THE PROBLEM OF PROBLEMS IN HIGH-SCHOOL CHEMISTRY
If all the students in beginning courses in chemistry were mathematicians, the average chemistry teacher would have a much easier task. Many students with little or no foundation in algebra find their way into classes in chemistry. Obviously, the course must then be made largely "descriptive" or else a means of presenting chemical mathematics in an extremely simple form must be devised, since there is never time to go back and secure the groundwork in algebra that is desired. This places the conscientious teacher in a difficult position. I am quite convinced of two things-that any method of science teaching which neglects the mathematical side is not scientific (that was Lord Kelvin's idea, too) and that most of the essential problems that appear in beginning courses in chemistry can be reduced to a few type forms which do not require any great amount of mathematical ability. It is my purpose here to present, without any special claim to originality, some methods which I have found useful in simplifying a few of my students' difficulties. Usually one class period is sufficienttime to polish up a few of the tools which the student has almost forgotten-the equations of algebra, the halfdozen or so of axioms of elementary algebra and geometry, and the laws of proportion. A few minutes, early in the course, may well be spent in discussing the metric system and its advantages. This brief review of fundamentals is well worth the time it takes. , High-school students meeting the gas laws of Boyle and of Charles for the first time are always more or less bewildered. f have found the following device useful in showing that in a system of three dependent variables, one factor cannot vary without effecting a change in a t least one other. Take a meter stick and call each end and the middle the three variables. Then hold the middle of the stick stationary and show that both ends must move if one of them does, or either end may be held stationary and the other end cannot be moved without moving the middle. Finally, show that all three variables may change simultaneously, and each either up or down. The illustration may be further applied directly to the gas laws by calling the ends Pressure and Volume, respectively, and the center, Temperature. Since in Boyle's Law the temperature must remain constant, hold the center stationary and observe that if P is raised, V is lowered-inverse variation of P and V with T constant. In Charles' Law P is constant, so hold that end still and observe that as T rises, V also rises-direct variation of V and T with P constant. And if the point called V is held still, P and T move up or down together-direct variation with V constant. The above demonstration takes but a few minutes and even the least mathematically minded members of the class can see reason in what might easily be mere words; and a way is opened up for a mathematical statement of the problem. 355
356
JOURNAL OF CHEMICAL EDUCATION
FBBEUARY, 1930
Most high-school texts will develop the gas law equations and leave the student with three formulas to remember. I have found i t advantageous to combine them into one formula which applies in all cases. But there are always some students who object to having the formula, PVT' = P'V'T, thrust upon them bodily: they want to know why i t is so. Some of them are satisfied by seeing i t developed in the following manner: since T varies jointly with P and V , i t varies as their product, i. e.,
T : T ::P : P V r ; whence PVT'
=
P'V'T.
Those to whom joint variations remains a mystery are usually placated in another way. Write the three formulas thus, and apply the axiom of multiplication: Boyle's Law, Charles' Law
P V = P'V' PT' = P ' T VT' = V'T (PVT')' = (P'V'T)¶; whence, by applying another axiom, we havePVTf = P'V'T
{
The practicality of the formula becomes evident when i t is pointed out that i t becomes the formula of Boyle's Law by making T constant, and of Charles' Law by making P or V constant. A few exercises in making correct substitutions in the formula soon make the application of the gas laws clear. Similar methods for clarifying other types of chemical calculations can be readily developed, and are well worth the time and effort that they require. Most of the problems that the beginning student will meet can be readily solved by applying the laws of proportion; and I am of the opinion that there will be less confusion if the problems are all approached in as nearly the same manner as possible. A few examples will show how a number of different problems can be reduced to one type which can be solved in four main steps which are always the same. These steps are: (1) form an equation which states the fundamental fact given or implied in the problem, (2) form a similar equation with X in the place of the quantity sought, (3) form a proportion involving the four quantities concerned, and (4) solve the resulting algebraic equation. The following examples will illustrate what I mean. ( a ) Find the weight of 50 liters of oxygen measured under standard conditions. (1) 1 liter 01 = 1.429 g. (2) 50 liters Oz = X g.
(3) 1:50::1.429:X (4) X = 50 X 1.429 = 71.45 g.
(b) Find the volume occupied by 73 grams of hydrogen under standard conditions. (1) 1 liter HZ= 0.08987 g. (2) X liters HS = 73 g. X = 812
+
(3) l : X ::0.08987:73 (4) 0.08987X = 73 liters
VOL.7, NO.2
PROBLEM OF PROBLEMS I N CHEMISTRY
357
(6). What weight of KC103 is required to produce 12 grams of oxygen? From the percentage composition of KC1O8,
(1) 100 g. KCIOs = 39.2 g. O1 12 g. 0, (2) X g. KClOs
(3) 1OO:X ::39.2:12 (4) 39.2X = 1200
--
X = 30.6 g.
(d). What volume of hydrogen can be prepared by the reaction of 10 grams of zinc with hydrochloric a d d ? What weight of hydrogen? ZnCL Hz, From the equation, Zn 2HC1-
-
+
+
(1) 65.37 g. Zn = 22.4 liters (2 g.) of HP (3) 65.37:10 ::22.4:X(2:Y) (2) 10 g. Zn X liters (Y g.) of HI (4) 65.371 = 224 65.37Y = 20 X = 3.43 liters Y = 0.306 g.
(e). How many cubic centimeters of N / 4 HC1 will juit react with 10 grams of NaHC03? Hz0 CO,, From the equation, HC1 NaHC03 --+NaCl
+
+
--
1000 cc. N/4 HCl 84/4 or 21 g. NaHCOi (2) X cc. N/4 HCI 10 g. N ~ H C O I X = 476 cc. (1)
--
+
+
(3) 1000:X ::21:lO (4) 21X = 10,000
The number of examples that might be given is unlimited. The chief advantage that I can see in this method is the uniformity of procedure, whether the problem deals in grams, mols, volumes, or normalities, so that the inexperienced student can always know how to start. I have tried the plan with good success in my classes, and 1 would welcome any criticisms or suggestions from any readers of THISJOURNAL,^^^ how to simplify further one of the hardest spots in teaching chemistry in high schools.
Turquoise Mined in Prehistoric California. Extensive mining operations were carried on near Victoriville, California, many centuries ago, M. J. Rogers, who has'recently completed a study of the mines in the Mohave Sink area for the San Diego museum, has discovered. There are indications of over two hundred mines in this region which Mr. Rogers believes were abandoned many centuries ago. The material mined was the semiprecious stone turquoise, which was so extensively used in prehistoric times hy the Indians in Arizona, New Mexico, Mexico, Yucatan, Peru, and Chile, where it was employed in mosaic inlay. The shields and masks from prehistoric times in Mexico are the finest examples of inlay mosaics known from the western hemisphere. The prehistoric mines were all of the open-pit type excavated through solid rock, by primitive stone tools still found near the mines, to a depth of six or eight feet. During the centuries since the mines were abandoned the pits have filled with loose rock and sand, thus preserving these ancient workings from destruction. There is evidence to show that the early Pueblans from eastern areas operated the mines, and it seems probable that they were abandoned because of the distance from water, which today is only to he had during part of the year a t a distance of several miles.-Science Service
