DONNABOGNER Hutchinson Seniw High School Hutchinson. KS 67501
Unit Basis-A Neglected Problem-Solving Technique George J. Beichl St. Joseph's University, Philadelphia, PA 19131
T h e maior stumbline block for students of eeneral chemistry is pro61em s o l v i n g ~ t u d e n t soon s realizethat this facet of the subject must he mastered if thev are to obtain respectable grades. However, given a choice, they will opt for the easiest route to obtain their objective. One of these "easy" routes is to memorize the solution of type problems illustrated in the text or to memorize a formula and plug quantities into these formulas. Another easy route is the factor-label method, which enables one to solve chemistry problems without understanding the problem. This was perceptively analyzed in a recent article hy Navidi and Baker.' In other words, it is possible to become competent a t solving prohlems without appreciating the reason for the steps taken, i.e., without learnine to think. There is an approach that, once learned, can serve as a framework for the logical solution of the vast majority of the problems encounter& in general chemistry. 1ti s a systematic application of common sense known as the unit basis In wlving problems, no matter ~ , h a system t is employed, student.; must first be taught to he systematic and to label the dimensions of all quantities. I h e n s i o n a l analysis, also known as the factor-label method, is a sine qua non for any successful approach to problem solving. It is useful in convertingone unit toanother, e.g., kilometers tomiles asshown below 1in. 1ft 1 mi 1000 m lOOcm 2.0km X----xX-X---= 1.2 mi 1 km 1m 2.54 em 12 in. 5280ft However, the cancellation of units is not a sufficient condition to guarantee a correct solution to more complex prohlems.2 T h e unit basis approach t o problem solving analyzes t h e question in such a way that the student clearly sees what data are needed to solve the prohlem. This is illustrated in the following examples. Problem 1 How many grams of Oz can be obtained from the decompoaitionof 50.0 g of KNOB?
146
Journal of Chemical Education
a) Restate the question but substitute one unit of material in place of the amount specified in the question. Thus, how many grams of O2 can be obtained from the decomposition of 1.00 g of KNOB?It is readily apparent that if this quantity were known, the answer to the
original question would he obtained by multiplying this quantity by 50.0 g of KNO, b) To obtain the quantity that would represent the weight af Oz obtainable from 1.W g of KNOBa relationship between two quantities must be obtained. That relationship is the weight of 0 2 obtainable from a given weight of KN03. The balanced chemical equation shows that 2 mol, or 202 g ofKN03, evolve 1mol, or 32.0 g of 02. C) The desired quantity, i.e., the grams of 0 2 evolved from the decomposition of 1 g of KN08, is obtained by setting up a fraction in which the denominator is the quantity of which we desire one unit, i.e., 32.0 g 0, 202 g KNOB It to note the units that students will eive if thev -~ is interestine ~.carrv. ~.~~ out theahow computation. Many will state that the answer is 0.158 g 0 ~Howewr, . when reminded that u n t r ~orr net er desrro)ed, they give the currect answer. U.158 g 0211 p, KNOJ. They soon realize tho1 ~
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Presented at the 188th American Chemical Society National meeting. Philadelphia. PA, August 28. 1984. Navidi, M. H.; Baker, A. D. J. Chem. Educ. 1984, 61, 522. A quiz contained the following problem: A 1.1O-g sample of quinone. CBH40,, was burned in excess oxygen in a calorimeter. The calorimeter contained 1.00 kg of water and had a heat capacity of 4.00 kJIoC (calorimeter constant). The temperature of the calorimeter and its contents increased from 20.00° to 23.4I0C. What is the AHfor the combustion of 1 moi of quinone? Among the solutions submitted was the following 3.41% 108 g quinone 4.00 kJ - 1339 kJ l0C 1.10 g quinone 1 mole
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which is dimensionally correct but ignores the absorption of heat by the water in the calorimeter.
that was the reason for making the denominator of the fraction, "grams of KNOZ". d) The setup for the problem then becomes
Weight of 0, obtainable from 1g of KN03 The factor-label approach would set the problem up an follows 1mol KNOB 1mol 0, 50.0 g KNO, X X 101 g KNO, 2 mol KNO,
iize their defihitions. One sueh tern that students can define but that they often do not understand is molarity. If students use the unit basis approach they will have to kilow that molarity is the number of millimoles of solute in one milliliter of solution (or the number of moles of solute in one liter of solution). This is illustrated in the followingproblem. Problem 3 Thirty-five mL of 3.50 M NaOH is diluted with water to a final volume of 400 mL. What is the molarity of the resultant solution? Restate the question so that it can be solved by unit basis, thus: "How many mmals of NaOH are there in 1mmol of the resulting solution?" To solve this we need to know the number of mmols of NaOH in a given volume. Since M is the number of millimoles of solute in each mL, 3.50 mmols X 35.0 mL = 123 mmals NaOH in the solution 1mL
Problem 2 How many oxygen atoms are there in a 100-g sample of ammonium dichromate (NH4)ZCrZ07 (FW = 252)? TOsolve this problem a) Restate thequestion,"Haw many oxygen atomsare there in a 1-g sample of ammonium dichromate (NHhCrz07? b) If this is found, the number of oxygen atoms in a 100-g sample would be 100 times as much, or (The number of oxygen atoms in 1 g of (NH4)&O7) X
(100 g (NH,),Cr,O,) = answer
To get an expression for the number of atoms of oxygen in 1g of (NH1)2Cr~07 it is obvious that one needs to know the number of atoms of oxygen in a given weight of the compound. In this case each formula weight of the compound, 252 g, contains 7 X 6.02 X loz3 atoms of oxygen. d) The correct setup then becomes C)
42'1
loZ0 atoms X 100 g of compound = 1.67 X 252 g compound
loz40 atoms
The number of 0 atoms in 1 g of compound The factor-label method of solving the problem sets the problem dp as follows (loo
1mole cpd
( 252 g cpd ) X
7 mol 0 (1 mol cpd)
(6'02
O)
= 1.67 X
lo2' atoms
Here the student learns to insert factors that will eventually convert "g of compound" to "atoms of 0" in a way similar to the changing of letters in solving anagrams. Another advantage of the unit basis method is that it forces students to understand the meaning of terms, not merely to memo-
Since these 123mmols of NaOH are still contained in 400mL of the diluted solution 123 mmol NaOH 400 mL
- 0.308 mmol NaOH = 0.308 M 1mL
Unit basis is really the made of thinking we employ in solving problems in our daily life. If we want to find out how much four oranges will eost when one dozen oranges sell for $3, we calculate the cost of one orange ($0.25) and then multiply by four oranges or X 4 oranges = $1 $3 12 oranges cost of 1orange
We do not ask, "BY what units must I multiply oranges to obtain dollars?" T h e application of unit basis t o the solution of chemical problems has many advantages. For example, it enables the students t o determine which experimental data are needed t o solve a problem. (Too often, students strive t o use all data given in t h e statement of a problem simply because they are given.) B u t t h e most important asset of the method is that, once it is mastered, it provides a logical system for solving t h a t requires a n understanding of the underlying phenomena. In other words, the students learn t o think: Acknowledgment I a m grateful t o J. V. Quagliano, who many years ago convinced me of the superiority of t h e unit basis method, and to E. L. Haenisch3 and G. N. Quam, of happy memory, whose "Basic General Chemistry in Outline Form" popularized the unit basis method.
Haenisch. E. L.: Quam. G. N. "Basic General Chemistry in Outline Form": Burgess: Minneapolis, 1953.
Volume 63 Number 2
February 1986
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