How to balance a chemical equation

Balancing a chemical equation ensures that the number of atoms for each element is equal on both the reactant and product sides, satisfying the law of conservation of mass. This method applies to any closed-system chemical reaction where the identities and formulas of all reactants and products are known.

The setup

Write the unbalanced skeletal equation. Place the chemical formulas of the reactants on the left and the products on the right, separated by a reaction arrow (\rightarrow). Do not alter the chemical formulas or their subscripts.

The steps

  1. Count the number of atoms of each element on both sides of the equation.
  2. Adjust the stoichiometric coefficients (the numbers placed in front of formulas) to balance the elements one at a time. Start with the most complex molecule or elements that appear in only one reactant and one product.
  3. Balance polyatomic ions as single units if they remain unchanged during the reaction.
  4. Balance diatomic elements (like O2O_2 or H2H_2) last.
  5. If a fractional coefficient is required to balance an element, multiply the entire equation by the denominator to clear the fraction.

Checking the result

Tally the final atom counts for each element on the left and right sides. Both sets of totals must match exactly. Ensure that the coefficients are in the lowest possible ratio of whole numbers. If all coefficients are divisible by a common integer, divide them to simplify.

Common errors

The most frequent error is changing the subscripts within a chemical formula to balance the atoms (e.g., changing H2OH_2O to H2O2H_2O_2). This changes the chemical identity of the substance. Only coefficients may be modified.

Worked example

Balance the following chemical equation for the combustion of ethane: C2H6+O2CO2+H2OC_2H_6 + O_2 \rightarrow CO_2 + H_2O

  1. Initial count: Left (C=2, H=6, O=2); Right (C=1, H=2, O=3).
  2. Balance Carbon: Place a coefficient of 2 before CO2CO_2. C2H6+O22CO2+H2OC_2H_6 + O_2 \rightarrow 2CO_2 + H_2O
  3. Balance Hydrogen: Place a coefficient of 3 before H2OH_2O. C2H6+O22CO2+3H2OC_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O
  4. Count Oxygen: Right side now has (2imes2)+(3imes1)=7(2 imes 2) + (3 imes 1) = 7 oxygen atoms.
  5. Balance Oxygen: Place a coefficient of 72\frac{7}{2} before O2O_2. C2H6+72O22CO2+3H2OC_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O
  6. Clear fraction: Multiply the entire equation by 2. 2C2H6+7O24CO2+6H2O2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O
  7. Final check: Left (C=4, H=12, O=14); Right (C=4, H=12, O=14). The equation is balanced.

FAQ

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References: OpenStax Chemistry 2e, Chapter 4: Stoichiometry of Chemical Reactions · Zumdahl Chemistry 10th Edition, Chapter 3: Stoichiometry

See also