
If your students are like mine, they are frequently confused and frustrated by stoichiometry problems. This year, I pioneered a different approach that has been much more successful than the traditional method. The method centers around the writing of mass balanced chemical equations, as opposed to mole balanced. I'll go through some example problems, mostly using ammonia synthesis.
A traditional balanced equation looks like this N[sub]2[/sub] + 3 H[sub]2 [/sub]> 2 NH[sub]3[/sub]
A simple gram to mole conversion makes the equation look like this 28 g N[sub]2[/sub] + 6 g H[sub]2[/sub] > 34 g NH[sub]3[/sub]
Now let's look at some increasingly complex problems that might show up in a chemistry book.
1. How many grams of hydrogen gas are needed to make 20 g of ammonia?
To solve this, students create a scale factor by dividing the mass of ammonia given in the problem by the mass of ammonia in the mass balanced equation. This creates the equation scale factor. Then rescale the equation by multiplying all masses with the scale factor.
20g NH[sub]3[/sub]/34 g NH[sub]3[/sub]=0.588
0.588*28 g N[sub]2[/sub] = 16.5 g N[sub]2[/sub]
0.588*6 g H[sub]2[/sub] = 3.5 g H[sub]2[/sub]
Rescaled equation 16.5g N[sub][size=2]2[/size][/sub] + 3.5 g H[sub][size=2]2[/size][/sub] > 20.0 g NH[sub][size=2]3[/size][/sub]
[size=2](Side note: students can catch errors by using conservation of massthe mass of products must equal the mass of reactants. Many of them appreciate this extra way to validate their work)[/size]
[size=2]The answer appears in the rescaled equation: 3.5 g of H[sub]2[/sub] are needed.[/size]
[size=2]2. If 15.1 g of nitrogen are reacted with 3.2 g of H[sub]2[/sub] what is the limiting reactant?[/size]
[size=2]To solve this, create two scale factors, one for each product. The smaller scale factor corresponds to the limiting reactant.[/size]
[size=2]15.1 g N[sub]2[/sub]/28.0 g N[sub]2[/sub] = .539[/size]
[size=2]3.2 g H[sub]2[/sub]/ 6.0 g H[sub]2[/sub] = .533[/size]
[size=2]Hydrogen's smaller scale factor indicates that it is the limiting reactant. This is fairly intuitive to the students as to why that is the case.[/size]
[size=2]3. If 15.1 g of nitrogen are reacted with 3.2 g of hydrogen, what is the theoretical yield, and how many grams of excess reactant remain?[/size]
[size=2]To solve this, first determine the limiting reactant, as described above. Then scale the reaction using the limiting reactant's scale factor.[/size]
[size=2]It was previously noted that hydrogen is the limiting reactant with a scale factor of 0.533[/size]
[size=2]0.533*28.0 g N[sub]2[/sub] =14.9 g N[sub]2[/sub][/size]
[size=2]0.533*34.0 g NH[sub]3[/sub] = 18.1 g NH[sub]3[/sub][/size]
Rescaled equation 14.9 g N[sub][size=2]2[/size][/sub] + 3.2g H[sub][size=2]2[/size][/sub] > 18.1 g NH[sub][size=2]3[/size][/sub]
[size=2][size=2]The theoretical yield shows up right in front of ammonia. It is 18.1 g NH[sub]3[/sub][/size][/size]
[size=2]The excess can be obtained by finding the difference in the given amount of nitrogen and the amount in the mass balanced eq.[/size]
[size=2]15.1 g N[sub]2[/sub]  14.9 g N[sub]2[/sub] = 0.2 g N[sub]2[/sub] excess.[/size]
[size=2]Percent yield is easy to obtain this way as well. As always, once you have the theoretical yield, percent yield is pretty easy.[/size]
[size=2]My students find this method much more intuitive and have been much more successful at solving stoichiometry problems with this approach. While this approach does not work for g > mol conversion questions (if 5.0 g of N[sub]2[/sub] are reacted, how many mol of hydrogen are needed?), such questions are typically contrived as baby steps to the full blown gram to gram type problems and not representative of actual problems solved by chemist and can be dispensed with. [/size]
