1.9.2 Balancing Chemical Equations

Adam Maltese; Joey Wu; and OpenStax

Learning Objectives

  1. Define chemical equation.
  2. Identify the parts of a chemical equation.
  3. Balance simple chemical equations.

Balancing Equations

This figure shows a balanced chemical equation followed below by a representation of the equation using space-filling models. The equation reads C H subscript 4 plus 2 O subscript 2 arrow C O subscript 2 plus 2 H subscript 2 O. Under the C H subscript 4, the molecule is shown with a central black sphere, representing a C atom, to which 4 smaller white spheres, representing H atoms, are distributed evenly around. All four H atoms are bonded to the central black C atom. This is followed by a plus sign. Under the 2 O subscript 2, two molecules are shown. The molecules are each composed of two red spheres bonded together. The red spheres represent O atoms. To the right of an arrow and under the C O subscript 2, appears a single molecule with a black central sphere with two red spheres bonded to the left and right. Following a plus sign and under the 2 H subscript 2 O, are two molecules, each with a central red sphere and two smaller white spheres attached to the lower right and lower left sides of the central red sphere. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms.

Figure 1. The reaction between methane and oxygen to yield carbon dioxide and water (shown at bottom) may be represented by a chemical equation using formulas (top).

When we see a chemical equation, it should be balanced, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, in Figure 1, both product species in the example reaction, CO2 and H2O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

(2 O atoms in 1 CO2 molecule) + (1 O atom in 2 H2O) = 4 O atoms

 

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:

CH4 + 2O2 -> CO2 + 2H2O

Element Reactants Products Balanced?
C 1 × 1 = 1 1 ×1 = 1 1 = 1, yes
H 4 × 1 = 4 2 × 2 = 4 4 = 4, yes
O 2 × 2 = 4 (1 ×2) + (2 × 1) = 4 4 = 4, yes

A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:

H2O -> H2 + O2 (unbalanced)

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

Element Reactants Products Balanced?
H 1 × 2 = 2 1 × 2 = 2 2 = 2, yes
O 1 × 1 = 1 1 × 2 = 2 1 ≠ 2, no

The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H2O to H2O2 would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H2O to 2.

2H2O -> H2 + O2 (unbalanced)

Element Reactants Products Balanced?
H 2 × 2 = 4 1 × 2 = 2 4 ≠ 2, no
O 2 × 1 = 2 1 × 2 = 2 2 = 2, yes

The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H2 product to 2.

2H2O -> 2H2 + O2 (unbalanced)
Element Reactants Products Balanced?
H 2 × 2 = 4 2 × 2 = 4 4 = 4, yes
O 2 × 1 = 2 1 × 2 = 2 2 = 2, yes

These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:

2H2O -> 2H2 + O2

Examples

Write a balanced equation for the reaction of molecular nitrogen (N2) and oxygen (O2) to form dinitrogen pentoxide (N2O5).

Solution

First, write the unbalanced equation.

N2 + O2 -> N2O5 (unbalanced)

Next, count the number of each type of atom present in the unbalanced equation.

Element Reactants Products Balanced?
N 1 × 2 = 2 1 × 2 = 2 2 = 2, yes
O 1 × 2 = 2 1 × 5 = 5 2 ≠ 5, no

Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O2 and N2O5 to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).

N2 + 5O2 -> 2N2O5 (unbalanced)

Element Reactants Products Balanced?
N 1 × 2 = 2 2 × 2 = 4 2 ≠ 4, no
O 5 × 2 = 10 2 × 5 = 10 10 = 10, yes

The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N2 to 2.

 

2N2 + 5O2 -> 2N2O5

Element Reactants Products Balanced?
N 2 × 2 = 4 2 × 2 = 4 4 = 4, yes
O 5 × 2 = 10 2 × 5 = 10 10 = 10, yes

The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.

It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation’s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C2H6) with oxygen to yield H2O and CO2, represented by the unbalanced equation:

C2H6 + O2 -> H2O + CO2  (unbalanced)

Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:

C2H6 + O2 -> 3H2O + 2CO2  (unbalanced)

This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O2 reactant to yield an odd number, so a fractional coefficient, 7/2, is used instead to yield a provisional balanced equation:

C2H6 + 7/2 O2 -> 3H2O + 2CO2 

A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:

2C2H6 + 7 O2 -> 6H2O + 4CO2

Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,

3N2 + 9H2 -> 6NH3

the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:

N2 + 3H2 -> 2NH3

Link to Learning

Use this interactive tutorial for additional practice balancing equations.

Exercises

  1. Balance: ___NaClO3 → ___NaCl + ___O2

  2. Balance: ___N2 + ___H2 → ___N2H4

  3. Balance: ___Al + ___O2 → ___Al2O3

  4. Balance: ___C2H4 + ___O2 → ___CO2 + ___H2O

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