Chemistry

How to balance chemical equations

Balancing a chemical equation means choosing whole-number coefficients so every element has the same number of atoms on both sides of the arrow. Atoms are not created or destroyed in a reaction, so the equation has to show that. You change only the coefficients in front of the formulas, never the subscripts inside them, because a new subscript makes a different substance.

Updated

On this page

Key ideas

  • Conservation of mass. The same atoms are on both sides of a reaction. They are only rearranged into new substances.
  • Coefficient vs subscript. In 3 H2O3\,\mathrm{H_2O} the 3 is a coefficient: three water molecules. The 2 is a subscript: two hydrogen atoms inside each molecule.
  • Counting atoms. Atoms of an element = coefficient × subscript (× the number outside any parentheses). In 3 H2O3\,\mathrm{H_2O} there are 3×2=63 \times 2 = 6 hydrogen atoms and 3×1=33 \times 1 = 3 oxygen atoms.
  • Smallest whole numbers. The final coefficients should have no common factor. A coefficient of 1 is not written.

A balanced equation reads like a recipe. This one says two hydrogen molecules react with one oxygen molecule to make two water molecules:

2 H2+O2→2 H2O\mathrm{2\,H_2 + O_2 \rightarrow 2\,H_2O}

Check it: hydrogen is 2×2=42 \times 2 = 4 on the left and 2×2=42 \times 2 = 4 on the right. Oxygen is 2 on the left and 2×1=22 \times 1 = 2 on the right.

A method that works every time

  1. Write the correct formula for every reactant and product. Do not touch these formulas again.
  2. Make an atom count for each side: list every element and how many atoms it has on the left and on the right.
  3. Balance elements that appear in only one formula on each side first. Save elements that stand alone, like O2\mathrm{O_2}, H2\mathrm{H_2} or Fe\mathrm{Fe}, for last, because you can change them without disturbing anything else.
  4. If a polyatomic ion such as NO3−\mathrm{NO_3^-}, SO42−\mathrm{SO_4^{2-}} or PO43−\mathrm{PO_4^{3-}} appears unchanged on both sides, balance it as one unit.
  5. If the last element needs a fraction (like 72 O2\tfrac{7}{2}\,\mathrm{O_2}), multiply every coefficient by the denominator.
  6. Recount every element. Then divide by any common factor so the coefficients are the smallest whole numbers.

Worked examples

Example 1: burning propane

Problem Balance the combustion of propane: C3H8+O2→CO2+H2O\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}

  1. Count atoms. Left: C 3, H 8, O 2. Right: C 1, H 2, O 3. Nothing matches yet.
  2. Carbon appears in one formula on each side. Put 3 in front of CO2\mathrm{CO_2} to get 3 carbon atoms on the right.
    C3H8+O2→3 CO2+H2O\mathrm{C_3H_8 + O_2 \rightarrow 3\,CO_2 + H_2O}
  3. Hydrogen: 8 on the left. Each water has 2, so put 4 in front of H2O\mathrm{H_2O}.
    C3H8+O2→3 CO2+4 H2O\mathrm{C_3H_8 + O_2 \rightarrow 3\,CO_2 + 4\,H_2O}
  4. Oxygen last, because O2\mathrm{O_2} stands alone. Right side: 3×2+4×1=103 \times 2 + 4 \times 1 = 10 oxygen atoms. Each O2\mathrm{O_2} has 2, so you need 5.
  5. Recount. C: 3 = 3. H: 8 = 8. O: 10 = 10. The coefficients 1, 5, 3, 4 share no common factor.
    C3H8+5 O2→3 CO2+4 H2O\mathrm{C_3H_8 + 5\,O_2 \rightarrow 3\,CO_2 + 4\,H_2O}

Answer C3H8+5 O2→3 CO2+4 H2O\mathrm{C_3H_8 + 5\,O_2 \rightarrow 3\,CO_2 + 4\,H_2O}, coefficients 1, 5, 3, 4

Atom count for the balanced propane equation
ElementLeft sideRight side
C1 × 3 = 33 × 1 = 3
H1 × 8 = 84 × 2 = 8
O5 × 2 = 103 × 2 + 4 × 1 = 10

Example 2: when you get a fraction

Problem Balance the formation of aluminum oxide: Al+O2→Al2O3\mathrm{Al + O_2 \rightarrow Al_2O_3}

  1. Aluminum: 2 on the right, so put 2 in front of Al.
    2 Al+O2→Al2O3\mathrm{2\,Al + O_2 \rightarrow Al_2O_3}
  2. Oxygen: 3 on the right, 2 per O2\mathrm{O_2} on the left. You need 32\tfrac{3}{2} of an O2\mathrm{O_2}.
    2 Al+32 O2→Al2O3\mathrm{2\,Al + \tfrac{3}{2}\,O_2 \rightarrow Al_2O_3}
  3. Fractions are not allowed in the final answer. Multiply every coefficient by 2.
    4 Al+3 O2→2 Al2O3\mathrm{4\,Al + 3\,O_2 \rightarrow 2\,Al_2O_3}
  4. Recount. Al: 4 = 2×22 \times 2. O: 3×2=63 \times 2 = 6 and 2×3=62 \times 3 = 6.

