Stoichiometry Calculator

Convert between moles and mass in chemical reactions using stoichiometric ratios, and identify the limiting reagent.

⚗️ Stoichiometry Calculator

Given moles of A, find moles of B using the balanced equation coefficients.

Convert grams of A to grams of B. Requires molar masses and stoichiometric coefficients.

Identify which reactant runs out first given available moles of both A and B.

Moles of B
Mole Ratio (A:B)

Reactant A
Product/Reactant B

Limiting Reagent
Excess Reagent

What Is a Stoichiometry Calculator?

Stoichiometry is the quantitative study of chemical reactions — specifically the relationships between the amounts of reactants consumed and products formed. The word comes from the Greek stoicheion (element) and metron (measure). Every balanced chemical equation is a stoichiometric statement: 2H₂ + O₂ → 2H₂O tells us that exactly two moles of hydrogen react with one mole of oxygen to produce two moles of water.

This calculator handles the three most common stoichiometry problems students and chemists encounter:

  1. Mole Ratio: Convert moles of one substance to moles of another using the equation's coefficients.
  2. Mass Calculation: Convert grams of a reactant to grams of a product (the full mole roadmap: g → mol → mol → g).
  3. Limiting Reagent: Determine which reactant runs out first and how much of the other remains in excess.

Understanding stoichiometry is essential for laboratory work, industrial process chemistry, pharmaceutical manufacturing, and environmental analysis — any field where chemical quantities must be precisely controlled or predicted.

Stoichiometry Formulas

All stoichiometry is built on three relationships:

CalculationFormulaWhen Used
Moles from massn = m / MMConverting grams to moles (MM = molar mass in g/mol)
Mole ration_B = n_A × (coef_B / coef_A)Converting moles of A to moles of B
Mass from molesm = n × MMConverting moles back to grams
Limiting reagentCompare n_A / coef_A vs n_B / coef_BSmallest ratio → limiting reagent
Theoretical yieldm_product = n_limiting × (coef_P / coef_L) × MM_PMax product from limiting reagent
Percent yield% = (actual / theoretical) × 100Measuring reaction efficiency

The universal stoichiometry roadmap: grams A → (÷ MM_A) → moles A → (× ratio) → moles B → (× MM_B) → grams B

📖 How to Use the Stoichiometry Calculator

Steps to Calculate

1
Select a mode: Mole Ratio (for moles-only problems), Mass Calculation (for grams-to-grams), or Limiting Reagent.
2
Enter the coefficients from your balanced chemical equation for reactant A and substance B.
3
Enter the known quantity: moles of A (Mole Ratio mode), mass and molar masses (Mass mode), or available moles of both A and B (Limiting Reagent mode).
4
Click Calculate to see the answer with complete step-by-step working.

Tip: Use the Molecular Weight Calculator to find molar masses from chemical formulas before doing a mass calculation.

Example Calculations

Example 1 — Mole Ratio (Combustion of Methane)

CH₄ + 2O₂ → CO₂ + 2H₂O — How many moles of O₂ react with 3 mol CH₄?

1
Coefficients: CH₄ = 1, O₂ = 2
2
Moles O₂ = 3 mol × (2/1) = 6 mol O₂
3
Also: moles CO₂ = 3 × (1/1) = 3 mol; moles H₂O = 3 × (2/1) = 6 mol
3 mol CH₄ requires 6 mol O₂ and produces 3 mol CO₂ and 6 mol H₂O.
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Example 2 — Mass Calculation (Synthesis of Water)

2H₂ + O₂ → 2H₂O — How many grams of water form from 10 g of H₂?

1
Molar mass H₂ = 2.016 g/mol; H₂O = 18.015 g/mol; ratio H₂:H₂O = 2:2 = 1:1
2
Moles H₂ = 10 g ÷ 2.016 g/mol = 4.96 mol
3
Moles H₂O = 4.96 × (2/2) = 4.96 mol → Mass = 4.96 × 18.015 = 89.4 g
10 g of H₂ produces 89.4 g of water (theoretical yield, assuming excess O₂).
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Example 3 — Limiting Reagent (Haber Process)

N₂ + 3H₂ → 2NH₃ — 4 mol N₂ and 9 mol H₂ available. Which is limiting?

