Stoichiometry Calculator: Mole Ratios, Limiting Reagent and Yield

Enter a reaction and known amount to calculate product mass, limiting reagent, excess reactant, theoretical yield, or percent yield. Equations are parsed and balanced automatically; calculations stay in your browser.

Enter the reaction

Use + between species and →, ->, =, or ⇌ as the arrow. Formula examples: Fe2(SO4)3, K4[Fe(CN)6], CuSO4·5H2O, NH4+(aq).
Balanced equation
Review or override coefficients
Enter a positive measured amount.
Optional amounts and molar-mass overrides

Add amounts for other reactants to find a limiting reagent. Purity adjusts the usable amount before conversion.

Optional gas conditions and actual yield

Keyboard shortcut: Ctrl/⌘ + Enter

Results

Enter an equation and amount, then calculate. Headline answers and full working will appear here.

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How it works

The calculator parses each formula, counts atoms, and automatically finds the smallest whole-number coefficients. Editable coefficient fields let you review or override that result; calculation stops if the override does not conserve every element.

It then converts each measured amount to moles. Mass uses molar mass, solution volume uses n = M × V, particles use Avogadro’s constant, and gases use the selected molar volume or PV = nRT. The balanced coefficients turn those moles into a reaction extent ξ.

  • Single-known mode: one known species sets the reaction extent, so every other species can be converted directly from the mole ratio.
  • Limiting-reagent mode: when two or more reactants have known amounts, the smallest value of n/coefficient is limiting and controls theoretical yield.
  • Percent yield: if you enter an actual product amount, percent yield is (actual ÷ theoretical) × 100%.

Method and data sources

  • Mole ratio: n_target = n_basis × (coefficient_target / coefficient_basis)
  • Mass conversion: m = n × M, where M is molar mass in g/mol
  • Particles: particles = n × 6.02214076×10²³
  • Ideal gas: V = nRT/P, with R = 0.082057366 L·atm·mol⁻¹·K⁻¹
  • Atomic weights: the CIAAW/IUPAC Abridged Standard Atomic Weights 2024 (based on the Atomic Weights 2021 report with the 2024 Gd, Lu, and Zr revisions). For radioactive elements without a standard atomic weight, the parser uses a conventional representative isotope mass number. Every value used is shown in the result.
  • Avogadro constant: exact SI value from the NIST/CODATA constants database.
  • Rounding: full floating-point precision is retained internally; displayed answers use up to six significant digits. Apply the significant-figure rule required by your measurements.
  • Notation: all 118 element symbols, parentheses, square brackets, hydrates, state symbols, and terminal charges are recognized. Isotopes, electrons, redox charge balancing, mixtures, equilibrium, side reactions, and multistep yield are outside scope.

Published verification cases

Calculator verification cases and expected answers
CaseInputExpected
Water formation4.00 g H2 + 32.0 g O2H2 limiting; 35.74 g H2O
Calcination100.0 g CaCO3about 56.03 g CaO
Solution precipitation25.00 mL of 0.1000 M AgNO30.3583 g AgCl

Worked stoichiometry examples

Mole-to-mole: ammonia

Problem: How many moles of NH3 form from 1.50 mol N2?

  1. Balance: N2 + 3H2 → 2NH3.
  2. Use the ratio: 1.50 mol N2 × (2 mol NH3 / 1 mol N2).
  3. Answer: 3.00 mol NH3.

Mass-to-mass: calcination

Problem: Find CaO from 100.0 g CaCO3.

  1. n(CaCO3) = 100.0 g / 100.086 g/mol = 0.9991 mol.
  2. The coefficient ratio CaO:CaCO3 is 1:1.
  3. m(CaO) = 0.9991 mol × 56.077 g/mol = 56.03 g CaO.

Limiting reagent and excess

Problem: React 4.00 g H2 with 32.0 g O2.

  1. H2 extent = (4.00/2.016)/2 = 0.9921 mol-reaction.
  2. O2 extent = (32.0/31.998)/1 = 1.0001 mol-reaction, so H2 is limiting.
  3. H2O = 2(0.9921)(18.015) = 35.74 g; O2 remaining ≈ 0.256 g.

