# Stoichiometry an Introduction

Stoichiometry is often the bane of a chemistry student’s existence. Unlike many lessons which can be learned and forgotten after the test, stoichiometry is a tool that is used constantly throughout the remainder of chemistry. If the skill is not learned, it makes difficult or impossible every concept that follows that requires it, and as a result, grades plummet. Teachers know this, and stress the importance of learning stoichiometry, but in case you missed it, here’s a chance to get to know everybody’s “favorite” aspect of chemistry.

Every balanced chemical equation functions as a recipe for a chemical reaction. If you’ve ever seen one, you know that there’s a mix of numbers and chemical formulas, something like this:

2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O

(Due to text formatting limitations, you’ll have to imagine that -> is an arrow and that the numbers immediately to the right of letters are subscripts.)

In the chemical equation, the numbers that precede each chemical formula (2, 13, 8, 10) are called coefficients, and represent the relative number of molecules that are used or produced by the reaction. These coefficients are much more important to stoichiometry than the actual chemicals involved in the reaction.

Suppose that in the reaction above there are 400 molecules of butane (C4H10). This is a very small amount, and burning it would only make the tiniest spark, but it is good for a first look at the strategy. If we do have 400 butane molecules, then we can calculate how many molecules of oxygen (O2) are needed to burn them all up in the combustion reaction above.

We know that 13 oxygens are required for every 2 butanes, giving a ratio of 13 / 2. (Note that the one we know goes on the bottom this is always going to be the case.) To find how many oxygens we need, you simply multiply the number of butanes by our ratio.

400 x 13 / 2 = 2600 oxygen molecules

Similarly, you can use ratios to see how many molecules of carbon dioxide (CO2) and water (H2O) will be produced.

400 x 8 / 2 = 1600 carbon dioxide molecules

400 x 10 / 2 = 2000 water molecules

You would have gotten the very same answer if you tried to find the answers using oxygen instead.

2600 x 8 / 13 = 1600 carbon dioxide molecules

2600 x 10 / 13 = 2000 water molecules

Such is the “magic” of stoichiometry. Right now, you may be thinking “well that was simple, why do students get so confused?” A very good question, but it does get more complex. In chemistry, we don’t usually count molecules outthe numbers would be enormous and difficult to work with. Instead, we count by moles, and that tends to be a place where students get lost. What is a mole? (There’s a whole article called “How big is a mole?” here on Helium.) A mole is a counting number. It is a very large number indeed, which means that it is excellent for keeping track of large numbers of molecules. (One mole represents 6.02 x 10^23 molecules, but that doesn’t actually matter for stoichiometry.) Don’t let the new name scare you, counting moles is the same as counting thousands, millions, or billions. We can have 1 mole, 4 moles, or even 5.38 moles. If you find yourself thinking of furry, burrowing pests, have no fear. Those moles behave exactly the same when it comes to math too.

Let’s go back to our equation.

2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O

Now suppose that we want to make 5.20 moles of carbon dioxide. Using ratios, we can calculate how many moles of butane and oxygen we need, and how many moles of water will also be produced. Remember, CO2 is the one we know, so it goes on the bottom of the ratio.

butane: 5.20 moles x 2 / 8 = 1.30 moles of butane

oxygen: 5.20 moles x 13 / 8 = 8.45 moles of oxygen

water: 5.20 moles x 10 / 8 = 6.50 moles of water

You should now see, the math works exactly the same, we just write in the word moles.

Work all stoichiometry problems in moles. If the problem is given in terms of grams, liters, or other measures, you must convert those to moles before multiplying by the ratio. If you aren’t sure how to convert, there are a few common conversions to and from moles at the end of the article.

You may also wish to further your knowledge of stoichiometry with the Helium articles:
“Finding the Limiting Reactant” and
“Understanding Percent Yield”

Converting to and from moles.
GRAMS
To convert grams to moles, divide by the molar mass of the chemical.
To convert moles to grams, multiply by the molar mass of the chemical.

LITERS (of a gas at STP)
To convert liters to moles, divide by 22.414 (the molar volume of any gas at STP)
To convert moles to liters, multiply by 22.414

LITERS (of a gas not at STP)
Use the Ideal Gas Law (or Van der Waal’s equation) to solve for moles or volume, as needed.

LITERS (of a liquid)
To change liters to moles, multiply the liters by the density to get grams, then convert grams to moles as above.
To convert moles to liters, first convert to grams (as above) and then divide by the density.

MOLARITY (and liters of solution)
To convert from molarity and liters to moles, multiply together the molarity and the liters of solution.
To convert from moles to molarity, divide the moles by the volume (in liters) of solution. Note that if you poured two solutions together, the final volume will be the sum of those volumes.