The Chemistry of Potato Guns

Chemistry was one of those subjects I was never that excited about. We did do experiments in chemistry but they were never all that captivating. I think I would have been much more motivated to learn chemistry if I got to blow stuff up. Explosive reactions are definitely a part of chemistry but for some reason my school was hesitant to allow students to play with those kinds of reactions.

Today I am going to try to educate you on some chemistry concepts I have had to brush up on to build my potato gun. Now before you click away and go look at something else on the internet, I will make it worth your while. At the end of this article will be a video of my potato gun and some slow motion footage of it blowing stuff up. You could skip to it… but I hope you won’t.

First, what is a potato gun? A potato gun is a cannon that shoots potatoes that can be built from plumbing parts you can buy at the hardware store. If you are interested in making one make sure to do your research! Improperly made potato guns are dangerous and can explode when fired.

Potato guns are typically powered by either pneumatic pressure supplied by a compressor or by a combustion reaction. My potato gun is a combustion type. Combustion guns often use either hairspray or propane for fuel. Typically you can measure out this fuel by spraying it into the combustion chamber for a few seconds and then sealing it up. I wanted to be a bit more precise. I chose butane as the fuel for my gun as it is the flammable component of many hairsprays and buying it pure makes it easier to measure. The container I got is a butane refill container for grill lighters.

Combustion is a reaction where a flammable compound is mixed with oxygen and burned to produce carbon dioxide and water . In my case the flammable compound is butane and I get the oxygen from the atmosphere. The chemical formula for butane combustion is as follows.

The reactants are on the left and the products are on the right. We don’t really care about the carbon dioxide or water that gets produced, what we do care about is the heat that is generated in the process. That heat increases the pressure in the combustion chamber and powers our cannon. In order to get the most power we need to balance the equation. This formula is unbalanced which means there are not the same amount of Carbon, Hydrogen, and Oxygen atoms on both sides. The left side has 4 Carbon atoms, 10 Hydrogen atoms and 2 Oxygen atoms while the right side has 1 Carbon atom, 2 Oxygen atoms, and 2 Hydrogen atoms. By balancing the equation we can determine what ratio of fuel to oxygen we need for a complete combustion reaction.

This is the balanced formula for complete butane combustion. On the left side we can see that for every 2 butane molecules we will need 13 oxygen molecules for a proper reaction. On the right side we see that a proper reaction will produce 8 carbon dioxide and 10 water molecules. Now that we know the optimal fuel ratio we have a different problem, how do we get the proper mixture into the potato gun?

Because the oxygen for our combustion reaction is supplied by the atmosphere we can’t control how much oxygen is in the combustion chamber, however we can control how much butane is injected. We really need to determine how much butane we need to inject to react with the oxygen already in the chamber.

At normal temperatures and pressure butane and oxygen are both gasses. Counting gas molecules one by one would take a long time so fortunately we can use the Ideal Gas Law to estimate how many will be in our potato gun.

The Ideal Gas Law is used to model ideal gasses. Ideal gasses are gasses with various assumptions made about them and have to follow the following rules:

1. Molecules in an ideal gas collide elastically, without any energy loss.
2. Gas particles have no volume and are infinitesimally small points.
3. Molecules do not have any intermolecular forces.
4. Molecules in an ideal gas are constantly in motion.

These rules mean that real world gasses are not ideal gasses but ideal gasses are easier to predict than real gasses. For instance, if water followed these rules it could not form a liquid at room temperature. Under certain circumstances, such as in our combustion chamber, ideal gas equations can be used to make reliable predictions about real gasses.

Lets break apart the Ideal Gas Law and figure out how we can use it to measure fuel for our cannon. The Ideal gas law takes the form

The variable is used to represent the pressure of the gas and is measured in Atmospheres () or Pascals (). For our calculation this will be set to atmospheric pressure.

The variable is used to represent the volume of the gas and is measured in Liters () or cubic meters (). For our calculation this will be the volume of our combustion chamber.

The variable is used to represent the substance of the gas. Substance measures the number of molecules in a sample. Because molecules are so small, instead of counting individual molecules, substance is measured in Moles (). 1 is equal to approximately or molecules. This number is known as Avogadros Number and the reason that specific number is used is so moles of substance can be easily converted between Atomic Mass Units () and Grams (). Because we want to determine the number of oxygen molecules in our combustion chamber this is the value we will be solving for.

The variable is used to represent the absolute temperature and is measured in Kelvin (). For our calculation this will be set to the outside temperature.

Finally, the variable is the Ideal Gas Constant which is used to relate the quantities of Pressure, Volume, Substance, and Temperature together and cancel out their units. Because of this, the exact value of the constant depends on the units being used for the other quantities.

Just to review

Variable	Quantity	Unit
	Pressure	Atmospheres () or Pascals ()
	Volume	Liters () or cubic meters ()
	Substance	Moles ()
	Ideal Gas Constant	Depents on other units
	Temperature	Kelvin ()

Now that we have gone over the Ideal gas law we can calculate the amount of oxygen in the chamber. First we gather our values

and then plug them into the Ideal Gas Law Equation.

While it may seem that is not very much it is actually equal to a quantity of molecules. This is still not quite what we are looking for because air is not pure oxygen. To specifically find the number of oxygen molecules we need to learn about another Gas Law, Dalton’s Law.

Dalton’s law will allow us to isolate the oxygen content of the air for use in the Ideal Gas Law equation. Dalton’s law states that the pressure of a gas is equal to the sum of the partial pressures of its components.

Air is nitrogen, oxygen, with the rest being trace compounds such as carbon dioxide, argon, water vapor, and more. In this case of the air’s pressure comes from nitrogen, comes from oxygen, and so on.

We can turn this around and determine the partial pressure of a component of a gas if we know the total pressure and the concentration of the component.

Because we know standard atmospheric pressure and the percentage of it that is oxygen

we can calculate the partial pressure of oxygen.

We can now use this partial pressure of oxygen with our other variables from before

and plug them into the Ideal Gas Law Equation.

This tells us that there are approximately 0.034 mol of oxygen in our combustion chamber when it is at standard atmospheric pressure and an outside temperature of . With this quantity we can determine how many moles of butane we need to inject into our combustion chamber for a proper reaction. Lets look at our balanced chemical formula from earlier.

We can extract a ratio of butane to oxygen from this formula that we can use to convert our moles of oxygen to moles of butane.

By multiplying our moles of oxygen by this conversion term we can determine how many moles of butane we need.

Now we know that we need 0.0052 mol of butane to react in our combustion chamber. The butane I bought is in a pressurized container. In order for it to be usable in the gun it needs to be vented into the chamber. Venting directly in would not give me the control I need to measure out specific volumes of butane. Instead I vent the chamber into a syringe. The pressure makes the volume of the syringe expand and when I release the butane container I have a specific volume of pure butane at standard air pressure and the outside air temperature. You have probably guessed by now that this is another problem that can be solved with the Ideal gas law. One last time lets gather our values

and plug them into the Ideal Gas Law Equation.

At last we have out final answer. 119.5 mL of butane at standard air pressure and temperature is precisely how much fuel is needed for my potato gun.

Finally, as promised, here is the footage of my potato gun firing.