Topics Covered

• The Mole and Molar Calculations
• Percent Mass
• Empirical and Molecular Formulas
• Chemical Reactions and the Balancing of Them
• Stoich Calculations
• Percent Yield

Stoich is short for stoichiometry. Don't be scared off by that name; I remembered in ninth grade when I heard that I thought I wasn't going to get through the year. It's a 13-letter name for something that's very easy, yet important part to becoming successful in chemistry. This is where the stuff becomes important to learn, because it's directly on the test.

The Mole and Molar Calculations

I am surprised to hear that this is the point where some chemistry students lose all hope; trying to figure out what a mole is. I admit, I was confused too, but keeping this in mind made life a lot easier: a mole is simply the measure of how many. A similar unit is a dozen, and since everyone knows what this is, I'll use this first.

Let's say you have a dozen eggs. What does this mean? It just means you have twelve of something. It doesn't say how much it weighs, or how much space it occupies, just how much. A dozen siege tanks doesn't have the same mass nor volume to a dozen specks of dust. There's just twelve of each.

Moles are the same concept. One mole of something has 6.022 x 10²³ objects in it. It's a lot, but atoms are small, so it's a good unit in chem. 6.022 x 10²³ is N_A, called Avogadro's number. It's a pretty big number. One mole of seconds is 4 million times as long as the age of the earth. One mole of meters is over 600 times the diameter of the Milky Way. For our purposes, it's pretty good.

It's important to understand the relation of amu's and moles. One mole of something that weighs one amu is one gram. In other words, 1 g = 6.022 x 10²³ amus. So one mole of hydrogen (H2) would weigh 1.0079 x 2 = 2.0158 grams. One mole of carbon is exactly 12 grams; that's how they derived Avogadro's number.

Often it's important to find out how many moles is some given quantity of mass. You can use dimensional analysis (DA) for this. Let's say you want to know how many moles is 36 grams of water, and also how many molecules are in it. First you must find the molar mass of water (how many amus it weighs). Water is H₂O, that means 2 of hydrogen and one of oxygen. So the adding of those masses is:

1.008 x 2 + 16.00 = 18.00 amu.

Now comes the DA part.

So it's 2 moles. And since 1 mole is N_A molecules, there are 2 x N_A = 1.2 x 10²⁴ molecules.

The key to solving these problems is getting the g/mol (molar mass, or amus) of the substance, and then using that in a DA setup to convert back and forth between grams and moles. One amu is one g/mol.

Example:

Example:

Percent Mass

This is just another way to show how much of something is in a compound. It's how much of the substance is made of one particular element, the percent by mass (I bet you could figure that out by the title). So if there is 60 g of something, and in that compound there is 12 g of carbon, then you would say the percent mass is 20% (12 g / 60 g x 100%).

Note that this number can't be directly obtained from the moles. Percent moles is not the percent mass too. You must convert or die.

Example:

Go try some more percent mass practice problems now!

Empirical and Molecular Formulas

The molecular formula is the type that we've always been using: H₂O, C₆H₁₂O₆, C₂H₄, and all the like. It's how many atoms of each element are present in one molecule of the substance. The empirical formula is the reduced form of this. It's the ratio of atoms. For H₂O, there's 2 hydrogens for every oxygen. So the empirical formula is the same. However in C₆H₁₂O₆, there is a ratio of 6 : 12 : 6. That can be reduced, so the ratio is 1 : 2 : 1. Therefore the empirical formula is CH₂O. Likewise, the ratio 2 : 4 in the last one can be reduced to 1 : 2. The empirical for it is is CH₂.

Example:

It's important to remember that the molecular formula is a multiple of the empirical formula in both subscripts and molar masses.

Chemical Equations and Balancing Them

A chemical reaction is when one or more molecules rearrange their atoms to form new molecules. Here is an example:

CH₄ + O₂ ---> CO₂ + H₂O

You can see this visually below:

		+
	Carbon Dioxide		Water

The things on the left side are called the reactants, and the things on the right are called the products. The reactants are generally transformed into the products until one of the reactants runs out, or the system reaches equilibrium (a balance).

