Methane and Oxygen

A guide for essentially using math to burn your hands.

So, you want to burn stuff in chemistry...

Well before doing that, you need to know what stoichiometry is. In a nutshell, it's everyone's worst nightmare: math and science meeting. Jokes aside, stoichiometry is the calculation of mass, moles, the conversion of them, to find the relationship between products and reactants.
Stoichiometry: Chemistry for Massive Creatures - Crash Course Chemistry #6

And yes, I know you want the video to burn things

Methane Bubbles Combustion Reaction

Wait, what?

When methane and oxygen combine, they create a type of reaction known as combustion. The first thing you want to do is to balance out the equation (and if you don't know how, there will be a video below to demonstrate it.)
With a simple Google search, we can determine that methane is CH4, while oxygen is O2 because it is a diatomic molecule. If added together, the equation will look like this:


CH4 + O2 --->


Now that's just the reactants side, so buckle up since it's about to get heated in here. Since it's a combustion problem, the products will always be CO2 and H2O. So now combined, it should look like this:


CH4 + O2 ---> CO2 + H2O


But wait, there's more! The equation is right, but it's not balanced yet. To balance it out, we need to add coefficients in front of every reactant and product (if needed.) For this particular problem, we add a 2 in front of CH4, a 3 in front of O2, a 2 in front of CO2, and a 2 in front of H2O.

2CH4 + 3O2 ---> 2CO2 +2H2O


Why? Well now, there's 2 carbons on both sides, 8 hydrogens on both sides, and 6 oxygens on both sides. That makes a balanced equation. Of course when we add in their states of matter it'd look a lot like this (even though it's not necessary for this equation.)


2CH4 (g) + 3O2 (g) ---> 2CO2 (g) + 2H2O (l)




If we were to put those letters and numbers in English, it would be called tetrahydridocarbon and oxygen yields carbon dioxide and dihydrogen monoxide.

A Full Guide on Balancing Equations

Balancing Chemical Equations Practice Problems

Oh no, calculations!

Time for the part that haunts me in my dreams. Before doing anything, such as crying for instance, it's always useful to find the molar mass of each element. Simply look at your periodic table and find the mass of the element.


C = 12.011

H = 1.008

O = 15.999


So if we have one carbon and 4 hydrogens on the reactant side, and one carbon, 2 oxygens, 2 hydrogens, plus another oxygen, our mathematical equation would look like this:


(12.011 + 1.008(4)) + (15.999(2)) ---> (12.011 + 15.999(2)) + (1.008(2) + 15.999)


I'll go ahead and spare you of the calculations and tell you that the total mass on the reactants side equals 48.041g/mol, while the products side equals 62.024g/mol.

There's no escape. (mole to mole conversions)

Given that the formula for mole to mole conversions is Mole A (from given) times Mole B (Coefficient) divided by Mole A (coefficient), we can calculate it out.


For example, if I had 3.07 moles of CH4, I would place that under Mole A (from given). Mole B would be the coefficient of the product you're trying to solve for. I'll be solving for H2O, which means the coefficient would be 2. Check and check. The next step is to find the coefficient in front of CH4, which is 2.

3.07 moles CH4 | 2 moles H2O

----it'sablank------| 2 moles CH4

Multiply the top numbers, divide by the bottom, excluding the blank space because it's blank. The number you should get is 3.07 moles of H2O. Not all your problems will cancel out, but coincidentally, mine did.

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There's no escape (part 2, featuring mass to mass conversions)

We did moles, now we need mass. Up next is the longer equation that you need to memorize.


Mass A (given) | Mole A (always 1) | Mole B (coefficient) | Molar Mass B (periodic table)
----------------------| Molar Mass A (periodic table) | Mole A ( coefficient ) | Mole B (always 1)


So say I had 12.1 grams of CH4, and I was solving for CO2. Simply applying prior knowledge, my table would look like this:



12.1g CH4 | 1 mole CH4 | 2 mole CO2 | 44.009g CO2
----------------| 16.043g CH4| 2 mole CH4 | 1 mole CO2


The mass I got was a whopping 33.74g of CO2.

Now what? (limiting and excess reactant + a bonus)

Now we get to calculate for the limiting and excess reactant (yay.) The excess is exactly what it sounds like: leftover/extra components to your lab/experiment. The limiting factor is the item that is scarce, or it limits how much you can make of the product. It's solved basically the same way as the mass to mass conversions, except you have 2 givens, rather than one. But worry not, you only have to do 2x the work now.


So if I had 12.3g of CH4 and 12.3g of O2, I'd just solve two different mass to mass conversion problems.


12.3g CH4 | 1 mole CH4 | 2 mole CO2 | 44.009g CO2
----------------| 16.043g CH4| 2 mole CH4| 1 mole CO2


Oh, what a surprise, it was the same problem as above (for this example at least.) Which means the answer is the same: 33.74g of CO2

Now for oxygen...

12.3g O2 | 1 mole O2 | 2 mole CO2 | 44.009g CO2
--------------| 31.998g O2 | 3 mole O2 | 1 mole CO2

The mass of CO2 in this problem is 11.28g of CO2.

So when we had CH4 as a reactant, it yielded more CO2 than O2, meaning CH4 is the excess factor and oxygen is the limiting factor.


This also ties in with the theoretical amount, which is the amount of a product you would get in an optimal situation, where nothing goes wrong. But just as life does, it rarely turns out this way. The theoretical amount is the same as the excess amount.

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Why is life always ruining my dreams? (percent yield)

This is probably the simplest thing I've written on this article yet. To get the percent yield, all you have to do is divide the actual yield (a.k.a. the real-life-hit-me-like-a-ton-of-bricks yield) of your product by the theoretical yield (a.k.a. my-dreams-were-happy yield,) then multiply by a 100.

Seriously, that's it.


If my actual yield was 11.15g of CO2, then my equation is:

(11.15/11.28) * 100 = 98.84%

So close. I was 2 percent off of the theoretical yield (although I'd blame it on my lab partners.) But it's okay, it could've been from wrong measurements, or the losing of products, or an excess of reactants. The possibility of messing up is endless, but there's only one reason how you'd be perfect.

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But hey, it's finally over! (Or is it?)

Congrats, you made it to the end! No more stoichiometry for a while! Unless you're Mr. Miller and/or happen to be reviewing for a final exam...
Hey, at least have some slightly-relevant pictures of chemistry puns to make you feel better. Now if you'll excuse me, I need to go cry in a corner. Good luck, have fun!
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