A chemical equation is a recipe written in formulas, and like any recipe it has to account for every ingredient. Atoms are not created or destroyed in a chemical reaction; they are only rearranged. A balanced equation shows the same number of atoms of each element on both sides.

Balancing follows the same loop every time (Figure 1): count the atoms, change a coefficient, count again, until every element matches with the smallest whole numbers.

1. startFormulaswrite them correctly; never change them
2. count the atoms on each side
3. Counteach element, both sides
4. change one coefficient
5. Recountrepeat until every element matches
6. divide by any common factor
7. answerBalancedlowest whole numbers

Change coefficients, never subscripts.Balance H and O last.

Learning Objectives

After this lesson, you should be able to:

1. Balance a chemical equation by inspection.
2. Check a balanced equation by counting atoms of each element.

Key Ideas

1. Atoms Are Conserved

Hydrogen burns in oxygen to form water. The unbalanced equation $\mathrm{H_2} + \mathrm{O_2} \rightarrow \mathrm{H_2\!O}$ has 2 O atoms on the left but only 1 on the right. Two water molecules need two hydrogen molecules:

Now each side has 4 H atoms and 2 O atoms (Figure 2). Because each atom keeps its mass, a balanced equation also conserves mass.

Key idea: balanced means the same number of atoms of every element on both sides of the arrow.

2. Change Coefficients, Never Subscripts

A coefficient is the number written in front of a formula. It multiplies the whole formula: $2\,\mathrm{H_2\!O}$ means two water molecules, 4 H and 2 O atoms in all. A subscript is part of the formula itself. Changing $\mathrm{H_2\!O}$ to $\mathrm{H_2\!O_2}$ does not make more water; it makes hydrogen peroxide, a different substance (Figure 3).

A formula with no coefficient has a coefficient of 1. The final coefficients should be the smallest whole numbers that balance the equation: $4\,\mathrm{H_2} + 2\,\mathrm{O_2} \rightarrow 4\,\mathrm{H_2\!O}$ balances, but every coefficient can be divided by 2.

Key idea: to balance, change only coefficients, and finish with the lowest whole numbers.

3. A Strategy That Works

• Start with an element that appears in only one formula on each side, often a metal or carbon.
• Keep a polyatomic group such as NO₃, SO₄ or PO₄ together when it appears unchanged on both sides (Figure 4).
• Leave H and O, and any element that appears as the free element (O₂, Al), for last.
• If you need half a molecule, such as $\frac{7}{2}\,\mathrm{O_2}$, double every coefficient.
• Recount every element at the end, and divide by any common factor.

Try it: balance an equation

Use the + and − buttons to set the coefficients. The atom counts on each side update as you go; the simulator tells you when every element matches in the smallest whole numbers.

Worked Example 1: Aluminum and Oxygen

Problem: Aluminum metal reacts with oxygen to form aluminum oxide. Balance the equation $\mathrm{Al} + \mathrm{O_2} \rightarrow \mathrm{Al_2\!O_3}$.

Step 1. Count the atoms of each element in the unbalanced equation.

• Al: 1 on the left, 2 on the right
• O: 2 on the left, 3 on the right

Step 2. Balance O first, using the least common multiple of 2 and 3. Six O atoms on each side need 3 $\mathrm{O_2}$ and 2 $\mathrm{Al_2\!O_3}$:

Step 3. Balance Al. The right side now has $2 \times 2 = 4$ Al atoms, so put 4 in front of Al:

Step 4. Check every element.

• $\mathrm{Al}:\ 4$ on the left, $2 \times 2 = 4$ on the right
• $\mathrm{O}:\ 3 \times 2 = 6$ on the left, $2 \times 3 = 6$ on the right

Final Answer: 4 Al + 3 O₂ → 2 Al₂O₃

The coefficients 4, 3 and 2 share no common factor, so they are the lowest whole numbers (Figure 5).

Your turn: find a coefficient

Same method as Worked Example 1: balance the equation, then read the coefficient the question asks for. Remember that no number means 1.

Worked Example 2: Burning Ethane

Problem: Ethane burns completely in oxygen: $\mathrm{C_2\!H_6} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2\!O}$. Balance the equation and find the sum of the coefficients.

Step 1. Balance C. 2 C atoms on the left need 2 $\mathrm{CO_2}$.

Step 2. Balance H. 6 H atoms on the left need 3 $\mathrm{H_2\!O}$:

Step 3. Balance O last. The right side has $2 \times 2 + 3 \times 1 = 7$ O atoms. That would take $\dfrac{7}{2}\,\mathrm{O_2}$, half a molecule, so double every coefficient:

Step 4. Check every element, then add the coefficients.

