Chemical equation balancer

Balancing an equation means finding positive whole numbers ${\nu }_j$ so that every element, and the total charge, is the same on both sides. Write one row per element (and one for charge) and one column per species, with the atom counts of reactants positive and of products negative. The coefficients are then a vector in the nullspace of this matrix:

$$A\cdot \nu =0,\text{with }{\nu }_j>0\text{ whole numbers}$$

The tool reduces $A$ to reduced row echelon form by Gauss–Jordan elimination using exact fractions (arbitrary-precision integers), the matrix method described by Blakley (1982). If the nullspace has dimension 1, its basis vector, multiplied by the least common multiple of the denominators and divided by the greatest common divisor, gives the smallest whole-number coefficients. If its dimension is 0, no balance exists (usually an element appears on one side only, or a formula is wrong). If it is 2 or more, the equation combines independent reactions and has infinitely many balances (for example $\mathrm{H}{\phantom{A}}_2+\mathrm{O}{\phantom{A}}_2⟶\mathrm{H}{\phantom{A}}_2\mathrm{O}+\mathrm{H}{\phantom{A}}_2\mathrm{O}{\phantom{A}}_2$); the tool says so rather than picking one. A zero coefficient means that species cannot take part, and a negative one that it belongs on the other side.

By hand for $\mathrm{Fe}+\mathrm{O}{\phantom{A}}_2⟶\mathrm{Fe}{\phantom{A}}_2\mathrm{O}{\phantom{A}}_3$: Fe: $a=2c$; O: $2b=3c$. Take $c=2$: $a=4$, $b=3$, so $4\mathrm{Fe}+3\mathrm{O}{\phantom{A}}_2⟶2\mathrm{Fe}{\phantom{A}}_2\mathrm{O}{\phantom{A}}_3$. Redox reactions need the charge row too: for $\mathrm{MnO}{\phantom{A}}_4{\phantom{A}}^{-}+\mathrm{Fe}{\phantom{A}}^{2+}+\mathrm{H}{\phantom{A}}^{+}$ $⟶\mathrm{Mn}{\phantom{A}}^{2+}+\mathrm{Fe}{\phantom{A}}^{3+}+\mathrm{H}{\phantom{A}}_2\mathrm{O}$ the charge balance $-1+2b+c=2+3b$ forces $b=5$.

Reading formulas. Element symbols are case-sensitive. A charge goes at the end: Fe3+ and Fe^3+ both mean $\mathrm{Fe}{\phantom{A}}^{3+}$. For a polyatomic ion, a single trailing digit is a subscript (NH4+ is $\mathrm{NH}{\phantom{A}}_4{\phantom{A}}^{+}$, MnO4- is $\mathrm{MnO}{\phantom{A}}_4{\phantom{A}}^{-}$), and when two or more digits come before the sign the last is the charge (SO42- is $\mathrm{SO}{\phantom{A}}_4{\phantom{A}}^{2-}$); writing ^ (SO4^2-) is never ambiguous. The equation is typeset with the mhchem package for MathJax.

Sources: Blakley GR. Chemical equation balancing: a general method which is quick, simple, and has unexpected applications. J Chem Educ 1982;59:728–734. Atomic weights: Prohaska T et al. Standard atomic weights of the elements 2021 (IUPAC Technical Report). Pure Appl Chem 2022;94:573–600, doi:10.1515/pac-2019-0603.

Potassium permanganate and hydrochloric acid. Type KMnO4 + HCl = KCl + MnCl2 + H2O + Cl2. The element matrix has five rows (K, Mn, O, H, Cl) and six columns, so its nullspace has dimension 1, and the balancer returns

$$\begin{array}{rl}& 2\mathrm{KMnO}{\phantom{A}}_4+16\mathrm{HCl}⟶2\mathrm{KCl}\\ & +2\mathrm{MnCl}{\phantom{A}}_2+8\mathrm{H}{\phantom{A}}_2\mathrm{O}+5\mathrm{Cl}{\phantom{A}}_2\end{array}$$

Mass balance per mole of reaction:

$$\begin{array}{rl}\text{reactants: }& 2\times 158.032+16\times 36.458\\ & =899.393\mathrm{g}\\ \text{products: }& 2\times 74.548+2\times 125.838\\ & +8\times 18.015+5\times 70.900\\ & =899.393\mathrm{g}\end{array}$$

Other results from the tool

The LaTeX for the first example is \ce{2 KMnO4 + 16 HCl -> 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2}; paste it into the LaTeX tool to change its colour or size.