Chemical Education Today
Letters What Should We Teach about Significant Figures? There is a plethora of papers about significant figures. There is too little time in introductory chemistry to discuss them exhaustively, so what should we teach? Where an “official” recommendation exists, we should not deviate from it without compelling reason. The ASTM recommendations (1) should therefore be followed except where this leads to the dangerous error— 0.98 × 1.05 = 1.0 1.05 ÷ 0.98 = 1.1 —giving less precision to the answer than to the data, which has an uncertainty of only about 0.01/0.98 or about 1%, whereas the answer shows uncertainty 0.1/1.1 or 0.1/1.0 or about 10%. To avoid this error, the ASTM recommendation 5.3.4.2 The rule for multiplication and division is that the product or quotient shall contain no more significant digits than are contained in the number with the fewest significant digits used in the multiplication or division
should be followed by the caution (2) unless the least precise factor in the calculation begins with a 1 and the product or quotient begins with 5, 6, 7, 8, or 9 (or vice versa) then give the quantity beginning with 1 one more significant figure than the other quantity.
For example 1.09 ÷ 0.86 = 1.26744. The ASTM method rounds this to 1.3, but the answer begins with 1 so one more significant figure is required to represent the probable precision of the answer, that is, 1.27.
Rounding The ASTM rule 5.4.1.3 is that When the first digit discarded is exactly 5, followed only by zeros, the last digit retained should be rounded upward if it is an odd number, but no adjustment made if it is an even number.
There has been discussion of whether this convention is the best possible (3–7). Muranaka (6 ) tells us that always rounding 5 upwards is recommended practice in Japan: arguments for it are given by Guare (3) and arguments against it have been given by Midden (4). The arguments on both sides have merit, but the effect on the precision of resulting calculations is too tiny to pursue. Few teachers of introductory chemistry will follow the discussion to its conclusion, so will rely on authority or consensus. In the USA the ASTM rule, quoted above, should be followed. In Japan, they should follow whatever authority provides the “shi-sha, go-nyu” rule (6 ), which translates to “drop 4, enter 5.” Other countries may have other authorities. It doesn’t matter enough to be an issue. Literature Cited 1. 2. 3. 4. 5. 6. 7.
1997 Annual Book of ASTM Standards, Vol. 03-05, 525-536, section 5.3. Fields, L. D.; Hawkes, S. J. J. Coll. Sci. Teach. 1986, 16(1), 30. Guare, C. J. J. Chem. Educ. 1991, 68, 818; 1998, 75, 971. Midden, W. R. J. Chem. Educ. 1997, 74, 405; 1998, 75, 971. Sykes, R. M. J. Chem. Educ. 1998, 75, 970. Muranaka, K. J. Chem. Educ. 1998, 75, 970. Rustad, D. J. Chem. Educ. 1998, 75, 970. Stephen J. Hawkes Department of Chemistry Oregon State University Corvallis, OR 97331-4003
Corrections An error appears in my paper “How Mathematics Figures in Chemistry: Some Examples” (J. Chem. Educ. 1999, 76, 258–267). With respect to the widths of parabolas discussed in Problem 3, shallow wells should be characterized as having the a parameter larger than 1 (a > 1) and deep wells should be characterized as having the a parameter between 0 and 1 (0 < a < 1). The words “deep” and “shallow” appearing in the paragraph following eq 17 are incorrectly transposed.
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❖❖❖ In the article, “Pushing the Rainbow: Frontiers in Color Chemistry; Light and Color in Chemistry” (J. Chem. Educ. 1999, 76, 737–746), the section on Exploring Color and Periodic Properties with Light-Emitting Diode Lasers should have stated that the new, intense blue LEDs contain aluminum, gallium, indium, and nitrogen (not phosphorus).
It uses scanning tunneling microscope (STM) data to fly over atomic surfaces.
