In the Classroom
How to Say How Much: Amounts and Stoichiometry Addison Ault Department of Chemistry, Cornell College, Mount Vernon, IA 52314-1098;
[email protected] Part 1 My purpose in the first part of this paper is to offer a concise and consistent pictorial representation of the ways by which chemists describe an amount of material. I use this representation in my introductory chemistry course and offer it here as a recommendation to both teachers and students. Although the central unit for the chemical amount is the mole, chemists must measure and express amounts in several other ways: grams of a pure substance, milliliters of a pure liquid, liters of a solution, liters of a gas at standard and at nonstandard conditions, and number of particles. These different expressions of “how much” can be interconverted by conversion factors, as indicated in Figure 1. The conversion factors, in the ovals, are density (g/mL), molar mass (g/mol), Avogadro’s number (particles/mol), concentration (mol/L), molar volume of a gas at standard conditions (22.4 L/mol), and the “adjustment factors”, the factors by which the volume of a gas must be adjusted to take into account deviations from the standard conditions of pressure (1 atm) and temperature (273 K). In my teaching I present this diagram a portion at a time, starting with MOLES in the middle and adding grams and its conversion factor (molar mass), milliliters of a liquid and its conversion factor (density), and particles and its conversion factor (Avogadro’s number). When we get to solutions, I redraw the figure, adding liters of a solution and its conversion factor (concentration), and when we get to gases I redraw the diagram again, adding liters of a gas at the standard conditions of temperature and pressure (STP) and its conversion factor (22.4 L/mol, the molar volume of any gas at STP) and liters of a gas not at STP. I point out later that one can go directly from liters of a gas not at STP to moles via PV = nRT. I also tell my students that if the units used in the statement of the problem are not represented on the diagram, the
first step is to reexpress the given amount in terms of the units of the diagram. Thus if the mass is in kilograms, pounds, or ounces, convert to grams; if the volume of a pure liquid is not in milliliters, convert to milliliters; if the volume of a solution is not in liters, convert to liters, etc. If concentrations are expressed as molality or percent by weight, the conversions are not so easy. In my opinion, however, such problems should not be included in the first semester of general chemistry. Finally, if the “answer” is to be expressed in units that are not represented in the diagram, reexpress the calculated mass or volume in the specified units as the last step in the problem.
Amounts:
density
Conversion Factors:
mL of a pure liquid
grams
molar mass
N
liters of a solution
MOLES
particles
22.4 L/mole liters of a gas not at STP
concentration
liters of a gas at STP
P,T adjustments
Figure 1. Ways to express amount and the conversion factors by which the various ways of expressing amount can be interconverted.
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In the Classroom
Part 2 My purpose in the second part of this paper is to present a systematic way to create a visual representation or “map” for the solution of the typical stoichiometry problems discussed in general chemistry. Once you explicitly lay out the path to the solution, you can do the actual calculations easily and accurately. You can apply this approach to mole-to-mole and gram-to-gram calculations (or any combination of these), and to limiting-reagent and percent-yield problems. You can also extend the basic approach to reactions that involve solutions or gases and to titration problems. The advantage here is that you approach all stoichiometry problems as variations on a central theme and subdivide the problems into the same types of elementary steps.
Mole-to-Mole Problems The first example is a simple problem that will illustrate the general idea. Example 1. Calculate the number of moles of nitrogen gas needed to produce 5.0 moles of ammonia. First, write the balanced equation: N2
+ 3H2
moles
+ 3H2
2.5
2NH3
mole–to–mole
5
Gram-to-Gram Problems Example 2. Calculate the number of grams of ammonia gas that can be formed by the reaction of 10 grams of nitrogen gas with an excess of hydrogen gas. Since grams are specified in this problem, the map for this problem, shown below the balanced equation, will contain a “grams line” in addition to the moles line. Again, we have assumed that all of the nitrogen has been converted to ammonia.
moles
moles
8.5
10 molar mass N2 0.36
grams; initial
molar mass NH3 0.72
+ 3H2
10 M.M. N2
moles; initial
0.36
2NH3
10 M.M. H2
0
5.0
0
moles; final
0
mole –to– mole 3.9 M.M. H2
grams; final
0
7.8
0.72 M.M. NH3 12.1
Since the initial number of moles of hydrogen is more than three times the initial number of moles of nitrogen, the nitrogen must be the limiting reagent, and its complete consumption would provide 0.72 moles or 12.1 grams of ammonia. I’ve included a few more numbers that can be used to show that while moles are not conserved, grams are conserved.
Extension to Amount of Gas Expressed by Volume at STP Example 5. Calculate the number of liters of ammonia gas at standard conditions of temperature and pressure that can be formed by the reaction of 10 grams of nitrogen gas with an excess of hydrogen gas. N2 grams
12.1
moles
molar mass NH3 0.72
liters of gas, STP
mole–to–mole
mole–to–mole
12.1
Limiting Reagent Problems Example 4. Ten grams of nitrogen gas and 10 grams of hydrogen are reacted together to give ammonia. How many grams of ammonia can be formed? This time we suspect that some of one of the starting materials will be left over at the end of the reaction. We therefore provide two mole lines: moles initial and moles final, and two gram lines: grams initial and grams final.
