Chapter 8: Stoichiometry of Chemical Reactions

# 8.7 Quantitative Chemical Analysis

### Learning Outcomes

• Describe the fundamental aspects of titrations and gravimetric analysis
• Perform stoichiometric calculations using typical titration and gravimetric data

In the eighteenth century, the strength (actually the concentration) of vinegar samples was determined by noting the amount of potassium carbonate, $\ce{K2CO3}$, which had to be added, a little at a time, before bubbling ceased. The greater the weight of potassium carbonate added to reach the point where the bubbling ended, the more concentrated the vinegar.

We now know that the effervescence that occurred during this process was due to reaction with acetic acid, $\ce{CH3CO2H}$, the compound primarily responsible for the odor and taste of vinegar. Acetic acid reacts with potassium carbonate according to the following equation:

$2{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\text{(}aq\text{)}+{\text{K}}_{2}{\text{CO}}_{3}\text{(}s\text{)}\rightarrow{\text{KCH}}_{3}{\text{CO}}_{3}\text{(}aq\text{)}+{\text{CO}}_{2}\text{(}g\text{)}+{\text{H}}_{2}\text{O}\text{(}l\text{)}$

The bubbling was due to the production of $\ce{CO2}$.

The test of vinegar with potassium carbonate is one type of quantitative analysis—the determination of the amount or concentration of a substance in a sample. In the analysis of vinegar, the concentration of the solute (acetic acid) was determined from the amount of reactant that combined with the solute present in a known volume of the solution. In other types of chemical analyses, the amount of a substance present in a sample is determined by measuring the amount of product that results.

## Titration

The described approach to measuring vinegar strength was an early version of the analytical technique known as titration analysis. A typical titration analysis involves the use of a buret (Figure 8.7.1) to make incremental additions of a solution containing a known concentration of some substance (the titrant) to a sample solution containing the substance whose concentration is to be measured (the analyte). The titrant and analyte undergo a chemical reaction of known stoichiometry, and so measuring the volume of titrant solution required for complete reaction with the analyte (the equivalence point of the titration) allows calculation of the analyte concentration.

The equivalence point of a titration may be detected visually if a distinct change in the appearance of the sample solution accompanies the completion of the reaction. The halt of bubble formation in the classic vinegar analysis is one such example, though, more commonly, special dyes called indicators are added to the sample solutions to impart a change in color at or very near the equivalence point of the titration.

Equivalence points may also be detected by measuring some solution property that changes in a predictable way during the course of the titration. Regardless of the approach taken to detect a titration’s equivalence point, the volume of titrant actually measured is called the end point. Properly designed titration methods typically ensure that the difference between the equivalence and end points is negligible. Though any type of chemical reaction may serve as the basis for a titration analysis, the three described in this chapter (precipitation, acid-base, and redox) are most common. Additional details regarding titration analysis are provided in the chapter on acid-base equilibria.

### Example 8.7.1: Titration Analysis

The end point in a titration of a 50.00-mL sample of aqueous $\ce{HCl}$ was reached by addition of 35.23 mL of 0.250 M $\ce{NaOH}$ titrant. The titration reaction is:

$\text{HCl}\text{(}aq\text{)}+\text{NaOH}\text{(}aq\text{)}\rightarrow\text{NaCl}\text{(}aq\text{)}+{\text{H}}_{2}\text{O}\text{(}l\text{)}$

What is the molarity of the $\ce{HCl}$?

Show Solution

As for all reaction stoichiometry calculations, the key issue is the relation between the molar amounts of the chemical species of interest as depicted in the balanced chemical equation. The approach outlined earlier is followed, with additional considerations required, since the amounts of reactants are provided and requested are expressed as solution concentrations.

For this exercise, the calculation will follow the following outlined steps:

The molar amount of $\ce{HCl}$ is calculated to be:

$35.23\cancel{\text{mL NaOH}}\times \dfrac{1\cancel{\text{L}}}{1000\cancel{\text{mL}}}\times \dfrac{0.250\cancel{\text{mol NaOH}}}{1\cancel{\text{L}}}\times \dfrac{\text{1 mol HCl}}{1\cancel{\text{mol NaOH}}}=8.81\times {10}^{-3}\text{mol HCl}$

Using the provided volume of $\ce{HCl}$ solution and the definition of molarity, the HCl concentration is:

