Stoichiometry Calculator
Convert between moles and mass in chemical reactions using stoichiometric ratios, and identify the limiting 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:
- Mole Ratio: Convert moles of one substance to moles of another using the equation's coefficients.
- Mass Calculation: Convert grams of a reactant to grams of a product (the full mole roadmap: g → mol → mol → g).
- 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:
| Calculation | Formula | When Used |
|---|---|---|
| Moles from mass | n = m / MM | Converting grams to moles (MM = molar mass in g/mol) |
| Mole ratio | n_B = n_A × (coef_B / coef_A) | Converting moles of A to moles of B |
| Mass from moles | m = n × MM | Converting moles back to grams |
| Limiting reagent | Compare n_A / coef_A vs n_B / coef_B | Smallest ratio → limiting reagent |
| Theoretical yield | m_product = n_limiting × (coef_P / coef_L) × MM_P | Max product from limiting reagent |
| Percent yield | % = (actual / theoretical) × 100 | Measuring 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
- Select a mode: Mole Ratio (for moles-only problems), Mass Calculation (for grams-to-grams), or Limiting Reagent.
- Enter the coefficients from your balanced chemical equation for reactant A and substance B.
- 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).
- 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.
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).
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.
❓ 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.