# 8.1 Writing and Balancing Chemical Equations

## Learning Objectives

By the end of this section, you will be able to:

• Derive chemical equations from narrative descriptions of chemical reactions
• Write and balance chemical equations in molecular formats

The preceding chapter introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH4) and two diatomic oxygen molecules (O2) to produce one carbon dioxide molecule (CO2) and two water molecules (H2O). The chemical equation representing this process is provided in the upper half of Figure 8.1a, with space-filling molecular models shown in the lower half of the figure.

This example illustrates the fundamental aspects of any chemical equation:

1. The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.
2. The substances generated by the reaction are called products, and their formulas are placed on the right sight of the equation.
3. Plus signs (+) separate individual reactant and product formulas, and an arrow (→) separates the reactant and product (left and right) sides of the equation.
4. The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.

It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the relative numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (Figure 8.1b). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:

• One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.
• One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.
• One mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide molecules and 2 moles of water molecules.

## Balancing Equations

The chemical equations described above are 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

$(1 \;\text{CO}_2 \;\text{molecule} \times \frac{2 \;\text{O atoms}}{\text{CO}_2 \;\text{molecule}}) + (2\;\text{H}_2\text{O molecule} \times \frac{1 \;\text{O atom}}{\text{H}_2\text{O molecule}}) = 4 \;\text{O atoms}$

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:

$\text{CH}_4 + 2 \text{ O}_2 \longrightarrow \text{CO}_2 + 2\text{ H}_2\text{O}$
Tally Chart for Reaction of Methane and Oxygen to Yield Carbon Dioxide and Water
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:

$\text{H}_2\text{O} \longrightarrow \text{H}_2 + \text{O}_2 \;(\text{unbalanced})$

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

Tally Chart for Decomposition of Water to Yield Hydrogen and Oxygen #1
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.

$2 \text{ H}_2\text{O} \longrightarrow \text{H}_2 + \text{O}_2 \;(\text{unbalanced})$
Tally Chart for Decomposition of Water to Yield Hydrogen and Oxygen #2
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.

$2 \text{ H}_2\text{O} \longrightarrow 2 \text{ H}_2 + \text{O}_2 \;(\text{balanced})$
Tally Chart for Decomposition of Water to Yield Hydrogen and Oxygen #3
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:

$2 \text{ H}_2\text{O} \longrightarrow 2 \text{ H}_2 + \text{O}_2$

### Example 8.1a

#### 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.

$\text{N}_2 + \text{O}_2 \longrightarrow \text{N}_2 \text{O}_5 \;(\text{unbalanced})$

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

Tally Chart for Reaction of Nitrogen and Oxygen to Form Dinitrogen Pentoxide #1
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).

$\text{N}_2 + 5 \text{ O}_2 \longrightarrow 2 \text{ N}_2\text{O}_5 \;(\text{unbalanced})$
Tally Chart for Reaction of Nitrogen and Oxygen to Form Dinitrogen Pentoxide #2
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.

$2 \text{ N}_2 + 5 \text{ O}_2 \longrightarrow 2 \text{ N}_2 \text{O}_5$
Tally Chart for Reaction of Nitrogen and Oxygen to Form Dinitrogen Pentoxide #3
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.

### Exercise 8.1a

Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)

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:

$\text{C}_2 \text{H}_6 + \text{O}_2 \longrightarrow \text{H}_2 \text{O} + \text{C} \text{O}_2 \;(\text{unbalanced})$

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

$\text{C}_2 \text{H}_6 + \text{O}_2 \longrightarrow 3\text{ H}_2 \text{O} + 2\text{ C} \text{O}_2 \;(\text{unbalanced})$

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:

$\text{C}_2 \text{H}_6 + \frac{7}{2}\text{O}_2 \longrightarrow 3\text{ H}_2 \text{O} + 2\text{ C} \text{O}_2 \;$

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

$2\text{ C}_2 \text{H}_6 + 7\text{ O}_2 \longrightarrow 6\text{ H}_2 \text{O} + 4\text{ C} \text{O}_2 \;$

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,

$3\text{ N}_2 + 9\text{ H}_2 \longrightarrow 6\text{ N} \text{H}_3$

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:

$\text{N}_2 + 3\text{ H}_2 \longrightarrow 2\text{ N} \text{H}_3$

## Additional Information in Chemical Equations

The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include s for solids, l for liquids, g for gases, and aq for substances dissolved in water (aqueous solutions, as introduced in the preceding chapter). These notations are illustrated in the example equation here:

$2\text{ Na}(s) + 2\text{ H}_2 \text{O}(l) \longrightarrow 2\text{ NaOH}(aq) + \text{H}_2(g)$

This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).

Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation’s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (Δ) over the arrow.

$\text{CaCO}_3(s) \;\xrightarrow{\Delta} \; \text{CaO}(s) + \text{CO}_2(g)$

### Endothermic vs. Exothermic Reactions

Sometimes ΔH for a chemical reaction is displayed which may be positive or negative. The number is assumed to be positive if it has no sign; a + sign can be added explicitly to avoid confusion. A chemical reaction that has a positive ΔH is said to be endothermic while a chemical reaction that has a negative ΔH is said to be exothermic.

What does it mean if the ΔH of a process is positive? It means that the system in which the chemical reaction is occurring is gaining energy. If one considers the energy of a system as being represented as a height on a vertical energy plot, the enthalpy change that accompanies the reaction can be shown in Figure 8.1c. It displays the energy of the reactants with some energy, and the system increases its energy as it goes to products. The products are higher on the vertical scale than the reactants. Endothermic, then, implies that the system gains, or absorbs, energy.

An opposite situation exists for an exothermic process, as shown in Figure 8.1d. If the ΔH of a reaction is negative, the system is losing energy, so the products have less energy than the reactants, and the products are lower on the vertical energy scale than the reactants are. Exothermic, then, implies that the system loses, or gives off, energy.

Source: “Endothermic vs. Exothermic Reactions” adapted by Adrienne Richards from Chapter 7: Enthalpy and Chemical Reactions In Introductory Chemistry: 1st Canadian Edition by David W. Ball and Jessica A. Key, licensed under CC BY NC SA 4.0.

### Exercise 8.1b

Practice using the following PhET simulation: Balancing Chemical Equations

Activity source: Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC-BY-4.0

### Links to Interactive Learning Tools

1. $2\text{ NH}_4 \text{NO}_3 \longrightarrow 2\text{ N}_2 + \text{O}_2 + 4\text{ H}_2\text{O}$