Balance Equation Calculator

A chemical equation is balanced when every element has the same number of atoms on both sides, and the total charge matches too. That requirement is the law of conservation of mass written in symbols: atoms are rearranged by a reaction, never created or destroyed.

Balancing means choosing coefficients — the whole numbers in front of each formula:

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

Only coefficients may change. Subscripts are part of a substance's identity: turning $\mathrm{H_2O}$ into $\mathrm{H_2O_2}$ balances the oxygen but replaces water with hydrogen peroxide, which is a different reaction entirely.

What balancing does not tell you. The coefficients say nothing about how fast the reaction goes, whether it goes at all, or what conditions it needs. They also assume the products are already known and correctly written — balancing cannot invent a product formula for you.

Why it matters. Every stoichiometric calculation reads its mole ratio off these coefficients, so an unbalanced equation makes every mass, volume and yield derived from it wrong.

Balancing by inspection

1. Start with the element that appears in the fewest formulas — usually a metal or carbon — and leave elements that appear alone (like $\mathrm{O_2}$ or $\mathrm{H_2}$) for last, since a lone element can absorb any leftover.
2. Balance polyatomic ions as a unit when they survive the reaction intact; treating $\mathrm{SO_4}$ as one item saves several steps.
3. Balance hydrogen next, oxygen last.
4. Clear fractions at the end by multiplying everything by the common denominator.
5. Recount every element in the final equation.

The algebraic method

For a stubborn equation, assign a variable to each coefficient and write one equation per element:

$a\,\mathrm{Al} + b\,\mathrm{HCl} \to c\,\mathrm{AlCl_3} + d\,\mathrm{H_2}$

gives $\mathrm{Al}: a = c$, $\mathrm{H}: b = 2d$, $\mathrm{Cl}: b = 3c$. The system is underdetermined by one degree of freedom — set the simplest variable to 1, solve, then scale to the smallest set of whole numbers. This always works, including for redox equations where inspection stalls.

Checking the answer

Tabulate atom counts on each side, and confirm the coefficients share no common factor. $2\mathrm{H_2} + \mathrm{O_2} \to 2\mathrm{H_2O}$ is balanced; $4\mathrm{H_2} + 2\mathrm{O_2} \to 4\mathrm{H_2O}$ is balanced but not in lowest terms.

• Changing a subscript to make the count work. That changes the substance. Only the coefficients out front are adjustable.
• Balancing oxygen or hydrogen first. These appear in the most formulas, so fixing them early guarantees you will undo the work later.
• Leaving a fractional coefficient. $\mathrm{Fe} + \tfrac{3}{2}\mathrm{O_2} \to \mathrm{Fe_2O_3}$ is a legitimate intermediate step, but multiply through to reach whole numbers.
• Not reducing to lowest terms. Coefficients with a common factor are conventionally divided down.
• Forgetting charge in an ionic equation. Net charge must balance as well as atoms.
• Missing atoms inside brackets. $\mathrm{Ca(OH)_2}$ contributes two oxygens and two hydrogens.
• Assuming there is always a solution. If no set of coefficients works, the products themselves are usually written wrongly.