Solution stoichiometry

Problem: Find AgCl from 25.00 mL of 0.1000 M AgNO3 with excess NaCl.

  1. n = 0.1000 mol/L × 0.02500 L = 0.002500 mol.
  2. AgNO3:AgCl is 1:1.
  3. m = 0.002500 mol × 143.32 g/mol = 0.3583 g AgCl.

Gas volume

Problem: At the same temperature and pressure, find NH3 gas from 5.00 L N2 and excess H2.

  1. N2 + 3H2 → 2NH3.
  2. Equal-condition gas volumes follow the mole ratio.
  3. 5.00 L × 2/1 = 10.0 L NH3.

Percent yield

Problem: If the water experiment above produces 34.2 g H2O, find percent yield.

  1. Theoretical yield = 35.74 g H2O.
  2. Percent yield = (34.2 g / 35.74 g) × 100.
  3. Answer: approximately 95.7%.
Common mistake: do not compare reactant grams directly. Convert each to moles and divide by its coefficient. The smallest raw mole amount is not necessarily limiting because the coefficients can differ.

Frequently asked questions

Does the calculator balance equations automatically?

Yes. Enter reactants and products with an arrow; the calculator parses the formulas and finds the smallest whole-number coefficients. You can override coefficients, but calculation is blocked unless every element is balanced.

How does the limiting reagent calculation work?

Each known reactant amount is converted to moles and divided by its stoichiometric coefficient. The smallest reaction extent identifies the limiting reagent, which then sets the theoretical product yield.

What is the difference between limiting and excess reagent?

The limiting reagent has the smallest available moles divided by its coefficient and runs out first. Any other measured reactant is excess; its remaining amount is the initial amount minus the amount consumed.

What is the difference between theoretical and actual yield?

Theoretical yield is the maximum product predicted from stoichiometry. Actual yield is the measured product. Percent yield equals actual divided by theoretical, multiplied by 100%.

How do I enter a solution using molarity?

Choose solution mL or solution L for a known species, then enter its molarity. The calculator uses n = M × V in litres before applying the mole ratio.

How is reagent purity used?

Enter the assay or purity percentage in Optional amounts and overrides. The stated amount is multiplied by purity divided by 100 before it is converted to moles.

How many significant figures does the answer have?

The calculator keeps full precision internally and displays up to six significant digits. Round the final answer to the least precise measured input required by your course or laboratory method.

What happens when reactants are co-limiting?

If reaction extents agree within the calculator tolerance, all tied reactants are reported as co-limiting and each has approximately zero remaining.

Can the equation contain multiple products?

Yes. The balancer and target selector support multiple products. Choose the product or other species you want in the Find species field.

Are hydrates, state symbols, brackets, and ions supported?

The parser supports parentheses, square brackets, hydrate dots such as CuSO4·5H2O, state symbols such as (aq), and terminal ionic charges. Use a caret for multi-digit charges, for example SO4^2-. Electrons and half-reaction charge balancing are not supported.

Can this solve multistep reactions?

Not in one calculation. Solve each balanced step separately and carry the moles or yield from one step into the next. The calculator does not model equilibrium, side reactions, or kinetics.

Quick stoichiometry facts

Coefficients are mole ratios

A balanced equation does not directly compare grams. It compares particles or moles, which is why molar-mass conversion matters.

Core rule

Limiting reagent sets the ceiling

The reactant that runs out first fixes the maximum theoretical amount of every product in the reaction.

Yield control

Gas volumes can mirror moles

For ideal gases at the same temperature and pressure, equal-volume relationships follow directly from Avogadro’s law.

Gas shortcut

Percent yield can exceed 100% on paper

If experimental samples are wet, impure, or not fully dried, an apparent yield above 100% often flags a measurement issue rather than impossible chemistry.

Lab sanity check

Molar mass links the bench to the equation

Balanced equations speak in moles, while chemists often weigh grams. Molar mass is the bridge that connects those two worlds.

Mass to moles

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