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AP Chemistry Balancing Equations

A five-step method, four worked examples, a particle-count check, and six practice problems with answers.

Almost every calculation in the course starts from a balanced equation, so a balancing slip quietly ruins the stoichiometry, limiting reactant, and equilibrium work that follows. Most guides stop at simple molecular equations. AP Chemistry also asks for equations with polyatomic ions, net ionic equations that must balance in charge as well as atoms, and particulate diagrams that must match the coefficients, so this page covers all three. Where a topic here also appears in the Unit 4 review, that page gives the course context.

A Five-Step Method for Balancing Any Chemical Equation

  1. Write the correct formula for every substance. Get the formulas right first. Once the skeleton equation is written, you never change a subscript; only coefficients change.
  2. Make an atom ledger. List each element with its count on the left and on the right. If a polyatomic ion such as SO42− or OH− appears unchanged on both sides, count it as one unit.
  3. Balance in a sensible order. Start with an element that is in only one substance on each side. Leave oxygen, hydrogen, and any element standing alone (like O2 or Fe) for last, because changing them disturbs less.
  4. Change one coefficient at a time, and recount. If you get stuck at a half, use a fraction, then multiply every coefficient to clear it.
  5. Check three things. Each element (or ion) is equal on both sides, the charge is equal if the equation has ions, and the coefficients are the smallest whole numbers.

Worked Example 1: Balancing a Reaction With an Odd Number of Oxygen Atoms

Balance: Fe + O2 → Fe2O3

Ledger: left Fe 1, O 2; right Fe 2, O 3.
Oxygen is the problem: O2 always supplies an even number of atoms, and Fe2O3 an odd one. Find the least common multiple of 2 and 3, which is 6. Use 3 O2 (6 O) and 2 Fe2O3 (6 O).
Now iron: 2 Fe2O3 contains 4 Fe, so put 4 Fe on the left.
Balanced: 4 Fe + 3 O2 → 2 Fe2O3. Check: Fe 4 = 4, O 6 = 6. ✓

Worked Example 2: Balancing a Combustion Reaction, Including a Fractional Coefficient

Balance the combustion of ethane: C2H6 + O2 → CO2 + H2O

In combustion, balance carbon first, then hydrogen, and oxygen last.
Carbon: 2 C on the left, so 2 CO2.
Hydrogen: 6 H on the left, so 3 H2O.
Oxygen: the right side now has 2 × 2 + 3 = 7 O atoms. O2 supplies 2 per molecule, so the coefficient is 7/2: C2H6 + 7/2 O2 → 2 CO2 + 3 H2O.
Clear the fraction: multiply every coefficient by 2.
Balanced: 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O. Check: C 4 = 4, H 12 = 12, O 14 = 8 + 6 = 14. ✓

Worked Example 3: Balancing a Reaction With Polyatomic Ions Treated as Units

Balance: Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + CaSO4

Sulfate and hydroxide both survive intact, so the ledger lists Al, Ca, SO4, and OH rather than S and O separately.
Left: Al 2, SO4 3, Ca 1, OH 2. Right: Al 1, OH 3, Ca 1, SO4 1.
Aluminum: 2 on the left, so 2 Al(OH)3 (giving 6 OH on the right).
Sulfate: 3 on the left, so 3 CaSO4 (giving 3 Ca on the right).
Calcium and hydroxide: 3 Ca needs 3 Ca(OH)2, which supplies 6 OH to match the 6 on the right.
Balanced: Al2(SO4)3 + 3 Ca(OH)2 → 2 Al(OH)3 + 3 CaSO4. Check: Al 2 = 2, SO4 3 = 3, Ca 3 = 3, OH 6 = 6. ✓

Counting sulfate as a unit turns a messy four-element problem into a two-step one. Reactions like this appear again when you write net ionic equations.