$\begin{array}{l}\\ M=\dfrac{\text{mol HCl}}{\text{L solution}}\\ M=\dfrac{8.81\times {10}^{-3}\text{mol HCl}}{\text{50.00 mL}\times \frac{\text{1 L}}{\text{1000 mL}}}\\ M=0.176M\end{array}$

Note: For these types of titration calculations, it is convenient to recognize that solution molarity is also equal to the number of millimoles of solute per milliliter of solution:

$M=\dfrac{\text{mol solute}}{\text{L solution}}\times \dfrac{\frac{{10}^{3}\text{mmol}}{\text{mol}}}{\frac{{10}^{3}\text{mL}}{\text{L}}}=\dfrac{\text{mmol solute}}{\text{mL solution}}\text{.}$

Using this version of the molarity unit will shorten the calculation by eliminating two conversion factors:

$\dfrac{35.23\text{mL NaOH}\times \frac{0.250\text{mmol NaOH}}{\text{mL NaOH}}\times \frac{1\text{mmol HCl}}{1\text{mmol NaOH}}}{50.00\text{mL solution}}=0.176M\text{HCl.}$

## Gravimetric Analysis

A gravimetric analysis is one in which a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from the other components of the sample. Mass measurements of the sample, the isolated analyte, or some other component of the analysis system, used along with the known stoichiometry of the compounds involved, permit calculation of the analyte concentration. Gravimetric methods were the first techniques used for quantitative chemical analysis, and they remain important tools in the modern chemistry laboratory.

The required change of state in a gravimetric analysis may be achieved by various physical and chemical processes. For example, the moisture (water) content of a sample is routinely determined by measuring the mass of a sample before and after it is subjected to a controlled heating process that evaporates the water. Also common are gravimetric techniques in which the analyte is subjected to a precipitation reaction of the sort described earlier in this chapter. The precipitate is typically isolated from the reaction mixture by filtration, carefully dried, and then weighed (Figure 8.7.2). The mass of the precipitate may then be used, along with relevant stoichiometric relationships, to calculate analyte concentration.

### Example 8.7.2: Gravimetric Analysis

A 0.4550-g solid mixture containing $\ce{MgSO4}$ is dissolved in water and treated with an excess of $\ce{Ba(NO3)2}$, resulting in the precipitation of 0.6168 g of $\ce{BaSO4}$.

${\text{MgSO}}_{4}\text{(}aq\text{)}+\text{Ba}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}\rightarrow{\text{BaSO}}_{4}\text{(}s\text{)}+\text{Mg}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}$

What is the concentration (percent) of $\ce{MgSO4}$ in the mixture?

Show Solution

The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of $\ce{BaSO4}$ and $\ce{MgSO4}$ through their stoichiometric factor. Once the mass of $\ce{MgSO4}$ is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.

The mass of $\ce{CaSO4}$ that would yield the provided precipitate mass is

$0.6168\cancel{{\text{g BaSO}}_{4}}\times \dfrac{1\cancel{{\text{mol BaSO}}_{4}}}{233.43\cancel{{\text{g BaSO}}_{4}}}\times \dfrac{1\cancel{{\text{mol MgSO}}_{4}}}{1\cancel{{\text{mol BaSO}}_{4}}}\times \dfrac{120.37{\text{g MgSO}}_{4}}{1\cancel{{\text{mol MgSO}}_{4}}}=0.3597{\text{g MgSO}}_{4}$

The concentration of $\ce{MgSO4}$ in the sample mixture is then calculated to be

$\begin{array}{l}\\ {\text{percent MgSO}}_{4}=\dfrac{{\text{mass MgSO}}_{4}}{\text{mass sample}}\times 100\%\\ \dfrac{\text{0.3597 g}}{0.4550 g}\times 100\%=69.91\%\end{array}$

The elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product (Figure 8.7.3). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.

### Example 8.7.3: Combustion Analysis

Polyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a $0.00126$ g sample of polyethylene yields $0.00394$ g of $\ce{CO2}$ and $0.00161$ g of $\ce{H2O}$. What is the empirical formula of polyethylene?

Show Solution

The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:

${\text{C}}_{x}{\text{H}}_{y}\text{(}s\text{)}+{\text{excess O}}_{2}\text{(}g\text{)}\rightarrow x{\text{CO}}_{2}\text{(}g\text{)}+\dfrac{y}{2}{\text{H}}_{2}\text{O}\text{(}g\text{)}$

Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts $x$ and $y$ are needed.

