D. Harvey - Modern Analytical Chemistry (794078), страница 10
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These stoichiometric relationships provide thebasis for many analytical calculations. Consider, for example, the problem of determining the amount of oxalic acid, H2C2O4, in rhubarb. One method for this analysis uses the following reaction in which we oxidize oxalic acid to CO2.2Fe3+(aq) + H2C2O4(aq) + 2H2O(l) → 2Fe2+(aq) + 2CO2(g) + 2H3O+(aq) 2.2The balanced chemical reaction provides the stoichiometric relationship betweenthe moles of Fe3+ used and the moles of oxalic acid in the sample being analyzed—specifically, one mole of oxalic acid reacts with two moles of Fe3+.
As shown in Example 2.6, the balanced chemical reaction can be used to determine the amount ofoxalic acid in a sample, provided that information about the number of moles ofFe3+ is known.1400-CH02 9/8/99 3:48 PM Page 21Chapter 2 Basic Tools of Analytical ChemistryEXAMPLE 2.6The amount of oxalic acid in a sample of rhubarb was determined by reactingwith Fe3+ as outlined in reaction 2.2.
In a typical analysis, the oxalic acid in10.62 g of rhubarb was extracted with a suitable solvent. The completeoxidation of the oxalic acid to CO2 required 36.44 mL of 0.0130 M Fe3+. Whatis the weight percent of oxalic acid in the sample of rhubarb?SOLUTIONWe begin by calculating the moles of Fe3+ used in the reaction0.0130 mol Fe3+× 0.03644 L = 4.737 × 10 –4 mol Fe3+LThe moles of oxalic acid reacting with the Fe3+, therefore, is4.737 × 10 –4 mol Fe3+ ×1 mol C2 H 2 O4= 2.369 × 10 –4 mol C 2 H 2 O42 mol Fe3+Converting moles of oxalic acid to grams of oxalic acid2.369 × 10 –4 mol C 2 H 2 O4 ×90.03 g C2 H 2 O4= 2.132 × 10 –2 g oxalic acidmol C 2 H 2 O4and converting to weight percent gives the concentration of oxalic acid in thesample of rhubarb as2.132 × 10 –2 g C 2 H 2 O4× 100 = 0.201% w/w C 2 H 2 O410.62 g rhubarbIn the analysis described in Example 2.6 oxalic acid already was present in thedesired form.
In many analytical methods the compound to be determined must beconverted to another form prior to analysis. For example, one method for the quantitative analysis of tetraethylthiuram disulfide (C10H20N2S4), the active ingredient inthe drug Antabuse (disulfiram), requires oxidizing the S to SO2, bubbling the SO2through H2O2 to produce H2SO4, followed by an acid–base titration of the H2SO4with NaOH. Although we can write and balance chemical reactions for each of thesesteps, it often is easier to apply the principle of the conservation of reaction units.A reaction unit is that part of a chemical species involved in a reaction.
Consider, for example, the general unbalanced chemical reactionA + B → ProductsConservation of reaction units requires that the number of reaction units associatedwith the reactant A equal the number of reaction units associated with the reactantB. Translating the previous statement into mathematical form givesNumber of reaction units per A × moles A= number of reaction units per B × moles B2.3If we know the moles of A and the number of reaction units associated with A andB, then we can calculate the moles of B.
Note that a conservation of reaction units,as defined by equation 2.3, can only be applied between two species. There are fiveimportant principles involving a conservation of reaction units: mass, charge, protons, electron pairs, and electrons.211400-CH02 9/8/99 3:48 PM Page 2222Modern Analytical Chemistry2C.1 Conservation of MassThe easiest principle to appreciate is conservation of mass.
Except for nuclear reactions, an element’s total mass at the end of a reaction must be the same as that present at the beginning of the reaction; thus, an element serves as the most fundamental reaction unit. Consider, for example, the combustion of butane to produce CO2and H2O, for which the unbalanced reaction isC4H10(g) + O2(g) → CO2(g) + H2O(g)All the carbon in CO2 comes from the butane, thus we can select carbon as a reaction unit. Since there are four carbon atoms in butane, and one carbon atom inCO2, we write4 × moles C4H10 = 1 × moles CO2Hydrogen also can be selected as a reaction unit since all the hydrogen in butaneends up in the H2O produced during combustion. Thus, we can write10 × moles C4H10 = 2 × moles H2OAlthough the mass of oxygen is conserved during the reaction, we cannot applyequation 2.3 because the O2 used during combustion does not end up in a singleproduct.Conservation of mass also can, with care, be applied to groups of atoms.
Forexample, the ammonium ion, NH4+, can be precipitated as Fe(NH4)2(SO4)2 ⋅ 6H2O.Selecting NH4+ as the reaction unit gives2 × moles Fe(NH4)2(SO4)2 · 6H2O = 1 × moles NH4+2C.2 Conservation of ChargeThe stoichiometry between two reactants in a precipitation reaction is governed bya conservation of charge, requiring that the total cation charge and the total anioncharge in the precipitate be equal. The reaction units in a precipitation reaction,therefore, are the absolute values of the charges on the cation and anion that makeup the precipitate. Applying equation 2.3 to a precipitate of Ca3(PO4)2 formed fromthe reaction of Ca2+ and PO43–, we write2 × moles Ca2+ = 3 × moles PO43–2C.3 Conservation of ProtonsIn an acid–base reaction, the reaction unit is the proton.
