573. Derivation of Molecular Formulas -
Recall that empirical formulas are symbols representing the relative numbers of a compound's elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in a previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.

Molecular formulas are derived by comparing the compound's molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If the molecular (or molar) mass of the substance is known, it may be divided by the empirical formula mass to yield the number of empirical formula units per molecule (n):

$$
\frac{\text { molecular or molar mass }\left(\mathrm{amu} \text { or } \frac{\mathrm{g}}{\mathrm{~mol}}\right)}{\text { empirical formula mass }\left(\mathrm{amu} \text { or } \frac{\mathrm{g}}{\mathrm{~mol}}\right)}=n \text { formula units/molecule }
$$

The molecular formula is then obtained by multiplying each subscript in the empirical formula by $n$, as shown by the generic empirical formula $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}}$ :

$$
\left(A_{x} B_{y}\right)_{n}=A_{n x} B_{n y}
$$

For example, consider a covalent compound whose empirical formula is determined to be $\mathrm{CH}_{2} \mathrm{O}$. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one 0 atom). If the compound's molecular mass is determined to be 180 amu , this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:


573. Derivation of Molecular Formulas -
$$
\frac{180 \mathrm{amu} / \mathrm{molecule}}{30 \frac{\mathrm{amu}}{\text { formula unit }}}=6 \text { formula units } / \mathrm{molec} \text {. }
$$

Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:

$$
\left(\mathrm{CH}_{2} \mathrm{O}\right)_{6}=\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}
$$

Note that this same approach may be used when the molar mass ( $\mathrm{g} / \mathrm{mol}$ ) instead of the molecular mass (amu) is used. In this case, one mole of empirical formula units and molecules is considered, as opposed to single units and molecules.


575. Determination of the Molecular Formula for Nicotine -
Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains $74.02 \% \mathrm{C}, 8.710 \% \mathrm{H}$, and $17.27 \% \mathrm{~N}$. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?


576. Solution -
Determining the molecular formula from the provided data will require comparison of the compound's empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound's empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:

$$
\begin{aligned}
& (74.02 \mathrm{~g} \mathrm{C})\left(\frac{1 \mathrm{~mol} \mathrm{C}}{12.01 \mathrm{~g} \mathrm{C}}\right)=6.163 \mathrm{~mol} \mathrm{C} \\
& (8.710 \mathrm{~g} \mathrm{H})\left(\frac{1 \mathrm{~mol} \mathrm{H}}{1.01 \mathrm{~g} \mathrm{H}}\right)=8.624 \mathrm{~mol} \mathrm{H} \\
& (17.27 \mathrm{~g} \mathrm{~N})\left(\frac{1 \mathrm{~mol} \mathrm{~N}}{14.01 \mathrm{~g} \mathrm{~N}}\right)=1.233 \mathrm{~mol} \mathrm{~N}
\end{aligned}
$$

Next, calculate the molar ratios of these elements relative to the least abundant element, N .

$$
\begin{gathered}
6.163 \mathrm{~mol} \mathrm{C} / 1.233 \mathrm{~mol} \mathrm{~N}=5 \\
8.264 \mathrm{~mol} \mathrm{H} / 1.233 \mathrm{~mol} \mathrm{~N}=7 \\
1.233 \mathrm{~mol} \mathrm{~N} / 1.233 \mathrm{~mol} \mathrm{~N}=1 \\
\frac{1.233}{1.233}=1.000 \mathrm{~mol} \mathrm{~N} \\
\frac{6.163}{1.233}=4.998 \mathrm{~mol} \mathrm{C} \\
\frac{8.624}{1.233}=6.994 \mathrm{~mol} \mathrm{H}
\end{gathered}
$$


576. Solution -
The C-to- N and H -to- N molar ratios are adequately close to whole numbers, and so the empirical formula is $\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{~N}$. The empirical formula mass for this compound is therefore $81.13 \mathrm{amu} /$ formula unit, or $81.13 \mathrm{~g} / \mathrm{mol}$ formula unit.

Calculate the molar mass for nicotine from the given mass and molar amount of compound:

$$
\frac{40.57 \mathrm{~g} \text { nicotine }}{0.2500 \mathrm{~mol} \text { nicotine }}=\frac{162.3 \mathrm{~g}}{\mathrm{~mol}}
$$

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

$$
\frac{162.3 \mathrm{~g} / \mathrm{mol}}{81.13 \frac{\mathrm{~g}}{\text { formula unit }}}=2 \text { formula units } / \text { molecule }
$$

Finally, derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

$$
\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{~N}\right)_{2}=\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2}
$$


577. Check Your Learning -
What is the molecular formula of a compound with a percent composition of $49.47 \% \mathrm{C}, 5.201 \% \mathrm{H}, 28.84 \% \mathrm{~N}$, and $16.48 \% \mathrm{O}$, and a molecular mass of 194.2 amu ?


578. Answer: -
$\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}$


579. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the fundamental properties of solutions
- Calculate solution concentrations using molarity
- Perform dilution calculations using the dilution equation

Preceding sections of this chapter focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures-samples of matter containing two or more substances physically combined-are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet's atmosphere determines its ability to sustain aerobic
life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an "alloy") determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure 6.8). This section will describe one of the most common ways in which the relative compositions of mixtures may be quantified.

FIGURE 6.8 Sugar is one of many components in the complex mixture known as coffee. The amount of sugar in a given amount of coffee is an important determinant of the beverage's sweetness. (credit: Jane Whitney)


580. Solutions -
Solutions have previously been defined as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. A more thorough treatment of solution properties is provided in the chapter on solutions and colloids, but provided here is an introduction to some of the basic properties of solutions.

The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution.

A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration).

Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity ( $\boldsymbol{M}$ ) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter ( 1 L ) of the solution:

$$
M=\frac{\text { mol solute }}{\mathrm{L} \text { solution }}
$$


582. Calculating Molar Concentrations -
A $355-\mathrm{mL}$ soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?


583. Solution -
Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L :

$$
M=\frac{\text { mol solute }}{\mathrm{L} \text { solution }}=\frac{0.133 \mathrm{~mol}}{355 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}}=0.375 \mathrm{M}
$$


584. Check Your Learning -
A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL ?


587. Deriving Moles and Volumes from Molar Concentrations -
How much sugar (mol) is contained in a modest sip ( $\sim 10 \mathrm{~mL}$ ) of the soft drink from Example 6.8?


588. Solution -
Rearrange the definition of molarity to isolate the quantity sought, moles of sugar, then substitute the value for molarity derived in Example 6.8, 0.375 M :

$$
\begin{gathered}
\qquad \begin{array}{c}
M=\frac{\text { mol solute }}{\mathrm{L} \text { solution }} \\
\text { mol solute }
\end{array}=M \times \mathrm{L} \text { solution } \\
\text { mol solute }=0.375 \frac{\mathrm{~mol} \text { sugar }}{\mathrm{L}} \times\left(10 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}\right)=0.004 \mathrm{~mol} \text { sugar }
\end{gathered}
$$


589. Check Your Learning -
What volume (mL) of the sweetened tea described in Example 6.8 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?


590. Answer: -
80 mL


592. Calculating Molar Concentrations from the Mass of Solute -
Distilled white vinegar (Figure 6.9) is a solution of acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, in water. A 0.500 - L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?

FIGURE 6.9 Distilled white vinegar is a solution of acetic acid in water.


593. Solution -
As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. Since the mass of solute is provided instead of its molar amount, use the solute's molar mass to obtain the amount of solute in moles:

$$
\begin{gathered}
M=\frac{\text { mol solute }}{\mathrm{L} \text { solution }=} \frac{25.2 \mathrm{~g} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \times \frac{1 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}}{60.052 \mathrm{~g} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}}}{0.500 \mathrm{~L} \text { solution }}=0.839 \mathrm{M} \\
\begin{array}{c}
M=\frac{\text { mol solute }}{\mathrm{L} \text { solution }}=0.839 \mathrm{M} \\
M=\frac{.0839 \text { mol solute }}{1.00 \mathrm{~L} \text { solution }}
\end{array}
\end{gathered}
$$


594. Check Your Learning -
Calculate the molarity of 6.52 g of $\mathrm{CoCl}_{2}(128.9 \mathrm{~g} / \mathrm{mol})$ dissolved in an aqueous solution with a total volume of 75.0 mL .


595. Answer: -
0.674 M


597. Determining the Mass of Solute in a Given Volume of Solution -
How many grams of NaCl are contained in 0.250 L of a $5.30-\mathrm{M}$ solution?