Answer 4 Al+3 O2→2 Al2O3\mathrm{4\,Al + 3\,O_2 \rightarrow 2\,Al_2O_3}, coefficients 4, 3, 2

Example 3: keep polyatomic ions together

Problem Balance the precipitation reaction: Pb(NO3)2+KI→PbI2+KNO3\mathrm{Pb(NO_3)_2 + KI \rightarrow PbI_2 + KNO_3}

  1. The nitrate ion NO3\mathrm{NO_3} appears unchanged on both sides, so count it as one unit. Left: 2 nitrate groups. Right: 1.
  2. Put 2 in front of KNO3\mathrm{KNO_3}. Now nitrate is 2 = 2, but potassium is 1 on the left and 2 on the right.
    Pb(NO3)2+KI→PbI2+2 KNO3\mathrm{Pb(NO_3)_2 + KI \rightarrow PbI_2 + 2\,KNO_3}
  3. Put 2 in front of KI. That fixes potassium (2 = 2) and iodine (2 = 2) in one move.
    Pb(NO3)2+2 KI→PbI2+2 KNO3\mathrm{Pb(NO_3)_2 + 2\,KI \rightarrow PbI_2 + 2\,KNO_3}
  4. Recount. Pb 1 = 1, N 2 = 2, O 6 = 6, K 2 = 2, I 2 = 2.

Answer Pb(NO3)2+2 KI→PbI2+2 KNO3\mathrm{Pb(NO_3)_2 + 2\,KI \rightarrow PbI_2 + 2\,KNO_3}, coefficients 1, 2, 1, 2

Example 4: a harder combustion

Problem Balance the combustion of ethane: C2H6+O2→CO2+H2O\mathrm{C_2H_6 + O_2 \rightarrow CO_2 + H_2O}

  1. Carbon: put 2 in front of CO2\mathrm{CO_2}. Hydrogen: 6 on the left, so put 3 in front of H2O\mathrm{H_2O}.
    C2H6+O2→2 CO2+3 H2O\mathrm{C_2H_6 + O_2 \rightarrow 2\,CO_2 + 3\,H_2O}
  2. Oxygen on the right: 2×2+3×1=72 \times 2 + 3 \times 1 = 7. That needs 72 O2\tfrac{7}{2}\,\mathrm{O_2}.
  3. Multiply everything by 2 to clear the fraction.
    2 C2H6+7 O2→4 CO2+6 H2O\mathrm{2\,C_2H_6 + 7\,O_2 \rightarrow 4\,CO_2 + 6\,H_2O}
  4. Recount. C: 4 = 4. H: 12 = 12. O: 14 = 8 + 6.

Answer 2 C2H6+7 O2→4 CO2+6 H2O\mathrm{2\,C_2H_6 + 7\,O_2 \rightarrow 4\,CO_2 + 6\,H_2O}, coefficients 2, 7, 4, 6

Common mistakes and how to fix them

  • Changing a subscript. Turning H2O\mathrm{H_2O} into H2O2\mathrm{H_2O_2} makes hydrogen peroxide, a different substance. Fix: only write numbers in front of formulas.
  • Forgetting the parentheses. Ca(NO3)2\mathrm{Ca(NO_3)_2} has 2 nitrogen atoms and 6 oxygen atoms, not 1 and 3. Fix: multiply everything inside the parentheses by the number outside.
  • Balancing oxygen first in a combustion. Oxygen shows up in two products, so it keeps changing. Fix: do C, then H, then O last.
  • Leaving a fraction or a common factor. 72\tfrac{7}{2} and 4, 6, 2 are not final answers. Fix: clear fractions, then divide by any common factor (4, 6, 2 becomes 2, 3, 1).
  • Not recounting. Fixing one element often breaks another. Fix: recount every element after the last change.

