# Basic Stoichiometry

Stoichiometry is a division of chemistry that involves proportions and relationships between reacting elements and compounds. Stoichiometry can take several forms. Three very common forms of stoichiometry are reaction stoichiometry, composition stoichiometry, and gas stoichiometry.

Pronounce stoichiometry as "stoy-kee-ah-met-tree," if you want to sound like you know what you are talking about, or "stoyk," if you want to sound like a real geek. Stoichiometry is just a five dollar idea dressed up in a fifty dollar name. You can compare the amounts of any materials in the same chemical equation using the formula weights and the coefficients of the materials in the equation.

The most common type of stoichiometry studied in introductory chemistry is reaction stoichiometry (more commonly known as balancing equations). This type of stoichiometry involves finding coefficients of elements in chemical equations. This provides a way to quantify (using a ratio) the relationship between elements in a chemical reaction. Most of the time, we assume that a chemical reaction goes to completion. This means that all the reactants are used up to make as much product as possible. Sometimes, however, a reaction only happens part way; it doesn't go to completion. In that case, you'll be given a percent yield, and will have to use that to figure out the ratio of elements that react. For most of our work here, we'll assume that the reactions go to completion.

## Balancing Equations

There are a few rules for balancing equations. First, the purpose of balancing an equation is to get the same ratio of all elements involved. This is in accordance witht the Law of Conservation of Energy, which says that matter (energy) is neither created nor destroyed. Therefore, we'd have to have the same relative amounts in the reactants as we do the products. As mentioned before, this relies on correcting the ratios between elements. Balancing equations is like balancing a scale-you have to have the same ratio on both sides of the yields arrow. Here is a basic skeleton to guide you in balancing equations.

1. Balance the metals.

2. Balance any uncommon elements.

3. Balance oxygen (O). Make sure that elements that you've already balanced are still balanced. If not, you may have to go back and double the ratio you started with.

4. Balance hydrogen (H). Make sure that all elements are balanced. If they are, you're done. If they are not, identify which ones are out of balance, balance them, and then go back and check that O and H are balanced as well.

We'll give you a couple of examples of how to do this.

### Balancing a simple equation:

**___ H _{2} + ___ O_{2} <---> ___ H_{2}O**

First, balance oxygen. We note that the ratio of oxygen is 2:1, so we know we have to double the compound on the right (1) in order to equalize the ratio. Now, our equation looks like this:

**___ H _{2} + _1_ O_{2} <---> _2_ H_{2}O**

We can see now that there are two oxygen on the left, and two on the right. The
coefficient of H_{2}O means that there are not just 2 H_{2}, but
also 2 O. So, our ratio of oxygen is now 2:2.

Next, we have to look at balancing hydrogen. We look at our hydrogen ratio and see that it is 2:4 (note that coefficients are multiplied by subscripts to discern the total number of molecules-in this case, the coefficient 2 is multiplied by the subscript 2 in order to get 4 hydrogen molecules). Now, there are two hydrogen molecules on the reactant side, and 4 on the product side. In order to even this out, we would put a coefficient of 2 in front of the hydrogen on the reactant side, which means we'd be multiplying 2 times 2 in order to get 4 on the reactant side as well.

Now the equation is completely balanced. There are 2 oxygen molecules on each side, and 4 hydrogen molecules on each side.

### Balancing an intermediate equation:

Now, let's try a slightly more complicated example. Here's the original equation:

**___ Zn + ___ HCl <---> ___ ZnCl _{2} + ___ H_{2}**

First, we balance Zn, because it is a metal! We look at the zinc ratio and see that it is 1:1. That's great, it means we don't have to do anything with it!

Next, we look at Cl because it is a nonmetal that is not O or H. The Cl ratio is 1:2, which means that we need to increase (double) the number of Cl molecules on the reactant side. To do this, we place a coefficient of 2 in front of HCl, so the new equation looks like this:

**Zn + _ 2_ HCl <---> ZnCl_{2} + ___ H_{2}**

Now, we just have to make sure the hydrogen molecules balance. We look at the reactants' side and see that there are two molecules of hydrogen. Then, we look at the products' side and see that hydrogen has a subscript of 2, which means that there are also two molecules of hydrogen in the products' side. Since this is balanced, we know that our final equation is:

**Zn + 2 HCl <---> ZnCl _{2} + H_{2}**

Notice that in this equation, if the coefficient of an element or compound is one (1), we do not write it in the final equation. This is because we assume that any element or compound without a written coefficient has a coefficient of one.