598. Solution -
The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 6.9:

$$
\begin{gathered}
\qquad M=\frac{\text { mol solute }}{\mathrm{L} \text { solution }} \\
\text { mol solute }=M \times \mathrm{L} \text { solution } \\
\text { mol solute }=5.30 \frac{\mathrm{~mol} \mathrm{NaCl}}{\mathrm{~L}} \times 0.250 \mathrm{~L}=1.325 \mathrm{~mol} \mathrm{NaCl}
\end{gathered}
$$

Finally, this molar amount is used to derive the mass of NaCl :

$$
1.325 \mathrm{~mol} \mathrm{NaCl} \times \frac{58.44 \mathrm{~g} \mathrm{NaCl}}{\mathrm{~mol} \mathrm{NaCl}}=77.4 \mathrm{~g} \mathrm{NaCl}
$$


599. Check Your Learning -
How many grams of $\mathrm{CaCl}_{2}(110.98 \mathrm{~g} / \mathrm{mol})$ are contained in 250.0 mL of a $0.200-M$ solution of calcium chloride?


600. Answer: -
$5.55 \mathrm{~g} \mathrm{CaCl}_{2}$

When performing calculations stepwise, as in Example 6.11, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 6.11, the molar amount of NaCl computed in the first step, 1.325 mol , would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If the guard digit had not been retained, the final calculation for the mass of NaCl would have been 77.1 g , a difference of 0.3 g .

In addition to retaining a guard digit for intermediate calculations, rounding errors may also be avoided by performing computations in a single step (see Example 6.12). This eliminates intermediate steps so that only the final result is rounded.


602. Determining the Volume of Solution Containing a Given Mass of Solute -
In Example 6.10, the concentration of acetic acid in white vinegar was determined to be 0.839 M . What volume of vinegar contains 75.6 g of acetic acid?


603. Solution -
First, use the molar mass to calculate moles of acetic acid from the given mass:

$$
\mathrm{g} \text { solute } \times \frac{\text { mol solute }}{\mathrm{g} \text { solute }}=\text { mol solute }
$$

Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:

$$
\text { mol solute } \times \frac{\mathrm{L} \text { solution }}{\text { mol solute }}=\mathrm{L} \text { solution }
$$

Combining these two steps into one yields:

$$
\begin{gathered}
\mathrm{g} \text { solute } \times \frac{\text { mol solute }}{\mathrm{g} \text { solute }} \times \frac{\mathrm{L} \text { solution }}{\text { mol solute }}=\mathrm{L} \text { solution } \\
75.6 \mathrm{~g} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\frac{\mathrm{~mol} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}}{60.05 \mathrm{~g}}\right)\left(\frac{\mathrm{L} \text { solution }}{0.839 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}}\right)=1.50 \mathrm{~L} \text { solution }
\end{gathered}
$$


604. Check Your Learning -
What volume of a $1.50-M \mathrm{KBr}$ solution contains 66.0 g KBr ?


605. Answer: -
0.370 L


606. Dilution of Solutions -
Dilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 6.10).

FIGURE 6.10 Both solutions contain the same mass of copper nitrate. The solution on the right is more dilute because the copper nitrate is dissolved in more solvent. (credit: Mark Ott)

Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution, a solution of lesser concentration may be prepared.
For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents.

A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the number of moles of solute in a solution $(n)$ is equal to the product of the solution's molarity $(M)$ and its volume in liters $(L)$ :

$$
n=M L
$$

Expressions like these may be written for a solution before and after it is diluted:

$$
\begin{aligned}
& n_{1}=M_{1} L_{1} \\
& n_{2}=M_{2} L_{2}
\end{aligned}
$$

where the subscripts " 1 " and " 2 " refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution, $n_{1}=n_{2}$. Thus, these two equations may be set equal to one another:

$$
M_{1} L_{1}=M_{2} L_{2}
$$


606. Dilution of Solutions -
$$
M_{1} L_{1}=M_{2} L_{2}
$$

This relation is commonly referred to as the dilution equation. Although this equation uses molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used as long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:

$$
C_{1} V_{1}=C_{2} V_{2}
$$

where $C$ and $V$ are concentration and volume, respectively.


607. LINK TO LEARNING -
Use the simulation (http://openstax.org/l/16Phetsolvents) to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation.


609. Determining the Concentration of a Diluted Solution -
If 0.850 L of a $5.00-M$ solution of copper nitrate, $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?


610. Solution -
The stock concentration, $C_{1}$, and volume, $V_{1}$, are provided as well as the volume of the diluted solution, $V_{2}$. Rearrange the dilution equation to isolate the unknown property, the concentration of the diluted solution, $C_{2}$ :

$$
\begin{gathered}
C_{1} V_{1}=C_{2} V_{2} \\
C_{2}=\frac{C_{1} V_{1}}{V_{2}}
\end{gathered}
$$

Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L ), the diluted solution's concentration is expected to be less than one-half 5 M . This ballpark estimate will be compared to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:

$$
C_{2}=\frac{0.850 \mathrm{~L} \times 5.00 \frac{\mathrm{~mol}}{\mathrm{~L}}}{1.80 \mathrm{~L}}=2.36 M
$$

This result compares well to our ballpark estimate (it's a bit less than one-half the stock concentration, 5 M ).


611. Check Your Learning -
What is the concentration of the solution that results from diluting 25.0 mL of a $2.04-\mathrm{M}$ solution of $\mathrm{CH}_{3} \mathrm{OH}$ to 500.0 mL ?


612. Answer: -
$0.102 \mathrm{MCH}_{3} \mathrm{OH}$


614. Volume of a Diluted Solution -
What volume of 0.12 MHBr can be prepared from $11 \mathrm{~mL}(0.011 \mathrm{~L})$ of 0.45 M HBr ?


615. Solution -
Provided are the volume and concentration of a stock solution, $V_{1}$ and $C_{1}$, and the concentration of the resultant diluted solution, $C_{2}$. Find the volume of the diluted solution, $V_{2}$ by rearranging the dilution equation to isolate $V_{2}$ :

$$
\begin{gathered}
C_{1} V_{1}=C_{2} V_{2} \\
V_{2}=\frac{C_{1} V_{1}}{C_{2}}
\end{gathered}
$$

Since the diluted concentration $(0.12 M)$ is slightly more than one-fourth the original concentration $(0.45 M)$,
the volume of the diluted solution is expected to be roughly four times the original volume, or around 44 mL . Substituting the given values and solving for the unknown volume yields:

$$
\begin{gathered}
V_{2}=\frac{(0.45 M)(0.011 \mathrm{~L})}{(0.12 M)} \\
V_{2}=0.041 \mathrm{~L}
\end{gathered}
$$

The volume of the $0.12-M$ solution is $0.041 \mathrm{~L}(41 \mathrm{~mL})$. The result is reasonable and compares well with the rough estimate.


616. Check Your Learning -
A laboratory experiment calls for $0.125 \mathrm{M} \mathrm{HNO}_{3}$. What volume of $0.125 \mathrm{M} \mathrm{HNO}_{3}$ can be prepared from 0.250 L of $1.88 \mathrm{M} \mathrm{HNO}_{3}$ ?


617. Answer: -
3.76 L


619. Volume of a Concentrated Solution Needed for Dilution -
What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH ?


620. Solution -
Given are the concentration of a stock solution, $C_{1}$, and the volume and concentration of the resultant diluted solution, $V_{2}$ and $C_{2}$. Find the volume of the stock solution, $V_{1}$ by rearranging the dilution equation to isolate $V_{1}$ :

$$
\begin{gathered}
C_{1} V_{1}=C_{2} V_{2} \\
V_{1}=\frac{C_{2} V_{2}}{C_{1}}
\end{gathered}
$$

Since the concentration of the diluted solution $0.100 M$ is roughly one-sixteenth that of the stock solution (1.59 $M$ ), the volume of the stock solution is expected to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:

$$
\begin{gathered}
V_{1}=\frac{(0.100 M)(5.00 \mathrm{~L})}{1.59 M} \\
V_{1}=0.314 \mathrm{~L}
\end{gathered}
$$

Thus, 0.314 L of the $1.59-M$ solution is needed to prepare the desired solution. This result is consistent with the rough estimate.


621. Check Your Learning -
What volume of a $0.575-M$ solution of glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$, can be prepared from 50.00 mL of a $3.00-\mathrm{M}$ glucose solution?


