	Equation	Moles of product per mole of reagent	Product from 100 g of the reagent
Lime burning	CaCO3 → CaO + CO2	1	56.03 g
Ammonia synthesis	N2 + 3H2 → 2NH3	2	121.59 g
Hydrogen combustion	2H2 + O2 → 2H2O	2	112.60 g
Thermite welding	Fe2O3 + 2Al → 2Fe + Al2O3	2	69.94 g
Propane combustion	C3H8 + 5O2 → 3CO2 + 4H2O	3	299.40 g
Sodium sulfate	2NaOH + H2SO4 → Na2SO4 + 2H2O	1/2	177.56 g
Nitric acid	3NO2 + H2O → 2HNO3 + NO	2/3	91.31 g

The last column is computed the same way the calculator computes it, so it agrees with the panel above by construction. Read the third column before using any of these: it is the number you would type into the mole ratio field, and it is written as a fraction where the answer is not a whole number — three moles of nitrogen dioxide give two of nitric acid, so the ratio is two thirds, not 0.6667. The fourth column shows how much the ratio matters: 100 grams of sodium hydroxide is 2.5 moles of it, and the equation turns that into 1.25 moles of sodium sulfate — which weighs 177.56 grams, because the product is a heavier molecule than the reagent it came from.

Formula

Theoretical yield = (mass of limiting reagent ÷ its molar mass) × mole ratio × molar mass of product

m: The mass of the limiting reagent you are starting from, in grams. It is the one reagent that runs out first — everything else is in excess
M(reagent): The molar mass of the limiting reagent, in grams per mole. It is summed from the standard atomic weights of the formula you type
n(reagent): The moles of limiting reagent: the mass divided by its molar mass. If you already know the moles, enter them and the mass field can stay empty
n(product): The moles of product, which is the reagent's moles multiplied by the mole ratio. That ratio is read off the coefficients of the balanced equation and entered by you
M(product): The molar mass of the product, in grams per mole. Multiplying the product's moles by it is the last step, and the answer is a mass

Use it whenever a stoichiometry problem asks how much product a given amount of starting material can give — before the experiment, to plan the scale, or after it, to have the number that percent yield is measured against. It is also the page for checking a recipe you found: enter the amounts it calls for and see whether the stated product mass is consistent with the equation. The two things it will not do for you are balance the equation and decide which reagent runs out first; both are inputs here, and the reference table below shows several ratios so you can see how much they differ.

Worked examples

Ammonia from nitrogen, the textbook first case
1. Molar mass of N2: 2 × 14.007 = 28.014 grams per mole
2. Moles of nitrogen: 28.014 ÷ 28.014 = 1 mol
3. N2 + 3H2 → 2NH3 gives two moles of ammonia per mole of nitrogen: 1 × 2 = 2 mol
4. Molar mass of NH3: 14.007 + 3 × 1.008 = 17.031 grams per mole
5. Mass of ammonia: 2 × 17.031 = 34.062 g
Every number here is exact by construction, which is why it makes a good first case: the mass you started with is the molar mass of nitrogen, so the moles come out at exactly 1. The hydrogen is not entered at all — the page is told that nitrogen is the limiting reagent, and it does not ask how much hydrogen was there. That decision is yours to make and yours to get right.
Lime from limestone, where the ratio is one to one
1. Molar mass of CaCO3: 40.078 + 12.011 + 3 × 15.999 = 100.086 grams per mole
2. Moles of limestone: 100.086 ÷ 100.086 = 1 mol
3. CaCO3 → CaO + CO2 is one to one, so one mole of lime
4. Molar mass of CaO: 40.078 + 15.999 = 56.077 grams per mole
5. Mass of lime: 1 × 56.077 = 56.077 g
The mass drops by 44.009 grams, which is the carbon dioxide driven off by the kiln — 100.086 in, 56.077 of solid out. That difference is worth noticing, because it is the whole point of calcination and the reason the answer is smaller than the starting mass. Nothing is lost in the arithmetic; CO2 is a product here, not a loss.
Entering moles directly, when that is what you have
1. The mass field is left empty and the moles are given as 1 mol of methane
2. CH4 + 2O2 → CO2 + 2H2O burns one mole of methane into two moles of water
3. Moles of water: 1 × 2 = 2 mol
4. Molar mass of water: 2 × 1.008 + 15.999 = 18.015 grams per mole
5. Mass of water: 2 × 18.015 = 36.03 g
One mole of methane weighs 16.043 grams, and this case never needs that number — the moles went in directly, so the reagent's molar mass is reported for reference rather than used. Entering both is allowed and is the safer habit: 1 mol against 16.043 grams agree, and a disagreement of more than a percent means one of the two was typed wrong.

