Chemical Equation Balancer

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balance combustion equation

How to balance a combustion equation (methane, propane, ethanol, glucose)

In complete combustion carbon becomes CO2 and hydrogen becomes H2O, and only oxygen takes any effort. Type the fuel into the tool and check the coefficients, or use the rule below to get them by hand.

The oxygen rule in complete burning

For a hydrocarbon CxHy, x carbon atoms become x molecules of CO2 and y hydrogen atoms become y/2 molecules of H2O. The oxygen needed is the number of O atoms in the products, 2x + y/2, divided by 2, that is x + y/4 molecules of O2. For methane, CH4, that gives 1 + 1 = 2. For propane, C3H8, it gives 3 + 2 = 5. When the result is a fraction, multiply everything by 2: butane, C4H10, needs 4 + 2.5 = 6.5 O2 and becomes 2C4H10 + 13O2 → 8CO2 + 10H2O; octane, C8H18, needs 12.5 and becomes 2C8H18 + 25O2 → 16CO2 + 18H2O.
If the molecule already carries oxygen, it counts in your favor and the O2 from the air has to supply less. For CxHyOz the value is x + y/4 - z/2. Ethanol, C2H6O, uses 2 + 1.5 - 0.5 = 3 O2, and glucose, C6H12O6, uses 6 + 3 - 3 = 6 O2, which means each mole of sugar consumes exactly 6 mol of oxygen and returns 6 mol of CO2 and 6 mol of water, in burning as in cellular respiration. By mass, 1 g of propane (44.097 g/mol) needs 5 x 31.998 / 44.097 = 3.63 g of O2, and that proportion is what the Stoichiometry tab computes for you.
With too little oxygen the burning is incomplete: part of the carbon leaves as carbon monoxide, CO, or as soot, C, instead of CO2. Methane burning only as far as CO gives 2CH4 + 3O2 → 2CO + 4H2O, with 1.5 O2 per fuel molecule instead of 2. The tool balances whichever scheme you type, but it does not decide which one occurs, because that depends on how much oxygen is available. If you write CO2 and CO together among the products, the equation has two independent solutions and the tool says so instead of picking one.

Combustion equations balanced by the tool

Methane, natural gas

Input: CH4 + O2 -> CO2 + H2O
Expected output: CH4 + 2O2 -> CO2 + 2H2O

x + y/4 = 1 + 1 = 2 mol of O2 per mol of CH4.

Propane, cooking gas

Input: C3H8 + O2 -> CO2 + H2O
Expected output: C3H8 + 5O2 -> 3CO2 + 4H2O

3 + 2 = 5 mol of O2, and 4 H2O come from the 8 hydrogens.

Butane, the lighter

Input: C4H10 + O2 -> CO2 + H2O
Expected output: 2C4H10 + 13O2 -> 8CO2 + 10H2O

6.5 O2 per molecule forces every coefficient to be doubled.

Octane, in gasoline

Input: C8H18 + O2 -> CO2 + H2O
Expected output: 2C8H18 + 25O2 -> 16CO2 + 18H2O

12.5 O2 per molecule, doubled as well.

Ethanol, with oxygen in the fuel

Input: C2H5OH + O2 -> CO2 + H2O
Expected output: C2H5OH + 3O2 -> 2CO2 + 3H2O

x + y/4 - z/2 = 2 + 1.5 - 0.5 = 3.

Incomplete combustion of methane

Input: CH4 + O2 -> CO + H2O
Expected output: 2CH4 + 3O2 -> 2CO + 4H2O

The carbon product is CO, and oxygen drops from 2 to 1.5 per CH4.

Full tool FAQ

How do you balance a chemical equation?

Count the atoms of each element on both sides and adjust the coefficients (never the subscripts) until the counts match. By hand this is done by trial; the tool solves the exact linear system. For Fe + O2 → Fe2O3 the answer is 4Fe + 3O2 → 2Fe2O3, with 4 Fe atoms and 6 O atoms on each side.

Why does an equation need to be balanced?

How does balancing by trial and error work?

How do you balance equations with ions?

How do you balance a redox reaction?

What does "more than one independent solution" mean?

Why does the tool say a substance is on the wrong side or takes no part?

Can I type the equation with coefficients?

How do you find the limiting reagent and the yield?

Does the tool predict the products of a reaction?

Frequently asked questions

How can I know how much O2 to use without trying values?

For CxHyOz the count is x + y/4 - z/2 molecules of O2 per fuel molecule. If it comes out as a half, double everything. For pentane, C5H12: 5 + 3 = 8, so C5H12 + 8O2 → 5CO2 + 6H2O.

Why does butane show coefficients 2 and 13?