622. Answer: -
0.261 L


623. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Define the concentration units of mass percentage, volume percentage, mass-volume percentage, parts-permillion (ppm), and parts-per-billion (ppb)
- Perform computations relating a solution's concentration and its components' volumes and/or masses using these units

The previous section introduced molarity, a very useful measurement unit for evaluating the concentration of solutions. However, molarity is only one measure of concentration. This section will describe some other units of concentration that are commonly used in various applications, either for convenience or by convention.


624. Mass Percentage -
Earlier in this chapter, percent composition was introduced as a measure of the relative amount of a given element in a compound. Percentages are also commonly used to express the composition of mixtures, including solutions. The mass percentage of a solution component is defined as the ratio of the component's mass to the solution's mass, expressed as a percentage:

$$
\text { mass percentage }=\frac{\text { mass of component }}{\text { mass of solution }} \times 100 \%
$$

Mass percentage is also referred to by similar names such as percent mass, percent weight, weight/weight percent, and other variations on this theme. The most common symbol for mass percentage is simply the percent sign, \%, although more detailed symbols are often used including \%mass, \%weight, and (w/w)\%. Use of these more detailed symbols can prevent confusion of mass percentages with other types of percentages, such as volume percentages (to be discussed later in this section).
Mass percentages are popular concentration units for consumer products. The label of a typical liquid bleach bottle (Figure 6.11) cites the concentration of its active ingredient, sodium hypochlorite ( NaOCl ), as being $7.4 \%$. A 100.0-g sample of bleach would therefore contain 7.4 g of NaOCl .

FIGURE 6.11 Liquid bleach is an aqueous solution of sodium hypochlorite ( NaOCl ). This brand has a concentration of $7.4 \% \mathrm{NaOCl}$ by mass.


625. Calculation of Percent by Mass -
A $5.0-\mathrm{g}$ sample of spinal fluid contains $3.75 \mathrm{mg}(0.00375 \mathrm{~g})$ of glucose. What is the percent by mass of glucose in spinal fluid?


626. Solution -
The spinal fluid sample contains roughly 4 mg of glucose in 5000 mg of fluid, so the mass fraction of glucose should be a bit less than one part in 1000, or about $0.1 \%$. Substituting the given masses into the equation defining mass percentage yields:

$$
\% \text { glucose }=\frac{3.75 \mathrm{mg} \text { glucose } \times \frac{1 \mathrm{~g}}{1000 \mathrm{mg}}}{5.0 \mathrm{~g} \text { spinal fluid }}=0.075 \%
$$

The computed mass percentage agrees with our rough estimate (it's a bit less than 0.1\%).
Note that while any mass unit may be used to compute a mass percentage ( $\mathrm{mg}, \mathrm{g}, \mathrm{kg}, \mathrm{oz}$, and so on), the same unit must be used for both the solute and the solution so that the mass units cancel, yielding a dimensionless ratio. In this case, the solute mass unit in the numerator was converted from mg to g to match the units in the denominator. Alternatively, the spinal fluid mass unit in the denominator could have been converted from g to mg instead. As long as identical mass units are used for both solute and solution, the computed mass percentage will be correct.


627. Check Your Learning -
A bottle of a tile cleanser contains 135 g of HCl and 775 g of water. What is the percent by mass of HCl in this cleanser?


628. Answer: -
14.8\%


630. Calculations using Mass Percentage -
"Concentrated" hydrochloric acid is an aqueous solution of $37.2 \% \mathrm{HCl}$ that is commonly used as a laboratory reagent. The density of this solution is $1.19 \mathrm{~g} / \mathrm{mL}$. What mass of HCl is contained in 0.500 L of this solution?


631. Solution -
The HCl concentration is near $40 \%$, so a 100-g portion of this solution would contain about 40 g of HCl . Since the solution density isn't greatly different from that of water ( $1 \mathrm{~g} / \mathrm{mL}$ ), a reasonable estimate of the HCl mass in $500 \mathrm{~g}(0.5 \mathrm{~L})$ of the solution is about five times greater than that in a 100 g portion, or $5 \times 40=200 \mathrm{~g}$. In order to derive the mass of solute in a solution from its mass percentage, the mass of the solution must be known. Using the solution density given, convert the solution's volume to mass, and then use the given mass percentage to calculate the solute mass. This mathematical approach is outlined in this flowchart:

For proper unit cancellation, the $0.500-\mathrm{L}$ volume is converted into 500 mL , and the mass percentage is expressed as a ratio, $37.2 \mathrm{~g} \mathrm{HCl} / \mathrm{g}$ solution:

$$
500 \mathrm{~mL} \text { solution }\left(\frac{1.19 \mathrm{~g} \text { solution }}{\mathrm{mL} \text { solution }}\right)\left(\frac{37.2 \mathrm{~g} \mathrm{HCl}}{100 \mathrm{~g} \text { solution }}\right)=221 \mathrm{~g} \mathrm{HCl}
$$

This mass of HCl is consistent with our rough estimate of approximately 200 g .
Check Your Learning
What volume of concentrated HCl solution contains 125 g of HCl ?


632. Answer: -
282 mL


633. Volume Percentage -
Liquid volumes over a wide range of magnitudes are conveniently measured using common and relatively inexpensive laboratory equipment. The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is therefore often expressed as a volume percentage, $\% \mathrm{vol}$ or $(\mathrm{v} / \mathrm{v}) \%$ :

$$
\text { volume percentage }=\frac{\text { volume solute }}{\text { volume solution }} \times 100 \%
$$


635. Calculations using Volume Percentage -
Rubbing alcohol (isopropanol) is usually sold as a 70\%vol aqueous solution. If the density of isopropyl alcohol is $0.785 \mathrm{~g} / \mathrm{mL}$, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol?


636. Solution -
Per the definition of volume percentage, the isopropanol volume is $70 \%$ of the total solution volume. Multiplying the isopropanol volume by its density yields the requested mass:
$\left(355 \mathrm{~mL}\right.$ solution) $\left(\frac{70 \mathrm{~mL} \text { isopropyl alcohol }}{100 \mathrm{~mL} \text { solution }}\right)\left(\frac{0.785 \mathrm{~g} \text { isopropyl alcohol }}{1 \mathrm{~mL} \text { isopropyl alcohol }}\right)=195 \mathrm{~g}$ isopropyl alchol


637. Check Your Learning -
Wine is approximately $12 \%$ ethanol $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)$ by volume. Ethanol has a molar mass of $46.06 \mathrm{~g} / \mathrm{mol}$ and a density $0.789 \mathrm{~g} / \mathrm{mL}$. How many moles of ethanol are present in a $750-\mathrm{mL}$ bottle of wine?


638. Answer: -
1.5 mol ethanol


639. Mass-Volume Percentage -
"Mixed" percentage units, derived from the mass of solute and the volume of solution, are popular for certain biochemical and medical applications. A mass-volume percent is a ratio of a solute's mass to the solution's volume expressed as a percentage. The specific units used for solute mass and solution volume may vary, depending on the solution. For example, physiological saline solution, used to prepare intravenous fluids, has a concentration of $0.9 \%$ mass $/$ volume ( $\mathrm{m} / \mathrm{v}$ ), indicating that the composition is 0.9 g of solute per 100 mL of solution. The concentration of glucose in blood (commonly referred to as "blood sugar") is also typically expressed in terms of a mass-volume ratio. Though not expressed explicitly as a percentage, its concentration is usually given in milligrams of glucose per deciliter ( 100 mL ) of blood (Figure 6.12).

FIGURE 6.12 "Mixed" mass-volume units are commonly encountered in medical settings. (a) The NaCl concentration of physiological saline is $0.9 \%(\mathrm{~m} / \mathrm{v})$. (b) This device measures glucose levels in a sample of blood. The normal range for glucose concentration in blood (fasting) is around $70-100 \mathrm{mg} / \mathrm{dL}$. (credit a: modification of work by "The National Guard"/Flickr; credit b: modification of work by Biswarup Ganguly)


640. Parts per Million and Parts per Billion -
Very low solute concentrations are often expressed using appropriately small units such as parts per million (ppm) or parts per billion (ppb). Like percentage ("part per hundred") units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules.

The mass-based definitions of ppm and ppb are given here:

$$
\begin{aligned}
& \mathrm{ppm}=\frac{\text { mass solute }}{\text { mass solution }} \times 10^{6} \mathrm{ppm} \\
& \mathrm{ppb}=\frac{\text { mass solute }}{\text { mass solution }} \times 10^{9} \mathrm{ppb}
\end{aligned}
$$

Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm . Inline water filters are designed to reduce the concentration of fluoride and several other tracelevel contaminants in tap water (Figure 6.13).

