Balancing the Combustion of Methane
Balance the chemical equation for the complete combustion of methane: CH4 + O2 -> CO2 + H2O. Show the final atom counts to verify the equation is balanced.
Given
Unbalanced equation: CH4 + O2 -> CO2 + H2O
Reaction type: complete combustion of a hydrocarbon
Find
The smallest whole-number coefficients that balance the equation, plus verification.
Principle
The law of conservation of mass says atoms are neither created nor destroyed in a chemical reaction, so each element must appear the same number of times on both sides of the arrow. Balancing adjusts only coefficients (how many molecules), never subscripts (what each molecule is), because changing a subscript changes the substance itself.
Formula
Solution
Count atoms in the unbalanced equation. Left: C = 1, H = 4, O = 2. Right: C = 1, H = 2, O = 3. Carbon is balanced; hydrogen and oxygen are not.
Balance hydrogen: there are 4 H on the left, so place a 2 in front of H2O on the right: CH4 + O2 -> CO2 + 2H2O. Right side is now C = 1, H = 4, O = 4.
Balance oxygen: the right side now has 4 O atoms (2 from CO2, 2 from 2H2O). Place a 2 in front of O2 on the left: CH4 + 2O2 -> CO2 + 2H2O.
Verify. Left: C = 1, H = 4, O = 4. Right: C = 1, H = 2 x 2 = 4, O = 2 + 2 x 1 = 4. All three elements match, and all coefficients are the smallest whole numbers.
Unit check
Balancing is a pure counting check — no units are involved. The verification compares whole atom counts per element on both sides, which is exactly what the conservation law requires.
What this means physically
Chemically, this says one molecule of methane burns with two molecules of oxygen to make one molecule of carbon dioxide and two molecules of water — and the same ratio holds for moles. That ratio is the foundation of all combustion stoichiometry for natural gas.
Students often change subscripts (writing H4O instead of 2H2O) or balance oxygen first and end up with fractional coefficients. Always balance C and H before O in combustions, and only ever touch coefficients.