Hard ways and easy ways

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<ul><li><p>Wichita State University Wlchita. KS 67208 </p><p>Hard Ways and Easy Ways -R Lowell M. Schwartz Universit~ 01 M a s ~ a ~ h ~ ~ e n ~ B ~ ~ t o n Boston. MA 02125 </p><p>Instructors are all aware that chemical calculations often c3n he done different ways. Here are two examples, each of which featureionr "traditional"ap~roachthat requiressev- era1 steps and one "clever" methodthat is much shorter. </p><p>Example 1. A "LlmRlng Reagent" Problem 51.4 g of CaCz and 26.7 g of Hz0 react completely to form </p><p>products CzHz and c a ( 0 H ) ~ . Calculate the sum of the masses of C2H2, Ca(OHh, and excess reactant. The Hard Way </p><p>When treating this as a conventional limiting reagent problem, there are several strategies for finding the limiting reagent, but all take about the same amount of work. One of these is done here. (In order to avoid round-off errors. more digits are retained ih intermediate calculations than ale jus- tified by significant figure considerations.) 1. Balance the chemical equation </p><p>2. Calculate the formula weights of CaCz and H20 using an atomic weight tahle (64.102 and 18.015 g mol-', respectively). </p><p>3. Calculate the moles of CaCz and H20 (0.8018 and 1.4821 mol, respectively) </p><p>4. Calculate the moles of Hz0 required to react with all the CaC2 assuming that CsC2 is the limiting reagent (1.6036 mol). </p><p>5. Observe that only 1.4821 mol of Hz0 are available. Because there is insufficient H20 to react with all the CaCZ, Hz0 is the limiting reagent. </p><p>6. Calculate the moles of C2H2 and Ca(OH)2 produced (both 0.74105 mol). </p><p>7. Cnlculnre the formula weightsd C?H, and Ca(OH)? (26.027 and 74095g moll, rrsprcti\elyl. </p><p>8. Calrdnte thr mass of CIHI and Ca(OHJi produced (19.287 and 54.908 g, respectively). </p><p>9. Calculate the moles of CaC2 reacted (0.74105 mol). 10. Calculate the moles of excess CaC2 (0.06075 mol). 11. Calculate the mass of excess CaC2 (3.894 g). 12. Sum the mases of excess CaC? and of CQH~ and CdOHL </p><p>- - . .- (78.1 g). The Easy Way </p><p>Recognizing that this problem requires only the total mass of products and excess reactant. the Law of Consemation of ass dictates that this sum equals the original mass of reactants. These reactant masses are 51.4 and 26.7. Their sum is 78.1. </p><p>Example 2. A "Mixlng" Problem In beaker 1 we have T moles of pure liquid A. In heaker 2 </p><p>we have the same number Tmoles of pure liquid B. Species A and B are nonreactive and are miscible in all proportions. We transfer an uns~ecified nortion. sav n moles. of liauid B to beaker 1 and sudsequentiy transfer exactly n koles'of the contents of beaker 1 hack to beaker 2. Compare the mole </p><p>898 Journal of Chemical Education </p><p>fraction concentration of foreien snecies A in beaker 2 with the mole fraction concentration of'foreign species B in hea- ker 1. Is the concentration of A in heaker 2 neater than. less than, or equal to the concentration of B in ieaker l? </p><p>The Hard Way 1. After the first transfer, beaker 1 contains the following: </p><p>n moles of B T moles of A (T + n) total moles mole fraction B equal to nl(T + n) mole fraction A equal to T/(T + n) </p><p>2. Assuming complete mixingin beaker 1 after the first transfer, the second transfer carries n[TI(T+ n)] moles ofA back to beaker 2. </p><p>3. The total moles in heaker 2 has now been restored to T so the mole fraction of A in 2 is [nT/(T + n)]lT = nl(T + n), which we see is identical to the mole fraction of B in beaker 1. </p><p>The Easy Way Suppose after both transfers we end up with m moles of </p><p>foreign species A in beaker 2. Because heaker 2 started with Ttotalmoles and ends with Ttotal moles, this means that m moles of B must be displaced from heaker 2. These m moles of B end up in beaker 1. Thus both beakers contain mole fraction mlT of the foreign species. </p><p>Notice that the "Easy Way" does not make the assump- tion of complete mixing after the first transfer. Try solving this problem the "Hard Way" or any other way without making this assumption. </p><p>When these problems are given to a class of first-year chemistry students, most proceed the "Hard Way". Why? Because we teach a topic, such as limiting reagent stoichio- metry, and then seek to drive home the topic by assigning several homework problems dealing with limiting reagent calculations. E x a m ~ l e 1 above certainlv looks like a limiting reagent problem add so we are not surprised to see students taking that approach. Perhaps we should not expect more from students who are struggling with their first exposure to such a concept. But somewhere in the educational Drocess the transition must he made from "narrow" thinkingwithin a specific topic to "wide" thinking within the science as a whole. This transition is a difficult one and should he post- poned until a fair amount of chemistry has been taught. The two examples described above might he used by instructors to attempt the transition. My suggestion is to pose these problems late in the course and well after stoichiometry or mole fraction concentration problems have been treated. Encourage the class to search for alternative methods of solution and challenge them to calculate the answer using the fewest ~oss ihle steps. When the results are in and the " ~ a s y ways" are cornpired to the "Hard Ways", it should be clear why "wide" thinking is superior to "narrow" thinking. </p><p>Acknowledgment The author is thankful to Klaus Schultz of the University </p><p>of Massarhusetts/Amherst for showing him Example 2 in a slightly different guise. </p></li></ul>