applications and analogies A Concrete Analogy for Combustion Analysis Problems I. David Reingold Juniata College Huntingdon, PA 16652 hlanv beginning s r u d r n r s h i n ~ .;i p ~ a dt m 1 of 1 1 m ~ b l c wlth sroichonierry I think rhis i i a t least p a r t i n l l j hecause t h t w chtmic;d forn1ulus thnr we write a r e norhing hur lett e r s a n d n u m b e r s t o them. T h e l e t t e r s h a v e n o b h y s i c a l reality. Because the letters are n o t real, s t u d e n t s need not follow t h e r u l e s of c o m m o n s e n s e . I n s t e a d , o n e m u s t memorize t h e weird ways chemists h a v e devised for gett i n g r i g h t answers. To counter t h i s attitude. I frequently t r y to u s e real objects i n stoichiometry problems a n d h a v e h a d reasonable success w i t h food analogies. G i v e n a recipe c a l l i n g for sticks of butter, cups of flour, cups of sugar, e k . , how m a n y p o u n d s of cookies c a n y o u m a k e s t a r t i n g w i t h c e r t a i n quantities of ingredients? S t u d e n t s h a v e n o trouble recognizing that you m u s t first determine h o w m a n y cups of flour there are i n t h e weight you h a v e given t h e m , a n d that once you h a v e r u n o u t of b u t t e r t h e r e c a n b e n o m o r e cookies, n o m a t t e r how m u c h you h a v e of a n y t h i n g else. Of course, n o t everyone c a n t r a n s f e r t h i s rational approach t o chemicals, b u t for m a n y t h e analogy i s useful. I found i t m u c h more difficult t o come u p with a n analogy for combustion analysis problems, u s i n g r e a l life objects t o which s t u d e n t s c a n apply logic. Because I anticipate t h a t others m i g h t b e having t h e s a m e difficulty, I wish to present the solution I devised. Again, it's difficult t o evaluate the usefulness of t h e approach, h u t I t h i n k some s t u d e n t s found i t helpful. Considering a S h e r l o c k H o l m e s C a s e Sherloek Holmes and his friend Dr. John Watson arrived a t Scotland Yard headquarters. "I'm so glad to see you, Mr. Holmes," said Inspector Lestrade. "I sent for you because I have finally found a case even you can't solve." "We'll see about that," said Holmes. "Tell me the facts." 'You recall the robbery a t the jewelry factory last year? Of course vau do. A crate eontainine some eold. some diamonds. collect an its insurance." "Why not?" asked Holmes. 'You see, the erate had just arrived a t the company. They knew that it weighed 50 pounds (net), hut they had not yet opened it, so they did not know how much of that weight was gold, how much was diamonds, rubies, or emeralds. Since each of these items is worth quite a different amount per pound, nobody knows how much the crate was worth. The insurance company won't pay a large amount, arguing that the erate may have contained mostly the least valuable material, hut the company won't accept a small payment, arguing that the erate may have contained mostly the most valuable material. "Through brilliant detective work, our men have been able to trace the crate to the XYZ Jewelrv Co. Thev stole the crate
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Journal of Chemical Education
edited by
RONDELORENZO Middle Georgia College Cachran. GAS1014
without gems. They have confessed to the crime, but they too don't know the contents of the erate." "How is that possible?" asked Watson. "Elementary, my dear Watson," said Halmes. "Obviously they used the contents of the crate, along with their own gold, to make jewelry which has since been sold and cannot be retrieved." "Precisely," continued Lestrade. "All the diamonds were made into rings. I have one here. They shipped out 12 pounds of rings, all exactly like this one. All the rubies were made into broaches, all exactly like this one over here. And all the emeralds were made into earrings, like those." "How many pounds of brooches and earrings were shipped?" asked Holmes. Lestrade answered, "There were 35 pounds of brooches and 33 pounds of earrings." Holmes declared, "Ifyou will let me take the ring, the brooch and the earrings, I will salve this conundrum." "Gladly," said Lestrade, "No one in Scotland Yard has been able to do it." When they arrived home, Holmes unwrapped the jewelry and examined it carefully "Observe, Watson," he said, "this ring contains one diamond." He brought out the balance on which he usually weighed his cocaine. "The ring weighs 1.297 ounces and the diamond. . ." He carefullv mied it nut. ". . .