FIGURE 6.13 (a) In some areas, trace-level concentrations of contaminants can render unfiltered tap water unsafe for drinking and cooking. (b) Inline water filters reduce the concentration of solutes in tap water. (credit a: modification of work by Jenn Durfey; credit b: modification of work by "vastateparkstaff"/Wikimedia commons)


642. Calculation of Parts per Million and Parts per Billion Concentrations -
According to the EPA, when the concentration of lead in tap water reaches 15 ppb , certain remedial actions must be taken. What is this concentration in ppm? At this concentration, what mass of lead ( $\mu \mathrm{g}$ ) would be contained in a typical glass of water ( 300 mL )?


643. Solution -
The definitions of the ppm and ppb units may be used to convert the given concentration from ppb to ppm . Comparing these two unit definitions shows that ppm is 1000 times greater than $\mathrm{ppb}\left(1 \mathrm{ppm}=10^{3} \mathrm{ppb}\right)$. Thus:

$$
15 \mathrm{ppb} \times \frac{1 \mathrm{ppm}}{10^{3} \mathrm{ppb}}=0.015 \mathrm{ppm}
$$

The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. Since the volume of solution ( 300 mL ) is given, its density must be used to derive the corresponding mass. Assume the density of tap water to be roughly the same as that of pure water ( $\sim 1.00 \mathrm{~g} / \mathrm{mL}$ ), since the concentrations of any dissolved substances should not be very large. Rearranging the equation defining the ppb unit and substituting the given quantities yields:

$$
\begin{gathered}
\mathrm{ppb}=\frac{\text { mass solute }}{\text { mass solution }} \times 10^{9} \mathrm{ppb} \\
\text { mass solute }=\frac{\mathrm{ppb} \times \text { mass solution }}{10^{9} \mathrm{ppb}} \\
\text { mass solute }=\frac{15 \mathrm{ppb} \times 300 \mathrm{~mL} \times \frac{1.00 \mathrm{~g}}{\mathrm{~mL}}}{10^{9} \mathrm{ppb}}=4.5 \times 10^{-6} \mathrm{~g}
\end{gathered}
$$

Finally, convert this mass to the requested unit of micrograms:

$$
4.5 \times 10^{-6} \mathrm{~g} \times \frac{1 \mu \mathrm{~g}}{10^{-6} \mathrm{~g}}=4.5 \mu \mathrm{~g}
$$


644. Check Your Learning -
A 50.0-g sample of industrial wastewater was determined to contain 0.48 mg of mercury. Express the mercury concentration of the wastewater in ppm and ppb units.


645. Answer: -
9.6 ppm, 9600 ppb


646. Key Terms -
aqueous solution solution for which water is the solvent
Avogadro's number ( $\boldsymbol{N}_{\boldsymbol{A}}$ ) experimentally determined value of the number of entities comprising 1 mole of substance, equal to $6.022 \times$ $10^{23} \mathrm{~mol}^{-1}$
concentrated qualitative term for a solution containing solute at a relatively high concentration
concentration quantitative measure of the relative amounts of solute and solvent present in a solution
dilute qualitative term for a solution containing solute at a relatively low concentration
dilution process of adding solvent to a solution in order to lower the concentration of solutes
dissolved describes the process by which solute components are dispersed in a solvent
empirical formula mass sum of average atomic masses for all atoms represented in an empirical formula
formula mass sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass
mass percentage ratio of solute-to-solution mass
expressed as a percentage
mass-volume percent ratio of solute mass to solution volume, expressed as a percentage
molar mass mass in grams of 1 mole of a substance
molarity (M) unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution
mole amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of ${ }^{12} \mathrm{C}$
parts per billion (ppb) ratio of solute-to-solution mass multiplied by $10^{9}$
parts per million (ppm) ratio of solute-to-solution mass multiplied by $10^{6}$
percent composition percentage by mass of the various elements in a compound
solute solution component present in a concentration less than that of the solvent
solvent solution component present in a concentration that is higher relative to other components
volume percentage ratio of solute-to-solution volume expressed as a percentage


647. Key Equations -
$\% \mathrm{X}=\frac{\text { mass } \mathrm{X}}{\text { mass compound }} \times 100 \%$
$\frac{\text { molecular or molar mass }\left(\mathrm{amu} \text { or } \frac{\mathrm{g}}{\mathrm{mol}}\right)}{\text { empirical formula mass }\left(\mathrm{amu} \text { or } \frac{\mathrm{g}}{\mathrm{mol}}\right)}=n$ formula units/molecule
$\left(\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}}\right)_{\mathrm{n}}=\mathrm{A}_{\mathrm{nx}} \mathrm{B}_{\mathrm{ny}}$
$M=\frac{\text { mol solute }}{\mathrm{L} \text { solution }}$
$C_{1} V_{1}=C_{2} V_{2}$
Percent by mass $=\frac{\text { mass of solute }}{\text { mass of solution }} \times 100$
$\mathrm{ppm}=\frac{\text { mass solute }}{\text { mass solution }} \times 10^{6} \mathrm{ppm}$
$\mathrm{ppb}=\frac{\text { mass solute }}{\text { mass solution }} \times 10^{9} \mathrm{ppb}$


649. 1 Formula Mass -
The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula
mass (amu) and molar mass ( $\mathrm{g} / \mathrm{mol}$ ) for any substance are numerically equivalent (for example, one $\mathrm{H}_{2} \mathrm{O}$ molecule weighs approximately18 amu and 1 mole of $\mathrm{H}_{2} \mathrm{O}$ molecules weighs approximately 18 g).


650. Formulas -
The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound's percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound's empirical formula. The empirical formula mass of a covalent compound may be compared to the compound's molecular or molar mass to derive a molecular formula.


650. Formulas - 650.1. Molarity
Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given
amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.


650. Formulas - 650.2. Other Units for Solution Concentrations
In addition to molarity, a number of other solution concentration units are used in various applications. Percentage concentrations based on the solution components' masses, volumes, or both are useful for expressing relatively high concentrations, whereas lower concentrations are conveniently expressed using ppm or ppb units. These units are popular in environmental, medical, and other fields where mole-based units such as molarity are not as commonly used.


651. Exercises - 651.1. Formula Mass
1. What is the total mass (amu) of carbon in each of the following molecules?
(a) $\mathrm{CH}_{4}$
(b) $\mathrm{CHCl}_{3}$
(c) $\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{O}_{6}$
(d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$
2. What is the total mass of hydrogen in each of the molecules?
(a) $\mathrm{CH}_{4}$
(b) $\mathrm{CHCl}_{3}$
(c) $\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{O}_{6}$
(d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$
3. Calculate the molecular or formula mass of each of the following:
(a) $\mathrm{P}_{4}$
(b) $\mathrm{H}_{2} \mathrm{O}$
(c) $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$
(d) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ (acetic acid)
(e) $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ (sucrose, cane sugar)
4. Determine the molecular mass of the following compounds:
(a)

(b)

(c)

(d)

5. Determine the molecular mass of the following compounds:
(a)

(b)

(c)

(d)

6. Which molecule has a molecular mass of 28.05 amu ?
(a)

(b)

(c)


651. Exercises - 651.2. Determining Empirical and Molecular Formulas
7. What information is needed to determine the molecular formula of a compound from the empirical formula?
8. Calculate the following to four significant figures:
(a) the percent composition of ammonia, $\mathrm{NH}_{3}$
(b) the percent composition of photographic fixer solution "hypo," $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$
(c) the percent of calcium ion in $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$
9. Determine the following to four significant figures:
(a) the percent composition of hydrazoic acid, $\mathrm{HN}_{3}$
(b) the percent composition of TNT, $\mathrm{C}_{6} \mathrm{H}_{2}\left(\mathrm{CH}_{3}\right)\left(\mathrm{NO}_{2}\right)_{3}$
(c) the percent of $\mathrm{SO}_{4}{ }^{2-}$ in $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$
10. Determine the percent ammonia, $\mathrm{NH}_{3}$, in $\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{3}$, to three significant figures.
11. Determine the percent water in $\mathrm{CuSO}_{4} \bullet 5 \mathrm{H}_{2} \mathrm{O}$ to three significant figures.
12. Determine the empirical formulas for compounds with the following percent compositions:
(a) $15.8 \%$ carbon and $84.2 \%$ sulfur
(b) $40.0 \%$ carbon, $6.7 \%$ hydrogen, and $53.3 \%$ oxygen
13. Determine the empirical formulas for compounds with the following percent compositions:
(a) $43.6 \%$ phosphorus and $56.4 \%$ oxygen
(b) $28.7 \% \mathrm{~K}, 1.5 \% \mathrm{H}, 22.8 \% \mathrm{P}$, and $47.0 \% \mathrm{O}$


