Stoichiometry

Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products. Chemical reactions in aqueous solution that involve ionic reactants or products may be represented more realistically by complete ionic equations and, more succinctly, by net ionic equations. A balanced chemical equation may be used to describe a reaction’s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.

31.1 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, total ionic, and net ionic formats.

An earlier chapter of this text 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 31.1, with space-filling molecular models shown in the lower half of the figure.

Figure 31.1

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).

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.

This example illustrates the fundamental aspects of any chemical equation:

The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.
The substances generated by the reaction are called products, and their formulas are placed on the right side of the equation.
Plus signs (+) separate individual reactant and product formulas, and an arrow $(⟶)$ separates the reactant and product (left and right) sides of the equation.
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 31.2). 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.

Figure 31.2

Regardless of the absolute numbers of molecules involved, the ratios between numbers of molecules of each species that react (the reactants) and molecules of each species that form (the products) are the same and are given by the chemical reaction equation.

This image has a left side, labeled, “Mixture before reaction” separated by a vertical dashed line from right side labeled, “Mixture after reaction.” On the left side of the figure, two types of molecules are illustrated with space-filling models. Six of the molecules have only two red spheres bonded together. Three of the molecules have four small white spheres evenly distributed about and bonded to a central, larger black sphere. On the right side of the dashed vertical line, two types of molecules which are different from those on the left side are shown. Six of the molecules have a central red sphere to which smaller white spheres are bonded. The white spheres are not opposite each other on the red atoms, giving the molecule a bent shape or appearance. The second molecule type has a central black sphere to which two red spheres are attached on opposite sides, resulting in a linear shape or appearance. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms. On each side of the dashed line, twelve red, three black, and twelve white spheres are present.

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

(1 {CO}_{2} molecule \times \frac{2 O atoms}{{CO}_{2} molecule}) + (2 H_{2} O molecules \times \frac{1 O atom}{H_{2} O 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:

{CH}_{4} + 2 O_{2} ⟶ {CO}_{2} + 2 H_{2} O

Element	Reactants	Products	Balanced?
C	1 $\times$ 1 = 1	1 $\times$ 1 = 1	1 = 1, yes
H	4 $\times$ 1 = 4	2 $\times$ 2 = 4	4 = 4, yes
O	2 $\times$ 2 = 4	(1 $\times$ 2) + (2 $\times$ 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:

H_{2} O ⟶ H_{2} + O_{2} (unbalanced)

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

Element	Reactants	Products	Balanced?
H	1 $\times$ 2 = 2	1 $\times$ 2 = 2	2 = 2, yes
O	1 $\times$ 1 = 1	1 $\times$ 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 H_{2} O ⟶ H_{2} + O_{2} (unbalanced)

Element	Reactants	Products	Balanced?
H	2 $\times$ 2 = 4	1 $\times$ 2 = 2	4 ≠ 2, no
O	2 $\times$ 1 = 2	1 $\times$ 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 H_{2} O ⟶ 2 H_{2} + O_{2} (balanced)

Element	Reactants	Products	Balanced?
H	2 $\times$ 2 = 4	2 $\times$ 2 = 4	4 = 4, yes
O	2 $\times$ 1 = 2	1 $\times$ 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 H_{2} O ⟶ 2 H_{2} + O_{2}

Example 31.1

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.

N_{2} + O_{2} ⟶ N_{2} O_{5} (unbalanced)

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

Element	Reactants	Products	Balanced?
N	1 $\times$ 2 = 2	1 $\times$ 2 = 2	2 = 2, yes
O	1 $\times$ 2 = 2	1 $\times$ 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).

N_{2} + 5 O_{2} ⟶ 2 N_{2} O_{5} (unbalanced)

Element	Reactants	Products	Balanced?
N	1 $\times$ 2 = 2	2 $\times$ 2 = 4	2 ≠ 4, no
O	5 $\times$ 2 = 10	2 $\times$ 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 N_{2} + 5 O_{2} ⟶ 2 N_{2} O_{5}

Element	Reactants	Products	Balanced?
N	2 $\times$ 2 = 4	2 $\times$ 2 = 4	4 = 4, yes
O	5 $\times$ 2 = 10	2 $\times$ 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. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)

Answer

2 {NH}_{4} {NO}_{3} ⟶ 2 N_{2} + O_{2} + 4 H_{2} O

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:

C_{2} H_{6} + O_{2} ⟶ H_{2} O + {CO}_{2} (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:

C_{2} H_{6} + O_{2} ⟶ 3 H_{2} O + 2 {CO}_{2} (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_{2} H_{6} + \frac{7}{2} O_{2} ⟶ 3 H_{2} O + 2 {CO}_{2}

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

2 C_{2} H_{6} + 7 O_{2} ⟶ 6 H_{2} O + 4 {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 N_{2} + 9 H_{2} ⟶ 6 {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:

N_{2} + 3 H_{2} ⟶ 2 {NH}_{3}

Link to Learning

Use this interactive tutorial for additional practice balancing equations.

