Chemical Equation Balancer

A balanced chemical equation has the same number of each type of atom on both sides of the reaction arrow. This reflects the law of conservation of mass: matter is neither created nor destroyed in a chemical reaction. Balancing equations correctly is a prerequisite for any stoichiometric calculation in chemistry.

How to Use the Chemical Equation Balancer

1. Enter the unbalanced equation using standard element symbols and subscript numbers (e.g., H2 + O2 -> H2O).
2. Separate reactants and products with an arrow (use -> or =).
3. Use a plus sign (+) to separate multiple reactants or products.
4. Click Balance to see the equation with correct stoichiometric coefficients.
5. Check the atom count table to verify each element balances on both sides.

The Formula

Balancing works by finding the smallest set of whole-number coefficients that makes the atom count equal on both sides for every element.

The algebraic method assigns variables (a, b, c...) as coefficients to each compound, then writes simultaneous equations, one per element. Solving the system gives the coefficients. For example, for aH2 + bO2 -> cH2O: hydrogen gives 2a = 2c, oxygen gives 2b = c. Setting c=2 gives a=2, b=1, yielding 2H2 + O2 -> 2H2O.

The inspection (trial-and-error) method works by balancing one element at a time, starting with elements that appear in the fewest compounds, saving hydrogen and oxygen for last.

Real-World Example

Balance the combustion of methane: CH4 + O2 -> CO2 + H2O.

1. Carbon: 1 on left, 1 on right. Balanced already.
2. Hydrogen: 4 on left, 2 on right. Place coefficient 2 before H2O: CH4 + O2 -> CO2 + 2H2O.
3. Oxygen: 2 on left, 2 + 2 = 4 on right. Place coefficient 2 before O2: CH4 + 2O2 -> CO2 + 2H2O.
4. Verify: C=1, H=4, O=4 on both sides. Balanced.

The balanced equation is CH4 + 2O2 -> CO2 + 2H2O.

Common Types of Chemical Reactions

Chemical equations fall into several general categories. Combination reactions join two or more substances into one product (A + B -> AB). Decomposition reactions break one compound into simpler products. Single displacement reactions have one element replacing another in a compound. Double displacement reactions involve two compounds exchanging ions. Combustion reactions involve a fuel reacting with oxygen to produce carbon dioxide and water. Knowing the reaction type often provides a shortcut for predicting products before balancing.

Frequently Asked Questions

Why must coefficients be whole numbers? Atoms exist in discrete whole-number quantities, so the coefficients representing moles of substances must also be whole numbers. Fractional coefficients are sometimes used as an intermediate step when balancing complex equations, but the final answer should always be scaled to the smallest whole-number ratio.

Can I change subscripts to balance an equation? No. Subscripts are part of the chemical formula and changing them changes the substance itself. For example, H2O and H2O2 are completely different compounds (water and hydrogen peroxide). You may only change the coefficients in front of each formula.

What is a net ionic equation? A net ionic equation removes spectator ions (ions that appear unchanged on both sides) to show only the species that actually participate in the reaction. For example, the reaction of hydrochloric acid with sodium hydroxide shows Na+ and Cl- as spectator ions; the net ionic equation is simply H+ + OH- -> H2O.

How do I balance redox equations? Redox (oxidation-reduction) equations are best balanced using the half-reaction method. Separate the equation into oxidation and reduction half-reactions, balance atoms and charges in each, then multiply the half-reactions by factors that equalise the electron transfer before adding them together.

Also try these free tools:

• Molar Mass Calculator
• Periodic Table Lookup
• Density Calculator

A second equation, balanced and then checked against the masses

The methane example above balances in three steps. Propane takes the same route and gives more room to see why the order of the steps matters.

Balance C3H8 + O2 -> CO2 + H2O.

1. Carbon: 3 on the left, 1 on the right. Put a coefficient of 3 on carbon dioxide.
2. Hydrogen: 8 on the left, 2 on the right. Put a coefficient of 4 on water.
3. Oxygen: 3 x 2 + 4 x 1 = 10 on the right, so oxygen gas needs a coefficient of 5 on the left.
4. Verify: C = 3, H = 8, O = 10 on both sides.

The balanced equation is C3H8 + 5O2 -> 3CO2 + 4H2O. Atom counting proves the coefficients are right. Molar mass proves the same thing a second way, and the second proof is the one that catches a coefficient that balances atoms but was written against the wrong formula.

