4.1 Writing and Balancing Chemical Equations - Chemistry 2e | OpenStax (2024)

Balancing Equations

The chemical equation described in section 4.1 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

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

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:

$${\text{CH}}_4+2{\text{O}}_2⟶{\text{CO}}_2+2{\text{H}}_2\text{O}$$

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:

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

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

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.

$$\mathbf{2}{\text{H}}_2\text{O}⟶{\text{H}}_2+{\text{O}}_2\text{(unbalanced)}$$

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

$$2{\text{H}}_2\text{O}⟶\mathbf{2}{\text{H}}_2+{\text{O}}_2\text{(balanced)}$$

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

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

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:

$${\text{C}}_2{\text{H}}_6+{\text{O}}_2⟶{\text{H}}_2\text{O}+{\text{CO}}_2\text{(unbalanced)}$$

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

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

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:

$${\text{C}}_2{\text{H}}_6+\frac{7}{2}{\text{O}}_2⟶3{\text{H}}_2\text{O}+2{\text{CO}}_2$$

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

$$2{\text{C}}_2{\text{H}}_6+7{\text{O}}_2⟶6{\text{H}}_2\text{O}+4{\text{CO}}_2$$

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,

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

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:

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

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:

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

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.

$${\text{CaCO}}_3(s)\stackrel{\text{Δ}}{⟶}\text{CaO}(s)+{\text{CO}}_2(g)$$

Other examples of these special conditions will be encountered in more depth in later chapters.

Equations for Ionic Reactions

Given the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of CaCl₂ and AgNO₃ are mixed, a reaction takes place producing aqueous Ca(NO₃)₂ and solid AgCl:

$${\text{CaCl}}_2(aq)+2{\text{AgNO}}_3(aq)⟶\text{Ca}({\text{NO}}_3){}_2(aq)+2\text{AgCl}(s)$$

This balanced equation, derived in the usual fashion, is called a molecular equation because it doesn’t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed hom*ogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:

$$\begin{array}{}\\ \\ {\text{CaCl}}_2(aq)⟶{\text{Ca}}^{\mathrm{2+}}(aq)+2{\text{Cl}}^{\text{−}}(aq)\\ 2{\text{AgNO}}_3(aq)⟶2{\text{Ag}}^{\text{+}}(aq)+2{\text{NO}}_3{}^{\text{−}}(aq)\\ \text{Ca}({\text{NO}}_3){}_2(aq)⟶{\text{Ca}}^{\mathrm{2+}}(aq)+2{\text{NO}}_3{}^{\text{−}}(aq)\end{array}$$

Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, s.

Explicitly representing all dissolved ions results in a complete ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:

$${\text{Ca}}^{\mathrm{2+}}(aq)+2{\text{Cl}}^{\text{−}}(aq)+2{\text{Ag}}^{\text{+}}(aq)+2{\text{NO}}_3{}^{\text{−}}(aq)⟶{\text{Ca}}^{\mathrm{2+}}(aq)+2{\text{NO}}_3{}^{\text{−}}(aq)+2\text{AgCl}(s)$$

Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, Ca²⁺(aq) and ${\text{NO}}_3{}^{\text{−}}(aq).$ These spectator ions—ions whose presence is required to maintain charge neutrality—are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation:

$$\begin{array}{c}\sout{{\text{Ca}}^{\mathrm{2+}}(aq)}+2{\text{Cl}}^{\text{−}}(aq)+2{\text{Ag}}^{\mathrm{+}}(aq)+\sout{2{\text{NO}}_3{}^{\text{−}}(aq)}⟶\sout{{\text{Ca}}^{\mathrm{2+}}(aq)}+\sout{2{\text{NO}}_3{}^{\text{−}}(aq)}+2\text{AgCl}(s)\\ 2{\text{Cl}}^{\text{−}}(aq)+2{\text{Ag}}^{\text{+}}(aq)⟶2\text{AgCl}(s)\end{array}$$

Following the convention of using the smallest possible integers as coefficients, this equation is then written:

$${\text{Cl}}^{\text{−}}(aq)+{\text{Ag}}^{\text{+}}(aq)⟶\text{AgCl}(s)$$

This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of Cl⁻ and Ag⁺.