JAMES G. TRAYNHAM' Denison University, Granville, Ohio
A N EXAMINATION
of the 25-year Cumulative Index to JOURNAL reveals a number of contributions dealing with the solution of problemsused in general chemistry. The authors of most of these articles emphasize the need for sound reasoning on the part of students solvine ~roblems: vet this emuhasis is missing from m a n y x t h e modern textbooks i s well as from tge separate books on chemical arithmetic itself. The presentation in these hooks convinces one that the emphasis is more on correct numerical answers than on the teaching of chemical principles. Thus, in the preface of one of the books now being advertised widely, the author states his opinion that simple proportion is the best approach for the beginning student and then uses that method wherever applicable. With little practice, a student who understands little chemistry can arrive a t a correct numerical answer to a problem that lends itself to a simple ratio approach. But the final answers t o most of these problems are incidental. Examinations from several colleges illustrate this point when the student is asked to give "a set-up that will allow one to calculate a correct answer; you need not perform the arithmetic."
THIS
FORMULA APPROACH
Texts often give a mathematical formula to fit each type of problem. Thus in the chapter on gases we find PV - = - P'V'
T
T'
a neat little formula that will permit anyone who can recognize which pressure, volume, and temperature are given first t o solve for an unknown after a change has taken place, whether he has ever heard of the laws governing the behavior of gases or not. A formula relating density, pressure, and temperature is slightly different and t o the student may appear unrelated or perhaps confusing. Another formula, Mol. wt. =
1.86 PC X
g. eolute -X 1000 g. water
will permit one to compute the molecular weight of a non-electrolyte in solution. The errors attending the transposition of this formula to allow one to calculate, for example, the freezing-point lowering are well known to teachers of general chemistry. Other formulas for other types of problems are available. Note, however, that a different formula must be memorized for each type of problem, and even then there is no assurance
that the student has learned the fundamental chemical principles. It must be admitted that the poorer, unthinking student can perhaps obtain correct numerical answers more often by the formula approach than by any other method. But these answers are not what we seek. REASONING APPROACH
Our concern then is with a simple, consistent approach t o problems that will emphasize above all a knowledge and understanding of the chemical principles involved. Several illustrative examples follow. An attempt has been made to give the reasoning process that might he used by a student in the course. We know that many departments of chemistry use simiiar methods and that these methods have been outlined by previous authors for one type of problem or another. A variety of types are included here t o emphasize the versatility of the method. A single approach-one of reasoning from fundamental principles or definitions-suffices for a11 problems used in first-year college chemistry. Calculation of Simplest Formula (1) From ex~erimentaldata: Problem: When 1.025 g. bf copper ma; heated with sulfur in a crucible, the product of the reaction was found to weigh 1.283 g. w h a t is the simplest formula for the copper sulfide formed? Solution: The difference in weight between the product and the copper is the weight of the sulfur in the sample. Therefore 1.025 g. of copper has combined with 0.258 g. of sulfur and g. Cu = the number of gams of
0.258g.S
Cu oomhining with 1 g. of S
Multiplying this fraction by 32 g. S/g. at. wt. S gives the grams of copper combining with 1 gram atomic i , e., weight of '025 g' Cu 0.258 g. S
32 g' = grams of copper per gram atomic 1 g. at. wt. S weight of sulfur
Finally, dividing by the grams of copper in 1 gram atomic weight of copper gives the number of g r a atomic weights of copper comhiningwith 1gram atomic weight of sulfur, or the atoms of copper Per atom of sulfur. The complete set-UPand solution is 1.025 g. cu 0.268 g. s
32 g. s I g. at. wt.
s
1 g. at. wt. Cu 63.54 g. cu 2 g. at. wt. Cu or 2 atoms Cu 1 atom 9 1 g. at. wt. S
and the is Cu2S. (2) From percentage composition: Problem: A'