### Balancing an advanced equation:

Let's try one last problem-this one is fairly difficult, so we'll go through it slowly.

**___ Pb(NO _{3})_{2} + ___ AlCl_{3} <---> ___PbCl_{2}
+ ___Al(NO_{3})_{3}**

From the onset, we notice that there's something different about this equation than
previous equations. You'll see that polyatomic ions are written with parenthesis.
This is because many polyatomic ions have subscripts naturally, like the one above
(NO_{3}). However, when balancing equations, we only look at the number
of ions needed to balance on each side of the equation. Therefore, instead of balancing
the nitrogen (N) in NO_{3}, and then balancing the oxygen (O) separately,
we balance them together, as one ion (NO_{3}). Sometimes, in order to be
in agreement with the compound, you'll see another subscript on the outside of the
parentheses, which indicates that more than one ion was needed to make the compound
neutral. If you need to change the amount of an ion, you'll still work with the
coefficients to get the equation to balance out.

Now that we've recalled the rule about polyatomic ions, we can begin to balance our equation. First, we balance Pb, because it is a metal. We look and see that the Pb ratio is 1:1, which means that we do not need to add any coefficients. Our equation still looks like this:

**Pb(NO _{3})_{2} + AlCl_{3} <---> PbCl_{2} + Al(NO_{3})_{3}**

Next, we'd look at Al, our second metal. (Please note that there is not a specific order in which to balance when you have more than one metal; however, we find it in best practice to start with the first metal listed in the reactants and then move on to the second metal listed, and so on). After looking at the Al ratio, which we find to also be 1:1, we note that this element is also balanced, so no action needs to be taken with it. So, our equation still looks like this:

**Pb(NO _{3})_{2} + AlCl_{3} <---> PbCl_{2} + Al(NO_{3})_{3}**

Now, we begin to look at the anions. Again, there is no specific order in which
you should balance them, but best practice says start with the anion of the first
compound in the reactants. In this problem, that anion is NO_{3}. Notice
that there are two molecules of NO_{3} on the reactants' side and three
molecules of NO_{3} on the products' side. This is not going to be resolved
by simply doubling one side, because 2 doubled is 4, and 3 doubled is 6, neither
of which give us the same number on each side. However, if we list the multiples
of 2 and 3, we find that the least common multiple between 2 and 3 is 6. Therefore,
we need each side to have 6 molecules of NO_{3}. In order to do this, we
need to find a number for each side of the equation that we can multiply by the
number that's already there in order to get 6. In other words, the reactants' side
has 2. What can we multiply by 2 to get 6? We know the answer to that question is
3. Therefore, our coefficient of NO_{3} on the reactants' side is 3. However,
we cannot just put the 3 in front of NO_{3}, we need it to be in front of
the compound containing NO_{3}. So, we're going to put the coefficient 3
in front of the Pb, like this: 3 Pb(NO_{3})_{2}. We do the same
thing with the products' side. We see that the subscript is 3, so we know we need
to coefficient to be 2 in order to produce 6 NO_{3} molecules to balance
both sides. Again, we put the coefficient in front of the compound, like this: 2
Al(NO_{3})_{3}. Our total equation so far looks like this:

**3 Pb(NO _{3})_{2} + AlCl_{3} <---> PbCl_{2} + 2 Al(NO_{3})_{3}**

Now, your NO_{3} ratio is 6:6. Perfect! But, don't forget to go back and
make sure everything you've balanced previously is still balanced. When we look
back, we see that now our Pb ratio is 3:1, which is unbalanced. We need to make
the ratio 3:3, so we look at the PbCl_{2} on the products' side. The coefficient
is 1, but we need it to be 3 to match the reactants' side, so we change it from
1 to 3. Now, we have 3 PbCl_{2} on the products' side. Our equation right
now looks like this:

**3 Pb(NO _{3})_{2} + AlCl_{3} <---> 3 PbCl_{2} + 2
Al(NO_{3})_{3}**

Now, Pb and NO_{3} are both balanced, so we should go back and check Al.
We look at the reactants' side and see that the coefficient on Al is 1. We look
at the products' side and see that the coefficient of Al is 2. To balance this,
we simply change the coefficient on the reactants' side to 2. Now, the equation
looks like this:

**3 Pb(NO _{3})_{2} + 2 AlCl_{3} <---> 3 PbCl_{2} +
2 Al(NO_{3})_{3}**

We have now balanced Pb, NO_{3}, and Al. Last, we need to balance Cl. We
look at Cl and see that the ratio is 6:6 (the reactants side has a coefficient of
2 and a subscript of 3, which multiply to give us 6. The products' side has a coefficient
of 3, and a subscript of 2, which also multiply to give us 6). Therefore, we don't
need to do anything to balance the Cl.