2NH3
0.36
2NH3
The fractional yield, (8.5 grams/12.1 grams) times 100 is the percent yield, 70%.
+ 3H2
Percent Yield Problems Example 3. The reaction of 10 grams of nitrogen gas with an excess of hydrogen gas gives 8.5 grams of ammonia. What is the percent yield of ammonia? In this problem we suspect that not all of the nitrogen has been converted to ammonia. We proceed, however, as in Example 2 and calculate the gram amount of ammonia that should be formed if all of the nitrogen is converted to ammonia (the theoretical yield, grams theo.), and compare this to 1348
grams; theo.
10 molar mass N2
N2
+ 3H2
N2
The balanced equation indicates that the mole-to-mole conversion factor for reexpression of 5.0 moles of ammonia as its equivalent in nitrogen is (1 mole nitrogen/2 moles ammonia). The result of the calculation is shown in italics in the box under nitrogen on the “mole line”. We have assumed that all of the nitrogen has been converted to ammonia.
grams
N2 grams; actual
2NH3
You solve the problem by reexpressing the given amount of ammonia as its equivalent in moles of nitrogen for the process represented by the balanced equation. The “map” below the balanced equation represents the solution. N2
the gram amount of ammonia actually formed (the actual yield, grams actual) as indicated here:
+ 3H2
2NH3
10 molar mass N2 0.36
mole–to–mole
0.72 22.4 liters/mole 16
Example 5 is just like Example 2 except for the requirement that the amount of ammonia, the product, is to be expressed as liters of gas at STP.
Extension to Amount of Gas Expressed by Volume Not at STP If the problem requires that the amount of a gas be expressed as liters of gas at nonstandard condition, the adjustment from standard to nonstandard conditions can be made at the end, as shown in Example 6.
Journal of Chemical Education • Vol. 78 No. 10 October 2001 • JChemEd.chem.wisc.edu
In the Classroom
Example 6. Calculate the number of liters of ammonia gas at 25 °C and 0.8 atm. that can be formed by the reaction of 10 grams of nitrogen gas with an excess of hydrogen gas. N2 grams
+ 3H2
2NH3
10 molar mass N2
moles
0.36
mole–to–mole
liters of gas, STP
0.72 22.4 liters/mole 16 conditions
liters of gas, conditions
21.8
Thus the volume of 16 liters at standard conditions is “adjusted” to the volume that 16 liters would occupy at 25 °C and 0.8 atmospheres by multiplying 16 liters by (298/273) and by (1.0/0.8), which gives 21.8 liters.
Extension to Problems That Involve a Solution Example 7 illustrates the extension of this approach to a problem in which one of the reagents is used as a solution. Example 7. What volume of 0.500 molar aqueous silver nitrate is required to completely precipitate the chloride ion present in a solution prepared by dissolving 0.750 grams of sodium chloride in 50 mL of water? How many grams of solid silver chloride will be produced? The approach to this problem is mapped out under the balanced equation. AgNO3(aq)
+ NaCl(aq)
AgCl(s)
0.750 molar mass N2
grams
moles
0.0128
liters
0.0256
m–to–m
conc. AgNO3
0.0128
m–to–m
+ NaNO3(aq)
1.839 molar mass NH3 0.0128
Notice how easily you can add to the map the second part of a two-part question.
Extension to a Typical Titration Problem Example 8 illustrates the application of this approach to a typical titration problem. Example 8. Titration of 30.00 mL of an oxalic acid solution required 38.0 mL of 0.500 molar sodium hydroxide. Find the molarity of the oxalic acid solution. The insight here is to realize that the unknown in this problem is a conversion factor, not an amount. Thus the unknown will be represented by a vertical line rather than a box, and the values in the two boxes connected by the line will define the conversion factor. In this case the boxes are a mole box and a volume box. 2NaOH(aq)
+ H2C2O4(aq)
liters
0.03880 0.500 M
0.03000
moles
0.0190
0.0095
m–to–m
Na2C2O4(aq)
+ 2H2O(l)
conc. oxalic acid (0.0317 M )
Thus the molarity of the oxalic acid solution is (0.0095 moles)/(0.0300 liter) = 0.0317 moles per liter.
Comments To emphasize the similarities between problems I insist that masses be expressed as grams, chemical amounts as moles, volumes of liquids as milliliters, volumes of solutions and gases as liters, and concentrations as molarity. If the given or requested amounts are in units different from these, the unit conversions are done as preliminary steps, as in Example 7 (mL converted to L), or as a final step. When writing on the blackboard I systematically write all conversion factors, the arrows that represent them, and the corresponding calculations in yellow; everything else is written in white.
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