651. Exercises - 651.2. Determining Empirical and Molecular Formulas
(b) $28.7 \% \mathrm{~K}, 1.5 \% \mathrm{H}, 22.8 \% \mathrm{P}$, and $47.0 \% \mathrm{O}$
14. A compound of carbon and hydrogen contains $92.3 \% \mathrm{C}$ and has a molar mass of $78.1 \mathrm{~g} / \mathrm{mol}$. What is its molecular formula?
15. Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of $99 \mathrm{~g} / \mathrm{mol}$. Analysis of a sample shows that it contains $24.3 \%$ carbon and $4.1 \%$ hydrogen. What is its molecular formula?
16. Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: $28.03 \% \mathrm{Mg}, 21.60 \% \mathrm{Si}, 1.16 \% \mathrm{H}$, and $49.21 \% \mathrm{O}$. The molar mass for chrysotile is $520.8 \mathrm{~g} / \mathrm{mol}$.
17. Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers:
(a) Lucite (Plexiglas); $59.9 \% \mathrm{C}, 8.06 \% \mathrm{H}, 32.0 \% \mathrm{O}$
(b) Saran; $24.8 \%$ C, $2.0 \%$ H, $73.1 \%$ Cl
(c) polyethylene; $86 \% \mathrm{C}, 14 \% \mathrm{H}$
(d) polystyrene; $92.3 \% \mathrm{C}, 7.7 \% \mathrm{H}$
(e) Orlon; $67.9 \%$ C, $5.70 \% \mathrm{H}, 26.4 \% \mathrm{~N}$


651. Exercises - 651.2. Determining Empirical and Molecular Formulas
(e) Orlon; $67.9 \%$ C, $5.70 \% \mathrm{H}, 26.4 \% \mathrm{~N}$
18. A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of $75.95 \% \mathrm{C}, 17.72 \% \mathrm{~N}$, and $6.33 \% \mathrm{H}$ by mass with a molar mass of about $240 \mathrm{~g} / \mathrm{mol}$. Determine the molecular formula of the dye.


651. Exercises - 651.3. Molarity
19. Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L .
20. What information is needed to calculate the molarity of a sulfuric acid solution?
21. A $200-\mathrm{mL}$ sample and a $400-\mathrm{mL}$ sample of a solution of salt have the same molarity. In what ways are the two samples identical? In what ways are these two samples different?
22. Determine the molarity for each of the following solutions:
(a) 0.444 mol of $\mathrm{CoCl}_{2}$ in 0.654 L of solution
(b) 98.0 g of phosphoric acid, $\mathrm{H}_{3} \mathrm{PO}_{4}$, in 1.00 L of solution
(c) 0.2074 g of calcium hydroxide, $\mathrm{Ca}(\mathrm{OH})_{2}$, in 40.00 mL of solution
(d) 10.5 kg of $\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}$ in 18.60 L of solution
(e) $7.0 \times 10^{-3} \mathrm{~mol}$ of $\mathrm{I}_{2}$ in 100.0 mL of solution
(f) $1.8 \times 10^{4} \mathrm{mg}$ of HCl in 0.075 L of solution
23. Determine the molarity of each of the following solutions:
(a) 1.457 mol KCl in 1.500 L of solution
(b) 0.515 g of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in 1.00 L of solution
(c) 20.54 g of $\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$ in 1575 mL of solution
(d) 2.76 kg of $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ in 1.45 L of solution
(e) 0.005653 mol of $\mathrm{Br}_{2}$ in 10.00 mL of solution


651. Exercises - 651.3. Molarity
(e) 0.005653 mol of $\mathrm{Br}_{2}$ in 10.00 mL of solution
(f) 0.000889 g of glycine, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{2}$, in 1.05 mL of solution
24. Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$, used for intravenous injection?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
25. Consider this question: What is the mass of solute in 200.0 L of a $1.556-M$ solution of KBr ?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
26. Calculate the number of moles and the mass of the solute in each of the following solutions:
(a) 2.00 L of $18.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$, concentrated sulfuric acid
(b) 100.0 mL of $3.8 \times 10^{-6} \mathrm{M} \mathrm{NaCN}$, the minimum lethal concentration of sodium cyanide in blood serum
(c) 5.50 L of $13.3 \mathrm{MH}_{2} \mathrm{CO}$, the formaldehyde used to "fix" tissue samples
(d) 325 mL of $1.8 \times 10^{-6} \mathrm{M} \mathrm{FeSO}_{4}$, the minimum concentration of iron sulfate detectable by taste in drinking water
27. Calculate the number of moles and the mass of the solute in each of the following solutions:
(a) 325 mL of $8.23 \times 10^{-5} \mathrm{M} \mathrm{KI}$, a source of iodine in the diet
(b) 75.0 mL of $2.2 \times 10^{-5} \mathrm{MH}_{2} \mathrm{SO}_{4}$, a sample of acid rain


651. Exercises - 651.3. Molarity
(b) 75.0 mL of $2.2 \times 10^{-5} \mathrm{MH}_{2} \mathrm{SO}_{4}$, a sample of acid rain
(c) 0.2500 L of $0.1135 \mathrm{M}_{2} \mathrm{CrO}_{4}$, an analytical reagent used in iron assays
(d) 10.5 L of $3.716 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$, a liquid fertilizer
28. Consider this question: What is the molarity of $\mathrm{KMnO}_{4}$ in a solution of 0.0908 g of $\mathrm{KMnO}_{4}$ in 0.500 L of solution?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
29. Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl ?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
30. Calculate the molarity of each of the following solutions:
(a) 0.195 g of cholesterol, $\mathrm{C}_{27} \mathrm{H}_{46} \mathrm{O}$, in 0.100 L of serum, the average concentration of cholesterol in human serum
(b) 4.25 g of $\mathrm{NH}_{3}$ in 0.500 L of solution, the concentration of $\mathrm{NH}_{3}$ in household ammonia
(c) 1.49 kg of isopropyl alcohol, $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}$, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol
(d) 0.029 g of $\mathrm{I}_{2}$ in 0.100 L of solution, the solubility of $\mathrm{I}_{2}$ in water at $20^{\circ} \mathrm{C}$
31. Calculate the molarity of each of the following solutions:
(a) 293 g HCl in 666 mL of solution, a concentrated HCl solution


651. Exercises - 651.3. Molarity
31. Calculate the molarity of each of the following solutions:
(a) 293 g HCl in 666 mL of solution, a concentrated HCl solution
(b) $2.026 \mathrm{~g} \mathrm{FeCl}_{3}$ in 0.1250 L of a solution used as an unknown in general chemistry laboratories
(c) $0.001 \mathrm{mg} \mathrm{Cd}^{2+}$ in 0.100 L , the maximum permissible concentration of cadmium in drinking water
(d) $0.0079 \mathrm{~g} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{SNO}_{3}$ in one ounce ( 29.6 mL ), the concentration of saccharin in a diet soft drink.
32. There is about 1.0 g of calcium, as $\mathrm{Ca}^{2+}$, in 1.0 L of milk. What is the molarity of $\mathrm{Ca}^{2+}$ in milk?
33. What volume of a $1.00-M \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}$ solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M ?
34. If 0.1718 L of a $0.3556-M \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}$ solution is diluted to a concentration of 0.1222 M , what is the volume of the resulting solution?
35. If 4.12 L of a $0.850 \mathrm{M}-\mathrm{H}_{3} \mathrm{PO}_{4}$ solution is be diluted to a volume of 10.00 L , what is the concentration of the resulting solution?
36. What volume of a $0.33-M \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M ?
37. What is the concentration of the NaCl solution that results when 0.150 L of a $0.556-\mathrm{M}$ solution is allowed to evaporate until the volume is reduced to 0.105 L ?


651. Exercises - 651.3. Molarity
37. What is the concentration of the NaCl solution that results when 0.150 L of a $0.556-\mathrm{M}$ solution is allowed to evaporate until the volume is reduced to 0.105 L ?
38. What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?
(a) 1.00 L of a $0.250-\mathrm{M}$ solution of $\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}$ is diluted to a final volume of 2.00 L
(b) 0.5000 L of a $0.1222-\mathrm{M}$ solution of $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}$ is diluted to a final volume of 1.250 L
(c) 2.35 L of a $0.350-M$ solution of $\mathrm{H}_{3} \mathrm{PO}_{4}$ is diluted to a final volume of 4.00 L
(d) 22.50 mL of a $0.025-M$ solution of $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ is diluted to 100.0 mL
39. What is the final concentration of the solution produced when 225.5 mL of a $0.09988-M$ solution of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ is allowed to evaporate until the solution volume is reduced to 45.00 mL ?
40. A $2.00-\mathrm{L}$ bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl . What is the molarity of the solution?
41. An experiment in a general chemistry laboratory calls for a $2.00-M$ solution of HCl . How many mL of 11.9 $M \mathrm{HCl}$ would be required to make 250 mL of 2.00 M HCl ?
42. What volume of a $0.20-M \mathrm{~K}_{2} \mathrm{SO}_{4}$ solution contains 57 g of $\mathrm{K}_{2} \mathrm{SO}_{4}$ ?


