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How to balance chemical equations

To balance a chemical equation, change only the coefficients, the numbers in front of each formula, until every element has the same number of atoms on both sides. Never change the subscripts inside a formula, because that changes the substance. Balance the elements that appear in only one compound first, leave hydrogen and oxygen until last, and finish by recounting every atom.

Why equations must balance

In a chemical reaction, atoms are rearranged into new substances, but no atoms are created or destroyed. This is the law of conservation of mass: the total mass of the products equals the total mass of the reactants. If you burn a candle in a sealed jar, the jar weighs the same afterwards.

A chemical equation is a record of that rearrangement, so it has to show the same number of each kind of atom on the left, the reactants, as on the right, the products. An unbalanced equation describes a reaction that cannot happen.

Coefficients and subscripts

A subscript is the small number inside a formula. In H₂O, the 2 says each water molecule has two hydrogen atoms and one oxygen atom. Subscripts are fixed by the substance's identity: change H₂O to H₂O₂ and you have hydrogen peroxide, a different chemical.

A coefficient is the large number in front of a formula. In 3H₂O, the 3 means three water molecules. To count atoms, multiply the coefficient by each subscript: 3H₂O contains 3 × 2 = 6 hydrogen atoms and 3 × 1 = 3 oxygen atoms. Coefficients are the only numbers you change when balancing; a missing coefficient means 1.

The step-by-step method

This method works for most equations you will meet in middle and high school:

Worked example: burning methane

Methane, CH₄, is the main gas in natural gas. When it burns, it reacts with oxygen to make carbon dioxide and water. This is the pattern for every hydrocarbon burning completely: hydrocarbon plus oxygen gives carbon dioxide plus water.

ProblemBalance CH4+O2→CO2+H2O

  1. C:1 and 1,H:4 and 2,O:2 and 3

    Count first. Carbon balances, but hydrogen and oxygen do not. The right side has 2 oxygen atoms in CO₂ plus 1 in H₂O, which is 3.

  2. CH4+O2→CO2+2H2O

    Hydrogen: 4 on the left, so put a 2 in front of H₂O to get 2 × 2 = 4 on the right.

  3. O right:2+2×1=4

    Recount oxygen. The right side now has 2 in CO₂ and 2 in the two water molecules, so 4.

  4. CH4+2O2→CO2+2H2O

    The left has only O₂, so 2O₂ gives 2 × 2 = 4 oxygen atoms. Final check: C 1 and 1, H 4 and 4, O 4 and 4.

Answercoefficients 1,2,1,2

Worked example: iron rusting

Iron reacts with oxygen to form iron(III) oxide, Fe₂O₃, a major part of rust. Here oxygen has 2 atoms on one side and 3 on the other, so the trick is to find the lowest common multiple, 6.

ProblemBalance Fe+O2→Fe2O3

  1. Fe:1 and 2,O:2 and 3

    Count first. Neither element balances yet.

  2. Fe+3O2→2Fe2O3

    Oxygen comes in 2s on the left and 3s on the right. Make 6 on each side: 3O₂ has 3 × 2 = 6, and 2Fe₂O₃ has 2 × 3 = 6.

  3. Fe right:2×2=4

    The coefficient 2 also doubled the iron on the right, to 4 atoms.

  4. 4Fe+3O2→2Fe2O3

    Put a 4 in front of Fe. Final check: Fe 4 and 4, O 6 and 6.

Answercoefficients 4,3,2

Worked example: burning propane

Propane, C₃H₈, is the fuel in many gas grills. It follows the same pattern as methane, so use the same order: carbon, then hydrogen, then oxygen.

ProblemBalance C3H8+O2→CO2+H2O

  1. C3H8+O2→3CO2+H2O

    Carbon: 3 atoms on the left, so put a 3 in front of CO₂.

  2. C3H8+O2→3CO2+4H2O

    Hydrogen: 8 atoms on the left, and each water has 2, so 4H₂O gives 4 × 2 = 8.

  3. O right:3×2+4×1=10

    Count the oxygen on the right: 6 in the carbon dioxide and 4 in the water.

  4. C3H8+5O2→3CO2+4H2O

    5O₂ gives 5 × 2 = 10 oxygen atoms on the left. Final check: C 3 and 3, H 8 and 8, O 10 and 10.

Answercoefficients 1,5,3,4

When oxygen will not come out even

Sometimes the oxygen needed is an odd number, which O₂ cannot supply with a whole number. Burning ethane, C₂H₆, is the classic case. Balancing carbon and hydrogen gives C₂H₆ + O₂ → 2CO₂ + 3H₂O, with 2 × 2 + 3 = 7 oxygen atoms on the right.

Use a fraction for a moment: 7/2 O₂ supplies 7 oxygen atoms. Then multiply every coefficient by 2 to clear the fraction: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. Check: C 4 and 4, H 12 and 12, O 14 and 14.

Common mistakes

Most wrong answers come from one of these:

How to check your work

Always finish with a full atom inventory. Write each element down the page with two columns, reactants and products, and fill in every count from scratch rather than trusting your earlier working. Every row must match.

Then check the coefficients have no common factor and that every formula is exactly as it was given. To practise, start with combustion of hydrocarbons such as methane, propane and butane, then move to reactions with polyatomic ions, like Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl, where SO₄ moves as one unit.

Questions

What are the steps to balance a chemical equation?

Write correct formulas, count each element's atoms on both sides, adjust coefficients one element at a time, balancing hydrogen and oxygen last, then reduce to the smallest whole numbers and recount.

Why can't you change subscripts when balancing equations?

Subscripts define the substance. Changing them turns one chemical into another, such as water into hydrogen peroxide, so the equation would describe a different reaction.

Why do chemical equations need to be balanced?

Atoms are not created or destroyed in a reaction, so mass is conserved. A balanced equation shows the same atoms before and after.

What is the balanced equation for burning methane?

CH₄ + 2O₂ → CO₂ + 2H₂O. Each side has 1 carbon, 4 hydrogen and 4 oxygen atoms.

Can a coefficient be a fraction?

It can be a temporary step, but final answers normally use the smallest whole numbers. Multiply every coefficient by the denominator to clear the fraction.