Worked Example 4: Balancing Both Atoms and Charge in a Net Ionic Equation

Balance: Ag+(aq) + Cu(s) → Ag(s) + Cu2+(aq)

Atoms: one Ag and one Cu on each side, so the atoms already match.
Charge: the left side has a total charge of +1 and the right side has +2, so the equation is not balanced.
Put a 2 in front of both silver species: 2 Ag+ gives +2 on the left, and 2 Ag keeps the silver atoms equal.
Balanced: 2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq). Check: Ag 2 = 2, Cu 1 = 1, charge +2 = +2. ✓

An equation that balances in atoms but not in charge is still wrong. The electrons transferred here are what the half-reaction method tracks explicitly.

Checking a Balanced Equation With a Particulate Diagram

A balanced equation also describes how many particles react. The diagram below turns N2 + 3 H2 → 2 NH3 into particles, and a quick count confirms the balance: 2 nitrogen atoms and 6 hydrogen atoms appear in each box, and the molecule counts follow the 1 : 3 : 2 ratio of the coefficients.

Checking a Balanced Equation With a Particle Count: N₂ + 3 H₂ → 2 NH₃ N N H H H H H H N H H H N H H H 1 N₂ + 3 H₂ 2 NH₃ N atoms: 2    H atoms: 6 N atoms: 2    H atoms: 6 Atoms are conserved: 2 N and 6 H on each side, and the particle ratio 1 : 3 : 2 matches the coefficients.

Exam questions use this in both directions: draw the products for a given reactant picture, or pick the picture that matches a balanced equation. Count atoms by element, not molecules, since they can be rearranged. The conventions for drawing these correctly are in the particulate diagrams guide, and the picture changes when one reactant runs out, as covered in limiting reactants.

Six Balancing Practice Problems With Answers

Try each one with the five steps before opening the answer.

  1. Al + O2 → Al2O3
    Show answer4 Al + 3 O2 → 2 Al2O3
  2. C4H10 + O2 → CO2 + H2O
    Show answer2 C4H10 + 13 O2 → 8 CO2 + 10 H2O (oxygen: 26 = 16 + 10)
  3. Na3PO4 + CaCl2 → Ca3(PO4)2 + NaCl
    Show answer2 Na3PO4 + 3 CaCl2 → Ca3(PO4)2 + 6 NaCl (treat PO4 as a unit)
  4. Fe2O3 + CO → Fe + CO2
    Show answerFe2O3 + 3 CO → 2 Fe + 3 CO2
  5. Al(s) + Cu2+(aq) → Al3+(aq) + Cu(s)
    Show answer2 Al + 3 Cu2+ → 2 Al3+ + 3 Cu (charge +6 = +6)
  6. C6H12O6 + O2 → CO2 + H2O
    Show answerC6H12O6 + 6 O2 → 6 CO2 + 6 H2O (oxygen: 6 + 12 = 12 + 6)

Common Equation Balancing Mistakes

A balanced equation is the starting line for every amount calculation, the stoichiometry ratios included. If you want to see how these skills feed into your overall score, the AP Chem Score Calculator lets you test a few scenarios.

Frequently Asked Questions

Why can you change coefficients but not subscripts when balancing?

Subscripts are part of a substance's identity: H2O is water, H2O2 is hydrogen peroxide. Changing a subscript changes what the substance is. Coefficients only change how many of that substance are present, which is exactly what balancing adjusts.

How do you balance an equation that contains polyatomic ions?

If a polyatomic ion such as SO4 2- or OH- appears unchanged on both sides, balance it as a single unit instead of counting its individual atoms. Count the whole ion on each side and adjust coefficients until the ion counts match.

Can a balanced equation have fractional coefficients?

As an intermediate step, yes: fractions such as 7/2 O2 are a convenient way to balance combustion reactions. The final equation is normally written with the smallest whole-number coefficients, so multiply every coefficient by the denominator to clear the fraction.

What does it mean for an ionic equation to be balanced in charge?

The total charge on the reactant side must equal the total charge on the product side, in addition to equal numbers of each kind of atom. For example, 2 Ag+ + Cu -> 2 Ag + Cu2+ has a total charge of +2 on both sides.

Why must an equation be balanced before doing stoichiometry?

The mole ratios used in stoichiometry come directly from the coefficients of the balanced equation. Using an unbalanced equation gives wrong ratios and therefore wrong amounts of product or reactant.

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