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 of the arrow. Balance the element that appears in the fewest formulas first, leave hydrogen and oxygen for last, then clear any fraction and reduce to the smallest whole numbers.
This guide covers the inspection method on easy reactions, the polyatomic-ion shortcut, the fraction trick that combustion of ethane and butane needs, and the algebraic method for reactions where guessing goes in circles. Every balanced equation below comes with an atom count, so you can see the check as well as the answer.
What a Balanced Equation Means
atoms on the reactant side = atoms on the product side. Atoms of an element = coefficient × subscript, summed over every formula that contains it.Atoms are not created or destroyed in a chemical reaction; they are only rearranged into new substances. A balanced equation is that fact written in symbols, and it's why the total mass of the reactants equals the total mass of the products.
The equation N₂ + H₂ → NH₃ tells you what reacts and what forms, but not in what proportions. Count the atoms: the left side has 2 N and 2 H, and the right side has 1 N and 3 H. Neither element matches, so the equation is unbalanced.
Here is how the counting rule works on a formula with brackets. In 3Ca(OH)₂, the coefficient 3 multiplies everything, and the subscript 2 multiplies everything inside the bracket. That gives 3 Ca, 3 × 2 = 6 O and 3 × 2 = 6 H.
Changing a subscript to make the numbers work. Turning H₂O into H₂O₂ does balance the oxygen in H₂ + O₂ → H₂O, but H₂O₂ is hydrogen peroxide, a different substance. Subscripts are part of a substance's identity, as OpenStax Chemistry 2e section 4.1 explains, so only coefficients may change. The correct answer is 2H₂ + O₂ → 2H₂O.
How to Balance by Inspection, Step by Step
Inspection is organized trial and adjustment. It works for most equations you meet in a first chemistry course, and the steps below keep it from turning into random guessing.
- Write correct formulas
Get every reactant and product formula right first. Balancing a wrong formula gives a confident, wrong answer.
- Make an atom table
List each element with its count on the left and on the right. Treat a polyatomic ion that appears unchanged on both sides (such as
SO₄²⁻orPO₄³⁻) as a single unit. - Start with the element in the fewest formulas
Usually a metal or carbon. Elements that appear in only one formula on each side have only one coefficient to fix them, so they can't be undone later. Some teachers say to start with the most complex formula instead; that usually points to the same element.
- Leave H and O for last
They tend to appear in several formulas, and an element that stands alone (like
O₂orFe) is the easiest to adjust because changing it disturbs nothing else. - Clear fractions and reduce
A fractional coefficient is fine as a step. Multiply every coefficient by the denominator, then divide by any common factor so the set is the smallest whole numbers.
- Recount every element
Fill the atom table again from the final equation. This is the check, and it takes thirty seconds.
Balance N₂ + H₂ → NH₃.
Nitrogen: 2 on the left, 1 on the right. Put a 2 in front of NH₃: N₂ + H₂ → 2NH₃.
Hydrogen: the right now has 2 × 3 = 6 H. The left has 2 per H₂ molecule, so it needs 6 ÷ 2 = 3 molecules: N₂ + 3H₂ → 2NH₃.
Count: N is 2 and 2. H is 3 × 2 = 6 and 2 × 3 = 6.
N₂ + 3H₂ → 2NH₃Balancing with Polyatomic Ions
When a group such as phosphate (PO₄) or sulfate (SO₄) travels through the reaction intact, count it as one unit. You get fewer things to track, and you avoid counting the oxygen in the ion together with the oxygen in water, which is where most errors in these problems start.
Balance Ca(OH)₂ + H₃PO₄ → Ca₃(PO₄)₂ + H₂O.
Calcium appears once on each side: 1 on the left, 3 on the right. Put a 3 in front of Ca(OH)₂.
Phosphate as a unit: 1 on the left, 2 on the right. Put a 2 in front of H₃PO₄: 3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂O.
Now the leftover H and O. The left has 6 OH groups, and the 6 H from 2H₃PO₄. Each OH pairs with one H to make one water, so 6 waters: 3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O.
Full atom count. Ca: 3 and 3. P: 2 and 2. H: 6 + 6 = 12 and 6 × 2 = 12. O: 6 + 8 = 14 and 8 + 6 = 14.