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 Na (s) + 2 H_{2} O (l) ⟶ 2 NaOH (a q) + 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.

{CaCO}_{3} (s) \overset{Δ}{⟶} CaO (s) + {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:

{CaCl}_{2} (a q) + 2 {AgNO}_{3} (a q) ⟶ Ca ({NO}_{3})_{2} (a q) + 2 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 homogenously 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}{l} {CaCl}_{2} (a q) ⟶ {Ca}^{2+} (a q) + 2 {Cl}^{−} (a q) \\ 2 {AgNO}_{3} (a q) ⟶ 2 {Ag}^{+} (a q) + 2 {NO}_{3}^{−} (a q) \\ Ca ({NO}_{3})_{2} (a q) ⟶ {Ca}^{2+} (a q) + 2 {NO}_{3}^{−} (a q) \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:

{Ca}^{2+} (a q) + 2 {Cl}^{−} (a q) + 2 {Ag}^{+} (a q) + 2 {NO}_{3}^{−} (a q) ⟶ {Ca}^{2+} (a q) + 2 {NO}_{3}^{−} (a q) + 2 Ag​Cl (s)

Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, Ca²⁺(aq) and ${NO}_{3}^{−} (a q) .$ 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{matrix} {Ca}^{2+} (a q) + 2 {Cl}^{−} (a q) + 2 {Ag}^{+} (a q) + 2 {NO}_{3}^{−} (a q) ⟶ {Ca}^{2+} (a q) + 2 {NO}_{3}^{−} (a q) + 2 AgCl (s) \\ 2 {Cl}^{−} (a q) + 2 {Ag}^{+} (a q) ⟶ 2 AgCl (s) \end{matrix}

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

{Cl}^{−} (a q) + {Ag}^{+} (a q) ⟶ 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⁺.

Example 31.2

Ionic and Molecular Equations

When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced molecular, complete ionic, and net ionic equations for this process.

Solution

Begin by identifying formulas for the reactants and products and arranging them properly in chemical equation form:

{CO}_{2} (a q) + NaOH (a q) ⟶ {Na}_{2} {CO}_{3} (a q) + H_{2} O (l) (unbalanced)

Balance is achieved easily in this case by changing the coefficient for NaOH to 2, resulting in the molecular equation for this reaction:

{CO}_{2} (a q) + 2 NaOH (a q) ⟶ {Na}_{2} {CO}_{3} (a q) + H_{2} O (l)

The two dissolved ionic compounds, NaOH and Na₂CO₃, can be represented as dissociated ions to yield the complete ionic equation:

{CO}_{2} (a q) + 2 {Na}^{+} (a q) + 2 {OH}^{−} (a q) ⟶ 2 {Na}^{+} (a q) + {CO}_{3}^{2−} (a q) + H_{2} O (l)

Finally, identify the spectator ion(s), in this case Na⁺(aq), and remove it from each side of the equation to generate the net ionic equation:

\begin{matrix} {CO}_{2} (a q) + 2 {Na}^{+} (a q) + 2 {OH}^{−} (a q) ⟶ 2 {Na}^{+} (a q) + {CO}_{3}^{2−} (a q) + H_{2} O (l) \\ {CO}_{2} (a q) + 2 {OH}^{−} (a q) ⟶ {CO}_{3}^{2−} (a q) + H_{2} O (l) \end{matrix}

Check Your Learning

Diatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:

\begin{matrix} NaCl (a q) + H_{2} O (l) \overset{electricity}{\to} NaOH (a q) + H_{2} (g) + {Cl}_{2} (g) \end{matrix}

Write balanced molecular, complete ionic, and net ionic equations for this process.

Answer

2 NaCl (a q) + 2 H_{2} O (l) ⟶ 2 NaOH (a q) + H_{2} (g) + {Cl}_{2} (g) (molecular)

2 {Na}^{+} (a q) + 2 {Cl}^{−} (a q) + 2 H_{2} O (l) ⟶ 2 {Na}^{+} (a q) + 2 {OH}^{−} (a q) + H_{2} (g) + {Cl}_{2} (g) (complete ionic)

2 {Cl}^{−} (a q) + 2 H_{2} O (l) ⟶ 2 {OH}^{−} (a q) + H_{2} (g) + {Cl}_{2} (g) (net ionic)

Link to Supplemental Exercises

Supplemental exercises are available if you would like more practice with these concepts.

