A chemical equation is balanced when every element appears the same number of times on both sides of the arrow. That is the law of conservation of mass in symbols: atoms are rearranged in a reaction, never created or destroyed. You balance an equation only by changing the coefficients in front of the formulas – changing a subscript would turn one substance into a different one (H₂O into H₂O₂, for example).
Simple equations can be balanced by inspection, adjusting one element at a time and leaving hydrogen and oxygen until last. Larger ones – combustion of fuels, redox reactions with permanganate or dichromate – are easier with the algebraic method: write one unknown coefficient per species, write one equation per element (plus one for charge if ions are present), and solve the system for the smallest positive whole numbers. That is what this balancer does, using exact fractions rather than rounding.
Type the reactants and products separated by + signs. Brackets, hydrates (CuSO4·5H2O), ions (Fe3+, MnO4-, SO4^2-) and electrons (e-) are all understood, and state symbols such as (aq) or (s) are ignored. The result lists every coefficient, the mass of each species per reaction as written, and confirms that atoms, mass and charge balance.
How to use the Chemical Equation Balancer
- Reactants (left-hand side) — Separate species with ” + “, e.g. Fe2O3 + CO
- Products (right-hand side) — Ions: Fe3+, SO4^2-, MnO4-; electrons: e-
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
Each species becomes one column of an element-count matrix (negative on the product side); the coefficients are the smallest positive integer solution of that homogeneous system, found with exact fraction arithmetic. Molar masses use IUPAC standard atomic weights.
Worked example
For propane burning in oxygen, C3H8 + O2 → CO2 + H2O: carbon fixes 3 CO₂, hydrogen fixes 4 H₂O, and the products then hold 3 × 2 + 4 = 10 O atoms, so 5 O₂ are needed. Balanced: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O.
Mass check with standard atomic weights: 44.097 g + 5 × 31.998 g = 204.087 g of reactants and 3 × 44.009 g + 4 × 18.015 g = 204.087 g of products – mass is conserved.
Equations to try
| Type | Unbalanced | Balanced |
|---|---|---|
| Synthesis | N2 + H2 → NH3 | N₂ + 3 H₂ → 2 NH₃ |
| Combustion | C6H12O6 + O2 → CO2 + H2O | C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O |
| Thermite | Al + Fe2O3 → Al2O3 + Fe | 2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe |
| Precipitation | Pb(NO3)2 + KI → PbI2 + KNO3 | Pb(NO₃)₂ + 2 KI → PbI₂ + 2 KNO₃ |
| Redox (ionic) | MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O | MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O |
Tips
- Check the equation is balanced before using coefficients.
- Find the limiting reagent before calculating a yield.
- Keep temperatures in kelvin for rate and equilibrium equations.
- Everything is calculated in your browser.
Common mistakes to avoid
- Using the excess reagent to calculate theoretical yield.
- Including pure solids and liquids in equilibrium expressions.
- Entering temperatures in °C into the Arrhenius equation without converting.
Formula
For every element: Σ (coefficient × atoms in reactants) = Σ (coefficient × atoms in products) · for ions also Σ charge (left) = Σ charge (right)Worked example
Inputs — Reactants (left-hand side): C3H8 + O2; Products (right-hand side): CO2 + H2O.
Results — Balanced equation = C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O; Coefficients in order = 1 : 5 : 3 : 4; Coefficients and masses per reaction = Side Species Coefficient M (g/mol) Mass (g)
───────────────────────────────────────────────────────
reactant C₃H₈ 1 44.097 44.097
reactant O₂ 5 31.998 159.99
product CO₂ 3 44.009 132.027
product H₂O 4 18.015 72.06; Atom count check = Element Reactant side Product side Balanced
─────────────────────────────────────────────────
C 3 3 yes
H 8 8 yes
O 10 10 yes; Mass check = 204.087 g of reactants → 204.087 g of products – mass is conserved.
Key terms
- Coefficient
- The number in front of a formula giving how many formula units take part.
- Subscript
- The number inside a formula giving how many atoms of an element are in one formula unit.
- Law of conservation of mass
- Matter is neither created nor destroyed in a chemical reaction.
- Algebraic method
- Balancing by solving a system of linear equations, one per element (and one for charge).
Frequently asked questions
How do you balance a chemical equation?
Count the atoms of each element on both sides, then change coefficients (never subscripts) until every count matches. Balance elements that appear in only one compound per side first, and H and O last.
What is the balanced equation for propane combustion?
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. Check: 3 C, 8 H and 10 O on each side.
Why can't I change subscripts?
Subscripts define the substance. Changing H₂O to H₂O₂ balances nothing – it describes hydrogen peroxide instead of water.
How are ionic and redox equations balanced?
The charges must balance too. For MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O the balancer gives 1, 5, 8, 1, 5, 4 – a net charge of +17 on both sides.
What does "cannot be balanced" mean?
Either a formula is wrong or a species is missing (no set of positive coefficients works), or the equation is really two independent reactions added together. Check the formulas and the products.
Does the mass check prove the equation is right?
It proves the equation is balanced – both sides weigh the same per reaction as written – but not that the reaction actually happens as written.
Can I enter fractions or hydrates?
Hydrates such as CuSO4·5H2O (or CuSO4*5H2O) work. The balancer always returns the smallest whole-number coefficients.
What formula does the Chemical Equation Balancer use?
For every element: Σ (coefficient × atoms in reactants) = Σ (coefficient × atoms in products) · for ions also Σ charge (left) = Σ charge (right).