Practice: balance these

  1. Balance N2+H2→NH3\mathrm{N_2 + H_2 \rightarrow NH_3}. Which coefficients are correct?

    1. 1, 1, 2
    2. 1, 3, 2
    3. 2, 3, 2
    4. 1, 3, 1
    Show answer

    Answer: 1, 3, 2

    Nitrogen: 2 on the left, so put 2 in front of NH3\mathrm{NH_3}. That gives 6 hydrogen on the right, so put 3 in front of H2\mathrm{H_2}: N2+3 H2→2 NH3\mathrm{N_2 + 3\,H_2 \rightarrow 2\,NH_3}.

  2. Balance Fe+O2→Fe2O3\mathrm{Fe + O_2 \rightarrow Fe_2O_3}. Which coefficients are correct?

    1. 2, 3, 1
    2. 4, 3, 2
    3. 2, 1, 1
    4. 4, 6, 2
    Show answer

    Answer: 4, 3, 2

    Oxygen needs a multiple of 2 and of 3, so aim for 6: 3 O23\,\mathrm{O_2} and 2 Fe2O32\,\mathrm{Fe_2O_3}. That makes 4 iron atoms on the right, so put 4 in front of Fe.

  3. Balance the combustion of methane: CH4+O2→CO2+H2O\mathrm{CH_4 + O_2 \rightarrow CO_2 + H_2O}.

    1. 1, 2, 1, 2
    2. 1, 1, 1, 2
    3. 2, 3, 2, 4
    4. 1, 3, 1, 2
    Show answer

    Answer: 1, 2, 1, 2

    Carbon is already 1 = 1. Hydrogen: 4 on the left, so 2 H2O2\,\mathrm{H_2O}. Oxygen on the right is 2+2=42 + 2 = 4, so 2 O22\,\mathrm{O_2}.

  4. How many oxygen atoms are in 3 Ca(NO3)23\,\mathrm{Ca(NO_3)_2}?

    Show answer

    Answer: 18 oxygen atoms

    Each nitrate has 3 oxygen atoms, each formula unit has 2 nitrates, and there are 3 formula units: 3×2×3=183 \times 2 \times 3 = 18.

  5. Balance Na3PO4+CaCl2→Ca3(PO4)2+NaCl\mathrm{Na_3PO_4 + CaCl_2 \rightarrow Ca_3(PO_4)_2 + NaCl}.

    1. 1, 3, 1, 3
    2. 2, 3, 1, 6
    3. 2, 3, 1, 3
    4. 3, 2, 1, 6
    Show answer

    Answer: 2, 3, 1, 6

    Treat phosphate as a unit: 2 on the right, so 2 Na3PO42\,\mathrm{Na_3PO_4}. Calcium: 3 on the right, so 3 CaCl23\,\mathrm{CaCl_2}. Now there are 6 Na and 6 Cl on the left, so 6 NaCl6\,\mathrm{NaCl}.

Frequently asked questions

Why can't I change the subscripts to balance an equation?

A subscript is part of what the substance is. CO\mathrm{CO} is carbon monoxide, a poison gas, and CO2\mathrm{CO_2} is carbon dioxide. Changing a subscript would describe a different reaction. Coefficients only change how many molecules take part, so they are the only numbers you may adjust.

What do I do if I end up with a fraction?

Fractions are a normal middle step, especially for O2\mathrm{O_2} in combustion. Finish balancing with the fraction, then multiply every coefficient in the equation by the denominator. For example, 72\tfrac{7}{2} becomes 7 and every other coefficient doubles.

Do I need state symbols like (s), (l), (g) and (aq)?

They do not change the balancing, because atoms are counted the same way in any state. Many teachers and the AP exam expect them in final answers, though, because they tell you if a substance is a solid, liquid, gas or dissolved in water. Add them if your class uses them.

Is there a faster way for really hard equations?

Yes, the algebraic method. Give each coefficient a letter, write one equation per element (atoms left = atoms right), set one letter to 1 and solve. It always works, but for most homework problems the inspection method on this page is faster.

Sources

  1. OpenStax Chemistry 2e, 4.1 Writing and Balancing Chemical Equations, accessed October 1, 2026

Try asking Ducky

  • “Look at my worksheet. Is number 4 balanced, and if not, which element is off?”
  • “Walk me through balancing this combustion reaction one element at a time, but let me pick the coefficients.”
  • “My atoms match on both sides but my teacher marked it wrong. What did I miss?”

Free to start. The web app works in any browser, Chromebooks included; the Mac app can also draw on your real screen.

Try it free in your browserGet the Mac app
DuckyHelper
The AI tutor that shows you
FREE TO TRY
Try it free