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Over the last few years, I have been tossing out all of my step-by-step instructions for solving various problems in chemistry.

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In the past, I thought it would be helpful to provide my students with a series of steps for solving problems. The idea was simple: the students would memorize the steps and then carry them out to solve a variety of problems. But while this would work for routine problems, it ultimately failed whenever I changed the structure of the problem to no longer fit the steps of the procedure.

It became evident to me that my students were not really learning chemistry, but instead, simply memorizing a series of steps. Based on talking with my students and observing their work in class, I came to realize that students do not need steps; rather, they need to understand the chemistry.

For example, as my students worked out molar mass conversion problems, I often asked them questions about what the answer represented if they didn’t include a unit on their final answer. Most of the students could calculate the correct numerical value, but they didn’t actually understand what the number meant. They could tell me that the answer was 1.73, and while they were correct, some didn’t realize that the answer was 1.73 moles. Additionally, some of the students who wrote 1.73 moles didn’t understand what that truly meant. They could easily follow the steps, but failed to understand that the mole represented how many particles they had.

Have you ever had a student ask for help because they didn’t know what to do on a particular problem, but when you looked at their paper, they already had the right answer in an earlier step? This tells me that the student lacks a true understanding of what the numbers are telling them, and what the steps mean.

I came to realize that rather than teaching my students chemistry, I had essentially become a computer software developer writing code. My step-by-step procedure was the code and the students were the computers following an algorithm. Just like computers, when the students encountered a problem that no longer fits the code, they experienced a critical error. They would not know what to do because they had learned only “what to do,” not “how to do.” This process left the students feeling negative about their abilities, and they would quickly give up. For my part, I felt like I was not doing a good job, because so many students lacked the ability to solve the “more challenging” problems.

True problem-solving is not simply following a bunch of steps. It actually involves creating the steps you need to solve the problem at hand.

A change of focus

Recently, I have been focusing more on helping my students to develop an understanding of the fundamental concept, rather than simply handing out a series of steps. This approach has taken more time, and in the beginning, students were very resistant because they did not see the value in the process. One of the issues was that students had been trained to follow steps. So I tell my students that true problem-solving is not simply following a bunch of steps. It actually involves creating the steps you need to solve the problem at hand.

Example 1: Empirical formulas

In the past, I would define empirical formulas and discuss the relationship between the moles of atoms that make up the formulas. Then I would provide students with a problem, and go over the steps for solving for the empirical formulas, which looked something like this:

  1. Assume you have 100 g of the compound. Now you can convert the percentages into grams.
  2. Convert the masses from Step 1 into moles using the molar mass.
  3. Find the ratio of the moles from Step 2 by dividing each mole value by the smallest value.
  4. If your ratio of moles is not in whole numbers, multiple all the moles by the smallest number to achieve a whole number for each element in the compound.

Most chemistry textbooks include these steps, with slight variations — but I have found a number of challenges using them. First, the student will always try to convert the problem’s data to fit into Step 1. For example, given the problem shown in Figure 1, students will divide the mass of each element by the total mass of the compound, and then multiply by 100 to obtain a percentage so that they can start at Step 1.

A 10.00 g sample of a compound contains 4.00 g C, 0.667 g H, and 5.33 g O. Find the empirical formula.

Figure 1. Sample problem showing the empirical formula

Without being aware of it, most students go full-circle, moving from each element’s mass, to its percentage, and then back to its mass. While there is nothing inherently wrong with this method and it ultimately leads to the correct answer, it speaks volumes about the students’ thought processes and understanding of the concept. More importantly, it tells me that the student is not even thinking about the problem but instead, mindlessly going through the steps.

A combustion device was used to determine the empirical formula of a compound containing only carbon, hydrogen and oxygen. A 0.6349g sample of the unknown compound produces 1.395g of CO­2 and 0.4565g of H2O.Determine the empirical formula of the compound.

Figure 2. A more complex problem involving empirical formulas

Things can get even worse for students when they are only following the steps. For example, when I give students a problem such as that shown in Figure 2, most students who are only using the steps will give up or not even attempt such a problem, because it does not fit clearly into the steps that they have learned. Instead, a new set of steps needs to be memorized to solve the problem.

I now approach empirical formula problems by starting off with a discussion about how we interpret chemical formulas. What exactly do the subscript numbers mean? If we are trying to solve for an empirical formula, we first need to know what a formula is. By the end of the discussion, I want the students to know that the numbers represent the number of atoms in a single molecule of the compound and more importantly, that the numbers represent the number of moles of each atom in a mole of the compound. I use a lot of “for every” statements to get the idea across. For example, “For every individual mole of carbon dioxide, how many moles of carbon and oxygen atoms do we have?” We then discuss how every molecule of carbon dioxide always has the same ratio of moles of carbon and hydrogen.

