The chemical equation described above is 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, 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
$$ (1\, CO_2\, molecule\, \times \frac{2\, O\, atoms}{CO_2\, molecule})\, +\, (2\, H_2O\, molecules\, \times \frac{1\, O\, atom}{H_2O\, molecule})=\, 4\, O\, atoms $$
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
| Atom | 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)
| 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 (balanced)
| 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
Balancing Chemical Equations
Write a balanced equation for the reaction of molecular nitrogen (N2) and oxygen (O2) to form dinitrogen pentoxide.Solution
First, write the unbalanced equation.
N2 + O2 ⟶ N2O5 (unbalanced)
Next, count the number of each type of atom present in the unbalanced equation.
If you feel confident drag and drop the numbers to balance the equation below:
If you find this a little bit of a struggle read the following to help address the question:
Click to expand the hints:
N2 + O2 ⟶ N2O5 (unbalanced)
| 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.
Check Your Learning
Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water.
Answer:
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, $\frac{7}{2}$ is used instead to yield a provisional balanced equation: $$C_2H_6\,+\,\frac{7}{2}O_2\,⟶\,3H_2O\,+\,2CO_2$$
A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:
2C2H6 + 7O2 ⟶ 6H2O + 4CO2
Finally with regard to balanced equations, recall that convention dictates the 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
Use this interactive tutorial for additional practice balancing equations.