Bridging the Stoichiometry Gap
by Jeanne M. Gizara
Are your students likely to recoil at the very mention of stoichiometry problems - those sticky exercises in which students are asked to find the weight of a product given the weight of a reactant in a chemical equation?  In an effort to lessen student anxiety over such problems, we in the chemistry department at Columbia-Greene Community College developed a model called The Bridge Method of Stoichiometry.  The method can be applied to any problem involving the relationships between constituents in a chemical reaction, and even students with little prior knowledge of chemistry can use the model successfully. 

Bridging the Stoichiometry Gap
The Bridge Method uses a map model consisting of an imaginery bridge and several roads leading to and away from the bridge (see Figure 1).  Students use the map as a guide to the steps they must follow in order to solve a stoichiometry problem.  On the left side of the bridge is Known Country, the territory of the chemical for which information is given in the problem.  On the right side of the bridge is Unknown Country, the land of the chemical about which information is sought. 
The sample problem illustrates how the bridge is used. 
How many grams of nitrogen must react to form 51 grams of ammonia by the reaction: 

                                     N2 + 3 H2 --> 2 NH3  ? 

The first step is to identify on the map both the starting point and the destination. The starting point would be 51 grams of ammonia on the left side of the bridge, and the destination ( or answer to the problem) would be grams of nitrogen on the right.  First, starting with grams of ammonia, follow Formula Weight Road, to get to moles of ammonia. Second, cross over the Coefficients Bridge to reach moles of nitrogen. Third, turn on Formula Weight Road and proceed to grams of nitrogen. 

The road names are clues to the calculations involved in the problem.  To convert 51 grams of ammonia to moles of ammonia along Formula Weight Road, use the formula weight of ammonia, 17 grams per mole.  Dimensional analysis (factor label method) shows that the 51 grams of ammoia is divided by 17 grams per mole: 

                          51 g of NH3  X   1 mole of NH3
                                                     17 g of  NH3

The manipulation carried this far would give an answer in moles of ammonia.  Next,  cross Coefficients Bridge.  As the name suggests, the coefficients in the balanced equation is used in this step.  The balanced equation shows that 2 moles of NH3 are chemically equivalent to1 mole of N2.  In order for the units to cancel correctly, the conversion factor should be 1 mole of N2 / 2 moles of NH3.   So far, this is the setup: 

           51 g of NH3  X   1 mole of NH3  X      1 mole of N2
                                       17 g of  NH3         2 moles of NH3

and the answer would be in moles of nitrogen. Finally, Formula Weight Road is used to reach grams of nitrogen. (The formula weight of nitrogen is 28 grams per mole.)  The entire setup is therefore: 

    51 g of NH3  X   1 mole of NH3  X      1 mole of N2   X      28 g of N2
                                17 g of  NH3         2 moles of NH3       1 mole of  N2

Computing the answer gives you 42 grams of nitrogen. 

Solving other problems
The bridge model can also be used to solve problems in converting grams of one substance into molecules or atoms of another.  For example:  How many molecules of hydrogen would be released if 10 grams of magnesium react with unlimited hydrochloric acid?   Also, how many atoms of hydrogen is this?  The balanced equation is: 

                         Mg   +  2 HCl ----->  MgCl2  +  H2

In this problem, the path to follow is from grams of magnesium to moles of magnesium via Formula Weight Road, then over the Coefficients Bridge to moles of hydrogen and along Route 6.02 X 1023 to molecules of hydrogen.  After stopping at molecules of hydrogen, the journey continues down Subscript Street to atoms of hydrogen. Before the Bridge, use the formula weight of magnesium (24 grams) to get from grams of magnesium to moles.  The first part of the setup is: 

                         10 g of Mg  X  1 mole of Mg
                                                  24 g of Mg 

Next, Coefficients Bridge must be crossed. The balanced equation shows that 1 mole of magnesium is chemically equivalent to 1 mole of hydrogen, so 1 mole of H2 / 1 mole of Mg is now included in the setup: 
                         10 g of Mg  X  1 mole of Mg    X   1 mole H2
                                                  24 g of Mg           1 mole Mg 

Next, the map suggests that moles are converted into molecules via Route 6.02 X 1023 (a reminder that there are 6.02 X 1023 molecules per mole).  Dimensional analysis dictates that this conversion factor be included in the form: 

                                      6.02  X 1023 molecules NH3
                                                     1 mole NH3

The calculation thus becomes: 

    10 g of Mg  X  1 mole of Mg    X   1 mole H2  X  6.02  X 1023 molecules H2
                            24 g of Mg           1 mole Mg                  1 mole H2

and the answer is calculated to be 2.5 X 1023 molecules of hydrogen.  The second part of the problem asks for the number of atoms of hydrogen.  Subscript Street leads to atoms, meaning that the hydrogen will be used to convert from molecules to atoms of hydrogen.  The formula H2 denotes that there are two atoms of hydrogen to every molecule.  Thus, 

                   2.5 X 1023 molecules H2  X   2 atoms of H2
                                                               1 molecules H2

yields 5 X 1023 atoms of hydrogen. 

The Bridge Method can be used to solve many other problem variations.  As long as one substance in a balanced equation is expressed in moles, grams, molecules, or atoms, any other substance in the equation can be so converted.  The appeal of the method is that every stoichiometry problem is handled the same way. 

Once your students are familiar with the territory, you can also expand the map to more complicated problems (see Figure 2).  For example, if you start with the volume of a liquid sample, take Density Avenue to grams and then continue as before:  Volume in millilitres X Density (in grams/millilitre) = grams, would be the first part of the setup.  Or if the molarity of a solution and its volume are given, follow Litres Lane to get from molarity to moles.  The setup would begin with Molarity (moles/ litres) X volume in litres = moles. 

In the fall of 1980, 65 percent of students at Columbia-Greene who were tested on stoichiometry problems using the Bridge Method achieved a grade of A; only 8 percent failed.  Following remedial work, those who failed went on to demonstrate that they understood stoichiometry problems in a subsequent exam. 

As already stated, the Bridge Method is a confidence builder for students first starting out with stoichiometry problems. It dosen't take long, however, for the astute student to realize that the method is more than a gimmick, and that he or she is always converting to moles of known and then crossing the bridge to moles of unknown.  At that point, the student is no longer dependant upon drawing the map, and is able to solve subsequent problems using the dimensional analysis (factor label) method. 

N.B.  this article was originally published in The Science Teacher, April 1981