Chemistry

Chemistry Equation Balancer

Balance chemical equations and inspect stoichiometric coefficients on both sides.

Chemistry equation balancer

Balances atoms in neutral formula equations and shows step notes plus an atom-check table. Ionic charges, electrons, and acidic/basic redox half-reactions (H⁺, OH⁻, e⁻) are not supported yet.

Balanced equation

4Fe + 3O2 → 2Fe2O3

Reactants

  • 4 Fe
  • 3 O2

Products

  • 2 Fe2O3

Balancing steps

  1. Parse species: Fe, O2, Fe2O3.
  2. Identify elements to conserve: Fe, O.
  3. Solve for the smallest positive integer coefficients that balance every element.
  4. Balanced equation: 4Fe + 3O2 → 2Fe2O3.
  5. Verify atom counts match on both sides for each element.
ElementReactantsProducts
Fe44
O66

What is the Chemistry Equation Balancer?

A balanced chemical equation obeys the law of conservation of mass: every atom present among the reactants must appear among the products in equal numbers. Balancing is not merely a bookkeeping exercise—it produces the stoichiometric coefficients that govern all subsequent mole-ratio calculations in reaction stoichiometry, equilibrium problems, and thermochemical equations. An unbalanced equation implies atoms are created or destroyed, which violates fundamental physical law.

The standard balancing method in general chemistry is inspection: adjust coefficients in front of compound formulas until each element has the same count on both sides. Start with elements that appear in only one reactant and one product, then move to more complex cases involving polyatomic ions that may transfer intact (such as sulfate or nitrate groups). For redox reactions, the half-reaction method or oxidation-number method provides a systematic approach when inspection becomes unwieldy.

Balanced equations appear in virtually every chemistry context. Combustion analysis requires balancing the burning reaction to relate CO₂ and H₂O produced back to the original compound. Acid–base neutralization, precipitation, and gas-evolution reactions all begin with a correctly balanced equation. In thermochemistry, coefficients scale the enthalpy change: if ΔH for forming 1 mol of product is known, doubling the coefficient doubles the enthalpy.

Students frequently struggle with balancing because they attempt to change subscripts within formulas rather than adjusting coefficients. The subscripts in H₂O, for instance, define water's identity and must never be altered—only the coefficient in front may change. The Chemistry Equation Balancer on Online Science Tools is a free chemistry equation balancer that applies algorithmic balancing to valid chemical formulas, returns the smallest whole-number coefficients, and shows balancing steps with an atom-check table for practice and homework verification.

Correct balancing is the gateway to the Reaction Stoichiometry Calculator and the Equilibrium Calculator. Without accurate coefficients, limiting reagent predictions and ICE table stoichiometry are wrong from the start. Treat balancing as the first step in any multi-part quantitative chemistry problem, and use our balancer to confirm your handwritten work during homework and exam preparation.

  • Coefficients multiply entire formulas; subscripts within formulas are fixed
  • Polyatomic ions unchanged on both sides can be balanced as units
  • Redox equations may require the half-reaction method in acidic or basic medium
  • The smallest whole-number coefficient set is the convention for balanced equations

Mathematical / chemical formulas

Balancing is a constraint satisfaction problem: find integer coefficients cᵢ for each species such that the total atom count of every element is identical on the reactant and product sides.

General form:
  c₁·(species₁) + c₂·(species₂) + … → c₃·(species₃) + c₄·(species₄) + …

Conservation for each element X:
  Σ (cᵢ × atoms of X in speciesᵢ)_reactants
    = Σ (cⱼ × atoms of X in speciesⱼ)_products

Example:  CH₄ + O₂ → CO₂ + H₂O

  C:  1 = 1           ✓ (already balanced for carbon)
  H:  4 ≠ 2           → need 2·H₂O
  O:  2 ≠ 4           → need 2·O₂

Balanced:  CH₄ + 2O₂ → CO₂ + 2H₂O
  • Fractional coefficients during balancing should be cleared by multiplying the entire equation by the denominator.
  • For ionic equations in aqueous solution, charge must also balance in addition to atom count.
  • Combustion of hydrocarbons always produces CO₂ and H₂O; balance C first, then H, then O.

Step-by-step example: Balancing the Combustion of Ethanol

Balance the combustion equation for ethanol: C₂H₅OH + O₂ → CO₂ + H₂O. Verify atom conservation on both sides.

