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Analytical Chem istry - DePauw University

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Chapter 2 Basic Tools of <strong>Analytical</strong> <strong>Chem</strong><strong>istry</strong>25form before we can complete the analysis. For example, one method for thequantitative analysis of disulfiram, C 10 H 20 N 2 S 4 —the active ingedient inthe drug Anatbuse—requires that we convert the sulfur to H 2 SO 4 by firstoxidizing it to SO 2 by combustion, and then oxidizing the SO 2 to H 2 SO 4by bubbling it through a solution of H 2 O 2 . When the conversion is complete,the amount of H 2 SO 4 is determined by titrating with NaOH.To convert the moles of NaOH used in the titration to the moles of disulfiramin the sample, we need to know the stoichiometry of the reactions.Writing a balanced reaction for H 2 SO 4 and NaOH is straightforwardHSO ( aq) + 2NaOH( aq) → 2H O() l + NaSO ( aq )2 4 2 2 4but the balanced reactions for the oxidations of C 10 H 20 N 2 S 4 to SO 2 , andof SO 2 to H 2 SO 4 are not as immediately obvious. Although we can balancethese redox reactions, it is often easier to deduce the overall stoichiometryby using a little chemical logic.Example 2.7An analysis for disulfiram, C 10 H 20 N 2 S 4 , in Antabuse is carried out byoxidizing the sulfur to H 2 SO 4 and titrating the H 2 SO 4 with NaOH. Ifa 0.4613-g sample of Antabuse is taken through this procdure, requiring34.85 mL of 0.02500 M NaOH to titrate the H 2 SO 4 , what is the %w/wdisulfiram in the sample?So l u t i o nCalculating the moles of H 2 SO 4 is easy—first, we calculate the moles ofNaOH used in the titration4( 0. 02500 M) × ( 0. 03485 L) = 8.7125× 10− molNaOHand then we use the balanced reaction to calcualte the correspondingmoles of H 2 SO 4 .48. 7125× 10− molNaOH×1mol HSO2 4= × − 44. 356210 mol HSO2mol NaOH2 4We do not need balanced reactions to convert the moles of H 2 SO 4 tothe corresponding moles of C 10 H 20 N 2 S 4 . Instead, we take advantage of aconservation of mass—all the sulfur in C 10 H 20 N 2 S 4 must end up in theH 2 SO 4 ; thus41 molS4. 3562× 10− molH SO × ×2 4molH SO1mol C H NS10 20 2 44 molS2 4= 1. 0890× 10−4 molC H NS10 20N SS NSdisulfram2 4Here is where we use a little chemical logic!A conservation of mass is the essence ofstoichiometry.S

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