applications and analoqies Relative Atomic Mass and the Mole: A Concrete Analoav to Help Students Understand ~hese~bstraci Concepts Josefina Arce de Sanabia University of Puerto Rim Rio Piedras, PR 00931
The of the mole as: 'The amount of - - - - modem ~ - definition any substance that contains the same number of elementary particles as there are carbon atoms in exactly 12.000 g of the C isotope of carbon" describes the mole as a measurable auantitv of matter that contains a certain number of partiJes withokt mentioning Avogadro's number. Even todav there is still discre~ancvamong chemists about the ~ revolves unitUthemole should repieseit (I).T G confusion around the close relationship between the concepts of mole, molar mass, and the number 6.02 x low. The fact is that, in the vast majority . . of instances, chemists are able to establish stoichiometric relationships hetween masses and the amount of moles these represent, without need~ngto know the actual number of articles involved. In soite of this, many textbooks (2) prGer to give first a defdiion of the mole as: ~
1 mole = 6.02 x
loz3particles
There are three principal difficulties students have in the comprehension of the mole concept. They (1) do not have a clear idea of what relative atomic mass
means,
(2) wonder why chemiats'chose" such a strangenumber (6.02 x los) a8 their counting unit, and (3) have difficultv understandine that in different amounts of grams of twoelements ( for &le, l2'gaf c and 32 g of S)there are the same number of atoms.
The d~ficultyin understanding items a and c above has todo with the difficulty most high schml and first-year college students have ~ o & ~ r e h e n dthe i n ~mathematical concept of ratio.This mathematical concept must he understood fully in order to grasp the abstract chemical concepts of relative atomic mass and the molar mass. As in many other instances in the teachine of chemistnr it is in the application of a not-well-under&od mathematical concept to a n abstract chemical conceDt that eives students soecial difficulty. I t is very difficult'for &dents to understand that there are the same number of carbon atoms in 12 g of carbon as there are sulfur atoms in 32 g of sulfur, when the concept of atoms, the exact number in a mole (6.02 x los3), 'we have had difficulty in finding paper clips, or any other small obiect.. that have a uniform weiaht to the first decimal dace. When se enmg the .&,one inould we gh me same q& ty ieveral times usmg d fferenlsampes of tne same type to be sure tnat tne total y weignt s reprod~cole to a tenth of a gram. We have s~aessful used paper clips of different materials such as: all plastic, metallic of different sizes, and plasticcovered metal clips and have also used bail bearings of different sizes. We have found that the lightest clip should be selected to weigh approximately 0.2 g and fill the match boxes with 12 items each to obtain the best results.
~.~~ c
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edited by RONDELORENZO Middle Georgia College Cochran, GA31014
and the awesome magnitude of this quantity are all being discussed in detail for the first time. We believe the concept of ratio should be presented and discussed first with concrete examples, and later discuss the relative atomic mass scale as an-abstract a~olicationof this concept. In addition, the student should byexposed to calculations using number of moles without stressing the actual number, NA.This is the way most chemical calculations are done. Comprehension of the relative atomic mass and mole concepts can be improved if they are introduced with a concrete analogy. Although this analogy could be presented as a hands-on exercise where each student works individually, it is more appropriate to present it as an interactive demonstration with the direct guidance of the teacher. This concrete analogy for the establishment of a relative mass scale is based on Avogadro's hypothesis that equal volume of gases (at the same P and 2') contain the same number of particles. Exercise Three or four identical match boxes are each filled with an eoual number of DaDer c l i ~ of s a different size.' Each box i i then sealed so i h i t the actual number of paper clips cannot be observed directly. The student is told that there are the same number of paper ehps in each box hut that each box contains items of R different size. The student is encouraged to think about the possibility of determining the number of paper clips inside each box by weighing the boxes. The mass of the empty boxes (to the nearest tenth of a gram) is made available and the student is instructed to weigh each box to the nearest tenth of n gram and record the observations. The total mass of the paper cl~psinside each box is then determined bv subtractingfrom eachmeasurement the mass of the box. The recoyded results will appear something like this:
Contents of Box #I Contents of Box #2 Contents of Box #3
2.4 g 3.6 g 8.4 g
What does this information tells us? It should be clear that the boxes have different masses because the paper clips have different masses. The total mass in each box is equal to the mass of each paper clip multiplied by the (unknown) number of paper clips (n)in it. The mass of one paper clip cannot be calculated because the exact amount of DaDer c l i ~ sin each box is not known. Nevertheless. kn&vkg t h k there are the same number of paper clips (nj in each box. the ratio of the masses of two different oaoer clips can bk obtained from this data, because, as ihokn below, it is the same as the ratio of the total masses. mtal g of clips s2 - n total g of clips U l n
of one c l i p - g of one clip 62 ( g done clip l l ) g ofoneclipUl ((I
Now, the relative masnes of the paper clips could he obtained by dividing the total mass of each box by the mass of the lightest. Dimding each mass by 2.4 g gives unitless quantities that are themass ratios.
Volume 70 Number 3 March 1993
233
Relative Masses of Paper Clips
Clip
relative mass units
1
1
2 3
1.5 3.5
mass of a "clip-ole" 1.og 1.5g 3.5 g
This means that the mass of clip #2 is 1.5 times that of clip #1 and the mass of c l i ~ #3 is 3.5 times the mass of clio #I. We will now define
