Chemical Equation Balancer

How to balance a chemical equation

To balance a chemical equation, put a whole-number coefficient in front of each formula so that every element has the same number of atoms on both sides, and for ions the same total charge: Fe + O2 = Fe2O3 balances as 4Fe + 3O₂ → 2Fe₂O₃, with 4 iron atoms and 6 oxygen atoms on each side. In symbols, every element obeys Σ_j A_ij ν_j = 0, where A_ij is the number of atoms of element i in one formula of species j and ν_j is that species’ coefficient, counted negative for a reactant and positive for a product. Charge adds Σ_j z_j ν_j = 0, with z_j the charge on each species.

Only the coefficients change. A subscript is part of a formula, so changing one changes the substance: H₂O₂ is hydrogen peroxide, not a way of writing more water. Type an equation above, with or without coefficients, and the calculator returns the smallest whole numbers that balance it, the algebra that found them, and an atom count you can check line by line.

Typing an equation

Write each formula as you would on paper, Ca(OH)2 or CuSO4·5H2O, join the formulas with +, and put = or → between the reactants and the products. -> works too, and ⇌ or <=> marks an equilibrium, which the answer keeps. State symbols go last and are carried through, as in Zn(s) + Cu^2+(aq); write them in lower case, because (S) would be sulfur.

Write a charge after a caret: Fe^3+, SO4^2-, NH4^+, and for a complex ion after its bracket, [Cu(H2O)6]^2+. Braces and brackets work as well, Fe{3+} and Fe(3+), and so do pasted superscripts, Fe³⁺, and the older Fe+3 and Ca++. A bare sign is a charge of one after a letter or a bracket, Na+ and OH-, and after the single-digit subscript of a polyatomic ion, NO3- and NH4+. What the calculator refuses is a sign straight after the digit of a lone element, after two digits, or after a digit that follows a complex ion’s bracket, Fe3+, SO42- or [Cu(H2O)6]2+, because there the digit could be either the charge or a subscript, and the two readings are different ions. One common formula is ambiguous even with a single digit: VO2+ reads as VO₂⁺, so write the vanadyl ion, VO²⁺, as VO^2+. However the charge is typed, the answer writes it the way IUPAC recommends, number before sign, as in Al³⁺; IUPAC’s Green Book lists Al+3 and Al+++ as obsolete forms.

An electron is e-, or just e. Write iron(III) as Fe^3+ rather than Fe(III), which a formula parser reads as iron with three iodine atoms, so the calculator refuses it. Coefficients you type, whole, decimal or written as 1/2, are checked before they are replaced: the answer says whether yours balance and whether they are the smallest whole numbers.

Worked example: propane burning in oxygen

Balance C3H8 + O2 = CO2 + H2O. Give each species an unknown coefficient, a C₃H₈ + b O₂ → c CO₂ + d H₂O, and write one equation for each element, atoms on the left equal to atoms on the right:

• C: 3a = c
• H: 8a = 2d
• O: 2b = 2c + d

Three equations in four unknowns fix the ratios but not the size, so set one coefficient to 1. Taking d = 1 gives a = 1/4, b = 5/4 and c = 3/4. Multiplying by 4, the smallest number that clears the fractions, gives C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. The check is to count: 3 carbon, 8 hydrogen and 10 oxygen atoms on each side, the 10 on the right being 3 × 2 in the carbon dioxide and 4 × 1 in the water.

That is the working the calculator prints, line for line. It is also why the method never fails on an equation that has an answer: each element is one linear equation, and the coefficients are the solution of all of them together.

Balancing by inspection, and why the algebra always works

By hand, the quicker route is inspection. Balance first an element that appears in only one formula on each side, then work outwards, and leave hydrogen and oxygen, which usually turn up everywhere, until last. For Fe + H2O = Fe3O4 + H2, Fe₃O₄ needs 3 iron, so write 3Fe; its 4 oxygens need 4H₂O; and the 8 hydrogens in 4H₂O need 4H₂. The result, 3Fe + 4H₂O → Fe₃O₄ + 4H₂, is what the algebra gives too.

Inspection slows down when every element sits in several formulas. KMnO4 + HCl = KCl + MnCl2 + H2O + Cl2 balances as 2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂, which takes some trial and error by hand, and an equation such as K4Fe(CN)6 + KMnO4 + H2SO4 = KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O needs coefficients of 10, 122, 299, 162, 5, 122, 60, 60 and 188. The algebra does not care how many there are. It solves every element equation at once, in exact fractions rather than decimals, so 299 comes out as 299 and never as 298.99999.

In matrix form the element equations are Aν = 0. A, which the IUPAC Green Book calls the conservation or formula matrix, holds the atom counts with one row per element and one column per species, and the balanced coefficients are its null space. The number of independent reactions an equation contains is the number of species minus the rank of A. That number, with the signs of the solutions, is how the calculator knows whether an equation has one answer, none, or more than one (IUPAC, Quantities, Units and Symbols in Physical Chemistry, 3rd edition, 2007, section 2.10.1, p. 53, after R. A. Alberty, 1991).

Ionic equations and half-equations

An ionic equation has to balance charge as well as atoms, and the calculator treats charge as one more quantity to conserve. MnO4^- + Fe^2+ + H^+ = Mn^2+ + Fe^3+ + H2O balances as MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, the reaction in a permanganate titration of iron(II). The charge is +17 on each side: −1 + 10 + 8 on the left, and 2 + 15 on the right.

