swered at this mint is whether the students see the relation between the two formsof the equation and whether they are aware of the advantages and disadvantages of using each form. Toby F. Block W g i a Institute of Technoiosy Atlanta. GA 30332-0400
To the Editoc
There is, of course, merit in your assessment of the glass as being half-full while I view it as half-empty. The conceptual gas law problem is a good example of this difference. The answer chosen hy the greatest number of students may indeed reflect the fact that we discuss balloons and moveable pistons in class more often than rigid containers. ~, but..in mv admittedlv random and non-quantitative poll of why students chose that particular, most c~mmon,~answer. it was resDonses such as "the molecules were huddled together for warmth" and "the molecules are closer together in solid Hp than gaseous HZ" that convinced me that the students did not understand the kinetic molecular theory of gases whether applied to balloons or rigid containers. Regarding the conceptual stoichiometry question, I agree that answer d is partially correct but cannot go along with it being the best answer. We train our students to understand that a balanced equation represents the stoichiometryof the reaction and, like an algebraic equation, is reduced to lowest whole number ratios with like terms collected. Limiting reagent problems, in particular, rely on the students' grasp of the fact that nonstoichiometric quantities of reactants is a common occurrence that must be reconciled with the balanced equation. Chemists would not accept
-
The body altitudes of a tetrahedron intersect at the center of the figure, and this point is at a distance three-fourths from the vertices and one-fourth from the faces (see Proof). We are trying to find angle AMB = 0, is the bond angle of the tetrahedral molecule. By definition, t h e s i n e of this angle's supplement, i.e., cos (180 - 8), is HM/BTij: = (hl 4)(3hl4) = 113, as can be seen in triangle BHM, where h is the body altitude of the tetrahedron. Thus cos (180 - 8) = -cos 0 = 113and 0 = arccw (-113) = 109.47' or 109" 28'16". Proof
To prove that the altitudes of an equilateral triangle intersect at a point that is at a distance two-thirds from the vertices and one-third from the sides we draw:
In equilateral triangle ABC, altitude BE is also the angle hisector, and angle DBE = 30'. BDF is a special right (30°-60° right A), where the side opposi&to the= an&e is half the hypotenuse. Thus DF = 112 BF. But BF = AF, being the sidesadjacent to the 30° angles in A ABF, therefore DF = 112 AF, which was to he proved. To prove that the body altitudes of a tetrahedron meet at a point of distance three-fourths from the vertices and onefourth from the faces we draw:
A
302+ 8Hz 6H20+ 2Hz
as the best way to express the formation of water from its elements anv more than 3A 8B = 3ABp 2B is a finished algebraic refationship Sure, we want ourstudents to be ahle to count, and the overwhelmingly popular choice of d as an .~ answer proves that most can do just that, hut they do so without making the connection to what chemists mean the balanced equation to represent.
+
+
Barbara A. Sawrey Academic Cowdinator Univwsity of California. San Oiego La Jolla, CA 92093
Flndlng the Bond Angle To the Editor:
We offer an alternative method of "Findina the Bond Angle in a Tetrahedral-Shaped Molecule" to thethree indicated bv Kawa ( I ) . This alternate method is a purely mathematicafproof that does not require the use of three-dimensional models or additional justification that a tetrahedron can be inscrihed in a cube or that the center of the inscrihed tetrahedral molecule is coincident with the point of intersection of the body diagonals of the cube. (This assumption was the starting point of Kawa's proof.) Consider the tetrahedron drawn below:
Assume that the edge of the tetrahedron is of
unit length.
BH, the face altitude, measures 4312 units. BH, being twothirdsofthe face altitude (first part of Proof) measures 1 / 4 3 unit. AH. the hodv altitude, is equal to (AB2- mz)'", i.e., (1 - ( l l i ) ) E = 4-33, Now, if HM = x, = I; being equal to also measures - x. In right triangle BHM, BM2= H M Z or:
+
Jm
m- m,
m, m2
--
Therefore AH:HM = 3:1, which was to be proved. Literature Clted 1. Ksws.C. J.J Cham. Educ. 1988.65.884.
Resat Apak and izzet Tor Istanbul University Faculty of Engineering Vezneciler 34459, Istanbul. Turkey
970
Journal of Chemical Education