31.2 Reaction Stoichiometry

Learning Objectives

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

Explain the concept of stoichiometry as it pertains to chemical reactions
Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products
Perform stoichiometric calculations involving mass, moles, and solution molarity

A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry, a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.

The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, $\frac{3}{4}$ cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is

1 cup mix + \frac{3}{4} cup milk + 1 egg ⟶ 8 pancakes

If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is

24 pancakes \times \frac{1 egg}{8 pancakes} = 3 eggs

Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:

N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3} (g)

This equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:

\frac{2 {NH}_{3} molecules}{3 H_{2} molecules} or \frac{2 doz {NH}_{3} molecules}{3 doz H_{2} molecules} or \frac{2 mol {NH}_{3} molecules}{3 mol H_{2} molecules}

These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.

Example 31.3

Moles of Reactant Required in a Reaction

How many moles of I₂ are required to react with 0.429 mol of Al according to the following equation (see Figure 31.3)?

2 Al + 3 I_{2} ⟶ 2 {AlI}_{3}

Figure 31.3

Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)

This figure shows three photos with an arrow leading from one to the next. The first photo shows a small pile of iodine and aluminum on a white surface. The second photo shows a small amount of purple smoke coming from the pile. The third photo shows a large amount of purple and gray smoke coming from the pile.

Solution

Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is

\frac{3 mol I_{2}}{2 mol Al} .

The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor: This figure shows two pink rectangles. The first is labeled, “Moles of A l.” This rectangle is followed by an arrow pointing right to a second rectangle labeled, “Moles of I subscript 2.”

This figure shows two pink rectangles. The first is labeled, “Moles of A l.” This rectangle is followed by an arrow pointing right to a second rectangle labeled, “Moles of I subscript 2.”

\begin{array}{l} {mol I}_{2} & = 0.429 mol Al \times \frac{3 mol I_{2}}{2 mol Al} \\ = 0.644 mol I_{2} \end{array}

Check Your Learning

How many moles of Ca(OH)₂ are required to react with 1.36 mol of H₃PO₄ to produce Ca₃(PO₄)₂ according to the equation

3 Ca {(OH)}_{2} + 2 H_{3} {PO}_{4} ⟶ {Ca}_{3} {({PO}_{4})}_{2} + 6 H_{2} O?

Answer

2.04 mol

Example 31.4

Number of Product Molecules Generated by a Reaction

How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?

C_{3} H_{8} + 5 O_{2} ⟶ 3 {CO}_{2} + 4 H_{2} O

Solution

The approach here is the same as for Example 31.3, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro’s number.

The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:

\frac{3 mol {CO}_{2}}{1 mol C_{3} H_{8}}

Using this stoichiometric factor, the provided molar amount of propane, and Avogadro’s number,

This figure shows two pink rectangles. The first is labeled, “Moles of C subscript 3 H subscript 8.” This rectangle is followed by an arrow pointing right to a second rectangle labeled, “Moles of C O subscript 2.”

0.75 mol C_{3} H_{8} \times \frac{3 mol {CO}_{2}}{1 mol C_{3} H_{8}} \times \frac{6.022 \times 10^{23} {CO}_{2} molecules}{mol {CO}_{2}} = 1.4 \times 10^{24} {CO}_{2} molecules

Check Your Learning

How many NH₃ molecules are produced by the reaction of 4.0 mol of Ca(OH)₂ according to the following equation:

{({NH}_{4})}_{2} {SO}_{4} + Ca {(OH)}_{2} ⟶ 2 {NH}_{3} + {CaSO}_{4} + 2 H_{2} O

Answer

4.8 $\times$ 10²⁴ NH₃ molecules

These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.

Example 31.5

Relating Masses of Reactants and Products

What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)₂] by the following reaction?

{MgCl}_{2} (a q) + 2 NaOH (a q) ⟶ Mg {(OH)}_{2} (s) + 2 NaCl (a q)

Solution

The approach used previously in Example 31.3 and in Example 31.4 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:

This figure shows four rectangles. The first is shaded yellow and is labeled, “Mass of M g ( O H ) subscript 2.” This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, “Moles of M g ( O H ) subscript 2.” This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, “Moles of N a O H.” This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, “Mass of N a O H.”