First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as we learned earlier. An outline of this approach is given in the following flow chart:

$\begin{array}{l}\\ \text{mol C}=0.00394{\text{g CO}}_{2}\times \dfrac{1{\text{mol CO}}_{2}}{\text{44.01 g/mol}}\times \dfrac{1\text{mol C}}{1{\text{mol CO}}_{2}}=8.95\times {10}^{-5}\text{mol C}\\ \text{mol H}=0.00161{\text{g H}}_{2}\text{O}\times \dfrac{1{\text{mol H}}_{2}\text{O}}{18.02\text{g/mol}}\times \dfrac{2\text{mol H}}{1{\text{mol H}}_{2}\text{O}}=1.79\times {10}^{-4}\text{mol H}\end{array}$

The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The $\ce{H}$-to-$\ce{C}$ molar ratio is

$\dfrac{\text{mol H}}{\text{mol C}}=\dfrac{1.79\times {10}^{-4}\text{mol H}}{8.95\times {10}^{-5}\text{mol C}}=\dfrac{2\text{mol H}}{1\text{mol C}}$

and the empirical formula for polyethylene is $\ce{CH2}$.

### Key Concepts and Summary

The stoichiometry of chemical reactions may serve as the basis for quantitative chemical analysis methods. Titrations involve measuring the volume of a titrant solution required to completely react with a sample solution. This volume is then used to calculate the concentration of analyte in the sample using the stoichiometry of the titration reaction. Gravimetric analysis involves separating the analyte from the sample by a physical or chemical process, determining its mass, and then calculating its concentration in the sample based on the stoichiometry of the relevant process. Combustion analysis is a gravimetric method used to determine the elemental composition of a compound by collecting and weighing the gaseous products of its combustion.

### Try It

1. Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M $\ce{NaOH}$ to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain?
2. In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO3)2 solution. $2{\text{Cl}}^{-}\text{(}aq\text{)}+\text{Hg}{\text{(}\text{NO}3\text{)}}_{2}\text{(}aq\text{)}\rightarrow 2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}+{\text{HgCl}}_{2}\text{(}s\text{)}$ What is the $\ce{Cl-}$ concentration in a 0.25-mL sample of normal serum that requires 1.46 mL of 5.25 × 10-4 M $\ce{Hg(NO3)2}$(aq) to reach the end point?
3. A sample of gallium bromide, $\ce{GaBr2}$, weighing 0.165 g was dissolved in water and treated with silver nitrate, $\ce{AgNO3}$, resulting in the precipitation of 0.299 g $\ce{AgBr}$. Use these data to compute the %Ga (by mass) $\ce{GaBr2}$.
4. A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of $\ce{B2O3}$. What are the empirical and molecular formulas of the compound?
5. What volume of a 0.00945-M solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with a $\ce{H2SO4}$ concentration of 1.23 × 10-4 M. ${\text{H}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}+2\text{KOH}\text{(}aq\text{)}\rightarrow{\text{K}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O}\text{(}l\text{)}$
Show Selected Solutions
1. The balanced equation is ${\text{H}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}+2\text{NaOH}\text{(}aq\text{)}\rightarrow{\text{Na}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O}\text{(}l\text{)}$The steps to follow in solving this problem if we use volumes in milliliters are$\text{Volume NaOH}\rightarrow\text{mmol NaOH}\rightarrow{\text{mmol H}}_{2}{\text{SO}}_{4}\rightarrow{\text{M H}}_{2}{\text{SO}}_{4}$$\begin{array}{l} 1.7\cancel{\text{mL}}\times \frac{0.0811\text{mmol NaOH}}{\cancel{\text{mL}}}=0.138\text{mmol NaOH}\\ {\text{mmol H}}_{2}{\text{SO}}_{4}=0.138\cancel{\text{mmol NaOH}}\times \frac{1{\text{mmol H}}_{2}{\text{SO}}_{4}}{2\cancel{\text{mmol NaOH}}}=\text{0.069 mmol}\\ M{\text{H}}_{2}{\text{SO}}_{4}=\frac{0.069{\text{mmol H}}_{2}{\text{SO}}_{4}}{\text{20.0 mL}}=3.4\times {10}^{-3}M\end{array}$
2. In this exercise, the volume is left in units of milliliters and the number of moles is expressed in units of millimoles to compensate for the factor of 1000 difference between units. This technique is often useful in calculations. The steps involved in solving the problem are$\text{Volume Hg}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\rightarrow\text{mmol Hg}\text{(}{\text{NO}}_{3}\text{)}\rightarrow{\text{mmol Cl}}^{-}\rightarrow{M\text{Cl}}^{-}$$\text{mmol Hg}{\text{(}{\text{NO}}_{3}\text{)}}_{2}=1.46\cancel{\text{mL}}\times \text{(}8.25\times {10}^{-4}\text{mmol/}\cancel{\text{mL}}\text{)}=1.20\times {10}^{-3}\text{mmol}$${\text{mmol Cl}}^{-}=\left[1.20\times {10}^{-3}\cancel{\text{mmol Hg}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}\right]\times \frac{2{\text{mmol Cl}}^{-}}{1\cancel{\text{mmol Hg}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}}=2.41\times {10}^{-3}{\text{mmol Cl}}^{-}$