Once the students understand that chemical formulas are based on moles, that becomes our focus for solving empirical formula problems. Looking again at the problem in Figure 1, instead of giving them a series of steps to follow, I start by saying something like, “We need to find the empirical formula, so what exactly is a formula indicating, and what is it based on?” Eventually, someone says that an empirical formula is based on moles. I write on the board:

? moles of carbon
? moles of hydrogen
? moles of oxygen

Now our goal is to find the number of moles of each element, instead of trying to get the problem to fit a series of steps. Since the 4.00 g carbon, 0.667 g H, and 5.33 g O all come from the same 10.00 g sample compound, all we need to do is convert each element to its respective number of moles. After the conversions, we have:

4.00 g C = 0.333 moles C;
0.667 g H = 0.660 moles H;
5.33 g O = 0.333 moles O

At this point, I write on the board C0.333H0.660O0.333 — and the students look at me funny. I ask them if there is anything wrong with writing the formula this way, and one student will say that you can’t have a third of a carbon atom. I agree, and we review what the numbers mean. I want them to realize that there are twice as many hydrogens as there are carbon and oxygen, while there is an equal amount of carbon and oxygen. This logic will lead us to C12O1 as the empirical formula.

This approach to solving empirical formula problems takes more time to teach and is more difficult for students to understand than simply listing a series of steps on the board. Even so, I believe it is more direct and demonstrates a stronger understanding of the empirical formula concept. The biggest obstacle is that most students are not taught to problem-solve, and do not want to take the time to perform the required work. When they struggle on a problem, they often want me to show them the “trick” to solving it. I remind them that the “trick” to problem-solving is practice.

Example 2: Molecular conversions and stoichiometry

I also take this approach when I teach molar conversions and stoichiometry. When I first started teaching 20 years ago, I would teach these concepts by heavily relying on the mole/stoichiometry road map. You may have seen or even used one of these at some point. I thought it was amazing. It was how I learned to perform molar conversions in my first-year chemistry class in college. I would use it religiously — in fact, I would copy it on the top of all of my exams as soon as they were handed out.

I often tell my students this story to illustrate why I believe it is important for them to learn chemistry, and not just memorize it. One day while I was working on some stoichiometry problems, it hit me. I finally understood what the conversion factors meant and I became aware of why I was doing what I was doing. At the time, it was a big breakthrough for me. I immediately stopped using the mole/stoichiometry road map and didn’t see it again until I began student teaching in the classroom.

During my first few years of teaching, I didn’t think much about the mole/stoichiometry road map. I assumed it was a rite of passage and that every chemistry student needs to have that moment of clarity that I experienced.

Then I had my second awakening. Why am I teaching this with the hope that one day my students will have the same epiphany that I had? I knew there had to be another way to teach this material without relying on the road maps. I had a vague recollection of one of my professors in college using a table to teach stoichiometry. So, I dug out my old notes and, sure enough, there it was: the Before-Change-After (BCA) table.

I started playing around with using BCA tables in class, and found that my students really enjoyed using them. I began focusing on trying to get my students to understand what the mole really means by making connections to things that they were familiar with, like the concept of a dozen. I constantly told my students that the way a chemist uses a mole is no different than how we use the concept of a dozen to buy donuts.

I also started using phrases like, “Every time you have a one mole, you have exactly Avogadro’s number of particles,” or “Every time you weigh out 12.01 g of carbon, you have one mole of carbon atoms.”

I tried to share my ideas with the other teachers, but they were set in their ways. I’m not completely sure why, but no teachers in my district were interested in changing their methods of instruction. For years, I felt alone in the methods I had been developing. I was okay with that because I knew I was onto something when the students in my AP Chemistry class, who didn’t have me for Honors Chemistry, would all jump on board with my new method.

I never force my students to use a BCA table. I always give them the option to use whatever method makes the most sense to them. Based on the fact that almost all of my students end up using the BCA table, however, I think I’m on the right track. In recent years, I have been reading about teachers using BCA tables and what are now called “for every” statements. It has been very reassuring to know that I’m not alone and that there are other teachers out there who share my thoughts and feelings.

Example 3: Reactant problems

Now let’s look at how I taught limiting reactant problems in the past. To solve the problem shown in Figure 3, I would have my students follow these steps:

  1. Convert the masses for each reactant to the mass of the product you are trying to find. For example, convert the 10.0 g of hydrogen and the 10.0 g of oxygen into the mass of water using molar conversions.
  2. Now pick the reactant that produces the smaller mass of water to be the limiting reactant. For example, 10.0 g of hydrogen will produce 89.2 g water, and 64.0 g of oxygen will produce 72.1 g water. Therefore, the 64.0 g of oxygen must be the limiting reactant, because it produces the smallest mass of water.