1
Divide by coefficient: N₂: 4/1 = 4; H₂: 9/3 = 3
2
H₂ has the smaller ratio (3 < 4) → H₂ is the limiting reagent
3
H₂ consumed to consume all 9 mol: uses 9/3 = 3 mol N₂. Excess N₂ = 4 − 3 = 1 mol N₂
H₂ is limiting (9 mol available, needs 12 for 4 mol N₂). 1 mol N₂ remains in excess. NH₃ produced = 9 × (2/3) = 6 mol.
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❓ Frequently Asked Questions

What is stoichiometry?+
Stoichiometry is the quantitative study of chemical reactions — specifically the relationships between amounts of reactants consumed and products formed. Based on the law of conservation of mass and balanced chemical equations, stoichiometry allows chemists to predict exactly how much product a reaction will produce, how much reactant is needed, and which substance limits the reaction. It is fundamental to chemistry lab work, industrial manufacturing, and pharmaceutical synthesis.
What is a mole ratio?+
A mole ratio is the ratio of stoichiometric coefficients from a balanced chemical equation, used to convert between moles of different substances. For aA + bB → cC, the mole ratio of A to B is a:b; A to C is a:c. If you have n moles of A, moles of B used = n × (b/a), moles of C produced = n × (c/a). Mole ratios are exact — they come directly from the integer coefficients, not from measurements.
What is a limiting reagent?+
The limiting reagent is the reactant completely consumed first, which determines the maximum product yield. To find it: divide available moles of each reactant by its stoichiometric coefficient. The reactant with the smallest result is the limiting reagent. The other reactants are in excess — some remain after the reaction. Only the limiting reagent's amount is used to calculate theoretical yield.
How do you calculate theoretical yield?+
Theoretical yield is the maximum product mass assuming the limiting reagent is fully consumed. Steps: (1) Identify the limiting reagent and find its moles. (2) Apply the mole ratio: moles product = moles limiting × (coef product / coef limiting). (3) Convert to grams: mass = moles × molar mass. The actual yield from a real experiment is always ≤ theoretical due to losses, side reactions, or incomplete reaction.
How do you do a mass-to-mass stoichiometry problem?+
The three-step mole roadmap: (1) Convert mass of reactant A to moles: n_A = mass_A / MM_A. (2) Apply the mole ratio from the balanced equation: n_B = n_A × (coef_B / coef_A). (3) Convert moles of product B to grams: mass_B = n_B × MM_B. The mole ratio is the bridge — you cannot directly convert grams of one substance to grams of another without going through moles.
Why must the chemical equation be balanced before doing stoichiometry?+
The stoichiometric coefficients in a balanced equation reflect the law of conservation of mass — atoms are neither created nor destroyed. An unbalanced equation gives incorrect mole ratios, leading to wrong answers for amounts of reactants and products. Always balance the equation first by adjusting coefficients so the number of each type of atom is the same on both sides. The coefficients, not the subscripts in formulas, are what you use in stoichiometry calculations.
What is percent yield?+
Percent yield = (actual yield / theoretical yield) × 100%. It measures how efficiently a reaction produced the desired product. Less than 100% is expected due to: incomplete reactions (equilibrium), side reactions, losses during product isolation and purification, and measurement uncertainties. In industrial chemistry, achieving high percent yield is economically critical. A 70–90% yield is acceptable for complex organic syntheses; simple reactions may achieve 95–100%.
What is the difference between moles and grams in stoichiometry?+
Moles measure number (1 mole = 6.022 × 10²³ particles). Grams measure mass. Different substances have different molar masses, so 1 mole of H₂ (2 g/mol) = 2 g, while 1 mole of H₂O (18 g/mol) = 18 g. Stoichiometry works with moles (via mole ratios from balanced equations), so you must convert grams to moles before applying the mole ratio, then convert back to grams. Molar mass is the conversion factor between grams and moles.
Can stoichiometry be applied to solutions and gases?+
Yes. For solutions: use molarity (mol/L) to find moles from volume and concentration. For gases at STP (0°C, 1 atm), 1 mole of any ideal gas occupies 22.4 L. At other conditions, use PV = nRT (ideal gas law) to find moles from pressure, volume, and temperature. All stoichiometry roads lead through moles — the specific conversion from measured quantity (volume of solution, volume of gas, mass of solid) to moles depends on the state of matter.
How is stoichiometry used in real life?+
Stoichiometry is essential across many fields: (1) Pharmaceutical manufacturing — exact drug dosages and reaction yields must be controlled. (2) Industrial chemistry — Haber process for ammonia, steel production, and refining all use stoichiometric ratios to minimise waste. (3) Environmental science — calculating combustion products and pollutant concentrations. (4) Food science — balancing fermentation reactions. (5) Explosives and propellants — stoichiometric oxygen-fuel ratios ensure complete combustion for maximum energy release.
What is the mole concept and why is it important?+
The mole (symbol: mol) is the SI unit for amount of substance. 1 mole = 6.02214076 × 10²³ particles (Avogadro's number). It is important because atoms and molecules react in whole-number ratios, but individual atoms are too small to count directly. The mole bridges the atomic world and the laboratory world: you can measure grams on a balance, then convert to moles to count 'how many molecules' in a practical, usable way. Without the mole concept, quantitative chemistry would be impossible.

