# How to balance chemical equations

Canonical: https://duckyhelper.com/learn/chemistry/balancing-chemical-equations/
Updated: 2026-10-01

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.

## 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 \(3\,\mathrm{H_2O}\) 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 \(3\,\mathrm{H_2O}\) there are \(3 \times 2 = 6\) hydrogen atoms and \(3 \times 1 = 3\) 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:

$$
\mathrm{2\,H_2 + O_2 \rightarrow 2\,H_2O}
$$

Check it: hydrogen is \(2 \times 2 = 4\) on the left and \(2 \times 2 = 4\) on the right. Oxygen is 2 on the left and \(2 \times 1 = 2\) 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 \(\mathrm{O_2}\), \(\mathrm{H_2}\) or \(\mathrm{Fe}\), for last, because you can change them without disturbing anything else.
4. If a polyatomic ion such as \(\mathrm{NO_3^-}\), \(\mathrm{SO_4^{2-}}\) or \(\mathrm{PO_4^{3-}}\) appears unchanged on both sides, balance it as one unit.
5. If the last element needs a fraction (like \(\tfrac{7}{2}\,\mathrm{O_2}\)), 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: \(\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}\)

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 \(\mathrm{CO_2}\) to get 3 carbon atoms on the right.

   $$
   \mathrm{C_3H_8 + O_2 \rightarrow 3\,CO_2 + H_2O}
   $$
3. Hydrogen: 8 on the left. Each water has 2, so put 4 in front of \(\mathrm{H_2O}\).

   $$
   \mathrm{C_3H_8 + O_2 \rightarrow 3\,CO_2 + 4\,H_2O}
   $$
4. Oxygen last, because \(\mathrm{O_2}\) stands alone. Right side: \(3 \times 2 + 4 \times 1 = 10\) oxygen atoms. Each \(\mathrm{O_2}\) 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.

   $$
   \mathrm{C_3H_8 + 5\,O_2 \rightarrow 3\,CO_2 + 4\,H_2O}
   $$

Answer: \(\mathrm{C_3H_8 + 5\,O_2 \rightarrow 3\,CO_2 + 4\,H_2O}\), coefficients 1, 5, 3, 4

**Atom count for the balanced propane equation**

| Element | Left side | Right side |
| --- | --- | --- |
| C | 1 × 3 = 3 | 3 × 1 = 3 |
| H | 1 × 8 = 8 | 4 × 2 = 8 |
| O | 5 × 2 = 10 | 3 × 2 + 4 × 1 = 10 |

**Example 2: when you get a fraction**

Problem: Balance the formation of aluminum oxide: \(\mathrm{Al + O_2 \rightarrow Al_2O_3}\)

1. Aluminum: 2 on the right, so put 2 in front of Al.

   $$
   \mathrm{2\,Al + O_2 \rightarrow Al_2O_3}
   $$
2. Oxygen: 3 on the right, 2 per \(\mathrm{O_2}\) on the left. You need \(\tfrac{3}{2}\) of an \(\mathrm{O_2}\).

   $$
   \mathrm{2\,Al + \tfrac{3}{2}\,O_2 \rightarrow Al_2O_3}
   $$
3. Fractions are not allowed in the final answer. Multiply every coefficient by 2.

   $$
   \mathrm{4\,Al + 3\,O_2 \rightarrow 2\,Al_2O_3}
   $$
4. Recount. Al: 4 = \(2 \times 2\). O: \(3 \times 2 = 6\) and \(2 \times 3 = 6\).

Answer: \(\mathrm{4\,Al + 3\,O_2 \rightarrow 2\,Al_2O_3}\), coefficients 4, 3, 2

**Example 3: keep polyatomic ions together**

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

1. The nitrate ion \(\mathrm{NO_3}\) appears unchanged on both sides, so count it as one unit. Left: 2 nitrate groups. Right: 1.
2. Put 2 in front of \(\mathrm{KNO_3}\). Now nitrate is 2 = 2, but potassium is 1 on the left and 2 on the right.

   $$
   \mathrm{Pb(NO_3)_2 + KI \rightarrow PbI_2 + 2\,KNO_3}
   $$
3. Put 2 in front of KI. That fixes potassium (2 = 2) and iodine (2 = 2) in one move.

   $$
   \mathrm{Pb(NO_3)_2 + 2\,KI \rightarrow PbI_2 + 2\,KNO_3}
   $$
4. Recount. Pb 1 = 1, N 2 = 2, O 6 = 6, K 2 = 2, I 2 = 2.