${M\text{Cl}}^{-}=\frac{2.41\times {10}^{-3}\text{mmol}}{0.25}=9.6\times {10}^{-3}M$

3. The reaction is ${\text{GaBr}}_{3}\text{(}aq\text{)}+3{\text{AgNO}}_{3}\text{(}aq\text{)}\rightarrow 3\text{AgBr}\text{(}s\text{)}+\text{Ga}{\text{(}{\text{NO}}_{3}\text{)}}_{3}\text{(}aq\text{)}\text{.}$
Begin by considering the definition of mass percentage:$\%\text{Ga}=\frac{\text{g Ga}}{{\text{g GaBr}}_{3}}\times 100\%$
4. Calculate the mass of $\ce{B}$ in the 0.063-g sample of $\ce{B2O3}$. The difference of the mass of this boron and the 0.025-g sample of boron and hydrogen gives the mass of the hydrogen present. Determine the moles of $\ce{B}$ and $\ce{H}$ in the sample. Divide by the smaller number of moles to find the empirical formula. Divide the mass of the empirical formula into the assumed molecular mass of ~28 amu. That number multiplied by the subscripts of the empirical formula gives the molecular formula.
$\begin{array}{l}{\text{mass of B in B}}_{2}{\text{O}}_{3}=\frac{2\times 10.811{\text{g mol}}^{-1}\text{B}}{69.6202{\text{g mol}}^{-1}{\text{B}}_{2}{\text{O}}_{3}}\times 0.063{\text{g B}}_{2}{\text{O}}_{3}=\text{0.0196 g}\end{array}$mass of $\ce{H}$ in $\ce{B2O3}$ = 0.025 g $\ce{B}$ & $\ce{H}$ – 0.0196 g $\ce{B}$ = 0.0054 g $\ce{H}$
5. $\text{mass KOH}\rightarrow\text{mol KOH}\rightarrow{\text{mol H}}_{2}{\text{SO}}_{4}\rightarrow{\text{M H}}_{2}{\text{SO}}_{4}\rightarrow{\text{volume H}}_{2}{\text{SO}}_{4}$
$\text{mol KOH}=24.74\cancel{\text{g}}\text{KOH}\times \frac{1\cancel{\text{mol KOH}}}{39.0983\cancel{\text{g}}}\times \frac{1{\text{mol H}}_{2}{\text{SO}}_{4}}{2\cancel{\text{mol KOH}}}=\text{0.031638 mol}$
${\text{volume H}}_{2}{\text{SO}}_{4}=\frac{\text{0.031638 mol}}{\text{0.3446 M}}=\text{0.09181 L}$

## Glossary

analyte: chemical species of interest

buret: device used for the precise delivery of variable liquid volumes, such as in a titration analysis

combustion analysis: gravimetric technique used to determine the elemental composition of a compound via the collection and weighing of its gaseous combustion products

end point: measured volume of titrant solution that yields the change in sample solution appearance or other property expected for stoichiometric equivalence (see equivalence point)

equivalence point: volume of titrant solution required to react completely with the analyte in a titration analysis; provides a stoichiometric amount of titrant for the sample’s analyte according to the titration reaction

gravimetric analysis: quantitative chemical analysis method involving the separation of an analyte from a sample by a physical or chemical process and subsequent mass measurements of the analyte, reaction product, and/or sample

indicator: substance added to the sample in a titration analysis to permit visual detection of the end point

quantitative analysis: the determination of the amount or concentration of a substance in a sample

titrant: solution containing a known concentration of substance that will react with the analyte in a titration analysis

titration analysis: quantitative chemical analysis method that involves measuring the volume of a reactant solution required to completely react with the analyte in a sample