# Determining Stoichiometry of pretty much any fuel

When it comes down to it, you can burn so many different things to use as a fuel for your car. What do you need? Simple – Hydrogen and Carbon, generally, in a compound that work in conjunction with Oxygen. This mixture (called a hydrocarbon because it contains both hydrogen and carbon) combusts, which means that you’re burning it!

You may be a bit like me – when I started playing with cars, I didn’t know jack about fuels. Just that if I wanted to have fun on the weekend, I could find leaded race fuel at a few select gas stations in the area. Boy how times have changed! A few years ago, I started looking at chemical compounds with industrial availability to see if there was something generally available that would give me an upper hand as far as making power. Ultimately, I came up with a list of hydrocarbons I wanted to check out.

• Ethanol – C2H6O
• Methanol – CH4O
• Acetone – C3H6O
• MTBE – C5H12O
• ETBE – C6H14O
• Toluene – C7H8
• Xylene – C8H1O
• Nitromethane – CH3NO2

I pretty quickly realized that I needed to figure out how much of each of these compounds I would need to burn, which is where stoichiometry comes into play… Stoichiometry of a fuel is the ratio that you have to mix that fuel with oxygen in order to completely combust the fuel. That ratio for Gasoline is 14.7:1. It varies a lot depending on the fuel you’re using. The good news is that you can calculate the stoichiometric ratio, or AFR, for anything that’s a known compound. And the good news about these compounds are that they are known compounds. So how does that work?

## Chemistry

Yep, chemistry is how you figure this stuff out. Unfortunately, I never took any chemistry in school, so I had to figure it out through the power of the Internet. Let’s start with just one of these compounds, Ethanol, and we’ll work through that.

Here’s a quick science lesson: the complete combustion of a hydrocarbon produces Carbon Dioxide (CO2) and Water (H2O).

### Balancing the Equation

In order for combustion to happen, it must happen in a balanced manner. This means that you need to take a certain number of molecules of fuel and mix it with a certain amount of oxygen to create complete molecules of carbon dioxide and water. A typical equation for that looks like this:

C2H6O + 3 O2 -> 2 CO2 + 3 H2O

This is a balanced equation. If you’re really interested in how it all breaks down, you can check it out on Wolfram Alpha. I’ll summarize it here: C2H6O is Ethanol. O2 is Oxygen. One molecule of Ethanol mixes with three Oxygen molecules and through combustion, two CO2 molecules and three H2O molecules are created, along with a lot of heat, which you use to make horsepower!

### Weighing Mass

We’re not quite done yet, though. We also need to know weights for the various elements in our hydrocarbons, and the weight of Oxygen. This allows us to get the total weight of a single molecule of our compound and we’re going to use the weight of Oxygen to determine our total ratio. For this we can turn to the periodic table!

Carbon (C on the periodic table) weighs 12.011 g
Hydrogen (H on the periodic table) weighs 1.001 g
Oxygen (O on the periodic table) weighs 15.99 g

Our Ethanol contains:

• 2 Carbon atoms, weighing 24.022 g total
• 6 Hydrogen atoms, weighing 6.048 g total
• 1 Oxygen atom, weighing 15.99 g total

This makes for a total weight of 46.069g per Ethanol molecule. Next up, we can figure out our oxygen:compound ratio.

### Finding Ratios

We’re almost done, so bear with me! In this step, we need to figure out the oxygen:compound ratio. This allows us to then find the AFR of the compound. Going back to our balanced equation, we know that we need 1 parts of Ethanol to 3 parts of Oxygen. That equation looks like this:

oxygen:fuel = ([O2 ratio] * ([O2 weight] * 2) / ([Compound weight] * [compound ratio])

So, that means:

(3*(15.99*2)) / (46.069 * 1) = 2.084

our Oxygen:Fuel ratio is 2.084:1.

### Finding the AFR, Finally!

Last step here, I promise. I’m not going to go into too many details because there are lots of places where you can find this information, but normal air we all breathe is about 23.2% Oxygen as a percentage of mass. Using our oxygen:air ratio above, we can now figure out our total AFR given the following formula:

AFR = ([o:a ratio] * 100) / [percentage of oxygen in air]

So, that means:

(2.084 * 100) / 23.2 = 8.98275

The AFR for Ethanol is 8.98:1. Close enough!

### Wrapping it Up

Yes, this was a lot of jumping through hoops to find AFR for one particular fuel – Ethanol, but it demonstrates that you can, in fact, find AFR for any hydrocarbon.

So what can you do with this information? Personally, I use it to find an AFR for a fuel that I’m mixing with other fuels to come up with AFR equivalents. I’ll write up something about this in the future, because it’s fascinating to be able to make a new fuel that is a drop-in replacement for another fuel. I have a fuel that consists of a majority of Methanol, yet runs the exact same AFR as E85. So, I can tune for E85 and reap some of the benefits of Methanol. I’ll tell you about it later! 🙂