Answer: \(\mathrm{Pb(NO_3)_2 + 2\,KI \rightarrow PbI_2 + 2\,KNO_3}\), coefficients 1, 2, 1, 2

**Example 4: a harder combustion**

Problem: Balance the combustion of ethane: \(\mathrm{C_2H_6 + O_2 \rightarrow CO_2 + H_2O}\)

1. Carbon: put 2 in front of \(\mathrm{CO_2}\). Hydrogen: 6 on the left, so put 3 in front of \(\mathrm{H_2O}\).

   $$
   \mathrm{C_2H_6 + O_2 \rightarrow 2\,CO_2 + 3\,H_2O}
   $$
2. Oxygen on the right: \(2 \times 2 + 3 \times 1 = 7\). That needs \(\tfrac{7}{2}\,\mathrm{O_2}\).
3. Multiply everything by 2 to clear the fraction.

   $$
   \mathrm{2\,C_2H_6 + 7\,O_2 \rightarrow 4\,CO_2 + 6\,H_2O}
   $$
4. Recount. C: 4 = 4. H: 12 = 12. O: 14 = 8 + 6.

Answer: \(\mathrm{2\,C_2H_6 + 7\,O_2 \rightarrow 4\,CO_2 + 6\,H_2O}\), coefficients 2, 7, 4, 6

## Common mistakes and how to fix them

- **Changing a subscript.** Turning \(\mathrm{H_2O}\) into \(\mathrm{H_2O_2}\) makes hydrogen peroxide, a different substance. Fix: only write numbers in front of formulas.
- **Forgetting the parentheses.** \(\mathrm{Ca(NO_3)_2}\) 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.** \(\tfrac{7}{2}\) 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.

> **Tip: Stuck after three tries?**
>
> Put a 2 in front of the most complicated formula and rebalance from there. Most textbook equations work out with coefficients of 6 or less.

**Practice: balance these**

1. Balance \(\mathrm{N_2 + H_2 \rightarrow NH_3}\). Which coefficients are correct?
   A. 1, 1, 2
   B. 1, 3, 2
   C. 2, 3, 2
   D. 1, 3, 1

   Answer: 1, 3, 2. Nitrogen: 2 on the left, so put 2 in front of \(\mathrm{NH_3}\). That gives 6 hydrogen on the right, so put 3 in front of \(\mathrm{H_2}\): \(\mathrm{N_2 + 3\,H_2 \rightarrow 2\,NH_3}\).

2. Balance \(\mathrm{Fe + O_2 \rightarrow Fe_2O_3}\). Which coefficients are correct?
   A. 2, 3, 1
   B. 4, 3, 2
   C. 2, 1, 1
   D. 4, 6, 2

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

3. Balance the combustion of methane: \(\mathrm{CH_4 + O_2 \rightarrow CO_2 + H_2O}\).
   A. 1, 2, 1, 2
   B. 1, 1, 1, 2
   C. 2, 3, 2, 4
   D. 1, 3, 1, 2

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

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

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

5. Balance \(\mathrm{Na_3PO_4 + CaCl_2 \rightarrow Ca_3(PO_4)_2 + NaCl}\).
   A. 1, 3, 1, 3
   B. 2, 3, 1, 6
   C. 2, 3, 1, 3
   D. 3, 2, 1, 6

   Answer: 2, 3, 1, 6. Treat phosphate as a unit: 2 on the right, so \(2\,\mathrm{Na_3PO_4}\). Calcium: 3 on the right, so \(3\,\mathrm{CaCl_2}\). Now there are 6 Na and 6 Cl on the left, so \(6\,\mathrm{NaCl}\).

## Frequently asked questions

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

A subscript is part of what the substance is. \(\mathrm{CO}\) is carbon monoxide, a poison gas, and \(\mathrm{CO_2}\) 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 \(\mathrm{O_2}\) in combustion. Finish balancing with the fraction, then multiply every coefficient in the equation by the denominator. For example, \(\tfrac{7}{2}\) 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

- [OpenStax Chemistry 2e, 4.1 Writing and Balancing Chemical Equations](https://openstax.org/books/chemistry-2e/pages/4-1-writing-and-balancing-chemical-equations), accessed 2026-10-01

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