Balancing Chemical Equations Step by Step

What coefficients may change and subscripts may not

Each formula fixes the composition of one substance. H₂O always has two hydrogen atoms and one oxygen atom; changing it to H₂O₂ describes hydrogen peroxide, a different compound. Balancing is therefore done only with coefficients, the numbers in front of formulas, which say how many formula units of each substance take part.

A balanced equation has the same number of atoms of each element on both sides. Because atoms keep their mass, the total mass of reactants then equals the total mass of products, which makes a useful independent check. When ions appear, the sum of charges on the left must also equal the sum on the right.

Coefficients are written as the smallest whole numbers that balance the equation. 4 H₂ + 2 O₂ → 4 H₂O is balanced but not in lowest terms; dividing by 2 gives 2 H₂ + O₂ → 2 H₂O. A coefficient of 1 is left unwritten.

An inspection order that avoids going in circles

Inspection works best with a fixed order, because balancing one element can unbalance another you already settled. The order below postpones elements that appear in several formulas, and especially free elements such as O₂ or a metal, because changing a free element's coefficient disturbs nothing else.

The final tally matters most. List each element with its count on each side, multiplying every subscript by the coefficient and by any subscript outside parentheses. It takes a minute and catches nearly every slip.

• Write correct formulas for every substance, including diatomic elements (H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂), and do not change them afterward.
• Give the most complex formula a trial coefficient of 1.
• Balance elements that appear in only one reactant and one product.
• Balance polyatomic ions that pass through unchanged as single units.
• Balance hydrogen and oxygen next, and free elements last.
• Clear any fraction by multiplying every coefficient by its denominator, then reduce to the smallest whole numbers.
• Tally every element, and the total charge, on both sides.

Worked example: combustion of 2-propanol

2-Propanol (isopropyl alcohol), C₃H₈O, burns in oxygen to give carbon dioxide and water: C₃H₈O + O₂ → CO₂ + H₂O. Give C₃H₈O a coefficient of 1. Carbon appears in one reactant and one product, so 3 C requires 3 CO₂. Hydrogen likewise: 8 H requires 4 H₂O.

Oxygen comes last because O₂ is a free element. The right side now has 3 × 2 + 4 × 1 = 10 O atoms. The fuel already supplies 1 O, so O₂ must supply 9 atoms, a coefficient of 9/2. Multiplying everything by 2 gives 2 C₃H₈O + 9 O₂ → 6 CO₂ + 8 H₂O.

Tally: C 6 = 6; H 16 = 16; O 2 + 18 = 20 on the left and 12 + 8 = 20 on the right. A mass check using IUPAC standard atomic weights rounded to two decimals (C 12.01, H 1.01, O 16.00) agrees for the molar amounts shown: 2(60.11 g) + 9(32.00 g) = 408.22 g of reactants and 6(44.01 g) + 8(18.02 g) = 408.22 g of products.

A common slip is to forget the oxygen atom inside the fuel and write 5 O₂ for each C₃H₈O. The tally exposes it at once: 11 O atoms on the left and 10 on the right.

Polyatomic ions as single units

When a polyatomic ion such as sulfate, nitrate or phosphate appears unchanged on both sides, count it as one unit instead of counting its atoms separately. This shortens the bookkeeping and avoids the most frequent counting error, forgetting that a subscript outside parentheses multiplies everything inside: Ca(NO₃)₂ contains 2 N and 6 O.

Worked example: barium hydroxide reacts with phosphoric acid to give barium phosphate and water, Ba(OH)₂ + H₃PO₄ → Ba₃(PO₄)₂ + H₂O. Start with Ba₃(PO₄)₂. It contains 3 Ba, so use 3 Ba(OH)₂, and 2 phosphate units, so use 2 H₃PO₄.

For the water, think of it as H–OH: each molecule forms from one H of the acid and one OH of the base. The left side now carries 3 × 2 = 6 OH groups and 2 × 3 = 6 acidic H atoms, so 6 H₂O form: 3 Ba(OH)₂ + 2 H₃PO₄ → Ba₃(PO₄)₂ + 6 H₂O. Tally: Ba 3 = 3; P 2 = 2; H 6 + 6 = 12 and 6 × 2 = 12; O 6 + 8 = 14 and 8 + 6 = 14.

Use the unit approach only when the ion survives intact. If nitrate is converted to NO in a redox reaction, for example, nitrogen and oxygen must be counted separately.

Balancing charge in ionic equations

An ionic equation can balance atoms and still be wrong. Consider aluminum metal placed in a solution of copper(II) ions: Al(s) + Cu²⁺(aq) → Al³⁺(aq) + Cu(s). There is one Al and one Cu on each side, but the charge is 2+ on the left and 3+ on the right.

Each Al atom loses 3 electrons and each Cu²⁺ ion gains 2. The least common multiple is 6, so 2 Al react with 3 Cu²⁺: 2 Al(s) + 3 Cu²⁺(aq) → 2 Al³⁺(aq) + 3 Cu(s). Charge: 3(2+) = 6+ on the left and 2(3+) = 6+ on the right; atoms Al 2 = 2 and Cu 3 = 3.

Electrons do not appear in an overall equation, so a charge mismatch always means the equation is not yet balanced: a coefficient is wrong, or a species such as H⁺, OH⁻ or H₂O is missing. For more involved redox equations, particularly in acidic or basic solution, the half-reaction method is more reliable than inspection.

The algebraic method for stubborn equations

Some equations, often redox reactions in which one element ends up in several products, resist inspection. Assign a letter to each coefficient, write one equation per element, set one letter to 1, and solve. For potassium permanganate and hydrochloric acid: a KMnO₄ + b HCl → c KCl + d MnCl₂ + e Cl₂ + f H₂O.

Potassium gives a = c; manganese gives a = d; oxygen gives 4a = f; hydrogen gives b = 2f; chlorine gives b = c + 2d + 2e. Setting a = 1 gives c = 1, d = 1, f = 4 and b = 8, and then 8 = 1 + 2 + 2e, so e = 5/2. Multiplying by 2: 2 KMnO₄ + 16 HCl → 2 KCl + 2 MnCl₂ + 5 Cl₂ + 8 H₂O. Tally: K 2, Mn 2, O 8, H 16 and Cl 16 on each side.

If, after you set one letter to 1, the element equations still leave a coefficient undetermined, the equation has more than one independent balanced form. That usually means it combines two separate reactions, and extra chemical information, such as which species is oxidized and which is reduced in each half-reaction, is needed to choose the right one.

Balancing errors and the tally that catches them

Each of the errors below is caught by the same final tally, element by element and then charge. Make the tally a fixed last line of every equation, not an optional extra.

If the tally keeps failing for the same element, return to the inspection order and check whether that element was balanced too early.

• Changing a subscript to force balance, such as writing H₂O₂ for water. Compare every formula with the unbalanced equation you started from.
• Writing a diatomic element as single atoms, such as O instead of O₂.
• Forgetting to multiply through parentheses, so Ba₃(PO₄)₂ is counted as having 4 O instead of 8.
• Leaving fractions, or coefficients with a common factor, in the final answer.
• Balancing atoms but not charge in an equation with ions.
• Balancing a free element first and then having to redo it after every other change.