651. Exercises - 651.3. Molarity
42. What volume of a $0.20-M \mathrm{~K}_{2} \mathrm{SO}_{4}$ solution contains 57 g of $\mathrm{K}_{2} \mathrm{SO}_{4}$ ?
43. The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to $0.50 \mathrm{mg} / \mathrm{L}$. If an industry is discharging hexavalent chromium as potassium dichromate $\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right)$, what is the maximum permissible molarity of that substance?


651. Exercises - 651.4. Other Units for Solution Concentrations
44. Consider this question: What mass of a concentrated solution of nitric acid ( $68.0 \% \mathrm{HNO}_{3}$ by mass) is needed to prepare 400.0 g of a $10.0 \%$ solution of $\mathrm{HNO}_{3}$ by mass?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
45. What mass of a $4.00 \% \mathrm{NaOH}$ solution by mass contains 15.0 g of NaOH ?
46. What mass of solid $\mathrm{NaOH}(97.0 \% \mathrm{NaOH}$ by mass) is required to prepare 1.00 L of a $10.0 \%$ solution of NaOH by mass? The density of the $10.0 \%$ solution is $1.109 \mathrm{~g} / \mathrm{mL}$.
47. What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of $1.19 \mathrm{~g} \mathrm{~cm}^{-3}$ and contains $37.21 \% \mathrm{HCl}$ by mass?
48. The hardness of water (hardness count) is usually expressed in parts per million (by mass) of $\mathrm{CaCO}_{3}$, which is equivalent to milligrams of $\mathrm{CaCO}_{3}$ per liter of water. What is the molar concentration of $\mathrm{Ca}^{2+}$ ions in a water sample with a hardness count of $175 \mathrm{mg} \mathrm{CaCO}_{3} / \mathrm{L}$ ?
49. The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of $1.0 \mathrm{~g} / \mathrm{mL}$ and calculate the molarity of mercury in the stream.


651. Exercises - 651.4. Other Units for Solution Concentrations
50. In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM is observed, what is the concentration of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ in $\mathrm{mg} / \mathrm{dL}$ ?
51. A throat spray is $1.40 \%$ by mass phenol, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$, in water. If the solution has a density of $0.9956 \mathrm{~g} / \mathrm{mL}$, calculate the molarity of the solution.
52. Copper(I) iodide (CuI) is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in $1.00 \mathrm{lb}(454 \mathrm{~g})$ of table salt containing $0.0100 \%$ CuI by mass?
53. A cough syrup contains $5.0 \%$ ethyl alcohol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, by mass. If the density of the solution is $0.9928 \mathrm{~g} / \mathrm{mL}$, determine the molarity of the alcohol in the cough syrup.
54. D5W is a solution used as an intravenous fluid. It is a $5.0 \%$ by mass solution of dextrose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ in water. If the density of D5W is $1.029 \mathrm{~g} / \mathrm{mL}$, calculate the molarity of dextrose in the solution.
55. Find the molarity of a $40.0 \%$ by mass aqueous solution of sulfuric acid, $\mathrm{H}_{2} \mathrm{SO}_{4}$, for which the density is $1.3057 \mathrm{~g} / \mathrm{mL}$.


652. CHAPTER 7 <br> Stoichiometry of Chemical Reactions -
Figure 7.1 Many modern rocket fuels are solid mixtures of substances combined in carefully measured amounts and ignited to yield a thrust-generating chemical reaction. (credit: modification of work by NASA)


653. CHAPTER OUTLINE - 653.1. Writing and Balancing Chemical Equations
7.2 Classifying Chemical Reactions


653. CHAPTER OUTLINE - 653.2. Reaction Stoichiometry
7.4 Reaction Yields
7.5 Quantitative Chemical Analysis

INTRODUCTION Solid-fuel rockets are a central feature in the world's space exploration programs, including the new Space Launch System being developed by the National Aeronautics and Space Administration (NASA) to replace the retired Space Shuttle fleet (Figure 7.1). The engines of these rockets rely on carefully prepared solid mixtures of chemicals combined in precisely measured amounts. Igniting the mixture initiates a vigorous chemical reaction that rapidly generates large amounts of gaseous products. These gases are ejected from the rocket engine through its nozzle, providing the thrust needed to propel heavy payloads into space. Both the nature of this chemical reaction and the relationships between the amounts of the substances being consumed and produced by the reaction are critically important considerations that determine the success of the technology. This chapter will describe how to symbolize chemical reactions using chemical equations, how to classify some common chemical reactions by identifying patterns of reactivity, and how to determine the quantitative relations between the amounts of substances involved in chemical reactions-that is, the reaction stoichiometry.


654. 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 $\left(\mathrm{CH}_{4}\right)$ and two diatomic oxygen molecules $\left(\mathrm{O}_{2}\right)$ to produce one carbon dioxide molecule $\left(\mathrm{CO}_{2}\right)$ and two water molecules $\left(\mathrm{H}_{2} \mathrm{O}\right)$. The chemical equation representing this process is provided in the upper half of Figure 7.2, with space-filling molecular models shown in the lower half of the figure.

FIGURE 7.2 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 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 side of the equation.
3. Plus signs (+) separate individual reactant and product formulas, and an arrow $(\longrightarrow)$ 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.


654. LEARNING OBJECTIVES -
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 7.3). 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.
Mixture before reaction

Mixture after reaction

FIGURE 7.3 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.


655. 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, $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

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

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:

$$
\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}
$$


655. Balancing Equations -
$$
\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}
$$

| 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:

$$
\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2}+\mathrm{O}_{2} \quad \text { (unbalanced) }
$$

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 | $1 \times 1=1$ | $1 \times 2=2$ | $1 \neq 2$, no |


655. Balancing Equations -
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 $\mathrm{H}_{2} \mathrm{O}$ to $\mathrm{H}_{2} \mathrm{O}_{2}$ 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 $\mathrm{H}_{2} \mathrm{O}$ to 2 .

| $2 \mathrm{H}_{2} \mathrm{O}$ |  | $\longrightarrow \mathrm{H}_{2}+\mathrm{O}_{2}$ | (unbalanced) |
| :--- | :--- | :--- | :--- |
| Element | Reactants | Products | Balanced? |
| H | $\mathbf{2} \times 2=4$ | $1 \times 2=2$ | $4 \neq 2$, 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 $\mathrm{H}_{2}$ product to 2 .

| $2 \mathrm{H}_{2} \mathrm{O}$ |  | $\longrightarrow \mathrm{H}_{2}+\mathrm{O}_{2}$ | (balanced) |
| :--- | :--- | :--- | :--- |
| Element | Reactants | Products | Balanced? |
| H | $2 \times 2=4$ | $\mathbf{2} \times 2=4$ | $4=4$, 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:


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

$$
2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{H}_{2}+\mathrm{O}_{2}
$$


657. Balancing Chemical Equations -
Write a balanced equation for the reaction of molecular nitrogen $\left(\mathrm{N}_{2}\right)$ and oxygen $\left(\mathrm{O}_{2}\right)$ to form dinitrogen pentoxide.


658. Solution -
First, write the unbalanced equation.

$$
\mathrm{N}_{2}+\mathrm{O}_{2} \longrightarrow \mathrm{~N}_{2} \mathrm{O}_{5} \quad \text { (unbalanced) }
$$

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 | $1 \times 2=2$ | $1 \times 5=5$ | $2 \neq 5$, 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 $\mathrm{O}_{2}$ and $\mathrm{N}_{2} \mathrm{O}_{5}$ to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).

| $\mathrm{N}_{2}+\mathbf{5} \mathrm{O}_{2} \longrightarrow \mathbf{2} \mathrm{~N}_{2} \mathrm{O}_{5}$ |  | (unbalanced) |  |
| :--- | :--- | :--- | :--- |
| Element | Reactants | Products | Balanced? |
| N | $1 \times 2=2$ | $\mathbf{2} \times 2=4$ | $2 \neq 4$, no |
| O | $\mathbf{5} \times 2=10$ | $\mathbf{2} \times 5=10$ | $10=10$, yes |

The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant $\mathrm{N}_{2}$ to 2 .