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Letters Determination of Percent Oxygen in Air I read with what I believe is justified skepticism the article “A Simplified Determination of Percent Oxygen in Air” in the January 1998 issue of J. Chem. Educ., pp 58–59. I recall that the famous general chemistry lab of determining the percent oxygen in air by using a burning candle in a jar that is inverted in a dish of water had long ago been debunked. I read that the rise of water in the jar was not directly due to the consumption of oxygen in the air (oxygen is replaced by carbon dioxide in the combustion reaction) but from the contraction of expanded gases due to the heat of the candle flame and the condensing of produced water vapor as it is cooled ( J. Chem. Educ. 1963, 40, A477). The apparatus the author uses aptly stops the expansion of gases with the use of a plastic valve. I do question the implication that water is drawn into the flask on the basis of reduced pressure caused by the consumption of the oxygen. I venture to say that initially there is an increase in pressure in the apparatus due to an increase in temperature of the gases in the flask and the actual production of more moles of gaseous product. The reduction in pressure is due to the rapid cooling of the gases in the flask and condensing of the water vapor product of the combustion. Please consider the following. The balanced reaction of the combustion of ethanol is C2H 5OH(,) + 3O2(g) → 2CO2(g) + 3H2O(g) Assuming that there are about 12 drops of ethanol to a milliliter, 1 drop of ethanol would be approximately 0.0015 mol of ethanol (1 drop × 1 mL/12 drops × 0.81 g/mL × 1 mol/46 g = 0.0015 mol) Choosing the largest flask in the data table (310 mL) and the least amount of methanol (1 drop) and assuming a temperature of 20 °C and a pressure of 760 mmHg, O 2 is still the limiting reactant. Moles of air in flask n = PV/RT = 760 × 0.0310/(62.3 × 293) = 0.013. Moles of O2 in flask 20% of 0.013 mol of air = 0.0026 mol. From the stoichiometry, 0.0015 mol of ethanol needs 0.0045 mol of O2. O2 is the limiting reactant in this flask and all the previous ones in the data table. Assuming the limiting reactant O2 is completely consumed in the reaction and using the stoichiometry of the reaction, 5 mol of gaseous product are produced for every 3 mol of oxygen consumed.
Again using the 310-mL flask as an example for computation the following theoretical results are obtained. Before the reaction there are 0.013 mol of air in the flask. After the reaction there are 0.013 mol of air – 0.0026 mol of O 2 consumed + 0.0017 mol of CO2 produced + 0.0026 mol of H2O vapor = 0.015 mol of gaseous product.
When the water vapor condenses owing to the rapid decrease in temperature, about 0.0002 mol of the water remains as vapor (owing to water vapor pressure of 17.5 mmHg at 20 °C). Therefore, theoretically, before the reaction there are 0.013 mol of air in the vessel, and after the reaction there are 898
approximately 0.012 mol of gases remaining in the vessel. This is hardly enough for the 20% decrease in volume of the vessel. Since the difference in solubility in water between CO2 and O2 is of little effect, the explanation for the decrease in volume must be found elsewhere. It is due to the increase in temperature of the air in the inverted vessel as it is placed over the burning flame. Even a rapid placement of the vessel over the flame does not stop the escape of air from the flask as it is heated. If an apparatus could be devised to ignite the ethanol in the sealed vessel without the possibility of an explosive situation due to the mixing of ethanol vapors with the air in the vessel, it could easily be shown that there would be little water entering the flask. This would put an end to the idea that a simple combustion reaction can be used to show the % oxygen in air. The oxygen is replaced with CO2. I would appreciate any correspondence from the readers of JCE if I am wrong in these conclusion. Please email. Leonard Parsons Classical High School Providence, RI 02903-4093
[email protected] The author responds: Thank you for the letter about my paper that appeared in J. Chem. Educ. 1998, 75, 58–59. My comments are the following. 1. In basic, I agree that the reduction in pressure is due to the rapid cooling of the gases in the flask and condensing of the water vapor product of the combustion. And the following reaction must be considered: C2H5OH(,) + 3O2(g) → 2CO2(g) + 3H2O(g) According to this balanced reaction, the amount of ethanol is enough that oxygen may be completely consumed in the reaction with ethanol and cotton. Oxygen in the flask is the limiting reactant. Therefore, 2 mol of carbon dioxide and 3 mol of water vapor products are produced for every 3 mol of oxygen consumed. 2. The products are 0.0026 mol of water vapor (with condensation due to the rapid cooling in the flask changing it to 0.0026 mol of liquid water, which occupied only 0.0468 mL) and 0.0017 mol of carbon dioxide gas, which may be absorbed by water or may form a white precipitate of calcium carbonate (d = 2.71, 0.0627 mL) when the water is replaced with a clear solution of calcium hydroxide entering the flask. The total volumes of products, 0.0468 mL of water and 0.0627 mL of calcium carbonate, are very small and may be omitted. 3. To obtain accurate results, the following procedures must be obeyed: a. The short-neck round flask must be used. b. The vessel must be placed very rapidly over the flame to avoid the escape of air from the flask as it is heated. c. It is dangerous to ignite the ethanol directly in the sealed vessel because of the possibility of an explosion due to the mixing of ethanol vapor with air in the sealed vessel.
