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Stoichiometry Stuff
The chemics book sure made this explanation conceptually harder than it had to be.

Suppose you come home one day after school and your parents tell you that they have decided to paint the whole house, inside and outside, the same color, but they have priced having the work done and it will cost a gazillion dollars so they have decided to make it a family project. Your parents think it will be a fun thing to do together. (You, of course, know it would be more fun just to catch yourself on fire.) They will do the painting, but it is your job to get the paint -- not too much and not too little.  They will give you a new car if you get it right.  The paint stores are 100 miles away, and the whole distance is through Galleria-at-Christmas-like traffic, so you need to figure out how much paint you need, so you only have to make one trip.

You have no idea how much paint you need, so you call a paint store and tell them the project, and you ask how to tell how much paint you need.  The paint guy says "Measure the dimensions of the walls, and you need a gallon of paint for every 400 square feet".  [I made up that number, cuz I don't remember how much paint covers what.]

So you measure the walls, and because you live in a mansion, you come up with a kabillion square feet.  You do the math, and figure out how many gallons of paint you need, and you figure you are ready to go to the paint store.

But...
You happen to mention the project to a friend who went through the same thing last year with his parents (which is where your parents got the idea, because you have let them associate with the wrong peer group), and he says "It won't be that easy, because...."

Wrinkle #1: It takes only one coat to cover light colored walls, but it will take two coats to cover medium color walls, and it will take three coats to cover the dark walls you already have in the house.

Nuts; that means you have to go back to your notes and see how many square feet will need one coat, how many will need two coats, and how many will need three coats.  Easy enough to calculate, so you do it.  And then you call the paint store to ask them to get the paint ready; you already have got a paint swatch so you know what the name of the color is to tell them. Ready to rock and roll.  3000GT, here I come.

But...
Wrinkle #2: The paint store says you will need to mix it yourselves; but it is easy to do. It is just made up of three colors. Good.  It takes two gallons of one color for every gallon of the second color, and for every three gallons of the third color.  But he only has one of the colors and you will need to get the rest from a different store.  Okay.  He gives you their phone number.  He has however many gallons on hand of one of the colors you will need.  So now you have to calculate how much of each color you need.  You have already figured out how many total gallons you need, so now you go back to figure out how many gallons of each of these different colors you need to add together to make that many total gallons.  You know you can figure it out, but this is starting to become a drag.  But you sit down and figure it out, so you can get your new car.  You call the second paint store and tell the salesguy how many gallons you need of each of the other two colors.
But...
But...
Wrinkle #4: The third color is made up of three colors, and they mix in a really weird way, because when you put two of them together, instead of making twice the amount, they react in such a way that when you add one gallon of each, you only get one gallon of total stuff; they kind of dissolve into each other.  Then when you add the third color, it mixes so that one gallon of it mixes with one gallon of the combo you just made up, to give you one and a half gallons of the color you want.  He asks how many gallons of each color you want (and by the way, the third color only comes in a powder by weight, so that 8 ounces of it makes up 5 quarts of paint).  You are now making up your own new four-letter words and applying them to the union of your grandparents for having children who get such stupid ideas as painting their own house.  But you sit down and do the last calculations so you can order all the #!%#!! paint. Notice that none of this is difficult to understand how to do; it is just a matter of keeping straight all the information and all the measurements and units and then just doing the arithmetic.  It is a pain, but it is not intellectually taxing; it is just boring and tedious.  You are about ready to call all the paint stores.
But...
Your friend points out there is just one last thing to calculate.  Your parents will not be as good a painters as real painters, and they will waste about 13% of the paint because they will overlap more than they should, and they will spill some as well, and some of it will stick to the cans and to the roller pans and the rollers and brushes, etc.

So you factor that into your other quantities, order the paint, pick it up, deliver it, and get the keys to your new car.

The only difference between the above and chemistry is that the names of the things you are mixing are unfamiliar to you, the units are not all that comfortable to work with, and you don't get a car when you figure out the answer.  Minor differences.

The main thing you need to keep in mind, in order to work these problems in chemistry, is that understanding mols is the key, in two ways: (1) mols (and the periodic chart with the atomic masses on it) lets you convert back and forth between weights and relative number of atoms/molecules of the chemicals you are working with for the various reactions, and (2) it is these relative numbers that is important, because it is not how much chemicals by weight, but how many atoms that will combine with each other that is the crucial thing.  For example, when you make up water molecules out of hydrogen and oxygen, you need two hydrogen atoms for every oxygen atom to make one molecule of water; having twice the weight of hydrogen that you have of oxygen will not, by itself tell you what relative amount of water you will get, because you first have to figure out what the relative number of atoms of each will be in that weight ratio.

These problems are kind of like (1) if I told you there were 5 tons of men in one room, and 4 tons of women in another, and asked you how many couples that might make on average, you need some way to figure out many men there are on average in a ton of men, and how many average women it takes up to make a ton of women.  (Then we get into the "excess" stuff they talk about in the chemics book, because it may turn out that you have more women than men or more men than women, so the number of couples -- of mixed gender, in this case -- will be the same as the number of whichever group, men or women, is fewer.)  Or, a little more difficult, like (2) if I told you there were 50 pounds of men for every 27 pounds of women in a room and asked you to calculate how many mixed couples there were for every average size man.  If you knew the relative weight of women to men, you could figure it out, and all it would take is paying attention to your units, your given quantities, and the conversion factor of the average weights per man and per woman.