A half-equation includes its electrons. Cr2O7^2- + H^+ + e- = Cr^3+ + H2O gives Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Leave the electrons out and the atoms still balance while the charge cannot, which the calculator recognises: for Fe^3+ = Fe^2+ it reports the mismatch and shows the half-equation with electrons added, Fe³⁺ + e⁻ → Fe²⁺.

By hand, a redox equation is balanced in halves. Write the reduction and the oxidation separately; in each, balance the atoms other than oxygen and hydrogen, then oxygen with H₂O, hydrogen with H⁺ and the charge with electrons. Scale the two halves to the same number of electrons and add them, so that the electrons cancel. For the titration above the halves are MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and Fe²⁺ → Fe³⁺ + e⁻: five of the second carry the five electrons the first takes, and their sum is the titration equation. In alkaline solution, finish by adding as many OH⁻ to both sides as there are H⁺, so that each H⁺ and OH⁻ make water, and cancel the water that then appears on both sides.

All 42 half-reactions in the table of standard reduction potentials balance here as written, electrons included, and one of them shows why a half-equation is chemistry as well as counting. Ozone’s, O₃ + 2H⁺ + 2e⁻ → O₂ + H₂O, is balanced, but so are O₃ + 6H⁺ + 6e⁻ → 3H₂O and 2O₃ → 3O₂, and any mixture of them, so atoms and charge alone cannot fix the number of electrons. The table’s 2 belong to the reduction that gives oxygen gas.

When oxygen or hydrogen appears on one side only of an ionic equation, the usual fix in water is to add H₂O for the oxygen and H⁺ for the hydrogen, or OH⁻ and H₂O in alkaline solution, and the calculator suggests that. The charges of the common ions are in the polyatomic ions table.

When there is no answer, or more than one

Some equations cannot be balanced as written, and the calculator says why instead of forcing numbers onto them. An element on one side only is the plainest case: in NaCl = Na2O the chlorine has nowhere to go. A species can be on the wrong side: H2O2 = H2 + H2O balances only as H₂O₂ + H₂ → 2H₂O. Or a species can take no part, like the Cl₂ in NaOH + HCl = NaCl + H2O + Cl2, where the rest balances only with none of it.

Other equations balance in more than one way, because they are two or more reactions written as one. H2 + O2 = H2O + H2O2 is 2H₂ + O₂ → 2H₂O and H₂ + O₂ → H₂O₂ added together, and any mixture of the two is balanced, so the atoms cannot choose the coefficients. The calculator says so and lists the simplest reactions the equation is made of.

A case worth knowing is H2O2 + KMnO4 + H2SO4 = K2SO4 + MnSO4 + O2 + H2O. It contains a reaction of permanganate alone, 4KMnO₄ + 6H₂SO₄ → 2K₂SO₄ + 4MnSO₄ + 5O₂ + 6H₂O, and the decomposition of the peroxide, 2H₂O₂ → O₂ + 2H₂O, so both 5H₂O₂ + 2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5O₂ + 8H₂O and 7H₂O₂ + 2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 6O₂ + 10H₂O are balanced. The first is the one the half-equation method gives, because it assumes every oxygen molecule comes from a peroxide molecule being oxidised. That is chemistry the atom count cannot see.

What the calculator does not do

It balances the equation you give it. It does not predict products, and a balanced equation says only that the atoms add up, not that the reaction happens, goes to completion or runs at a useful rate. It does not assign oxidation states or split a redox reaction into half-equations either, though it balances any half-equation you type.

Formulas are read literally, by the same parser as the molar mass calculator. Abbreviations such as Ph or Et are not understood, and one that is also an element symbol is read as the element, so AcOH is actinium, oxygen and hydrogen rather than acetic acid. An isotope label such as 13C reads as a coefficient, Roman numerals are refused, and every element symbol has to start with a capital letter. The coefficients count moles or particles, never grams. To turn a mass of reactant into a mass of product, take the balanced equation to the limiting reagent calculator, and then to the percent yield calculator to compare the result with what you actually made.

Common mistakes

• Changing a subscript. Writing H₂O₂ where the product is water balances the hydrogen and oxygen at a stroke and describes a different substance. Change coefficients only.
• Writing an element as single atoms. Oxygen, hydrogen, nitrogen and the halogens exist as diatomic molecules. Fe + O = Fe2O3 balances as 2Fe + 3O → Fe₂O₃, but the oxygen that reacts is O₂, and written with it the equation balances as 4Fe + 3O₂ → 2Fe₂O₃.
• Miscounting inside brackets. A subscript after a bracket multiplies everything inside it, so Ca₃(PO₄)₂ holds 2 phosphorus and 8 oxygen atoms. Ca(OH)2 + H3PO4 = Ca3(PO4)2 + H2O balances as 3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O.
• Stopping at fractions or at a common factor. H2 + 1/2O2 = H2O and 4H2 + 2O2 = 4H2O both balance, but the convention is the smallest whole numbers, 2H₂ + O₂ → 2H₂O. Type yours in and the calculator says which you have.
• Forgetting the charge. In Fe^2+ + Cl2 = Fe^3+ + 2Cl^- the atoms balance, but the charge is +2 on the left and +1 on the right. Balanced, it is 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, with +4 on each side.
• Putting a species on both sides. A catalyst or a spectator ion written on both sides cancels out, so there is no coefficient to find for it. Leave it out of the equation.