For an acid, the number of reaction units is given by the number of protons that can be donatedto the base; and for a base, the number of reaction units is the number ofprotons that the base can accept from the acid. In the reaction between H3PO4and NaOH, for example, the weak acid H3PO4 can donate all three of its protons to NaOH, whereas the strong base NaOH can accept one proton. Thus,we write3 × moles H3PO4 = 1 × moles NaOHCare must be exercised in determining the number of reaction units associated with the acid and base.
The number of reaction units for an acid, for instance, depends not on how many acidic protons are present, but on how many1400-CH02 9/8/99 3:48 PM Page 23Chapter 2 Basic Tools of Analytical Chemistryof the protons are capable of reacting with the chosen base. In the reaction between H3PO4 and NH3H3PO4(aq) + 2NH3(aq)t HPO4–(aq) + 2NH4+(aq)a conservation of protons requires that2 × moles H3PO4 = moles of NH32C.4 Conservation of Electron PairsIn a complexation reaction, the reaction unit is an electron pair. For the metal, thenumber of reaction units is the number of coordination sites available for bindingligands.
For the ligand, the number of reaction units is equivalent to the number ofelectron pairs that can be donated to the metal. One of the most important analytical complexation reactions is that between the ligand ethylenediaminetetracetic acid(EDTA), which can donate 6 electron pairs and 6 coordinate metal ions, such asCu2+; thus6 × mole Cu2+ = 6 × moles EDTA2C.5 Conservation of ElectronsIn a redox reaction, the reaction unit is an electron transferred from a reducingagent to an oxidizing agent.
The number of reaction units for a reducing agent isequal to the number of electrons released during its oxidation. For an oxidizingagent, the number of reaction units is given by the number of electrons needed tocause its reduction. In the reaction between Fe3+ and oxalic acid (reaction 2.2), forexample, Fe3+ undergoes a 1-electron reduction. Each carbon atom in oxalic acid isinitially present in a +3 oxidation state, whereas the carbon atom in CO2 is in a +4oxidation state. Thus, we can write1 × moles Fe3+ = 2 × moles of H2C2O4Note that the moles of oxalic acid are multiplied by 2 since there are two carbonatoms, each of which undergoes a 1-electron oxidation.2C.6 Using Conservation Principles in Stoichiometry ProblemsAs shown in the following examples, the application of conservation principles simplifies stoichiometric calculations.EXAMPLE 2.7Rework Example 2.6 using conservation principles.SOLUTIONConservation of electrons for this redox reaction requires thatmoles Fe3+ = 2 × moles H2C2O4which can be transformed by writing moles as the product of molarity andvolume or as grams per formula weight.MFe 3 + × VFe 3 + =2 × g H 2 C 2 O4FW H 2 C 2 O4231400-CH02 9/8/99 3:48 PM Page 2424Modern Analytical ChemistrySolving for g H2C2O4 givesMFe 3 + × VFe 3 + × FW H 2 C 2 O4(0.0130 M)(0.03644 L)(90.03 g/mole)=22= 2.132 × 10 –2 g H2 C 2 O4and the weight percent oxalic acid is2.132 × 10 –2 g C 2 H 2 O4× 100 = 0.201% w/w C 2 H 2 O410.62 g rhubarbEXAMPLE 2.8One quantitative analytical method for tetraethylthiuram disulfide, C10H20N2S4(Antabuse), requires oxidizing the sulfur to SO2, and bubbling the resultingSO2 through H2O2 to produce H2SO4.
The H2SO4 is then reacted with NaOHaccording to the reactionH2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)Using appropriate conservation principles, derive an equation relating themoles of C 10 H 20 N 2 S 4 to the moles of NaOH. What is the weight percentC10H20N2S4 in a sample of Antabuse if the H2SO4 produced from a 0.4613-gportion reacts with 34.85 mL of 0.02500 M NaOH?SOLUTIONThe unbalanced reactions converting C10H20N2S4 to H2SO4 areC10H20N2S4 → SO2SO2 → H2SO4Using a conservation of mass we have4 × moles C10H20N2S4 = moles SO2 = moles H2SO4A conservation of protons for the reaction of H2SO4 with NaOH gives2 × moles H2SO4 = moles of NaOHCombining the two conservation equations gives the following stoichiometricequation between C10H20N2S4 and NaOH8 × moles C10H20N2S4 = moles NaOHNow we are ready to finish the problem.
Making appropriate substitutions formoles of C10H20N2S4 and moles of NaOH gives8 × g C10 H 20 N 2 S4= MNaOH × VNaOHFW C10 H 20 N 2 S4Solving for g C10H20N2S4 givesg C10 H 20 N 2S 4 =1× M NaOH × VNaOH × FW C10 H 20 N 2S 481(0.02500 M)(0.03485 L)(296.54 g/mol) = 0.032295 g C10 H 20 N 2 S481400-CH02 9/8/99 3:48 PM Page 25Chapter 2 Basic Tools of Analytical Chemistry25The weight percent C10H20N2S4 in the sample, therefore, is0.32295 g C10 H 20 N 2 S4× 100 = 7.001% w/w C10 H 20 N 2 S40.4613 g sample2D Basic Equipment and InstrumentationMeasurements are made using appropriate equipment or instruments. The array ofequipment and instrumentation used in analytical chemistry is impressive, rangingfrom the simple and inexpensive, to the complex and costly.
With two exceptions,we will postpone the discussion of equipment and instrumentation to those chapters where they are used. The instrumentation used to measure mass and much ofthe equipment used to measure volume are important to all analytical techniquesand are therefore discussed in this section.2D.1 Instrumentation for Measuring MassAn object’s mass is measured using a balance. The most common type of balanceis an electronic balance in which the balance pan is placed over an electromagnet(Figure 2.2).
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