658. Solution -
The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant $\mathrm{N}_{2}$ to 2 .

| $2 \mathrm{~N}_{2}+5 \mathrm{O}_{2} \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{5}$ |  |  |  |
| :--- | :--- | :--- | :--- |
| 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.


659. 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.)


660. Answer: -
$2 \mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow 2 \mathrm{~N}_{2}+\mathrm{O}_{2}+4 \mathrm{H}_{2} \mathrm{O}$

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 $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ with oxygen to yield $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{CO}_{2}$, represented by the unbalanced equation:

$$
\mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \quad \text { (unbalanced) }
$$

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

$$
\mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{O}_{2} \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{CO}_{2} \quad \text { (unbalanced) }
$$

This results in seven $O$ atoms on the product side of the equation, an odd number-no integer coefficient can be used with the $\mathrm{O}_{2}$ reactant to yield an odd number, so a fractional coefficient, $\frac{7}{2}$, is used instead to yield a
provisional balanced equation:

$$
\mathrm{C}_{2} \mathrm{H}_{6}+\frac{7}{2} \mathrm{O}_{2} \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{CO}_{2}
$$

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


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

$$
2 \mathrm{C}_{2} \mathrm{H}_{6}+7 \mathrm{O}_{2} \longrightarrow 6 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{CO}_{2}
$$

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,

$$
3 \mathrm{~N}_{2}+9 \mathrm{H}_{2} \longrightarrow 6 \mathrm{NH}_{3}
$$

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:

$$
\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}
$$


661. LINK TO LEARNING -
Use this interactive tutorial (http://openstax.org/l/16BalanceEq) for additional practice balancing equations.


662. 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:

$$
2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)
$$

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 ( $\Delta$ ) over the arrow.

$$
\mathrm{CaCO}_{3}(s) \xrightarrow{\Delta} \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)
$$

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


663. 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 $\mathrm{CaCl}_{2}$ and $\mathrm{AgNO}_{3}$ are mixed, a reaction takes place producing aqueous $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ and solid AgCl :

$$
\mathrm{CaCl}_{2}(a q)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{AgCl}(s)
$$

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:

$$
\begin{aligned}
& \mathrm{CaCl}_{2}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \\
& 2 \mathrm{AgNO}_{3}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \\
& \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)
\end{aligned}
$$

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:


663. Equations for Ionic Reactions -
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:

$$
\mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)+2 \mathrm{Ag}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+2 \mathrm{Ag} \mathrm{Cl}(s)
$$

Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, $\mathrm{Ca}^{2+}(a q)$ and $\mathrm{NO}_{3}{ }^{-}(a q)$. 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:

$$
\begin{gathered}
\mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)+2 \mathrm{Ag}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+2 \mathrm{AgCl}(s) \\
2 \mathrm{Cl}^{-}(a q)+2 \mathrm{Ag}^{+}(a q) \longrightarrow 2 \mathrm{AgCl}(s)
\end{gathered}
$$

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

$$
\mathrm{Cl}^{-}(a q)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{AgCl}(s)
$$

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 $\mathrm{Cl}^{-}$and $\mathrm{Ag}^{+}$.


665. 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.


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

$$
\mathrm{CO}_{2}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \quad \text { (unbalanced) }
$$

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

$$
\mathrm{CO}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
$$

The two dissolved ionic compounds, NaOH and $\mathrm{Na}_{2} \mathrm{CO}_{3}$, can be represented as dissociated ions to yield the complete ionic equation:

$$
\mathrm{CO}_{2}(a q)+2 \mathrm{Na}^{+}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{Na}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
$$

Finally, identify the spectator ion(s), in this case $\mathrm{Na}^{+}(a q)$, and remove it from each side of the equation to generate the net ionic equation:

$$
\begin{aligned}
\mathrm{CO}_{2}(a q)+2 \mathrm{Na}^{+}(a q)+2 \mathrm{OH}^{-}(a q) & \longrightarrow 2 \mathrm{Na}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \\
\mathrm{CO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) & \longrightarrow \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
\end{aligned}
$$


667. 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:

$$
\mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \xrightarrow{\text { electricity }} \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)
$$

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


668. Answer: -
$2 \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \quad$ (molecular)
$2 \mathrm{Na}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{Na}^{+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \quad$ (complete ionic)
$2 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \quad$ (net ionic)


669. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Define three common types of chemical reactions (precipitation, acid-base, and oxidation-reduction)
- Classify chemical reactions as one of these three types given appropriate descriptions or chemical equations
- Identify common acids and bases
- Predict the solubility of common inorganic compounds by using solubility rules
- Compute the oxidation states for elements in compounds

Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they exchange blows with their fists or feet, we say they are fighting. Faced with a wide range of varied interactions between chemical substances, scientists have likewise found it convenient (or even necessary) to classify chemical interactions by identifying common patterns of reactivity. This module will provide an introduction to three of the most prevalent types of chemical reactions: precipitation, acid-base, and oxidation-reduction.


670. Precipitation Reactions and Solubility Rules -
A precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement, double replacement, or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central role in many chemical analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of matter (see the last module of this chapter).

The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility, defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large solubilities are said to be soluble. A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble, and these are the substances that readily precipitate from solution. More information on these important concepts is provided in a later chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that
have been observed for many ionic compounds (Table 7.1).


670. Precipitation Reactions and Solubility Rules -
| Soluble Ionic Compounds | contain these ions | exceptions |
| :---: | :---: | :---: |
|  | $\mathrm{NH}_{4}{ }^{+}$
 group I cations:
 $\mathrm{Li}^{+}$
 $\mathrm{Na}^{+}$
 $\mathrm{K}^{+}$
 $\mathrm{Rb}^{+}$
 $\mathrm{Cs}^{+}$ | none |
|  | $\begin{aligned} & \mathrm{Cl}^{-} \\ & \mathrm{Br}^{-} \\ & \mathrm{I}^{-} \end{aligned}$ | compounds with $\mathrm{Ag}^{+}, \mathrm{Hg}_{2}{ }^{2+}$, and $\mathrm{Pb}^{2+}$ |
|  | $\mathrm{F}^{-}$ | compounds with group 2 metal cations, $\mathrm{Pb}^{2+}$, and $\mathrm{Fe}^{3+}$ |
|  | $\begin{aligned} & \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} \\ & \mathrm{HCO}_{3}^{-} \\ & \mathrm{NO}_{3}^{-} \\ & \mathrm{ClO}_{3}^{-} \end{aligned}$ | none |
|  | $\mathrm{SO}_{4}{ }^{2-}$ | compounds with $\mathrm{Ag}^{+}, \mathrm{Ba}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Hg}_{2}{ }^{2+}, \mathrm{Pb}^{2+}$ and $\mathrm{Sr}^{2+}$ |
| Insoluble Ionic Compounds | contain these ions | exceptions |
|  | $\begin{aligned} & \mathrm{CO}_{3}{ }^{2-} \\ & \mathrm{CrO}_{4}{ }^{2-} \\ & \mathrm{PO}_{4}{ }^{3-} \\ & \mathrm{S}^{2-} \end{aligned}$ | compounds with group 1 cations and $\mathrm{NH}_{4}{ }^{+}$ |


670. Precipitation Reactions and Solubility Rules -
|  | $\mathrm{OH}^{-}$ | compounds with group 1 cations and $\mathrm{Ba}^{2+}$ |


670. Precipitation Reactions and Solubility Rules -
TABLE 7.1

A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide:

$$
2 \mathrm{KI}(a q)+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{PbI}_{2}(s)+2 \mathrm{KNO}_{3}(a q)
$$

This observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts.

The net ionic equation representing this reaction is:

$$
\mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow \mathrm{PbI}_{2}(s)
$$

Lead iodide is a bright yellow solid that was formerly used as an artist's pigment known as iodine yellow (Figure 7.4). The properties of pure $\mathrm{PbI}_{2}$ crystals make them useful for fabrication of X-ray and gamma ray detectors.