• $\mathrm{C}:\ 2 \times 2 = 4$ on the left, $4$ on the right
• $\mathrm{H}:\ 2 \times 6 = 12$ on the left, $6 \times 2 = 12$ on the right
• $\mathrm{O}:\ 7 \times 2 = 14$ on the left, $4 \times 2 + 6 \times 1 = 14$ on the right

Final Answer: 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O; sum of coefficients 19

1. Skeleton$\mathrm{C_2\!H_6} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2\!O}$
2. Step 1C: 2 on the left
3. C balanced$\mathrm{C_2\!H_6} + \mathrm{O_2} \rightarrow 2\,\mathrm{CO_2} + \mathrm{H_2\!O}$
4. Step 2H: 6 on the left
5. H balanced$\mathrm{C_2\!H_6} + \mathrm{O_2} \rightarrow 2\,\mathrm{CO_2} + 3\,\mathrm{H_2\!O}$
6. Step 3O: 7 on the right needs half a molecule
7. O balanced, with a fraction$\mathrm{C_2\!H_6} + \frac{7}{2}\,\mathrm{O_2} \rightarrow 2\,\mathrm{CO_2} + 3\,\mathrm{H_2\!O}$
8. Step 3$\times 2$clear the fraction
9. Balanced$2\,\mathrm{C_2\!H_6} + 7\,\mathrm{O_2} \rightarrow 4\,\mathrm{CO_2} + 6\,\mathrm{H_2\!O}$sum of coefficients 19

Exam tip: in a combustion equation, balance C, then H, then O last, because O appears in both products.

Your turn: sum of the coefficients

Same method as Worked Example 2: balance, then add every coefficient, including the 1s.

Worked Example 3: Checking by Counting Atoms

Problem: Calcium hydroxide neutralizes phosphoric acid: $\mathrm{Ca(OH)_2} + \mathrm{H_3\!PO_4} \rightarrow \mathrm{Ca_3\!(PO_4)_2} + \mathrm{H_2\!O}$. Balance it and check the result by counting the atoms of every element.

Step 1. Balance Ca, then the phosphate group. $\mathrm{Ca_3\!(PO_4)_2}$ has 3 Ca and 2 $\mathrm{PO_4}$ groups, so take 3 $\mathrm{Ca(OH)_2}$ and 2 $\mathrm{H_3\!PO_4}$.

Step 2. Balance H with water. The left side has $3 \times 2 + 2 \times 3 = 12$ H atoms, so the right needs 6 $\mathrm{H_2\!O}$:

Step 3. Check: count the atoms of every element on both sides.

• $\mathrm{Ca}:\ 3 \times 1 = 3$ on the left, $1 \times 3 = 3$ on the right
• $\mathrm{O}:\ 3 \times 2 + 2 \times 4 = 14$ on the left, $1 \times 8 + 6 \times 1 = 14$ on the right
• $\mathrm{H}:\ 3 \times 2 + 2 \times 3 = 12$ on the left, $6 \times 2 = 12$ on the right
• $\mathrm{P}:\ 2 \times 1 = 2$ on the left, $1 \times 2 = 2$ on the right

Final Answer: 3 Ca(OH)₂ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 6 H₂O; 3 Ca, 14 O, 12 H and 2 P on each side

Count atoms in the balanced equation as coefficient × subscript. In $\mathrm{Ca_3\!(PO_4)_2}$, the 2 after the parenthesis also multiplies: 8 O atoms (Figure 7).

Your turn: count the atoms

Same check as Worked Example 3: balance, then count one element on either side as coefficient × subscript.

Common Mistakes

Common mistake: Changing a subscript. Writing H₂O₂ to fix the oxygen count makes hydrogen peroxide, not water. Only coefficients may change.

Common mistake: Stopping before every element balances. In Worked Example 2, 2 CO₂ + 3 H₂O fixes C and H, but O is still 2 on the left and 7 on the right. Recount every element after each change.

Common mistake: Not using the lowest whole numbers. 8 Al + 6 O₂ → 4 Al₂O₃ balances, but every coefficient can be divided by 2. The answer is 4, 3, 2.

Common mistake: Leaving out the 1s. A formula with no number in front has a coefficient of 1, and it counts. The coefficients in Worked Example 3 add up to 3 + 2 + 1 + 6 = 12, not 11.

Common mistake: Counting subscripts without coefficients. In Worked Example 3 the formulas alone show 6 O atoms on the left (2 + 4); the balanced equation has 3 × 2 + 2 × 4 = 14. Multiply every subscript by its coefficient.

Key Takeaway

Atoms are conserved, so a balanced equation has the same number of atoms of each element on both sides. Change only the coefficients, balance one element at a time with H and O last, and finish with the smallest whole numbers. Check by counting every element as coefficient × subscript.

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Atomic weights: IUPAC 2021 abridged standard atomic weights. The simulator and the practice problems are audited; see the lesson audit log.