A Recipe for Teaching Stoichiometry J e a n B. Umland Department of Natural Science, University of Houston-Downtown, Most students have made cupcakes, and introducing the subject into an introductory course evokes a pleasant response. No one has any difficulty telling you that if a recipe says that i t takes 2lI4 cups of flour and 2 eggs (plus other ingredients such as sugar, baking powder, shortening, milk, and flavoring) to make 30 cupcakes and if you go to the refrigerator you find you have only one egg you can only make half a recipe. You will need 1and CUPS of flour and will get 15 cupcakes; if you have 3 cups of flour in your cannister, you will have 1 and % CUPS leftover. Pointing out to students that baking cupcakes involves a verv c o m ~ l e xchemical reaction and that identical reasoning is used for solving stoichiometry prohlems ahout the much simoler reactions uiuallv carried out in chem lahand in the chemich industry seems tohelp them learn to work the stoichiometrv nroblems. The balanced equation for the reaction is used as column headings for a table which includes: (1) any formula masses needed (2) the "recipe" for the reaction in moles, the numbers in this row being the same as the coefficients in the balanced equation (3) the "recipe" for the reaction in grams, if needed, and (4) the fraction of a recipe of each reactant used or product formed. The method is best exnlained bv some examnles. The first example is based on the reaition ~
phosphorus trichloride assuming that an excess of phosphorus is present?" The table would look like the one shown in Table 1.Because 1.8206 times a"recipe"of PC4 is to he made, 1.8206 times a "recipe" of Clz or 1.8206 X 212.70 g Clz = 387.2 g Clz must be used. For simple problems like this, the method works as well as any other, but does not offer any particular advantage. It is in solving problems involving a limiting reactant, solutions of known molarity, gases, energy, and/or electrolysis that the "recipe"method requires less arithmetic than other methods thus saving time and eivine .. ..less chance of mistakes. For example, suppone rhe prohlem ro he solved is: Problem I rh): "If a mixture of 12.6 r: uf ohusnhoru; and 97.6 e of chlorine reacts to form phosphorus tiichloride, how many grams of PC13, P, and Clz will be present after reaction is complete, and how many kilojoules of heat will he given off? (AH:, is -639.2 kJ.)" This time the fractions of a recipe are 0.6878 for phosphorus and 0.4589 for chlorine. Because the fraction 0.4589 is smaller than the fraction 0.6878, chlorine is limiting and only 0.4589 of arecipe or 0.4589 X 274.64 g PC13 = 126.0 g PC13 can be formed. The 0.4589th of a recipe of chlorine will react with 0.4589th of a recipe of phosphorus. Since 0.6878th of a recipe of phosphorus was used, (0.6878 0.4589) = 0.2289 of a recipe of phosphorus or 0.2289 X 61.94 g P = 14.2 g P will be leftover. No chlorine will be left hecause chlorine is the limiting reactant and is all used up in the reaction. The 639.2 kJ can be regarded as a product of the reaction. Since 0.4589th of a recipe of heat is 0.4589 X (-639.2 kJ) = -293.3 kJ, 293.3 kJ of heat will be given off.
~
Suppose the question to be answered is: Problem 1(a): "How many grams of chlorine gas are needed to make 500.0 g of Presented at the Associated Chemistly Teachers of Texas Summer Conference. Houston, July 29, 1983. Tahle 1
Formula masses,amu "Recipe." m o l s ~
"Recipe," grams
Houston, TX 77002
Oat
30.97 2 (30.97 g P) 2molP-=61.94gP mo1 P
3 mol CI
(70.90 g Clz) mo1 CI.
= 212.70 g Clr
2 ma1 PCI.
(137.32 g PC14 moi PCI.
= 274.64 g PCll
Fraetlon "recipe" (a)
Table 2.
+
MnOAs) Formula mass, am" "Recipe." moles
"Recipe." grams
Fraction "recipe"
1036
Data for Problem 2
3MnCl&q)
4HCl(aq)
+ Clr(g) + 2H,qi)
66.94 1
4
1 mol Mn02 (86.94 9 MnO*)= 66,g4 mol MnO. 5.0 g M n 0 2 86.94 g Mn02
= 0.0575
Journal of Chemical Education
1
Mn02
(
)
11.6 ma1 HCI
0.0050 I"I,,
4 mol HCi
= 0.01450
When students get stuck on problems, I can almost always get them to figure the problems out for themselves by giving them cupcake problems requiring the same reasoning. A second example shows the application of the method to a problem involving solution stoichiometry and gas stoichiometry as well as a limiting reactant. Problem 2: "Chlorine gas is sometimes made in the laboratory hy heating manganese(1V) oxide with concentrated hydrochloric acid. The equation for the reaction is
How many liters of chlorine gas a t standard temperature and pressure will be formed from 5.0 g of M n G and 5.0 ml of 11.6 M HCI?" Table 2 includes only information needed in the problem.
Because the fraction 0.01450 is smaller than the fraction 0.0575, HCI is limiting and only 0.01450 X 1 mole Ch = 0.01450 mole Cln can he formed. The volume of 0.01450 mole of C12 a t STP can he calculated, either by means of the S T P relationship or by means of the ideal gas equation, to be 0.32 1. Problems involving the stoichiometry of electrolysis can be handled by using the relationship, 96,500 coulombs (or 1 faraday) = charge of 1mole of electrons. Students like the method because it gives them a procedure to follow and because they understand it. I like i t because the procedure is a logical rather than a "plug-the-numbers-ina-formula" one, and because I have found that more students are able to solve more types of stoichiometry problems quickly and correctly using it than by any other method I have tried.
Volume 61
Number 12
December 1984
1037