FIGURE 7.4 A precipitate of $\mathrm{PbI}_{2}$ forms when solutions containing $\mathrm{Pb}^{2+}$ and $\mathrm{I}^{-}$are mixed. (credit: Der Kreole/ Wikimedia Commons)


670. Precipitation Reactions and Solubility Rules -
The solubility guidelines in Table 7.1 may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium chloride will yield a solution containing $\mathrm{Ag}^{+}, \mathrm{NO}_{3}{ }^{-}$, $\mathrm{Na}^{+}$, and $\mathrm{Cl}^{-}$ions. Aside from the two ionic compounds originally present in the solutions, $\mathrm{AgNO}_{3}$ and NaCl , two additional ionic compounds may be derived from this collection of ions: $\mathrm{NaNO}_{3}$ and AgCl . The solubility guidelines indicate all nitrate salts are soluble but that AgCl is one of insoluble. A precipitation reaction, therefore, is predicted to occur, as described by the following equations:

$$
\begin{gathered}
\mathrm{NaCl}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{AgCl}(s)+\mathrm{NaNO}_{3}(a q) \quad \text { (molecular) } \\
\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s) \quad \text { (net ionic) }
\end{gathered}
$$


672. Predicting Precipitation Reactions -
Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction.
(a) potassium sulfate and barium nitrate
(b) lithium chloride and silver acetate
(c) lead nitrate and ammonium carbonate


673. Solution -
(a) The two possible products for this combination are $\mathrm{KNO}_{3}$ and $\mathrm{BaSO}_{4}$. The solubility guidelines indicate
$\mathrm{BaSO}_{4}$ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

$$
\mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow \mathrm{BaSO}_{4}(s)
$$

(b) The two possible products for this combination are $\mathrm{LiC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ and AgCl . The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

$$
\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)
$$

(c) The two possible products for this combination are $\mathrm{PbCO}_{3}$ and $\mathrm{NH}_{4} \mathrm{NO}_{3}$. The solubility guidelines indicate $\mathrm{PbCO}_{3}$ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

$$
\mathrm{Pb}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \longrightarrow \mathrm{PbCO}_{3}(s)
$$


674. Check Your Learning -
Which solution could be used to precipitate the barium ion, $\mathrm{Ba}^{2+}$, in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate?


675. Answer: -
sodium sulfate, $\mathrm{BaSO}_{4}$


676. Acid-Base Reactions -
An acid-base reaction is one in which a hydrogen ion, $\mathrm{H}^{+}$, is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion, and a full chapter is devoted to this topic later in the text.

For purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an acid is a substance that will dissolve in water to yield hydronium ions, $\mathrm{H}_{3} \mathrm{O}^{+}$. As an example, consider the equation shown here:

$$
\mathrm{HCl}(a q)+\mathrm{H}_{2} \mathrm{O}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)
$$

The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, $\mathrm{H}_{3} \mathrm{O}^{+}$ions are produced by a chemical reaction in which $\mathrm{H}^{+}$ions are transferred from HCl molecules to $\mathrm{H}_{2} \mathrm{O}$ molecules (Figure 7.5).

FIGURE 7.5 When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).


676. Acid-Base Reactions -
FIGURE 7.5 When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).

The nature of HCl is such that its reaction with water as just described is essentially $100 \%$ efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called strong acids, and HCl is one among just a handful of common acid compounds that are classified as strong (Table 7.2). A far greater number of compounds behave as weak acids and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars:

$$
\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)
$$

When dissolved in water under typical conditions, only about $1 \%$ of acetic acid molecules are present in the ionized form, $\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}$(Figure 7.6). (The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.)

(a)

(b)

FIGURE 7.6 (a) Fruits such as oranges, lemons, and grapefruit contain the weak acid citric acid. (b) Vinegars contain the weak acid acetic acid. (credit a: modification of work by Scott Bauer; credit b: modification of work by Brücke-Osteuropa/Wikimedia Commons)

Common Strong Acids


676. Acid-Base Reactions -
Common Strong Acids

| Compound Formula | Name in Aqueous Solution |
| :--- | :--- |
| HBr | hydrobromic acid |
| HCl | hydrochloric acid |
| HI | hydroiodic acid |
| $\mathrm{HNO}_{3}$ | nitric acid |
| $\mathrm{HClO}_{4}$ | perchloric acid |
| $\mathrm{H}_{2} \mathrm{SO}_{4}$ | sulfuric acid |

TABLE 7.2

A base is a substance that will dissolve in water to yield hydroxide ions, $\mathrm{OH}^{-}$. The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion-for example, NaOH and $\mathrm{Ca}(\mathrm{OH})_{2}$. Unlike the acid compounds discussed previously, these compounds do not react chemically with water; instead they dissolve and dissociate, releasing hydroxide ions directly into the solution. For example, KOH and $\mathrm{Ba}(\mathrm{OH})_{2}$ dissolve in water and dissociate completely to produce cations ( $\mathrm{K}^{+}$and $\mathrm{Ba}^{2+}$, respectively) and hydroxide ions, $\mathrm{OH}^{-}$. These bases, along with other hydroxides that completely dissociate in water, are considered strong bases.

Consider as an example the dissolution of lye (sodium hydroxide) in water:

$$
\mathrm{NaOH}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q)
$$


676. Acid-Base Reactions -
$$
\mathrm{NaOH}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q)
$$

This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield $\mathrm{Na}^{+}$and $\mathrm{OH}^{-}$ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases.
Unlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (Figure 7.7). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:

$$
\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)
$$

This is, by definition, an acid-base reaction, in this case involving the transfer of $\mathrm{H}^{+}$ions from water molecules to ammonia molecules. Under typical conditions, only about $1 \%$ of the dissolved ammonia is present as $\mathrm{NH}_{4}{ }^{+}$ ions.

FIGURE 7.7 Ammonia is a weak base used in a variety of applications. (a) Pure ammonia is commonly applied as an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of work by National Resources Conservation Service; credit b: modification of work by pat00139)

A neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base (but not water), and the products are often a salt and water


676. Acid-Base Reactions -
A neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base (but not water), and the products are often a salt and water

$$
\text { acid + base } \longrightarrow \text { salt + water }
$$

To illustrate a neutralization reaction, consider what happens when a typical antacid such as milk of magnesia (an aqueous suspension of solid $\mathrm{Mg}(\mathrm{OH})_{2}$ ) is ingested to ease symptoms associated with excess stomach acid ( HCl ):

$$
\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)
$$

Note that in addition to water, this reaction produces a salt, magnesium chloride.


678. Writing Equations for Acid-Base Reactions -
Write balanced chemical equations for the acid-base reactions described here:
(a) the weak acid hydrogen hypochlorite reacts with water
(b) a solution of barium hydroxide is neutralized with a solution of nitric acid


679. Solution -
(a) The two reactants are provided, HOCl and $\mathrm{H}_{2} \mathrm{O}$. Since the substance is reported to be an acid, its reaction with water will involve the transfer of $\mathrm{H}^{+}$from HOCl to $\mathrm{H}_{2} \mathrm{O}$ to generate hydronium ions, $\mathrm{H}_{3} \mathrm{O}^{+}$and hypochlorite ions, $\mathrm{OCl}^{-}$.

$$
\mathrm{HOCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{OCl}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)
$$

A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely.
(b) The two reactants are provided, $\mathrm{Ba}(\mathrm{OH})_{2}$ and $\mathrm{HNO}_{3}$. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide $\left(\mathrm{Ba}^{2+}\right)$ and the anion generated when the acid transfers its hydrogen ion $\left(\mathrm{NO}_{3}{ }^{-}\right)$.

$$
\mathrm{Ba}(\mathrm{OH})_{2}(a q)+2 \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)
$$

Check Your Learning
Write the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. (Hint: Consider the ions produced when a strong acid is dissolved in water.)


680. Answer: -
$\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)$


682. Stomach Antacids -
Our stomachs contain a solution of roughly 0.03 M HCl , which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, $\mathrm{CaCO}_{3}$. The reaction,

$$
\mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}(a q) \rightleftharpoons \mathrm{CaCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)
$$

not only neutralizes stomach acid, it also produces $\mathrm{CO}_{2}(g)$, which may result in a satisfying belch.
Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, $\mathrm{Mg}(\mathrm{OH})_{2}$. It works according to the reaction:

$$
\operatorname{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)
$$

The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that:

$$
\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)
$$


682. Stomach Antacids -
$$
\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)
$$

This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect. Several antacids have aluminum hydroxide, $\mathrm{Al}(\mathrm{OH})_{3}$, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.


684. Culinary Aspects of Chemistry -
Examples of acid-base chemistry are abundant in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. $\mathrm{NaHCO}_{3}$ is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter "rises." Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.

Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure 7.8). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a "sour" taste that we seem to enjoy.

FIGURE 7.8 A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.

Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.


685. LINK TO LEARNING -
Explore the microscopic view (http://openstax.org/l/16AcidsBases) of strong and weak acids and bases.