Stoichiometry Mastery Exams
The Stoichiometry Mastery Exam must be passed at the 85% level or better. It deals with equations and manipulations of chemical algebra based on the various conservation laws and definitions that introduce the course. The Exam is offered six times during the course. It must be completed successfully to obtain a passing grade in the course if a passing grade is otherwise justified by performance on exams, homeworks and quizzes. But the Mastery Exam score is not otherwise factored into the grade calculation. If the Mastery Exam requirement is not satisfied, the entire course must be repeated. (You do not receive an "incomplete" grade if all that remains is the Mastery Exam.) The types of calculations that can appear on the Exam are drawn from the following categories (which can serve as a preparation guide):
|atomic weights||balancing equations|
|molecular weights||reaction yields|
|isotopic composition||limiting reagents*|
|molecular formulas||dilution problems|
From a number of years of experience with
difficulties some students have in dealing with the
Stoichiometry Mastery Exams, I have concluded that there
is a small subset of pitfalls that can be avoided by
paying attention to the following hints. Learn to recognize
i.) a limiting reagent problem, and ii.) a mixture
For limiting reagent problems, the appropriate tactic is practice, practice, practice. There are lots of examples in the archived exams. Always keep in mind that balanced equations (which conserve atoms on each side of an equation) for reactions involve numbers. Read the equation. Then, if the following seems to work for you, re-read it in terms of "dozens". Finally, re-read it in terms of "moles", which is just a very, very large number. Once you recognize what's going on with moles, you can deal with the algebra of the problem by converting masses of materials to moles of materials and vice versa. Do practice work with the exams without answers.
For the mixture problems, a key word in the phrasing of the question is "mixture". Look for it. An example of the major stumbling block will serve well. Consider the combustion of methane represented by the balanced equation CH4 + 2O2 --> CO2 + 2H2O. Now separately, look at the combustion of ethane, 2C2H6 + 7O2 --> 4CO2 + 6H2O. When dealing with "a mixture of methane and ethane", you cannot work with CH4 + 2C2H6 + 9O2 --> 5CO2 + 8H2O because that unjustifiably assumes you have half the number of methanes as ethanes in the mixture. But the relative amounts of methane and ethane are what you are trying to solve for in the problem. There are an infinite number of combinations of the two original independent balanced equations that satisfy the combustion of methane and ethane; CH4 + 8C2H6 + 30O2 --> 17CO2 + 26H2O to pick a random example. That's also balanced. To continue, suppose you had exactly 25 grams of the mixture and were given information about how much water was produced. You would proceed by setting up the reaction of x grams of methane according to its independent balanced equation, and also 25-x grams of ethane for its balanced equation. You can subsequently generate the amount of products in terms of x and equate that to the amount of a specific problem given in the wording of the question. See the many examples in the archived exams.
Also, in an equation such as that for oxidizing ethane 2C2H6 + 7O2 --> 4CO2 + 6H2O, the molecular weight of carbon dioxide is (12.01)+2(16.00)=44.01, not 4 times 44.01. The "four" is a coefficient, not part of the molecular weight calculation.