What is stoichiometry?

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction, based on the law of conservation of mass. It uses balanced chemical equations and molar masses to calculate how much of each substance reacts or is produced. The stoichiometric coefficients in a balanced equation give the mole ratios: for 2H₂ + O₂ → 2H₂O, two moles of H₂ react with one mole of O₂ to produce two moles of H₂O.

What is a mole ratio?

A mole ratio is the ratio of the stoichiometric coefficients from a balanced equation, used to convert between moles of different substances. For the reaction aA + bB → cC, the mole ratio of A to B is a:b. If you have n_A moles of A, the moles of B consumed = n_A × (b/a), and moles of C produced = n_A × (c/a). Mole ratios are exact — they come from the integer coefficients in the balanced equation.

What is a limiting reagent?

The limiting reagent (or limiting reactant) is the reactant that is completely consumed first in a reaction, thereby limiting how much product can be formed. The other reactant is the excess reagent — some of it is left over after the reaction. To find the limiting reagent: divide the available moles of each reactant by its stoichiometric coefficient; the reactant with the smallest result is the limiting reagent.

How do you calculate theoretical yield?

Theoretical yield is the maximum amount of product that could form if the limiting reagent is fully consumed with no losses. Steps: (1) Identify the limiting reagent. (2) Use the mole ratio to find moles of product: n_product = n_limiting × (coef_product / coef_limiting). (3) Convert to grams: mass = moles × molar mass. The actual experimental yield is always ≤ theoretical yield.

What is the mole-to-mole conversion method?

Mole-to-mole conversion uses the balanced equation's coefficients as the conversion factor. For N₂ + 3H₂ → 2NH₃: if you start with 4 mol H₂, moles of NH₃ = 4 × (2/3) = 2.67 mol. The steps are always: (1) Start with known moles of substance A. (2) Multiply by (coef_B / coef_A). (3) The result is moles of substance B. No molar masses are needed for mole-to-mole calculations.

How do you do a mass-to-mass stoichiometry calculation?

Mass-to-mass conversion: (1) Convert mass of A to moles: n_A = mass_A / MM_A. (2) Apply mole ratio: n_B = n_A × (coef_B / coef_A). (3) Convert moles of B to mass: mass_B = n_B × MM_B. Example: how many grams of H₂O form from 36 g of H₂? MM(H₂) = 2 g/mol, MM(H₂O) = 18 g/mol. Moles H₂ = 36/2 = 18 mol. Moles H₂O = 18 × (2/2) = 18 mol. Mass H₂O = 18 × 18 = 324 g.

What is percent yield in chemistry?

Percent yield = (actual yield / theoretical yield) × 100%. It measures the efficiency of a reaction. Less than 100% is expected due to: incomplete reactions (equilibrium not reached), side reactions producing other products, physical losses during isolation/purification, and measurement errors. A 70–90% yield is considered good for complex organic syntheses; simple inorganic reactions can achieve >95%.

What is molar mass and how is it used in stoichiometry?

Molar mass is the mass (in grams) of one mole of a substance — numerically equal to the molecular weight in atomic mass units (amu). It serves as the conversion factor between grams and moles: moles = mass / molar mass. Examples: H₂O = 18.015 g/mol; NaCl = 58.44 g/mol; CO₂ = 44.01 g/mol. In stoichiometry, you convert grams → moles → (via mole ratio) → moles of product → grams of product.

How do you find the excess reagent amount?

After identifying the limiting reagent, calculate how much of the excess reagent was consumed: moles consumed = moles of limiting reagent × (coef_excess / coef_limiting). Subtract from the initial moles: moles remaining = initial moles − moles consumed. Multiply by molar mass for grams remaining. This tells you how much excess reagent is left over after the reaction is complete.

What is the difference between reactants and products in stoichiometry?

In a chemical equation A + B → C + D, A and B are reactants (starting materials consumed) and C and D are products (substances formed). Stoichiometry applies equally to all: mole ratios connect any pair of substances. The arrow direction does not affect the ratios — you can calculate how many moles of any reactant react with any amount of another reactant or product using the same mole ratio approach.

What does 'balanced equation' mean and why is it essential for stoichiometry?

A balanced equation has the same number of each type of atom on both sides of the reaction arrow, satisfying the law of conservation of mass. Example: unbalanced H₂ + O₂ → H₂O; balanced 2H₂ + O₂ → 2H₂O. The integer coefficients in the balanced equation give the exact mole ratios for stoichiometry calculations. Using unbalanced coefficients gives incorrect mole ratios and wrong answers. Always balance the equation before doing any stoichiometry.