# 2.3: Stoichiometric Calculations

A balanced reaction, which gives the stoichiometric relationship between the moles of reactants and the moles of products, provides the basis for many analytical calculations. Consider, for example, an analysis for oxalic acid, H2C2O4, in which Fe3+ oxidizes oxalic acid to CO2.

$\ce{2Fe^{3+}}(aq) + \ce{H2C2O4}(aq) + \ce{2H2O}(l) \rightarrow \ce{2Fe^2+}(aq) + \ce{2CO2}(g) + \ce{2H3O+}(aq)$

The balanced reaction indicates that one mole of oxalic acid reacts with two moles of Fe3+. As shown in Example 2.6, we can use this balanced reaction to determine the amount of oxalic acid in a sample of rhubarb.

Example 2.6

The amount of oxalic acid in a sample of rhubarb was determined by reacting with Fe3+. After extracting a 10.62 g of rhubarb with a solvent, oxidation of the oxalic acid required 36.44 mL of 0.0130 M Fe3+. What is the weight percent of oxalic acid in the sample of rhubarb?

SOLUTION

We begin by calculating the moles of Fe3+ used in the reaction

$\mathrm{\dfrac{0.0130\: mol\: Fe^{3+}}{L} \times 0.03644\: L = 4.73{\color{Red} 7} \times 10^{-4}\: mol\: Fe^{3+}}$

The moles of oxalic acid reacting with the Fe3+, therefore, is

$\mathrm{4.73{\color{Red} 7}\times10^{-4}\: mol\: Fe^{3+} \times \dfrac{1\: mol\: H_2C_2O_4}{2\: mol\: Fe^{3+}} = 2.36{\color{Red} 8} \times 10^{-4}\: mol\: H_2C_2O_4}$

Converting the moles of oxalic acid to grams of oxalic acid

$\mathrm{2.36{\color{Red} 8}\times10^{-4}\: mol\: C_2H_2O_4 \times \dfrac{90.03\: g\: H_2C_2O_4}{mol\: H_2C_2O_4} = 2.13{\color{Red} 2}\times10^{-2}\: g\: H_2C_2O_4}$

and calculating the weight percent gives the concentration of oxalic acid in the sample of rhubarb as

$\mathrm{\dfrac{2.13{\color{Red} 2}\times10^{-2}\: g\: H_2C_2O_4}{10.62\: g\: rhubarb}\times100 = 0.201\%\: \textrm{w/w}\: H_2C_2O_4}$

Note: Oxalic acid, in sufficient amounts, is toxic. At lower physiological concentrations it leads to the formation of kidney stones. The leaves of the rhubarb plant contain relatively high concentrations of oxalic acid. The stalk, which many individuals enjoy eating, contains much smaller concentrations of oxalic acid.

Note

Note that we retain an extra significant figure throughout the calculation, rounding to the correct number of significant figures at the end. We will follow this convention in any problem involving more than one step. If we forget that we are retaining an extra significant figure, we might report the final answer with one too many significant figures. In this chapter we will mark the extra digit in red for emphasis. Be sure that you pick a system for keeping track of significant figures.

The analyte in Example 2.6, oxalic acid, is in a chemically useful form because there is a reagent, Fe3+, that reacts with it quantitatively. In many analytical methods, we must convert the analyte into a more accessible form before we can complete the analysis. For example, one method for the quantitative analysis of disulfiram, C10H20N2S4—the active ingedient in the drug Anatbuse—requires that we convert the sulfur to H2SO4 by first oxidizing it to SO2 by combustion, and then oxidizing the SO2 to H2SO4 by bubbling it through a solution of H2O2. When the conversion is complete, the amount of H2SO4 is determined by titrating with NaOH.

To convert the moles of NaOH used in the titration to the moles of disulfiram in the sample, we need to know the stoichiometry of the reactions. Writing a balanced reaction for H2SO4 and NaOH is straightforward

$\ce{H2SO4}(aq) + \ce{2NaOH}(aq) \rightarrow \ce{2H2O}(l) + \ce{Na2SO4}(aq)$

but the balanced reactions for the oxidations of C10H20N2S4 to SO2, and of SO2 to H2SO4 are not as immediately obvious. Although we can balance these redox reactions, it is often easier to deduce the overall stoichiometry by using a little chemical logic.

Practice Exercise 2.4

You can dissolve a precipitate of AgBr by reacting it with Na2S2O3, as shown here.

$\ce{AgBr}(s) + \ce{2Na2S2O3}(aq) \rightarrow \ce{Ag(S2O3)2^3-}(aq) + \ce{Br-}(aq) + \ce{4Na+}(aq)$

How many mL of 0.0138 M Na2S2O3 do you need to dissolve 0.250 g of AgBr?

Example 2.7

An analysis for disulfiram, C10H20N2S4, in Antabuse is carried out by oxidizing the sulfur to H2SO4 and titrating the H2SO4 with NaOH. If a 0.4613-g sample of Antabuse is taken through this procdure, requiring 34.85 mL of 0.02500 M NaOH to titrate the H2SO4, what is the %w/w disulfiram in the sample?

SOLUTION

Calculating the moles of H2SO4 is easy—first, we calculate the moles of NaOH used in the titration

$\mathrm{(0.02500\: M)×(0.03485\: L) = 8.712{\color{Red} 5}\times10^{-4}\: mol\: NaOH}$

and then we use the balanced reaction to calcualte the corresponding moles of H2SO4.

$\mathrm{8.712{\color{Red} 5}\times10^{-4}\: mol\: NaOH \times \dfrac{1\:mol\: H_2SO_4}{2\: mol\: NaOH} = 4.356{\color{Red} 2}\times10^{-4}\: mol\: H_2SO_4}$

We do not need balanced reactions to convert the moles of H2SO4 to the corresponding moles of C10H20N2S4. Instead, we take advantage of a conservation of mass—all the sulfur in C10H20N2S4 must end up in the H2SO4(Here is where we use a little chemical logic! A conservation of mass is the essence of stoichiometry.); thus

$\mathrm{4.356{\color{Red} 2}×10^{-4}\: mol\: \ce{H2SO4} \times \dfrac{1\:mol\: S}{mol\: \ce{H2SO4}} \times \dfrac{1\:mol\: \ce{C10H20N2S4}}{4\: mol\: S} = 1.089{\color{Red} 0}\times10^{-4}\: mol\: \ce{C10H20N2S4}}$

or

$\mathrm{1.089{\color{Red} 0}×10^{-4}\: mol\: \ce{C10H20N2S4} × \dfrac{296.54\: g\: \ce{C10H20N2S4}}{mol\: \ce{C10H20N2S4}} = 0.03229{\color{Red} 3}\: g\: \ce{C10H20N2S4}}$

$\mathrm{\dfrac{0.03229{\color{Red} 3}\: g\: \ce{C10H20N2S4}}{0.4613\: g\: sample} × 100 = 7.000\%\: \textrm{w/w}\: \ce{C10H20N2S4}}$