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Stoichiometry Set-up Method Mark as Favorite (0 Favorites)

LESSON PLAN in Stoichiometry. Last updated October 17, 2016.

Lesson Plan

Summary

In this lesson, students will learn how to follow a process of visual cues in combination with a step-by-step problem solving method for different types of stoichiometric problems. This method can be particularly beneficial for students who struggle with completing multi-step calculations.

Grade Level

High school

Objectives

By the end of this lesson, students should be able to

  • Apply a specific problem solving method to successfully answer any stoichiometry problem.
  • Use dimensional analysis to complete a calculation.

Chemistry Topics

This lesson supports students’ understanding of

  • Stoichiometry
  • Dimensional Analysis
  • Molarity
  • Gas Laws

Time

Teacher Preparation: None

Lesson: 60 minutes per topic

Materials

  • Stoichiometry guides for the particular topic of focus.
  • Calculator
  • Periodic table

Safety

  • No safety considerations are needed for this activity.

Teacher Notes

  • The stoichiometry set-up method is very valuable for students who struggle to complete multi-step calculations, and dimensional analysis.
  • Use these as student handouts, or guides when introducing the method of calculating stoichiometry problems would be most beneficial.

Read an article about this lesson in the September 2015 issue of Chemistry Solutions.

Downloads 

Submitted by
Richard A. Samsa
Grove City High School
Grove City, Pennsylvania

For the Student

Lesson
 

Stoichiometry Problems

Steps for the stoichiometry set-up method

  1. Write a balanced chemical equation.
  2. Write what is given and what needs to be calculated (X) on the equation.
    1. Data in moles go under the equation.
    2. Data in other units go above the equation.
  3. Draw a path, using arrows, from the given amount to what needs to be calculated on your chemical equation.
    1. Separate vertical and horizontal arrows.
    2. Horizontal arrows must go under the equation.
  4. Perform the appropriate calculations along the path.

Putting the steps into practice

To show how this method is used to convert the given unit to the unit asked for in the problem, I will solve a typical stoichiometric problem using this mechanism. Here is a two-part sample problem:

If a 2.8 g sample of mercuric oxide is decomposed by heating:
(a) How many grams of mercury will be produced?
(b) How many moles of oxygen will be produced?

Steps for solving sample question part (a)

  1. Write the balanced chemical equation.

    2 HgO → 2 Hg +  O2

  2. Write what is given and what needs to be calculated on the equation (moles go under the equation all other units go above the equation)

    2.8 g           X g
    2 HgO   → 2 Hg    +  O2

  3. Draw a path from what is given to what needs to be calculated on your chemical equation (separating vertical and horizontal arrows and having horizontal arrows go under the equation).
  1. Perform the calculations along the path.  Here is where the arrows provide a map for the student to follow. Just like the animal mole lives underground, the unit of mole lives under the equation. So the first arrow directs the student to convert what they are given (the 2.8 g of mercuric oxide) to moles of mercuric oxide (under the HgO, which is where the moles live).

Corresponding calculation for the first arrow:

2.8 g   HgO  ×   1 mol HgO      =
  216.6 g HgO

The second arrow goes from under HgO to under Hg. Therefore, it asks the student to convert moles of HgO to moles of Hg.

Corresponding calculation for the second arrow:

2.8 g   HgO  ×   1 mol HgO      x 2 mol Hg      =
  216.6 g HgO 2 mol HgO

The last arrow goes from under Hg (from moles of Hg) to above Hg (to grams of Hg).

Corresponding calculation for the third arrow:

2.8 g   HgO  ×   1 mol HgO      x 2 mol Hg      x 200.6 g Hg = 2.6g of Hg
  216.6 g HgO 2 mol HgO 1 mol Hg

Steps for solving sample question part (b)

  1. Balanced chemical equation.

    2 HgO → 2 Hg +  O2

  2. The student would place X under the oxygen (because the student is asked to solve for moles of oxygen).

    2.8 g
    2  HgO → 2 Hg    +  O2
                                  X
    mol

  3. Draw a path from the given data to what needs to be calculated.
  1. Perform the calculations along the path. The first arrow shows the student to go from above the HgO (grams of HgO ) to under the HgO (moles of HgO).

Corresponding calculation for the first arrow:

2.8 g   HgO  ×   1 mol HgO      =
  216.6 g HgO

The second arrow shows the student to go from under the HgO (moles of HgO) to under the O (moles of O2).

Corresponding calculation for the second arrow:    

2.8 g   HgO  ×   1 mol HgO      x 1 mol O2    =  .0065 mol O2
  216.6 g HgO 2 mol HgO

Notice that the number of arrows corresponds to the number of conversion factors in the dimensional analysis.

 

Find additonal worksheets in the downloads box to use this method for:

  • Gas Laws
  • Molarity Problems
  • Electrolysis applications

Also find a summary of the set-up in the downloads box.

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