Using the abridged standard atomic weights published by the IUPAC Commission on Isotopic Abundances and Atomic Weights, carbon 12.011, hydrogen 1.008 and oxygen 15.999:

• Molar mass of propane: 3 x 12.011 + 8 x 1.008 = 36.033 + 8.064 = 44.097 g/mol
• Molar mass of oxygen gas: 2 x 15.999 = 31.998 g/mol
• Molar mass of carbon dioxide: 12.011 + 2 x 15.999 = 44.009 g/mol
• Molar mass of water: 2 x 1.008 + 15.999 = 18.015 g/mol
• Mass in: 44.097 + 5 x 31.998 = 44.097 + 159.990 = 204.087 g
• Mass out: 3 x 44.009 + 4 x 18.015 = 132.027 + 72.060 = 204.087 g

The two totals agree to the milligram at three decimal places, and they will agree for any correctly balanced equation because the coefficients move whole formula units from one side to the other. A coefficient typed against the wrong formula breaks the agreement at the second decimal, which is a far larger signal than a rounding difference.

A second dataset, with a metal instead of a hydrocarbon. Iron rusts to Fe2O3 when it is written in its simplest form: 4Fe + 3O2 -> 2Fe2O3. Iron has an atomic weight of 55.845 and Fe2O3 comes to 2 x 55.845 + 3 x 15.999 = 111.690 + 47.997 = 159.687 g/mol. The mass check reads 4 x 55.845 + 3 x 31.998 = 223.380 + 95.994 = 319.374 g in, against 2 x 159.687 = 319.374 g out.

Atom counts on both sides, before and after

The atom table is the cheapest check available and it catches the most common class of error, which is a coefficient placed on the correct formula but with the wrong value. Read it element by element.

The third row is the one worth studying. Doubling every coefficient of a balanced equation does keep it balanced, so 2C3H8 + 10O2 -> 6CO2 + 8H2O is correct. Doubling only the oxygen coefficient is not, and the oxygen count alone will not reveal that, because 10 and 10 still match. The carbon and hydrogen columns are what expose it. Balance the elements in the order the inspection method recommends, one element at a time, and finish by reading every column rather than the one that was being fixed.

A worked case where a fractional coefficient appears and then has to be scaled up: ethane burns to 2C2H6 + 7O2 -> 4CO2 + 6H2O. Balancing the carbon and hydrogen first leaves 3.5 oxygen molecules on the right, and doubling the whole equation clears the fraction. Atom counts: carbon 4 and 4, hydrogen 12 and 12, oxygen 14 and 14. The mass check gives 2 x 30.070 + 7 x 31.998 = 60.140 + 223.986 = 284.126 g in, against 4 x 44.009 + 6 x 18.015 = 176.036 + 108.090 = 284.126 g out.

Why the algebraic method always has an answer to give

Assigning a variable to each compound turns the problem into a system of linear equations, one per element. For propane with coefficients a, b, c and d on C3H8, O2, CO2 and H2O:

• Carbon gives 3a = c
• Hydrogen gives 8a = 2d, so d = 4a
• Oxygen gives 2b = 2c + d, which is 2b = 6a + 4a, so b = 5a

Every coefficient is now a multiple of a single free variable. Setting a = 1 gives b = 5, c = 3 and d = 4, the smallest whole-number solution. Any positive multiple of that vector also satisfies the equations, which is exactly the pattern that produced the doubling error in the table above. Balancing gives a one-dimensional family of answers, and the convention is to report the member with the smallest whole numbers.

Two consequences follow from the shape of the system. Where the solution family has one dimension and the entries are rational, a smallest whole-number solution always exists, so any equation that can be balanced at all can be balanced with integers. And where the system has no non-trivial positive solution, no set of coefficients will balance it, because the element counts on the two sides cannot be matched with whole formula units. Solvers that return an error on such an input are reporting a property of the equation, not a limitation of the method.

What the equation tells you, and what it does not

A balanced equation is an accounting statement about atoms. It carries no information beyond that, and the boundary is worth stating plainly.

• It says nothing about whether the reaction proceeds, how fast it goes, or which of several possible products dominates. Those questions belong to thermodynamics and kinetics, and they need data the equation does not contain.
• It balances atoms, not charge. The net ionic form of a reaction in solution is balanced separately, after spectator ions are removed.
• It does not split a redox reaction into half-reactions, which is the form needed to see the electron transfer. The half-reaction method balances atoms and charge together and then equalises the electrons.
• It does not cover nuclear equations, where mass number and atomic number are conserved and a coefficient is a count of nuclei rather than of atoms.
• It ignores state symbols, catalysts, temperature and pressure. Those belong to the conditions written alongside the equation.
• It does not convert between moles and grams. That step needs the molar masses, and the balancer works on the coefficients alone.

A note on the atomic weights used

The molar masses on this page use the abridged standard atomic weights published by the IUPAC Commission on Isotopic Abundances and Atomic Weights: hydrogen 1.008, carbon 12.011 and oxygen 15.999. The commission also publishes the full standard atomic weights, and for carbon and oxygen those are intervals rather than single numbers, [12.0096, 12.0116] for carbon and [15.99903, 15.99977] for oxygen, because the isotopic composition of a sample varies with where it came from. The single-value abridged figures are the ones used in teaching and in routine laboratory work, and they are what makes the two mass totals above agree exactly at three decimal places rather than approximately.