If these were things you really wanted to figure out, you could do them.  The stoichiometry problems just use the same principles, so they are just a matter of keeping straight your formulas of reaction (which give you the relative number of atoms involved in the various elements and compounds) and your conversions back and forth, where necessary, between the weights of atoms and molecules and the numbers of atoms and molecules.  And you have to keep in mind what you are doing as you go through the steps.  Just like in the paint problem above.  It is not intellectually difficult, just bookkeepingly difficult.

If you need a refresher on the concept of mols, it is at www.Garlikov.com/chemistry/mols.html

Let me just run through one fictitious example (using "elements" A, B, C, D, and E) with you, but if you get bogged down in trying to follow it, just forget about it, since this may be more trouble than it is worth:

Suppose we know that some amounts of A2B3 mixed with C(DE2)4 gives some amounts of A(DE2)2 and CB3 .
Since there are two atoms of A involved in the molecules of the ingredients and only one atom of A in the molecule of the product, we know we get two molecules of that product for each molecule in the ingredients.  Similarly with (DE2), there are twice as many atoms of D and E in the reactants as there are in the products.  So the formula showing the correct balanced proportions must be:

A2B3 + C(DE2)4 = 2A(DE2)2 + CB3
That way you start out, and end up with 2 atoms of A, 3 of B, 1 of C, 4 of D, and 8 of E.  All the atoms you began with are accounted for in the product.

Now suppose they give you 150 grams of A2B3 and 850 grams of C(DE2)4 and they want to know whether all the reactants will be used or whether there is excess, and they want to know how much product you get, along with whatever is not used of the original ingredients.

From this information alone, we cannot tell, because we do not know how many relative atoms or molecules of any of these elements or compounds we have.  In order to know that, we need to know the atomic masses of A, B, C, D, and E.

So suppose we have a chart that gives us that information:

 Element mass A 1 B 2 C 3 D 4 E 5

and we work out the number of mols of the original reactants.  One mol of A2B3 will then be twice the mass of A in grams plus 3 times the mass of B in grams, so that will be 2 + 6 = 8g.  A mol of C(DE2)4 will be the mass of C in grams, which is 3g, plus 4 times the mass of D in grams or (4 x 4 = 16g), plus 8 times the mass of E in grams (8 x 5 = 40), so it will be 3 + 16 + 40 = 59.

Looking at A2B3 , we have 150g and we know it is 8g/mol, so we can say: 150g x 1mol/8g = 18.75 mols.
Looking at C(DE2)4 , we have 850g given to us and it is 1 mol/59g, so we have 850g x 1mol/59g = 14.41 mols.

Since one molecule of A2B3 mixes with one molecule of C(DE2)4 in a reaction, we know that only 14.41 mols of the 18.75 mols of A2B3 can react, and the other 4.34 mols will be left over unreacted.

And since the balanced formula shows there are two molecules of A(DE2)2 for each molecule of  A2B3, C(DE2)4 , or CB3 and since you can use the chart to figure out the molecular masses of the compounds in the product, you can then figure out their total masses from the number of mols you have of each.  You have 2 x 14.41 mols of A(DE2)2 which is 28.82 mols of it.  And since each mol is 29g, you have 835.78g of it.  You have 14.41 mols of CB3 which has a mass of 9, so it weighs 9 x 14.41 = 129.69g.  And you had left over 4.34 mols of the original A2B3 , which at 8g/mol gives 34.72g of it.

One way of checking your answer is to make sure you have the same total mass at the end as you began with, since a chemical reaction of this sort should not create or destroy any total mass or any number of atoms.  We started out with 1000g of material, and we ended up with 34.72g + 129.69g + 835.78g, which is 1000.19g which is close enough for me since it is 1:00 in the morning.

I could have given you the problem with 150g of A2B3 and 14.4 mols of C(DE2)4 and it should still be one you could work.

The important things in doing all these problems are:

• (1) for you to be able to balance formulas, so you don't lose any atoms or any mass, and so you know the proportions of the number of molecules of each substance in the reaction
• (2) for you to be able to go back and forth between mols and masses for any element or compound in the reaction -- all the reactants and all the products
• (3) for you to understand that it is the relative number of molecules (which is the same as the relative number of mols) that react with each other that is important for determining excess amounts of reactants.
You should be able to see how all the following values were obtained, and how to go back and forth between mols and masses for each compound.  If you can do that, you understand this pretty well:

 compound mass atomic mass mols A2B3 150g 8 18.75 C(DE2)4 850g 59 14.41 A(DE2)2 835.78g 29 28.82 CB3 129.69g 9 14.41

I did this a bit too late at night to guarantee all the math (or even all the English); if you find any mistakes, let me know.
Rick Garlikov (Rick@Garlikov.com)

 This work is available here free, so that those who cannot afford it can still have access to it, and so that no one has to pay before they read something that might not be what they really are seeking.  But if you find it meaningful and helpful and would like to contribute whatever easily affordable amount you feel it is worth, please do do.  I will appreciate it. The button to the right will take you to PayPal where you can make any size donation (of 25 cents or more) you wish, using either your PayPal account or a credit card without a PayPal account.