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Lesson 2 · 25 min

Balancing chemical equations

Conservation of atoms is the only rule: balance by inspection, then write balancing as linear equations and watch the exact solution come out in smallest whole numbers.

In this lesson

  • Explain with conservation of mass (of atoms) why equations must balance
  • Distinguish coefficients from subscripts: balancing changes only coefficients
  • Balance common equations by inspection and check them by counting atoms
  • Write balancing as a system of linear equations and solve hard cases algebraically

In the 1770s Lavoisier heated tin in a sealed retort. The tin turned into a greyish calx and gained weight — yet the sealed vessel as a whole weighed exactly the same: the extra mass came from the air inside. He concluded the law of conservation of mass: in a reaction the total mass of the reactants equals the total mass of the products. Today we know why: a reaction only rearranges atoms; atoms are neither created nor destroyed. A chemical equation must have the same number of each kind of atom on both sides — that is what “balanced” means.

Predict first

Propane burns: CX3HX8+?OX2→3 COX2+4 HX2O\ce{C3H8} + \ce{?O2} \mathrel{\ce{->}} \ce{3CO2} + \ce{4H2O}. What number replaces the question mark?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Coefficients and subscripts

In 2 HX2O\ce{2H2O} the subscript 2 belongs to the formula — “each water molecule has two hydrogen atoms” — while the coefficient 2 in front means “two water molecules”. Balance by changing coefficients, never subscripts: turning HX2O\ce{H2O} into HX2OX2\ce{H2O2} might fix the oxygen, but that is hydrogen peroxide, a different substance.

Coefficients are ratios of particles, and therefore ratios of moles: 2 HX2+OX2→2 HX2O\ce{2H2} + \ce{O2} \mathrel{\ce{->}} \ce{2H2O} reads “two hydrogen molecules and one oxygen molecule give two water molecules”, and equally “2 mol of hydrogen and 1 mol of oxygen give 2 mol of water”. Coefficients are normally the smallest whole numbers.

Worked example

By inspection: iron to iron(III) oxide

Balance Fe+OX2→FeX2OX3\ce{Fe} + \ce{O2} \mathrel{\ce{->}} \ce{Fe2O3}.

Algebra: one equation per element

Some equations — redox ones especially — are hard to “see”. Give each coefficient a letter and write one conservation equation per element. For a Cu+b HNOX3→c Cu(NOX3)X2+d NO+e HX2Oa\,\ce{Cu} + b\,\ce{HNO3} \to c\,\ce{Cu(NO3)2} + d\,\ce{NO} + e\,\ce{H2O}:

  • Cu: a=ca = c
  • H: b=2eb = 2e
  • N: b=2c+db = 2c + d
  • O: 3b=6c+d+e3b = 6c + d + e

Four equations, five unknowns — the coefficients are fixed only as a ratio, so set one of them and solve. Ionic equations add one more equation for conservation of charge. The balancer’s kernel does exactly this: it writes the system as an integer matrix, finds its “nullspace” with exact fractions (never rounding decimals), and scales the result to the smallest positive whole numbers. If the solution is not unique, or only zero, the equation itself is at fault.

∑reactantsνj aij=∑productsνj aijfor every element i∑reactantsνjzj=∑productsνjzj\sum_{\text{reactants}} \nu_j\,a_{ij} = \sum_{\text{products}} \nu_j\,a_{ij} \quad \text{for every element } i \qquad \sum_{\text{reactants}} \nu_j z_j = \sum_{\text{products}} \nu_j z_j
Conservation of each element, and of charge.
νj\nu_j
stoichiometric coefficient of species j
aija_{ij}
atoms of element i in one formula unit of species j
zjz_j
charge of species j (0 for neutral molecules)
Conservation of atoms · coefficients · particle view

Equation balancer

Cu\ce{Cu}
HNOX3\ce{HNO3}
Cu(NOX3)X2\ce{Cu(NO3)2}
NO\ce{NO}
HX2O\ce{H2O}

Not balanced yet: H, N, O differ between the sides.

Atoms of each element: left vs right

  • Cu 1 : 1
  • H 1 : 2
  • N 1 : 3
  • O 3 : 8

Reactants Products

Particle view of one reaction event (current coefficients)
→CuNOOOCuNNOOOOOONOO

Algebra: one equation per element (a, b, c … are the coefficients)

  • Cu: a = c
  • H: b = 2e
  • N: b = 2c + d
  • O: 3b = 6c + d + e

The kernel solves the nullspace exactly and finds: a = 3, b = 8, c = 3, d = 2, e = 4

Atom counts, the balance check and the “balanced equation” are all computed: one equation per element (and charge), the exact nullspace of the coefficient matrix in BigInt fractions, scaled to the smallest positive integers. Molecule shapes in the particle view are schematic; atom colours follow the CPK convention and sizes scale with covalent radii.
Balance copper with dilute nitric acid yourself (the answer is 3, 8, 3, 2, 4), then try permanganate with hydrochloric acid. You can also type any equation — try H2 + O2 -> H2O + H2O2 and see why the kernel calls it “not unique”.

Concept check

Which equation is balanced?

Try the calculation

Balance the combustion of ethane with the smallest whole numbers: CX2HX6+OX2→COX2+HX2O\ce{C2H6} + \ce{O2} \mathrel{\ce{->}} \ce{CO2} + \ce{H2O}. What is the coefficient of OX2\ce{O2}?

Key takeaways

  • Reactions only rearrange atoms, so each element — and, in ionic equations, the charge — must balance.
  • Balance with coefficients, never subscripts; coefficients are ratios of particles and of moles.
  • By inspection: start from the most complex formula and elements that appear once, use least common multiples, then check and reduce.
  • By algebra: one conservation equation per element; a unique ratio is the answer, while no or many solutions mean the equation itself is at fault.
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Data and references

  1. A.-L. Lavoisier, Traité élémentaire de chimie (Paris, 1789).
  2. S. S. Zumdahl, S. A. Zumdahl, Chemistry, 10th ed. (Cengage, 2018), Ch. 3.
  3. R. Blakley, “Chemical equation balancing: a general method which is quick, simple, and has unexpected applications”, J. Chem. Educ. 59, 728 (1982) — the matrix (nullspace) method.