16 g Mg {(OH)}_{2} \times \frac{1 mol Mg {(OH)}_{2}}{58.3 g Mg {(OH)}_{2}} \times \frac{2 mol NaOH}{1 mol Mg {(OH)}_{2}} \times \frac{40.0 g NaOH}{mol NaOH} = 22 g NaOH

Check Your Learning

What mass of gallium oxide, Ga₂O₃, can be prepared from 29.0 g of gallium metal? The equation for the reaction is

4 Ga + 3 O_{2} ⟶ 2 {Ga}_{2} O_{3} .

Answer

39.0 g

Example 31.6

Relating Masses of Reactants

What mass of oxygen gas, O₂, from the air is consumed in the combustion of 702 g of octane, C₈H₁₈, one of the principal components of gasoline?

2 C_{8} H_{18} + 25 O_{2} ⟶ 16 {CO}_{2} + 18 H_{2} O

Solution

The approach required here is the same as for Example 31.5, differing only in that the provided and requested masses are both for reactant species.

This figure shows four rectangles. The first is shaded yellow and is labeled, “Mass of C subscript 8 H subscript 18.” This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, “Moles of C subscript 8 H subscript 18.” This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, “Moles of O subscript 2.” This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, “Mass of O subscript 2.”

702 g C_{8} H_{18} \times \frac{1 mol C_{8} H_{18}}{114.23 g C_{8} H_{18}} \times \frac{25 mol O_{2}}{2 mol C_{8} H_{18}} \times \frac{32.00 g O_{2}}{mol O_{2}} = 2.46 \times 10^{3} g O_{2}

Check Your Learning

What mass of CO is required to react with 25.13 g of Fe₂O₃ according to the equation

{Fe}_{2} O_{3} + 3 CO ⟶ 2 Fe + 3 {CO}_{2} ?

Answer

13.22 g

These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure 31.6 provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.

Figure 31.6

The flowchart depicts the various computational steps involved in most reaction stoichiometry calculations.

This flowchart shows 10 rectangles connected by double headed arrows. To the upper left, a rectangle is shaded lavender and is labeled, “Volume of pure substance A.” This rectangle is followed by a horizontal double headed arrow labeled, “Density.” It connects to a second rectangle which is shaded yellow and is labeled, “Mass of A.” This rectangle is followed by a double headed arrow which is labeled, “Molar Mass,” that connects to a third rectangle which is shaded pink and is labeled, “Moles of A.” To the left of this rectangle is a horizontal double headed arrow labeled, “Molarity,” which connects to a lavender rectangle which is labeled, “Volume of solution A.” The pink, “Moles of A,” rectangle is also connected with a double headed arrow below and to the left. This arrow is labeled “Avogadro’s number.” It connects to a green shaded rectangle that is labeled, “Number of particles of A.” To the right of the pink “Moles of A,” rectangle is a horizontal double headed arrow which is labeled, “Stoichiometric factor.” It connects to a second pink rectangle which is labeled, “Moles of B.” A double headed arrow which is labeled, “Molar mass,” extends from the top of this rectangle above and to the right to a yellow shaded rectangle labeled, “Mass of B.” A horizontal double headed arrow which is labeled, “Density” links to a lavender rectangle labeled, “Volume of substance B,” to the right. A horizontal double headed arrow labeled, “Molarity,” extends right to the of the pink “Moles of B” rectangle. This arrow connects to a lavender rectangle that is labeled, “Volume of substance B.” Another double headed arrow extends below and to the right of the pink “Moles of B” rectangle. This arrow is labeled “Avogadro’s number,” and it extends to a green rectangle which is labeled, “Number of particles of B.”

Chemistry in Everyday Life

Airbags

Airbags (Figure 31.7) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN₃. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN₃ to initiate its decomposition:

2 {NaN}_{3} (s) ⟶ 3 N_{2} (g) + 2 Na (s)

This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN₃ will generate approximately 50 L of N₂.

Figure 31.7

Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)

This photograph shows the inside of an automobile from the driver’s side area. The image shows inflated airbags positioned just in front of the driver’s and passenger’s seats and along the length of the passenger side over the windows. A large, round airbag covers the steering wheel.

Link to Supplemental Exercises

Supplemental exercises are available if you would like more practice with these concepts.

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Previous Citation(s)

Flowers, P., Neth, E. J., Robinson, W. R., Theopold, K., & Langley, R. (2019). Chemistry in Context. In Chemistry: Atoms First 2e. OpenStax. https://openstax.org/books/chemistry-atoms-first-2e/pages/7-introduction

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