"What does that look like to you, Watson?" "Afish," Watson replied. "This must he a clue? "Ah," said Halmes, "but what kind of fish is it?" "How should I know that?" exclaimed Watson. "Oh, Watson, you're s o stupid!" said Holmes. "What color are rubies?" "Of course," said Watson, "I should have known." "Now, the brooch weighs 2.629 ounces." Again he carefully pried out one of the gems. "And a ruby weighs 0.203 ounces. Now let us look a t an earring. This weighs 1.231 ounces and has three emeralds in it. Each emerald weighs. . .0.142 ounces. Now Watson, I want you to take these gems to a jewelry store and have them appraised. Also, find out the current price ofgold." Watson soon returned with his reoort. "The diamond is worth $320. The ruby is worth $37. he emerald is worth $22. And gold is going for $120 an ounce." "Excellent? exclaimed Holmes, "Send for Lestrade!" When Lestrade arrived, Halmes announced, "The insurance company owes ABC Jewelry Ca. $169,334." 'You are amazing," said Lestrade. "How did you do it?" Thinking about t h e C a s e 1. Haw did he do it? 2. What does this have to do with chemistry? 3. What kind of fish was it? 4. What did Holmes do wrong? If you were Holmes, what would you have said?
Answer 1 For the diamonds: 12 lb x 16 odlb = 192 oz 1 1.297 ozlring = 148 rings x 1diamondlring = 148 diamonds x ,110 ozldiamond = 16.3 oz of diamonds
For the rubies: Similarly, there are 1704 rubies weighing345.9 *-
For the emeralds: There are 1287 emeralds weighing
182.7oz
who don't want to know the answer immediately won't see i t by accident: b sfe ifssjoh, pg dpvstf! Answer 4
h t a l weight ofjewels = 16.3 + 345.9 + 182.7 = 544.9 oz h t a l weight ofbox = 50 lb x 16 oznb = 800 oz Weight of gold = 800 - 544.9 = 255.1 oz Value of box = (148 diamonds x $3201 + (1704 rubies x $371 + (1287 emeralds x $221 +(255.1az goldx $120/oz) =$47360+63048+28314+30612 = $169,334
Answer 2
Obviously, this is exactly the sort of calculation one must do in a combustion analysis. Given an unknown compound (box) containing C, H, (gemstones) and 0 (gold), we combust it in excess oxygen (gold) to make CO2 and Hz0 (jewelry). All the C's i n the C 0 2came from the unknown, all the H's in the H 2 0 came from the unknown, but the 0's in the C 0 2 and H 2 0 do not relate to the unknown, because we added extra. To determine the amount of 0 in the unknown we must add up the weight of C and H and subtract it from the total weight of the unknown. Answer 3
The answer to question 3 i s presented here with the letters shifted ahead in the alphabet by one, so that readers
The answer to question 4 was intended to be a n entry into a discussion of simificant fimres: since the weight of we can assume that the rings is presented>s "12 there is uncertainty in the last digit; i t might have been 11 or 13. Of course this throws our number of diamonds off by up to 12 and our value off by u p to $3840. Similar logic applies to the eamngs, the brooches, and the gold. To claim that there was 255.1 oz of gold is ludicrous, when the actual answer could have been anywhere from 225 to 285 oz! Thus the closest Holmes could have come to the value of the shipment is $170,000. My students found other diffculties. Several pointed out that the problem requires you to assume that all the diamonds weighed the same, and were worth the same, and likewise for the other jewels. This is true. but I antici~atedthat: notice the word "exactly" in the descriptions of the jewelry. (It was the only way I could think of to get around the problem.) Of course, in real analysis problems, not all the carbon atoms are the same either, but there are enough of them that we can get away with using average masses. One clever student pointed out that ounces and pounds a s applied to gold are different than when applied to other things. I haven't figured out a way to avoid that problem, but i t is unlikely to arise very often. If i t does, it provides an entry into a discussion of the importance of standardized weights and measures!
Volume 72 Number 3
March 1995
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