Journal of Chemical Education • Vol. 76 No. 7 July 1999 • JChemEd.chem.wisc.edu
Chemical Education Today
d. A fountain could easily be created and the results could be more accurate when the water is replaced with a clear solution of calcium hydroxide which is drawn into the flask based under the reduced the pressure caused by the consumption of the oxygen. Chin-Hsiang Fang Department of Chemistry National Kaohsiung Normal University Kaohsiung, Taiwan, R.O.C. 80264
[email protected] Editor’s Note For more discussion on this topic in this issue, see the article, “The Persistence of the Candle-and-Cylinder Misconception”, on page 914. ❖❖❖
The Arrhenius Equation Revisited The recent article by Carroll (1) seems to hover uneasily between two themes. It is difficult to decide whether it is primarily about the implications of the Arrhenius equation or about the manner in which this equation is sometimes described in general chemistry texts. As to the first of these themes, a very good exposition was given by Pratt (2) nearly thirty years ago, with the help of just one diagram. This showed that the rate constant k might be expected to rise in sigmoid fashion as a function of the temperature T, tending towards the constant A as T keeps increasing. Also, the point of inflection comes at a value of T equal to E a/2R. As Pratt explains, typical values of E a are such that the temperature where the point of inflection is to be expected is likely to be at several thousand Kelvin. For very few reaction systems would such a temperature be attainable. Taking logarithms of the Arrhenius equation we have:
Soda Bottle Orbital Models In the August issue of the Journal (1998, 75, 985) Samoshin described a method for producing molecular models using plastic soda bottles. He attached the screw caps to wooden or plastic tetrahedra, trigonal prisms, and cubes by using screws, glue, or wires. The method works quite well, though drilling holes in the bottle caps for the wires is somewhat tedious. A quicker connection can be made using the standard wooden ball-and-stick molecular models. Dowels are inserted into the black tetrahedral carbon atom and then into #3 one-hole stoppers, which are then tightly inserted into the bottles. (For greater stability Super Glue can be applied to the stoppers.) The bottles can be spray-painted with fastdrying (15 minutes) Krylon paint. The entire process takes less than 30 minutes. To form the σ bond as in ethane, use a couple of 2-liter bottles with the bottoms removed. If four vertical cuts are made in one bottle, it can be inserted into the other as shown in the photograph. To overcome pressure from the compressed air in the stoppered bottles, a small hole can be made in the bottom of each bottle using the heated point of a skewer. Louis H. Adcock Department of Chemistry University of North Carolina at Wilmington Wilmington, NC 28403-3297
[email protected] ln k = ln A – E a/RT This shows that ln k changes linearly, not with temperature, but with its reciprocal. Herein lies the weakness of the statement with which Carroll (1) commences his article. The rate constant is not an exponential function of T, but of 1/T: it decreases exponentially as 1/T increases. Perhaps this degree of inexactitude is acceptable in a textbook of general chemistry, on the grounds of simplicity, but it could hardly be excused in a textbook of physical chemistry or chemical kinetics. Literature Cited 1. Carroll, H. F. J. Chem. Educ. 1998, 75, 1186. 2. Pratt, G. L. Gas Kinetics; Wiley: London, 1969; p 21. S. R. Logan School of ABCS University of Ulster Coleraine, N. Ireland BT52 1SA
The author replies: I would like to thank Prof. Logan for letting me know that some of the material in my paper had already been discussed by Pratt. I wasn’t familiar with it. I think that Prof. Logan has missed the point of my paper. Even though the behavior of the Arrhenius equation is mathematically very clear, there are still misleading remarks or implications about it in some general chemistry texts. It was because of these remarks that I wrote my paper. For example, Olmsted and Williams say that “According to the Arrhenius equation, rate constants depend exponentially on temperature (T in K)” (1). McMurray and Fay state that “As the temperature increases, the distribution of collision energies broadens and shifts to higher energies (Figure 12.15), resulting in an exponential increase in the fraction of collisions that lead to products” (2). Zumdahl says “In fact, the fraction of effective collisions increases exponentially with temperature” (3). Gillespie et al. don’t explicitly say that the rate increases exponentially. Instead, they say “In contrast, the exp({E a/RT ) factor in the Arrhenius equation, increases rapidly with increase