  1. Count atoms on the left: C = 2, H = 6, O = 1 (in ethanol) + O₂.
  2. Balance carbon: place coefficient 2 in front of CO₂ → C₂H₅OH + O₂ → 2CO₂ + H₂O.
  3. Balance hydrogen: 6 H on left needs 3 H₂O → C₂H₅OH + O₂ → 2CO₂ + 3H₂O.
  4. Balance oxygen: right side has 4 + 3 = 7 O; left has 1 + 2×O₂, so 2x = 6, x = 3.
  5. Balanced equation: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O.
  6. Verify: C: 2 = 2, H: 6 = 6, O: 1 + 6 = 7 and 4 + 3 = 7. All atoms conserved.

Enter C2H5OH + O2 -> CO2 + H2O into the Chemistry Equation Balancer on Online Science Tools. The tool should return coefficients 1, 3, 2, 3 for ethanol, oxygen, carbon dioxide, and water respectively, matching your hand-balanced result. Use these coefficients in the Reaction Stoichiometry Calculator if you need to compute yields, or check individual molar masses with the Stoichiometry Calculator.

Frequently asked questions

Is this a balancing chemical equations calculator with steps?

Yes. Enter an equation such as C2H6 + O2 = CO2 + H2O and the balancer returns the balanced result, short step notes, and an element-by-element atom inventory so you can practice inspection balancing and check your work.

Can I change subscripts to balance an equation?

No. Changing a subscript alters the identity of the substance. Writing H₂O as H₂O₂ would mean hydrogen peroxide instead of water. The only permissible changes are coefficients—the numbers placed before a formula that multiply every atom in that formula. If you find yourself wanting to change a subscript, reconsider your product or reactant formulas instead.

What if the balancer gives fractional coefficients?

The Chemistry Equation Balancer returns the smallest whole-number ratio by default. If you encounter fractions during manual balancing, multiply the entire equation by the least common denominator to clear them. For example, if you obtain C₂H₄ + 3.5O₂ → 2CO₂ + 3H₂O, multiply everything by 2 to get 2C₂H₄ + 7O₂ → 4CO₂ + 6H₂O.

How do I balance redox reactions in acidic solution?

Split the reaction into oxidation and reduction half-reactions. Balance atoms other than O and H first, then balance O by adding H₂O and H by adding H⁺ (in acidic medium). Balance charge by adding electrons. Multiply each half-reaction so electrons lost equal electrons gained, then add the half-reactions and cancel species appearing on both sides. For a dedicated acidic/basic medium tool with atom and charge checks, use the Redox Equation Balancer.

Why does balancing matter for enthalpy calculations?

Enthalpy of reaction ΔH is reported per mole of reaction as written. If you double all coefficients, ΔH doubles. Thermochemical equations must be balanced so the stated ΔH corresponds to the correct mole ratio of reactants and products. Using an unbalanced equation leads to enthalpy values that are off by an integer factor, producing incorrect heat predictions in calorimetry problems.

Keep learning with more calculators and study guides on Online Science Tools.

Practice problems & worked examples

Practice alongside the equation balancer above. Each problem includes a full worked solution so you can check your reasoning step by step.

Practice problem 1

Balance combustion of ethane

Balance: C₂H₆ + O₂ → CO₂ + H₂O.

Show solution

Worked solution

  1. Carbon: put 2 CO₂.
  2. Hydrogen: 6 H ⇒ 3 H₂O.
  3. Oxygen: right side has 4 + 3 = 7 O atoms ⇒ (7/2) O₂; multiply all by 2.
  4. Result: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.

Answer: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Practice problem 2

Balance iron oxidation

Balance: Fe + O₂ → Fe₂O₃.

Show solution

Worked solution

  1. Put 2 Fe on left to match Fe₂O₃ iron count, then scale oxygen.
  2. Standard integer solution: 4Fe + 3O₂ → 2Fe₂O₃.

Answer: 4Fe + 3O₂ → 2Fe₂O₃

Practice problem 3

Polyatomic species

Balance: AgNO₃ + CaCl₂ → AgCl + Ca(NO₃)₂.

Show solution

Worked solution

  1. NO₃ and Cl transfer as groups.
  2. 2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂.

Answer: 2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂

Practice problem 4

Practice: propane combustion

Balance: C₃H₈ + O₂ → CO₂ + H₂O.

Show solution

Worked solution

  1. C → 3 CO₂; H → 4 H₂O.
  2. O atoms on right = 6 + 4 = 10 ⇒ 5 O₂.
  3. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

Answer: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Practice problem 5

Practice: aluminum + HCl

Balance: Al + HCl → AlCl₃ + H₂.

Show solution

Worked solution

  1. AlCl₃ needs 3 Cl ⇒ 3 HCl; H₂ then needs even H ⇒ multiply by 2.
  2. 2Al + 6HCl → 2AlCl₃ + 3H₂.

Answer: 2Al + 6HCl → 2AlCl₃ + 3H₂

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