Analytical Chem istry - DePauw University

Chapter 7 Collecting and Preparing Samples311Disadvantages include contamination from the flux and the crucible, andthe loss of volatile materials.Finally, we can decompose organic materials by dry ashing. In thismethod the sample is placed in a suitable crucible and heated over a flameor in a furnace. The carbon present in the sample oxidizes to CO 2 , andhydrogen, sulfur, and nitrogen leave as H 2 O, SO 2 , and N 2 . These gases canbe trapped and weighed to determine their concentration in the organicmaterial. Often the goal of dry ashing is to remove the organic material,leaving behind an inorganic residue, or ash, that can be further analyzed.7DSeparating the Analyte from InterferentsWhen an analytical method is selective for the analyte, analyzing samples isa relatively simple task. For example, a quantitative analysis for glucose inhoney is relatively easy to accomplish if the method is selective for glucose,even in the presence of other reducing sugars, such as fructose. Unfortunately,few analytical methods are selective toward a single species.In the absence of interferents, the relationship between the sample’ssignal, S samp , and the concentration of analyte, C A , isS= k C7.9samp A Awhere k A is the analyte’s sensitivity. If an interferent, is present, then equation7.9 becomesS = k C + kCsamp A A I I7.10where k I and C I are, respectively, the interferent’s sensitivity and concentration.A method’s selectivity for the analyte is determined by the relativedifference in its sensitivity toward the analyte and the interferent. If k Ais greater than k I , then the method is more selective for the analyte. Themethod is more selective for the interferent if k I is greater than k A .Even if a method is more selective for an interferent, we can use it todetermine C A if the interferent’s contribution to S samp is insignificant. Theselectivity coefficient, K A,I , which we introduced in Chapter 3, providesa way to characterize a method’s selectivity.KA,IkI= 7.11kSolving equation 7.11 for k I , substituting into equation 7.10, and simplifying,givesAS = k ( C + K × C ) 7.12samp A A A,I IAn interferent, therefore, does not pose a problem as long as the product ofits concentration and its selectivity coefficient is significantly smaller thanthe analyte’s concentration.In equation 7.9, and the equations thatfollow, you can replace the analyte’s concentration,C A , with the moles of analyte,n A when working with methods, such asgravimetry, that respond to the absoluteamount of analyte in a sample. In this casethe interferent also is expressed in termsof moles.