From time to time I do little pro bono lessons for second grade in math, physics, and chemistry. Recently I came across this question.

In a container is $2~\mathrm{mol}$ of $\ce{SO3}$ (sulfur trioxide). This compound under certain conditions breaks down into $\ce{SO2}$ (sulfur dioxide) and $\ce{O2}$ (oxygen). At the same those two resulting compounds react together back to form $\ce{SO3}$. In equilibrium we have $1.52~\mathrm{mol}$ of $\ce{SO3}$. How much of remaining resulting compounds do we get?

The answer: It is obvious that $0.48~\mathrm{mol}$ of $\ce{SO3}$ broke down into $0.48~\mathrm{mol}$ of $\ce{SO2}$ and $0.24~\mathrm{mol}$ of $\ce{O2}$.

The problem: A few of students cannot grasp how could we start with $2~\mathrm{mol}$ of some substance and end up with $2.24~\mathrm{mol}$ of other substances. I have tried to explain to them that $\mathrm{mol}$ is not a quantity as they are used to as in weight or volume, but that a $\mathrm{mol}$ is a number of molecules. I have tried to simplify an example using just 2 molecules of $\ce{SO3}$ and showed that the break down to 2 molecules $\ce{SO2}$ and 1 molecule of $\ce{O2}$. That was understandable. But as soon as we got back to moles: "How can we get more than we started with?"

What is the best way to explain this? Any experience? To me, it seems so obvious.

  • 1
    $\begingroup$ Try showing them with legos maybe, and use the dissolution of NaCl as an example. $\endgroup$
    – user1160
    May 15, 2014 at 8:40
  • 3
    $\begingroup$ This is an exceptionally well asked questions! Ball-and-Stick modelling by hand may be an option. If you do not want to purchase an available set you can use plasticine and matches. You can go from single molecules and double and double [and double]$_n$ and maybe they understand better this way. $\endgroup$ May 15, 2014 at 8:53
  • $\begingroup$ Before getting into the less trivial combustion-like reactions, have your students mastered displacement equilibrium? $\endgroup$ Sep 5, 2014 at 3:52
  • $\begingroup$ I should point out that I am not a teacher. I teach this in my sparetime to kids who have problems but are willing to do something about it. Therefore I am bound to the issue at hand before exams. $\endgroup$
    – Anderson
    Sep 12, 2014 at 16:39

4 Answers 4


The fundamental misunderstandings here are related to two concepts:

  1. Conservation (atoms are conserved, molecules are not)
  2. A mole is just a number (like the chemist's dozen)

I would start with conservation, then move on to moles.

Explaining why atoms are conserved, but not moles

For second-graders, hands-on activities are probably best, so molecular models or something similar would be a good approach. I like your example using two molecules - if you can get some models, you could try having groups of students start with two (or more - 4+ would make it easier to see) molecules of $\ce{SO3}$. Have them count the number of molecules and the number of each atom. Write the results down where everyone can see them. Then, have them do the "reaction" - let them take apart the molecules and put them back together to make$\ce{O2}$ and $\ce{SO2}$. Again, have them count the molecules and the atoms. Show them that the number of atoms stays the same (for each element and in total), but the number of molecules can change.

Once your students are comfortable with this idea, introduce the concept of moles as a way of counting very large numbers. This is the more difficult part, and in my experience is the "make or break" concept for most chemistry students.

A mole is just a very large number

Goal: The number of atoms in macroscopic samples is huge

You can motivate the discussion by calculating the number of molecules/atoms in very small samples. In my classes (college level) I use the number of molecules in one mL of hydrogen gas at STP, and show that if you counted one molecule per second, it would take longer than the age of the universe (a lot longer) to count them all. Sometimes I use the number of silicon atoms in a bag of sand and show them that it is larger (much larger) than the number of stars in the universe. Other examples would work as well - the idea is to get them thinking about the largest quantity that they can possibly imagine and show them that the number of atoms in even a small amount of matter is much larger than that.

Goal: A mole is just a number

Now you want to link this to moles. I start with the concept of a dozen. I ask things like "how many donuts are in a dozen?" "Ok, how many cookies?" "How many cats?" "How many people" until someone points out that it's always 12. Then I ask them why we say a dozen instead of 12. If they don't figure it out quickly on their own, I tell them that for bakers and their customers, it is a convenient way to count stuff. Then I tell them chemists also have a convenient way to count atoms and molecules, which is good because there are a lot of them! The number they use is called a mole, and similarly to how a dozen = 12 of anything, a mole = 6.022 x 10^23 of anything.