What mass of water is formed when react 10.0 g of hydrogen gas with 64.0 g of oxygen gas?

Figure 3. Sample problem involving reactants

Some teachers have their students convert the mass of one reactant to the other. For example, 10.0 g of hydrogen would require 79.2 g of oxygen. Since we don’t have enough oxygen, it must be the limiting reactant. Both methods work, but I found they put too much importance on the mass, and too little on the moles.

Moreover, many of my students were not truly understanding the concept. This became evident when students would tell me that the limiting reactant was the 72.1 g of water, and not the 64.0 g of oxygen. They were not thinking about what a limiting reactant was, and would circle the solution to the molar conversions because it was the calculated value. To address this tendency, before we perform any calculations, I always ask my students, which reactant do we have in a larger quantity? Inevitably, they say it is the oxygen because we have 64.0 g. I believe the key to understanding stoichiometry is the mole, not the mass.

Since moving to the BCA table method and increasing my emphasis on the mole, my students are able to form a deeper understanding of how the mole concept works. The first thing I tell my students to do is to remove the measurement by converting it to the moles. For example:

  • When working with measurements in grams, use molar mass to find the moles of each reactant
  • When working with gases using pressure, temperature, and volume, use the ideal gas law equation and find the moles.
  • When working with molarity and volume, use the molarity equation to find the number of moles of each reactant.

In short, if you focus on the moles, all limiting reactants problem become the same.

After we have calculated the moles of each reactant, we can now confidently say which one we have in a greater quantity. For example, 10.0 g hydrogen is equal to 4.95 moles of hydrogen and 64.0 g of oxygen is equal to 2.00 moles of oxygen. Since the mole measures quantity, we can clearly see that we have more of the hydrogen. So, based on that idea alone, the 2.00 moles of oxygen must be the limiting reactant. But we need to be careful, because we can’t just use the initial amounts to find the limiting reactant.

I take my students through this thought process because I want them to understand two things. First, the moles tell us how much we have. In order to find the limiting reactant, we need to know which reactant will run out. To determine that, we first need to know how much we are starting with. Second, if we are going to find out which will run out, we need to know how much we are using. This is what the balanced equation tells us: that every time I make two moles of water I need to react 1 mole oxygen with 2 moles of hydrogen. So, if I start with 4.95 moles of hydrogen and 2.00 moles of oxygen and I use 2 moles of hydrogen “for every” 1 mole of oxygen, which will run out first?

Some students can see intuitively how these ratios play out, while others need more guidance. For both groups, I also have them convert the moles of one reactant to the other using the molar ratio of the balanced equation. For example, I ask the students to find out how many moles of hydrogen we would need if we used up all the oxygen. We figure this out by converting the 2.00 moles of oxygen into 4.00 moles of hydrogen. I focus on what this means: “If we have 2.00 moles of oxygen and we want to convert it into water, then we need at least 4.00 moles of hydrogen. Do we have enough hydrogen?”

Yes, the students respond — we have an “excess” amount of hydrogen. In fact, we can see how much hydrogen is left over by subtracting the amount of hydrogen we reacted with the oxygen from the amount of hydrogen we started with. Next, I tell them to use the moles of the limiting reactant to find the moles of the product they are looking for in the problem. Finally, they finish the problem by converting the moles back into whatever measurement is required.

To me, this method seems more intuitive. I sometimes use this analogy: if I’m going to make as many simple cheese sandwiches as I can, I’m not going to count how many slices of bread to figure out how many sandwiches I can make, and then make the same calculation based on how many slices of cheese I have. Instead, I’m going to first count the slices of bread I have, and then figure out how many slices of cheese I need.

I also believe that the method of focusing on the moles helps the students to understand the mole concept at a deeper level, build their problem-solving skills, and increase their overall confidence. When I used to teach limiting reactants using the steps, I never would have thought to give the students a more challenging problem such as finding the total pressure or the concentration of each ion after the reaction has gone to completion. Now, my students routinely take on higher-level problems that don’t fit a formula. I can’t say that all of my students have mastered stoichiometry at this level, but most have. Based on classroom conversations, the quality of the students’ work, and their overall grades, I can confidently say that most of my students have a deeper understanding of the mole concept.

Seeing the big picture

One could argue that at the end of the day, my students are still following steps. I believe that the difference is that my students will create their own unique steps for each problem they solve. The solution to any problem requires a series of steps, but if you are only taught the steps, then you will never truly understand the concept.

You could easily teach grade schoolers that in order to multiply five times five, you type the following buttons on the calculator: “5”, “x”, “5” and “=” to get the answer. They may get the correct answer, but do they understand what it means to multiply. I believe that students will get more out of learning how to problem-solve, rather than just following steps.

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