Because one C4H10 molecule needs 6.5 molecules of O2, and there is no half molecule in an equation balanced with whole numbers. Multiplying everything by 2 solves it: 2 butane, 13 oxygen, 8 CO2 and 10 H2O.

What changes in incomplete combustion?

Carbon ends up as CO (or C) and O2 use drops. For propane, the path to CO is 2C3H8 + 7O2 → 6CO + 8H2O, with 3.5 O2 per molecule against 5 in complete burning. Write the scheme you want and the tool checks the atoms.

Next steps

Atom count, before and after the coefficients
Element	Before (all 1)	After	Check
C	3	1	3	3	✓
H	8	2	8	8	✓
O	2	3	10	10	✓

Atom count, before and after the coefficients

Element

Before (all 1)

After

Check

React.

Prod.

React.

Prod.

✓

The oxygen rule in complete burning

For a hydrocarbon CxHy, x carbon atoms become x molecules of CO2 and y hydrogen atoms become y/2 molecules of H2O. The oxygen needed is the number of O atoms in the products, 2x + y/2, divided by 2, that is x + y/4 molecules of O2. For methane, CH4, that gives 1 + 1 = 2. For propane, C3H8, it gives 3 + 2 = 5. When the result is a fraction, multiply everything by 2: butane, C4H10, needs 4 + 2.5 = 6.5 O2 and becomes 2C4H10 + 13O2 → 8CO2 + 10H2O; octane, C8H18, needs 12.5 and becomes 2C8H18 + 25O2 → 16CO2 + 18H2O.

If the molecule already carries oxygen, it counts in your favor and the O2 from the air has to supply less. For CxHyOz the value is x + y/4 - z/2. Ethanol, C2H6O, uses 2 + 1.5 - 0.5 = 3 O2, and glucose, C6H12O6, uses 6 + 3 - 3 = 6 O2, which means each mole of sugar consumes exactly 6 mol of oxygen and returns 6 mol of CO2 and 6 mol of water, in burning as in cellular respiration. By mass, 1 g of propane (44.097 g/mol) needs 5 x 31.998 / 44.097 = 3.63 g of O2, and that proportion is what the Stoichiometry tab computes for you.

With too little oxygen the burning is incomplete: part of the carbon leaves as carbon monoxide, CO, or as soot, C, instead of CO2. Methane burning only as far as CO gives 2CH4 + 3O2 → 2CO + 4H2O, with 1.5 O2 per fuel molecule instead of 2. The tool balances whichever scheme you type, but it does not decide which one occurs, because that depends on how much oxygen is available. If you write CO2 and CO together among the products, the equation has two independent solutions and the tool says so instead of picking one.

Combustion equations balanced by the tool

Methane, natural gas

Input: CH4 + O2 -> CO2 + H2O
Expected output: CH4 + 2O2 -> CO2 + 2H2O

x + y/4 = 1 + 1 = 2 mol of O2 per mol of CH4.

Propane, cooking gas

Input: C3H8 + O2 -> CO2 + H2O
Expected output: C3H8 + 5O2 -> 3CO2 + 4H2O

3 + 2 = 5 mol of O2, and 4 H2O come from the 8 hydrogens.

Butane, the lighter

Input: C4H10 + O2 -> CO2 + H2O
Expected output: 2C4H10 + 13O2 -> 8CO2 + 10H2O

6.5 O2 per molecule forces every coefficient to be doubled.

Octane, in gasoline

Input: C8H18 + O2 -> CO2 + H2O
Expected output: 2C8H18 + 25O2 -> 16CO2 + 18H2O

12.5 O2 per molecule, doubled as well.

Ethanol, with oxygen in the fuel

Input: C2H5OH + O2 -> CO2 + H2O
Expected output: C2H5OH + 3O2 -> 2CO2 + 3H2O

x + y/4 - z/2 = 2 + 1.5 - 0.5 = 3.

Incomplete combustion of methane

Input: CH4 + O2 -> CO + H2O
Expected output: 2CH4 + 3O2 -> 2CO + 4H2O

The carbon product is CO, and oxygen drops from 2 to 1.5 per CH4.

Full tool FAQ

How do you balance a chemical equation?

Why does an equation need to be balanced?

How does balancing by trial and error work?

How do you balance equations with ions?

How do you balance a redox reaction?

What does "more than one independent solution" mean?

Why does the tool say a substance is on the wrong side or takes no part?

Can I type the equation with coefficients?

How do you find the limiting reagent and the yield?

Does the tool predict the products of a reaction?

Frequently asked questions

How can I know how much O2 to use without trying values?

For CxHyOz the count is x + y/4 - z/2 molecules of O2 per fuel molecule. If it comes out as a half, double everything. For pentane, C5H12: 5 + 3 = 8, so C5H12 + 8O2 → 5CO2 + 6H2O.