1. Chapter 15 Advanced Problem Solving by Christopher G. Hamaker
Illinois State University
© 2014 Pearson Education, Inc. 1

2. Steps for Advanced Problems
Advanced problems involve more than one concept.
There are three steps to follow for these multi-concept problems:
Examine the problem to determine the unknown quantity.
Write down the given value related to the unknown quantity.
Plan a strategy that relates the given value to the unknown quantity.

3. Unit Analysis We have seen the unit analysis method previously.
For advanced problems, we need to use a more complex approach.
Keep the following points in mind when using the unit analysis method for advanced problems:
Many problems cannot be solved in a single unit analysis operation.
If the unknown quantity is a single unit (e.g., cm, g, mL), the given value should also be a single unit.

4. Unit Analysis Continued
A problem may include given values that are not relevant.
A problem may omit information that is necessary to obtain a solution.
If the calculation is based upon a chemical reaction, first write a balanced chemical equation for the reaction.
Before using your calculator, estimate an approximate answer.

5. Algebraic Analysis
Most problems are best solved using the unit analysis method; however some problems are not suited to that method.
Many problems can be solved using algebraic analysis.
An example is density problems.
mass
volume
density =

6. Algebraic Analysis Example
What is the mass of 22.5 cm3 of iron metal with a density of g/cm3? m
22.5 cm3
7.874 g/cm3 =
m = (22.5 cm3) ×
7.874 g
cm3
= 178 g

7. Concept Maps
A concept map is a diagram relating one or more chemical concepts.
It is a flow diagram outlining the steps that need to be taken in order to solve a problem.
An example is shown below for the problem “What is the volume in liters of 3.24 g of oxygen gas at standard conditions?”

8. Visualization
Visualization can help to understand concepts by forming mental pictures.
Although we cannot see atoms and molecules, we can visualize what a sample of a gas may look like as we change the conditions of the sample.

9. “Ballpark” Answers
Before using the calculator it is useful to estimate the answer.
This estimate is called a ballpark answer because it should be near the actual answer.
If the answer obtained for the calculator is significantly different than the estimated answer it can indicate an error in the calculation.

10. Concept Map – Mole Relationships
The concept map for mole relationships from Chapter 8 is shown below.

11. Mole Relationship Example
Calculate the volume and number of atoms of 2.50 g of neon gas at STP.
= mol Ne ×
2.50 g Ne
1 mol Ne
20.18 g Ne
= L Ne ×
22.4 L Ne
mol Ne
0.124 mol Ne ×
6.02 x 1023 atoms Ne
mol Ne
0.124 mol Ne
= 7.47 × 1022 atoms Ne

12. Concept Map – Stoichiometry

13. Stoichiometry Example
How many atoms of iron metal are produced from the reaction of 125 grams of aluminum metal with excess Fe2O3 in the thermite reaction?
Fe2O3 (s) + 2 Al (s) → 2 Fe (l) + Al2O3 (s)
The conversion goes:
g Al → mol Al → mol Fe → atoms Fe

14. Stoichiometry Example Continued×
125 g Al
1 mol Al
26.98 g Al
2 mol Fe
2 mol Al
= 4.63 mol Fe ×
4.63 mol Fe
6.02 x 1023 atoms Fe
mol Fe
= 2.79 × 1024 atoms Fe

15. Multiple-Reaction Stoichiometry
Often you may need to consider the stoichiometry of more than one chemical reaction.
Using the same concept map, apply it to each of the reactions in the sequence stepwise until you arrive at your final answer.

16. Multiple-Reaction Stoichiometry Example
What volume of sulfur trioxide gas is produced at STP from the reaction of 45.5 g of sulfur with excess oxygen according to the following equations?
S (s) + O2 (g) → SO2 (g)
2 SO2 (g) + O2 (g) → 2 SO3 (g)
The conversion goes:
g S → mol SO2 → mol SO3 → mol SO3 → volume SO3

17. Multiple-Reaction Stoichiometry Problem×
45.5 g S
1 mol S
32.07 g S
1 mol SO2
= 1.42 mol SO2
Reaction 2
1.42 mol SO2 ×
2 mol SO3
2 mol SO2
= 1.42 mol SO3

18. Multiple-Reaction Stoichiometry Problem
Use PV = nRT to get volume of SO3.
At STP, T = 273 K and P = 1.00 atm
V =
nRT . P
V =
(1.42 mol) ( atmL/molK)(273K)
(1.00 atm)
V = 31.8 L

19. Critical Thinking: Nanotechnology
Nanotechnology is the use of materials on the nanometer (10–9 m) scale. Typically devices are in the 1–100 nm scale.
A human hair is approximately 100,000 nm thick!
DNA molecules are nanoscale substances, with a diameter of about 1 nm.
Nanoparticles of TiO2 are used in sunscreens and of silver are used as antibacterial agents.

20. Chapter Summary Follow the three steps for advanced problem solving:
Examine the problem to determine the unknown quantity.
Write down the given value related to the unknown quantity.
Plan a strategy that relates the given value to the unknown quantity.

21. Chapter Summary Continued
When solving problems, estimate a ballpark answer to be sure that you have used your calculator correctly.
Most often the dimensional analysis method is best, but some simpler problems can be done using algebraic analysis.
Visualization is a useful tool for estimating answers.

22. Chapter Summary Continued
The use of concept maps for problems can help you to ensure that you have all the necessary steps for an advanced problem.
For stoichiometry involving more than one reaction, apply the stoichiometry concept map to each reaction in sequence until you arrive at your answer.
Practice doing problems!