3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂OThe ion shortcut only works while the group stays whole. In a reaction where nitrate turns into NO, as in Example 5 below, nitrogen and oxygen must be counted as separate atoms.
Combustion Reactions and the Fraction Trick
A hydrocarbon burning completely in oxygen always gives carbon dioxide and water. Balance C first, then H, then O last, because O₂ is a pure element and its coefficient changes nothing else. The catch: oxygen arrives in pairs, so when the product side needs an odd number of O atoms, the O₂ coefficient comes out as a half.
Balance C₂H₆ + O₂ → CO₂ + H₂O.
Carbon: 2 on the left, so 2CO₂.
Hydrogen: 6 on the left, 2 per water, so 3H₂O.
Oxygen on the right: 2 × 2 + 3 × 1 = 7 atoms. O₂ supplies 2 per molecule, so it needs 7 ÷ 2 = 7/2 molecules: C₂H₆ + 7/2 O₂ → 2CO₂ + 3H₂O.
Clear the half by doubling every coefficient: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. The numbers 2, 7, 4, 6 share no common factor, so they are already the smallest set.
Count. C: 2 × 2 = 4 and 4. H: 2 × 6 = 12 and 6 × 2 = 12. O: 7 × 2 = 14 and 4 × 2 + 6 = 14.
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂OA balanced equation states a ratio. 1 : 7/2 : 2 : 3 and 2 : 7 : 4 : 6 are the same ratio, just as 1/2 and 2/4 are the same number. Multiplying every coefficient by the same number keeps every element balanced. You can't double just one formula, though, because that changes the ratio.
A quick predictor: in CₓHᵧ combustion, the water coefficient is y/2. If the hydrogen count y is a multiple of 4 (as in methane CH₄ or propane C₃H₈), the oxygen total is even and no fraction appears. If y is even but not a multiple of 4 (ethane C₂H₆, butane C₄H₁₀, hexane C₆H₁₄), expect a half and expect to double.
A camping stove burns butane, C₄H₁₀, completely in air. Write the balanced equation, then find how many oxygen molecules react with 10 butane molecules and how many carbon dioxide molecules form.
Skeleton: C₄H₁₀ + O₂ → CO₂ + H₂O. Carbon gives 4CO₂; hydrogen gives 10 ÷ 2 = 5, so 5H₂O.
Oxygen on the right: 4 × 2 + 5 = 13, so O₂ needs 13/2. Double everything: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.
Count. C: 8 and 8. H: 20 and 20. O: 26 and 16 + 10 = 26.
The coefficients give the molecule ratio C₄H₁₀ : O₂ : CO₂ = 2 : 13 : 8. For 10 butane molecules, scale by 5: 13 × 5 = 65 O₂ and 8 × 5 = 40 CO₂.
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O; 65 O₂ molecules react, 40 CO₂ molecules formThat last step is the reason balancing matters: the coefficients are the ratios every stoichiometry calculation uses, whether you count molecules or moles.
The Algebraic Method for Hard Equations
Some equations resist inspection. Fixing one element breaks another, and you end up going around in circles. The algebraic method removes the guessing: give every coefficient a letter, write one equation per element, and solve. It is the same skill as solving a system of linear equations in three variables, with one extra freedom: because only the ratio matters, you may set one letter to 1.
Balance Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O.
Name the coefficients: a Cu + b HNO₃ → c Cu(NO₃)₂ + d NO + e H₂O.
One equation per element. Cu: a = c. H: b = 2e. N: b = 2c + d. O: 3b = 6c + d + e.
Set a = 1, so c = 1. Then N gives b = 2 + d, and H gives e = b/2 = (2 + d)/2.
Substitute into O: 3(2 + d) = 6 + d + (2 + d)/2. Multiply by 2: 12 + 6d = 12 + 2d + 2 + d, so 3d = 2 and d = 2/3.
Back-substitute: b = 2 + 2/3 = 8/3 and e = 4/3. Multiply every coefficient by 3: a = 3, b = 8, c = 3, d = 2, e = 4.
Count. Cu: 3 and 3. H: 8 and 4 × 2 = 8. N: 8 and 3 × 2 + 2 = 8. O: 8 × 3 = 24 and 3 × 6 + 2 + 4 = 24.