Goal: A mole is a HUGE number

Depending on how good the students are with scientific notation, you might want to actually write out that number of zeroes - it's an impressive demonstration! When there is time, I also like to demonstrate the scale by comparing it to different things as the number of zeroes grows. For example, I say (or have them say) "thousand," "million," "billion," "trillion," for every three zeroes. At around 11 zeroes I start comparing it to unimaginably large things. As examples:

  • The number of stars in the Milky Way galaxy is about 100 billion ($1 \times 10^{11}$).
  • The number of sheets of paper that would fit in a stack between Earth and the moon is about 4 trillion ($4 \times 10^{12}$).
  • The number that would fit between Earth and Mars (at their average distance) is about $2 \times 10^{15}$.
  • Age of the universe in seconds is $4 \times 10^{17}$.
  • There are $10^{22}$ stars in the universe.

I think that this is effective because it connects the concept of moles being huge and the number of atoms being huge to a bunch of astonishing facts - people tend to remember weird little surprising facts, and because of the way memories form, the moles stuff gets stored along with it, leading to an artificially created "intuitive" understanding of the size scale of moles and their relationship to counting atoms.

Now you need to tie it all together.

Goal: Moles are used to count atoms and molecules

I connect the concept of a mole being a very large "convenience" number used for counting to the stoichiometry and conservation concepts with a simple exercise. I write out a chemical equation, which in your case would be:

$$ \ce{2SO3 <=> 2SO2 + O2} $$

Underneath, I write the number of molecules of each, starting with the smallest. So, I would write...

$$ \begin{array}{|c|c|c|c|c|}\hline \ce{2SO3} & \ce{<=>} & \ce{2SO2} & \ce{O2} \\ \hline 2 \space \rm{molecules} & & 2 \space \rm{molecules} & 1 \space \rm{molecules} \\ \hline \end{array} $$

... and say "If we have two molecules of $\ce{SO3}$, we will end up with two molecules of $\ce{SO2}$ and one of $\ce{O2}$." Then I ask them: "What if I start with 10 molecules $\ce{SO3}$?" and let them figure it out, writing on the board as I go. Next, I ask

  • "What if I start with 12?"
  • "What about a dozen?"
  • "How about 100?"
  • "A billion?"
  • "$6.022 \times 10^{23}$?"
  • "A mole?" - at this point, most of them will have the idea
  • "10 moles?"

At the end, the board looks like this:

$$ \begin{array}{|c|c|c|c|c|}\hline \ce{2SO3} & \ce{<=>} & \ce{2SO2} & \ce{O2} \\ \hline 2 \space \rm{molecules} & & 2 \space \rm{molecules} & 1 \space \rm{molecules} \\ \hline 10\space \rm{molecules} & & 10\space \rm{molecules} & 5 \space \rm{molecules} \\ \hline 12\space \rm{molecules} & & 12\space \rm{molecules} & 6 \space \rm{molecules} \\ \hline 1\space \rm{dozen} & & 1\space \rm{dozen} & 0.5 \space \rm{dozen} \\ \hline 100 \space \rm{molecules} & & 100 \space \rm{molecules} & 50 \space \rm{molecules} \\ \hline 1 \times 10^9 \space \rm{molecules} & & 1 \times 10^9\space \rm{molecules} & 0.5 \times 10^9 \space \rm{molecules} \\ \hline 6.022 \times 10^{23} \space \rm{molecules} & & 6.022 \times 10^{23}\space \rm{molecules} & 3.011 \times 10^{23} \space \rm{molecules} \\ \hline 1 \space \rm{mole} & & 1 \space \rm{mole} & 0.5 \space \rm{mole} \\ \hline 10 \space \rm{mole} & & 10 \space \rm{mole} & 5 \space \rm{mole} \\ \hline \end{array} $$

At this point most everyone will "get it" - and the advantage of this method is that they sort of figure it out for themselves along the way, which means they will remember it more clearly. If anyone still struggles with the conservation part, you can have them calculate the number of each atom and each molecule for all or some of the examples above, in the same manner as at the beginning of the lesson.

  • $\begingroup$ Nice answer! However sometimes some professors may think that is wrong to say that a mole is just a number: Avogadro number is just a number, a mole is an Avogadro number of "atoms, molecules, ions, electrons, other particles, or specified groups of such particles." $\endgroup$
    – G M
    Sep 6, 2014 at 17:31
  • $\begingroup$ @GM, I know that IUPAC specifies elementary particles, and that in practice that is what moles are used for, but I would argue that "elementary entities" can be defined for anything, and therefore the definition is needlessly strict (and by trying to draw line, actually becomes more ambiguous). Also, I agree that it is important to understand that a mole is an abstract concept, while Avogadro's number is the constant derived from the definition of a mole - but it's not necessary to understand that at this stage, and it usually just confuses people when it is introduced too early. $\endgroup$
    – thomij
    Sep 6, 2014 at 22:04
  • $\begingroup$ @thomij Yes preservation of atoms. I knew that, I said that at the moment but did not emphasize it enought. Great tip for future. $\endgroup$
    – Anderson
    Sep 12, 2014 at 16:36