As another example, if you wanted to say that two molecules combine to form one, just use the coefficients. Just like in algebra, you can put numbers in front of the molecule to show how many of them are in the equation. Let's say 2 of Molecule A and 3 of Molecule B are needed to make 4 of Molecule C. The equation would look like:

2A + 3B ---> 4C

Another thing that's important to right down is what state each reactant and product is, by little subscript letters. The symbols are:

• Solid: (s)
• Liquid: (l) (When you write it by hand, it's a lowercase cursive l.)
• Gas: (g)
• Aqueous (means dissolved in water): (aq)

So the first equation should have really been written as:

CH_{4 (g)} + O_{2 (g)} ---> CO_{2 (g)} + H₂O _(g)

It's funny how they teach you to do that, but on the AP test on the Equation Writing section, they will not look for those symbols. But your teacher most certainly will, so get into the habit!

That equation is better, but it still is missing a key point in chemical reactions. Atoms cannot be created nor destroyed; they can only be put in different orders. If you look above, there are two oxygens (green) on the left and three oxygens on the right. Where did that extra oxygen come from? The key is to realize that these aren't one to one reactions. That is, they all don't have coefficients of one. So you have to balance.

Let's start with hydrogen. There's four on the left and two on the right, so multiply the molecule on the right by two:

CH_{4 (g)} + O_{2 (g)} ---> CO_{2 (g)} + 2H₂O _(g)

Now the hydrogens are balanced, but the oxygens aren't. Now we have 4 on right and 2 on left. Multiply atom with oxygen on left by two:

CH_{4 (g)} + 2O_{2 (g)} ---> CO_{2 (g)} + 2H₂O _(g)

If you count them, everything adds up. There are the same number of atoms on the left as on the right. That is the correct chemical equation.

Here is The Art of Balancing Instruction Manual, by none other than Takalah:

Example:

Stoichiometric Calculations

One thing to realize when doing any calculations is that moles and coefficients are interchangeable. Both mean number of particles, or multiples thereof. So in 2A + 3B ---> 4C, it either means 2 particles of A combine with 3 particles of B to make 4 particles of C, or 2 moles reacts with 3 moles to make 4 moles.

With that in mind, here is a simple problem.

Example:

The "oxygen in excess" is important. We're assuming that there's tons of oxygen, enough to supply whatever the hydrogen needs. If there was only .0000000000000000000001 moles of oxygen, 5.94 moles of water obviously can't form.

Let's try one more of this type before moving on to LRs.

Example:

In these last two problems, we always assumed that one of the reactants was in excess. This simplified things, but it's often not the case in many reactions. You have given amounts of each reactant. So how do you do this? Simple, just find out which out runs out first molewise. The one that does is called the limiting reactant, or just LR. One quick example:

Example:

Note that mole ratios are important. Let's say 2 Tak's become 1 Asdf. Then it becomes a little bit more than just finding the bigger number, but not very harder. You start with one, say, Tak. Assume that it is the LR. Then, according to the mole ratio, you would need at least half as much Asdf for Tak to become the LR. Do you have a half? Well, one-half of .109 moles is about .05 something, and you do have that much. So if Asdf isn't your LR, then Tak must be. (Of course if it was exactly equal to half, then you could use either one, there would be no LR, and everything gets used up.)

What happens if you started with looking at Asdf? Well you need twice as much Asdf's. Twice .0890 is .17 something. You don't have that much Tak, therefore Tak is your LR, no matter how you look at it.

Once you find the LR, you can go ahead and finish the problem just like the others. So what's the difference? Once you find the LR, you use that amount and throw away the other amount. So like in the above example, you use the .0890 value, and forget about the .109. Of course, if they ask you to refer to the original amount, such as how much reactant is left over, then you need to use it. But not for stoich.

Example:

Percent Yield

One little thing. You can test the above stuff in a lab, like measuring out the reactants, and see if you get the calculated amount of product. You probably won't get the exact number, but rather a number that's below it. There could be many reasons for it, such as maybe the reaction doesn't go to completion, or maybe there's impurities or something. The theoretical yield is how much you should get, according to stoich. The actual yield is how much you actually get, determined in a lab. The percent yield is the ratio of these two. The equation:

Yields are masswise. That is, how much you get is measured in kilograms, grams, mass.

Note that the percent yield will always be under 100%. The theoretical yield is the max you can get. (You're not going to get more substance than calculations predict.)

Now it's time for the big example, that ties together balancing equations, stoich, LRs, and percent yield.

Example:

Copyright ©1999 by Takalah. All rights reserved. This concepts portion was last updated on September 4, 1999