1 Present address: Louisiana State University, Baton Rouge, Louisiana.
82
zertain compound is found t o be composed of 32.45 per zent sodium, 22.55 per cent sulfur, and 45.10 per cent mygen. What is the simplest fnrmula of the subrtance? Solution: T o find the simplest formula, one must aszertain the relative ratios of atoms in the compound. Percentage ratios are numerically the same as weight ratios and these weight ratios may be used t o calculate the nnmljer of atoms of one element which are combined with any other element in the compound. Thus, arbijrarily choosing sulfur as the basis for calculation :either sodium or oxygen may be used similarly with :qua1 success) and following the same step-wise reasonng outlined in (1) above, we make the following calxlation: 32.45g.Na 32.06g.S 1g.at.wt.Na22.55g.S X l g . a t . w t . S X 3 . 0 0 g . N a 2 g. at. wt. Na 2 atoms Na 0" 1 g. a t wt. S 1 atom S
and 45.10g.0 32.06g.S 1g.at.wt.O2 2 . 5 5 g . S X 1 g . a t w t . S X 16.00g.0 4 g. at. wt. 0 or 4 atoms 0 1 g. at. wt. S 1 atom S
Thus, for every one atom of sulfur in the compound there are two atoms of sodium and four atoms of oxygen; the simplest formula is NanSOa. If oxygen had been chosen as the bmis for the calculation, the results would have been '/s
atom Na and I/, atom S 1 atom 0 1 atom 0
giving a formula of Na,,, SL,,O or, to use only whole atoms, NanS04. This approach seems to be more understandable to the student than the conventional one of dividing the weights of each element by the appropriate gram etomic weight and then manipulating the quotients into whole number ratios. Gas Laws. Because problems of this type have been treated extensively by other authors, a single example, ncluded for completeness here, should be sufficient. Problem: When the temperature of a certain quan;ity of gas is 20°C. and the pressure is 750 mm., the ~olumeis 2.0 liters. What n u t the pressure of the gas 3e if the volume is changed t o 2.2 liters and the temperature to 30°C.? Solution: The temperatures must be changed t o degrees absolute. An expression that will permit a correct solution is P = 750 mm. x temperature factor X volume factor where the temperature and volume factors are set up by .easoning from the perfect gas laws (or common sense). rhetemperature is increasing (293 t o 303' A,). There'ore, other factors remaining constant, the pressure nnst increase and the correct temperature factor is zreater than one or 303'A. ,1293 OA. The volume is also , ncreasing but this results i n a decrease in pressure; ~~~~~
~
~~
~
~~
~
the volume factor is therefore less than one or 2.0 1.12.2 1. The final expression then is P
=
303"A. 2.01. 750 mm. X 2 % " ~ . i.21. = 705mm'
Volumetric Titration: Problem: How many grams of sodium hydroxide are contained in 25.00 ml. of a solution if 23.00 ml. of a 0.103 N sulfuric acid solution is required for titration? Solution: Nomnality means gram equivalent weights per liter. Therefore a 0.103 N solution of sulfuric acid contains 0.103 g. eq. wt. Haoh 1 I. HBO,
Multiplying this factor by 0.023 1. of H2S04gives the number of gram equivalent weights of used, 2. e., 0.103 g. eq. wt. H 8 0 , 11. Hi304
0, 023
H2S04 =
- .
number of e. ea. wt. H8O. . . used
This number is equal t o the number of gram equivalent weights of NaOH used because 1 gram equivalent weight of any acid will neutralize 1 gram equivalent weight of any base. Finally, multiplying the number gram equivalent weights of N ~ O Hby the number of grms of N ~ O Hin grm equivalent weight of N ~ O H gives the number of grams of NaOH in the original solution-the answer to the question asked in the problem. The complete setup and solution then is 0.103 g. eq. wt. &So4 11. HnSO.
1 g. eq. wt. NaOH X 1 g. eq. wt. H8O4 40'00 g. NaOH = 0.0947 g. NaOH 1 g. eq. wt. NaOH 0.023
H2S04
Interpretation of Equations: Problem: Oxygen is produced by the decomposition of KCIOSaccording to the equation 2 Kc101
-
2 KC1
+ 3 0~
What weight of oxygen can be collected by the complete decomposition of 3.017 grams of KC103? Solution: The equation means that 3 mols of oxygen are produced from 2 mols of KCIOa. Only a fraction of a mol of KC103has been used, however. To find the number of mols of KC103 which were used, one can multiply the number of grams of KC103 used by the number of mols per gram of KC103,i . e., 3'017 g'
1 mol KCIOs = 122.6 g. KCIOI number of mols of KCIOs used
Since 3 mols of oxygen are produced from 2 mols of KCIOa, multiplying the number of mols of KCIOa used by 3 mols 02/2 mols KCIOSgives the number of mols of oxygen 3'017 g'
1 mol Kc101 122.6 g. Kclo3
3 mols O1 = 2 mols KClOs number of mols 0% formed
~
The number of grams of oxygen formed is obtained by
84
JOURNAL OF CHEMICAL EDUCATION
Several types of problems used in general chemistry are illustrated above. One can readily extend the method to other types. Reasoning and understanding 1 mol KCI08 3 mols 01 are emphasized always. A student must know the 3'017 g' 122.6 g. KClOt 2 mols KCIOs X meanings of words and clearly perceive the chemical 32 g' O' = 1.180 g. 0 2 formed 1 mol OP relationships between terms. Although the method in some cases may not give the numerical answer as quickly DISCUSSION as other methods avoiding this emphasis, the student The single objection to this method raised by chem- experiences no difficulty in going from one type of istry teachers of the author's acquaintance is that the problem to another. In every solution setup, the poorer student has difficulty reasoning through the unknown quantity is set equal to something; there steps, i. e., the poorer student does not understand are no transpositions necessary. Sliderule computathe chemistry. In order to prevent a serious morale tions are simplified and the inclusion of units allows the problem, resulting from a large proportion of the class student to make a quick check on the accuracy of his failing the course because of failure to be able to solve setup. In examinations where solution of the arithproblems, it has been suggested that an "easier" metic will be too time-consuming, the instructor can method be taught. However, this author suggests that quickly learn much, or more, about the student's the same difficulty can be met, if necessary, by placing knowledge of chemistry by examining a setup of this less emphasis on problems as such. That is, agility type as by looking for a numerical answer. with numerical problems will not be considered a Whenever we use problems in general chemistry, prerequisite for satinfactory completion of the course. let us use them to teach chemistry-not just arithmetic.
multiplying the number of mols of oxygen formed by the number of grams of oxygen in onemol.