This is indeed a advanced exercise so maybe first you should made more easier exercises. In this case the student need to know mole concept and equilibrium concept that are not directly linked, the problem here is more related to equilibrium concept. For the future he/she should begin to reasoning following these steps:

Write the balanced equation: $$\ce{2SO3 <=> 2SO2 + O2}$$ Then for simplify you could write it so, this is more easier to understand, always ask Can I write it so?: $$\ce{SO3 <=> SO2 + \frac{1}{2}O2}$$

Then you should ask him: Which is the difference between this notation: $\ce{SO3 -> SO2 + \frac{1}{2}O2}$ [eq. 1] and this $\ce{SO3 <=> SO2 + \frac{1}{2}O2}$ ? You should focus that in the first reaction all the reagent are transformed in the products in the second only a part. So you can show him what happen if this is not an equilibrium. For the first case you can write this table: \begin{array}{|c|c|c|c|c|}\hline &\mathrm{SO}_{3} & \mathrm{SO}_{2} & \mathrm{O}_2 \\ \hline \mathrm{Initial} & 2\ \mathrm{mol} & 0 &0 \\ \hline \mathrm{Variation} & \small{(0-2=)}-~2 & (2-0=)+2&(1-0=)+1 \\ \hline \mathrm{At\ the\ end} & 0~\mathrm{mol}& 2~\mathrm{mol}& 1~\mathrm{mol}\\ \hline \end{array} (Maybe write the initial row and let the student write the "At the end" row) At this point you can talk about equilibrium: (the variation row may lead to confusion so maybe you could write a 3 row table). You can tell him to fill something like this:

\begin{array}{|c|c|c|c|c|}\hline &\mathrm{SO}_{3} & \mathrm{SO}_{2} & \mathrm{O}_2 \\ \hline \mathrm{Initial} & 2\ \mathrm{mol} & 0 &0 \\ \hline \mathrm{Variation} & ? & ?&? \\ \hline \mathrm{At\ the\ end\ (equilibrium)} & 1.52\ \mathrm{mol} & ? & ?\\ \hline \end{array} At the end: \begin{array}{|c|c|c|c|c|}\hline &\mathrm{SO}_{3} & \mathrm{SO}_{2} & \mathrm{O}_2 \\ \hline \mathrm{Initial} & 2\ \mathrm{mol}& 0 &0 \\ \hline \mathrm{Variation} & -~(2-1.52\ \mathrm{mol}) & +~(2-1.52\ \mathrm{mol}) & +~\frac{(2-1.52\ \mathrm{mol})}{2} \\ \hline \mathrm{At\ the\ end\ (equilibrium)} & 1.52\ \mathrm{mol} & 2-1.52\ \mathrm{mol} & \frac{(2-1.52\ \mathrm{mol})}{2}\\ \hline \end{array} So the core concept: you have a fixed number of molecules one part is transformed in products one part is still reagent. So starting form 2 mol of reagents only a part of its can be transform in products. The goal is simply find this part (with a simple subtraction) and calculate the products using eq. 1.

Could be useful use metaphors, some times I use dozen of eggs. This could be dangerous please note dozen can be compare to Avogadro's number but not to mole in fact a mole is not simply a number is a "number of" because only refers to elementary entities (may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles, so not only molecules but neither orange or ping pong balls). With these exercise maybe you should NOT use metaphors: one misconception you should avoid is that it is a static equilibrium, tell him that as $SO_{3}$ molecules decompose they are formed but at equilibrium the rate is constant. This is a dynamic equilibrium!


Say you have a piece of bread. Lets refer to that piece of bread as one large crumb. Then you split this large crumb into 10 smaller crumbs.

Start: one large crumb

End: 10 smaller crumbs

More advanced example:

Assume you have a piece of bread, consisting of three different kinds of bread (three pieces of each type of bread) that you just squeezed together in your hand so that they would stick. Then, by some magical force, the large multicrumb is split into its constituents (the three different kinds of bread).

Start: 1 large multicrumb

End: $\ce{3Crumb_{type 1} + 3Crumb_{type 2} + 3Crumb_{type 3}}$


Currency transactions or making change can be viewed as examples of an equilibrium process. Maybe this example might help the students see your point. For example, $$\ce{4Q <=> 2Q + 3D + 4N}$$ $$\ce{4Q <=> 3Q + 25P}$$ $$\ce{4Q <=> 1DB}$$ $$\ce{Q=quarter, D=dime, P=penny, DB=dollar bill}$$

  • $\begingroup$ Very nice analogy, I'll take it. $\endgroup$
    – user43021
    Sep 12, 2018 at 17:27

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