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂ONotice why inspection struggles here: nitrogen ends up in two products, and only 2 of the 8 nitrogen atoms change into NO. The algebra handles that split without any insight on your part. For a refresher on substitution and elimination, see solving systems of equations. For bigger reactions, the same equations can be written as a matrix and row-reduced, as in the matrix operations guide.
Reactions like this one are oxidation-reduction (redox) reactions. Ionic redox equations in acidic or basic solution must also balance charge, and General Chemistry courses teach the half-reaction method for those (see OpenStax Chemistry 2e section 17.1). For neutral molecular equations like the one above, the algebraic method is enough.
Which Method Should You Use?
| Equation type | Method | Example |
|---|---|---|
| Few formulas, small numbers | Inspection, element in fewest formulas first | N₂ + 3H₂ → 2NH₃ |
| A polyatomic ion stays whole | Inspection, ion counted as one unit | 3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O |
| Hydrocarbon combustion | C, then H, then O; double if O₂ gets a half | 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O |
| An element splits between products, or inspection loops | Algebraic: one equation per element, set one letter to 1 | 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O |
| Ionic redox in acidic or basic solution | Half-reaction method (balances charge too) | Taught in AP and General Chemistry |
How to Check Your Answer
The atom table is the main check: one row per element, both sides equal. Then ask two more questions. Are the coefficients the smallest whole numbers? 4C₂H₆ + 14O₂ → 8CO₂ + 12H₂O is balanced, but every number shares a factor of 2, so it is not in simplest form. And did any formula change while you worked? Compare each formula with the original skeleton.
For a second, independent check, add up the masses. Using the IUPAC (CIAAW) abridged standard atomic weights H = 1.008, C = 12.011 and O = 15.999, the molar masses in Example 3 are C₂H₆ = 30.070 g/mol, O₂ = 31.998 g/mol, CO₂ = 44.009 g/mol and H₂O = 18.015 g/mol. Read the coefficients as moles, so mol × g/mol leaves grams; every value is kept to three decimal places, the precision of these atomic weights.
- Reactants:
2 mol × 30.070 g/mol + 7 mol × 31.998 g/mol = 60.140 g + 223.986 g = 284.126 g - Products:
4 mol × 44.009 g/mol + 6 mol × 18.015 g/mol = 176.036 g + 108.090 g = 284.126 g
The masses match, as conservation of mass says they must. An unbalanced equation such as C₂H₆ + O₂ → CO₂ + H₂O fails this test (62.068 g on the left against 62.024 g on the right), so the mass check catches mistakes the eye can miss.
Forgetting that a coefficient multiplies the whole formula. In 2Cu(NO₃)₂ there are 2 × 2 = 4 N and 2 × 2 × 3 = 12 O, not 2 and 6. Students who count only the subscripts inside the bracket get an atom table that looks balanced when it isn't.
Once you've balanced an equation by hand, you can check it against a step-by-step solution: Solver AI on iPhone and iPad lets you snap a photo of a chemistry question and see each step, so you can find exactly where your working went off. The chemistry solver page shows the kinds of problems it covers.
Practice Problems
Balance each equation, then check it with an atom table before looking at the answer.
Fe + O₂ → Fe₂O₃(rust). Answer:4Fe + 3O₂ → 2Fe₂O₃. Fe: 4 and 4. O: 6 and 6.Al + CuSO₄ → Al₂(SO₄)₃ + Cu. Answer:2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu. Al: 2 and 2. Cu: 3 and 3. SO₄: 3 and 3.C₃H₈ + O₂ → CO₂ + H₂O(propane). Answer:C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. C: 3 and 3. H: 8 and 8. O: 10 and 10. No fraction, because 8 is a multiple of 4.C₆H₁₄ + O₂ → CO₂ + H₂O(hexane). Answer:2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O. C: 12 and 12. H: 28 and 28. O: 38 and24 + 14 = 38.KMnO₄ + HCl → KCl + MnCl₂ + Cl₂ + H₂O(use the algebraic method). Answer:2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O. Setting the KMnO₄ coefficient to 1 givese = 5/2for Cl₂, so double. K: 2 and 2. Mn: 2 and 2. O: 8 and 8. H: 16 and 16. Cl: 16 and2 + 4 + 10 = 16.
Change coefficients, never subscripts. Start with the element in the fewest formulas, finish with H and O, double away any half, and prove the result with an atom table. When inspection loops, switch to one equation per element and solve.