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Letters in temperature and is responsible for a large increase in reaction rates” (4). I agree that all texts don’t state that there is an exponential increase of rate with temperature. But enough do, and I wrote my paper to simply remind my colleagues of the actual behavior of the Arrhenius equation. Literature Cited 1. Olmsted, J. III; Williams, G. M. Chemistry, the Molecular Science, 2nd ed.; Wm. C. Brown: Dubuque, IA, 1997; p 707. 2. McMurray, J.; Fay, R. C. Chemistry, 2nd ed.; Prentice Hall: Upper Saddle River, NJ, 1998; p 492 3. Zumdahl, S. S. Chemistry, 4th ed.; Houghton Mifflin: Boston, 1997; p 578 4. Gillespie, R. J., et al. Chemistry, 2nd ed.; Allyn and Bacon: Needham Heights, MA, 1986; p 877. Harvey F. Carroll Department of Physical Sciences Kingsborough Community College Brooklyn, NY 11235
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Equilibrium: A Teaching/Learning Activity I began receiving the Journal of Chemical Education in March 1998 and have thoroughly enjoyed each issue. As a teacher at the high school level, I find a good portion of each publication well beyond my expertise, and this serves to remind me just how diverse this subject of chemistry really is. However, I have also found that each edition has a number of articles, ideas, and laboratory possibilities which I have and will continue to implement in my classroom. I am indeed thankful that I chose to subscribe. I should like to make a comment about the excellent equilibrium analogy by Wilson in the September issue (J. Chem. Educ. 1998, 75, 1176), which is similar to an aquarium demonstration I have seen but better lends itself to a hands-on activity. In activity three, Wilson attempts to raise the temperature and achieve a new constant. At a higher temperature, both forward and reverse reaction rates are increased. However, it is the endothermic rate which is increased by a larger factor, causing the shift to be in the direction of the endothermic reaction. Thus, the transaction for both R and P must increase but one must increase by a larger amount. In the case given, if R increases from 1/2 to 3/4 (an increase of 50%) then P could increase from 1/4 to 5/16 (an increase of 25%). Equilibrium will again be realized with a new constant of 2.3 and exactly the same conclusion. This clarification in no way intended to diminish the excellent teaching strategy of the paper. It is one of many fine offerings suited nicely to chemistry teachers at the high school level. Lyle Sadavoy Middlefield Collegiate Institute Markham, ON, Canada L3S 3L5
900
❖❖ We have found the paper by Audrey H. Wilson very helpful for our teaching at both secondary school and firstyear university level. May we, however, make a few suggestions in order to further improve the outcome of this activity: 1. Attention should be called to the fact that the analogy used is appropriate to phase equilibria (eventually also to isomerization equilibria) but the essential aspects of chemical equilibrium in general are being met. 2. It could be emphasized that the transfer rates of matches between teams correspond to the microscopic rate constants for the forward and backward reactions (considered as elementary reactions) whose ratio is the equilibrium constant. 3. In order to reduce the effects of rounding (matches, toothpicks, buttons, etc.) 10 boxes of 40 matches each could be used, being sufficient to keep to the appropriate number of significant figures. In this manner, the expected value of (3/4)/(1/8) = 6 for the equilibrium constant in Activity 3 would not be denied by the final outcome 35/5 = 7 (written 35/5 = 6 in the discussion of the Activity). 4. The effect of temperature simulated in Activity 3 could be further developed, distinguishing clearly between endothermic and exothermic forward reactions. It should be taken into account that the effect of increasing temperature has the same sign for both the direct and the reverse reactions. The consequent increase in the rate constants, however, is greater for the transformation having higher activation energy— ln(k1/k 2) = E a[(1/T2) – (1/T1)]/R —that is, the forward reaction if the process is endothermic and the backward reaction if it is exothermic. Thus, the values 3/4 > 1/2 and 1/8 < 1/4 for the transfer rates in Activity 3 should be replaced by, for example, 0.90 >> 0.50 and 0.30 > 0.25 (equilibrium constant 0.90/0.30 = 3.0). João C. M. Paiva Secondary School Penacova, Portugal
[email protected] Victor M. S. Gil Department of Chemistry University of Coimbra Coimbra, Portugal
The author replies: Both the Sadavoy and the Paiva and Gil letters are correct and make important contributions to improving the content of the paper. I thank them for their interest and contribution. Audrey Wilson Department of Chemistry University of Wollongong NSW 2252, Australia
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