Limitations

The mole ratio is an input, not a calculation. This page parses chemical formulas, not equations, so it cannot balance one — a ratio entered wrongly produces a confidently wrong mass, and nothing on the page will flag it. Deciding which reagent is the limiting one is also left to you, because that needs the amount of every reactant, and only one is collected here. Molar masses come from a table of standard atomic weights, so a formula containing an element without one is refused rather than approximated, and a hydrate must be written with its dot (CuSO4·5H2O). The result is the theoretical yield only: it assumes the reaction goes to completion with no losses.

Frequently asked questions

How do I find the theoretical yield?: Three steps. Convert the limiting reagent's mass into moles by dividing by its molar mass. Multiply by the mole ratio — how many moles of product the balanced equation gives per mole of that reagent. Multiply by the product's molar mass to get grams. Starting from 28.014 grams of nitrogen, which is one mole, the ammonia synthesis gives two moles of ammonia at 17.031 grams per mole, so 34.062 grams.
Does this page balance the chemical equation for me?: No, and it is worth being clear about that. It reads chemical formulas — Fe2O3 is parsed into two iron atoms and three oxygens — but an equation with an arrow in it is a different object, and balancing one means solving a system of equations. So the mole ratio is a field you fill in. The reference table lists seven reactions with their ratios so you can see the shape of the answer: one to one, two to one, or two thirds.
How do I know which reagent is limiting?: Divide each reactant's moles by its coefficient in the balanced equation; the smallest result is the limiting reagent. That needs the amount of every reactant, and this page only collects one, so the identification is yours to make and to enter in the formula field. Getting it wrong does not produce an error — it produces a theoretical yield based on the wrong reagent, which is usually too high, and a percent yield that then looks correspondingly low.
Can I enter moles instead of grams?: Yes. The limiting reagent has a mass field and a moles field and either one is enough. Fill in both and the page converts the mass to moles and compares: if they differ by more than one percent it reports a contradiction rather than choosing. When both are given and they agree, the mass field is the one the answer is built on. Entering both is the habit that catches a molar mass copied wrong.
Why is the answer smaller than the mass I started with?: Because the reaction gives off something. Burning 100.086 grams of calcium carbonate leaves 56.077 grams of lime, and the missing 44.009 grams is carbon dioxide that went up the stack. The atoms are conserved but the product you are weighing is only part of what went in. When the product is heavier than the reagent, the reaction has added something from the air or from the other reactant, as the thermite reaction does with oxygen.

References

Standard Atomic Weights — abridged to four significant figures — Commission on Isotopic Abundances and Atomic Weights (CIAAW), IUPAC
SI Units — Amount of Substance: the mole — National Institute of Standards and Technology (NIST)
IUPAC Gold Book — amount of substance — International Union of Pure and Applied Chemistry (IUPAC)

Related calculators

Percent Yield CalculatorPercent yield from actual and theoretical yield.Molar Mass CalculatorMolar mass, plus how much of it is one element.Mole CalculatorMoles from mass, particles or molarity and volume.Grams to Moles CalculatorConvert grams to moles using molar mass, and back again.