edited by: JOHNJ. ALEXANDER University of Cincinnati Cincinnati. Ohio 45221
exam que~tionexchange Finding Largest Common Factors and Simplest Integer Ratios A Useful Analogy Applicable to the M i I I ~ n Oil Drop Experiment and for Calculating Empirical Formulas Roger S. Macomber
UnlveniV of Cincinnati Cincinnati. OH 45221 Two seemingly unrelated topics that are discussed in virtually every college-level general chemistry text are the Millikan oil drop experiment (for determining the charge of an electron) and the calculation of empirical formulas from elemental composition (by weight). Yet, in order to understand the logic behind the oil drop experiment, or to calculate empirical formulas, the student must first appreciate the concepts of largest common factors and simplest integer ratios. Most students first encounter these concepts during elementary school in the context of fractions (e.g., lowest common denominators). But, by the time they reach a college chemistry course, many students are so intimidated by the chemistw and physics that they have trouble applying concepts introduced years before. T o help them rediscover the concepts of largest common factor and simplest integer ratios, I often use the question below. I t involves data with which every student of mine is intimately familiar: an exam score histogram.
Acceptable Solution (a) Calculating the mean score requires only that we know the relatiue numher of students receiving each score. We can obtain this information simply hy measuring the height of each bar with a ruler
calibrated in any arbitrary units. In Table 1 the bar heights in millimeters are listed. The mean score is found hv multiolvina each score by its bar height, adding the products, then dividi&thL sum (11,775)by the total bar height (187.5).The meanscore is, thus, 62.8 pts. (h) To solve this part we need to make the same type of assumption used in the ail drop experiment (i.e., that electrons are quantized) and in calculating empirical formulas (i.e., that atoms are quantized); we must assume that students are also quantized, i.e., that we can only have an integer numher of them. No fractional students (or electrons or atoms) are possible. (At least this is true of my students!) Since the height of each har in the figure is proportional to (but not necessarily equal t o ) the number of students receiving that score, we have no way of knowing what actual number of students each bar represents. We do know, however, that each bar must represent on m l e g e r number of s~udmrs.The prohlem reduces to thci: what is the largest cummon foeror of all the bar height$,division by which a d d express all the har heights a+ intrgers? Table 1. Tabulated Data fmm the Flgure
Question
A multiple-choice midterm exam yields the bar-graph historrram shown in the fiaure. Nnte that the ordinate scale has h e k deliberately omitted. (a) Calculate the average (mean) numerical score on the exam. (h) What is the minimum numher that took the exam?
Scare
Bar Height, mm
100 90 80 70 60 50 40 30
15 22.5 15 22.5 45 30 15 22.5 187.5
Totair
Table 2.
F I R S T MIDTERM
Table 3.
The histogram for the mldterm exam.
42
Journal of Chemical Education
1500 2025 1200 1575 2700 1500 600 675 -
-
11775
Relatlve Bar Helghts in Table 1, Expressed as Declmal Multlple or Shortest Bar Helght Scare
Relative HeigM
100 90 60 70 60 50 40 30
15/15 = 1.0 22.5115 = 1.5 15115 = 1.0 22.5115 = 1.5 45/15 = 3.0 30115 = 2.0 15/15 = 1.0 22.5115 = 1.5
Relatlve Bar Helghts In Table 2, Expressed as Slmplest Integer Ratios Score
EXAM SCORE (POINTS)
Product
100 90 60 70 60 50 40 30
Relative Height 1.0 X 1.5 X 1.0 X 1.5 X 3.0 X 2.0 X 1.0 X 1.5 X
2 2 2 2 2 2 2 2
=2 =3 =2 =3 =6 =4 =2 =3
To find this largest common factor, we can begin by dividing each bar height hythe smallestone (15mm), since that mustrepresentat least one student. The resultsareshown inTable 2. At this point we are still left with some nonintegers, but we recognize that the decima1 part of all noninteger values are half-integers. Therefore, multiplieation of each bar height in Tahle 2 by 2 renders all bar heights as integers (Table 3). Thus, the largest common factor for all bar
heights in Tahle 1 is 1512 = 7.5.If the decimal parts of the values in Table 2 had been 0.33, we would have multiplied by 3, and so on, until all bar heights were reduced to integers. The sum (25)of the simplest integer ratios in Table 3 is the minimum number of students who took the exam. Of course, the actual number of students could be any integer multiple of this number.
Volume 68
Number 1 January 1991
43