That brings us to the end of balancing the reaction! Our final equation looks like this:

**3 Pb(NO _{3})_{2} + 2 AlCl_{3} <---> 3 PbCl_{2} +
2 Al(NO_{3})_{3}**

Lastly, let's consider the equation for the Haber reaction, the combination of nitrogen gas and hydrogen gas to make ammonia.

**N _{2} + 3 H_{2} —> 2 NH_{3}**

The formula for nitrogen is N2 and the formula for hydrogen is H2. They are both diatomic gases. The formula for ammonia is NH3. The balanced equation requires one nitrogen molecule and three hydrogen molecules to make two ammonia molecules, meaning that one nitrogen molecule reacts with three hydrogen molecules to make two ammonia molecules or one MOL of nitrogen and three MOLS of hydrogen make two MOLS of ammonia. Now we are getting somewhere. The real way we measure amounts is by weight (actually, mass), so 28 grams (14 g/mol times two atoms of nitrogen per molecule) of nitrogen and 6 grams of hydrogen (1 g/mol times two atoms of hydrogen per molecule times three mols) make 34 grams of ammonia. Notice that no mass is lost or gained, since the formula weight for ammonia is 17 (one nitrogen at 14 and three hydrogens at one g/mol) and there are two mols of ammonia made. Once you have the mass proportions, any mass-mass stoichiometry can be done by good old proportionation. What is the likelihood you will get just a simple mass-mass stoich problem on your test? You should live so long. Well, you should get ONE.

Rather than thinking in terms of proportions, think in mols and mol ratios, a much more general and therefore more useful type of thinking. A mol ratio is just the ratio of one material in a chemical equation to another material in the same equation. The mol ratio uses the coefficients of the materials as they appear in the balanced chemical equation. What is the mol ratio of hydrogen to ammonia in the Haber equation? 3 mols of hydrogen to 2 mols of ammonia. Easy. In the standard stoichiometry calculations you should know, ALL ROADS LEAD TO MOLS. You can change any amount of any measurement of any material in the same equation with any other material in any measurement in the same equation. That is powerful. The setup is similar to Dimensional Analysis, and the calculations can include portions of DA.

1. Start with what you know (GIVEN), expressing it as a fraction.

2. Use definitions or other information to change what you know to mols of that material.

3. Use the mol ratio to exchange mols of the material given to the mols of material you want to find.

4. Change the mols of material you are finding to whatever other measurement you need.

How many grams of ammonia can you make with 25 grams of hydrogen? (Practice your mol math rather than doing this by proportion. Check it by proportion in problems that permit it.)

You are given the mass of 25 grams of hydrogen. Start there.

25g H_{2}/1 Change to mols of hydrogen by the formula weight of hydrogen 1 mol of H_{2} = 2.0 g. (The 2.0 g goes in the denominator to cancel with the gram units in the material given.) Change mols of hydrogen to mols of ammonia by the mol ratio. 3 mols of hydrogen = 2 mols of ammonia. (The mols of hydrogen go in the denominator to cancel with the mols of hydrogen. You are now in units of mols of ammonia.) Convert the mols of ammonia to grams of ammonia by the formula weight of ammonia, 1 mol of ammonia = 17g. (Now the mols go in the denominator to cancel with the mols of ammonia.) Cancel the units as you go.

The math on the calculator should be the last thing you do. 2 5 ÷ 2 . 0 x 2 ÷ 3 x 1 7 = and the number you get (141.66667) will be a number of grams of ammonia as the units in your calculations show. Round it to the number of significant digits your instructor requires (often three sig. figs.) and put into scientific notation if required. Most professors suggest that scientific notation be used if the answer is over one thousand or less than a thousandth. The answer is 142 grams of ammonia.

The calculator technique in the preceding paragraph illustrates a straightforward way to do the math. If you include all the numbers in order as they appear, you will have less chance of making an error. Many times students have been observed gathering all the numbers in the numerator, gathering all the numbers in the denominator, presenting a new fraction of the collected numbers, and then doing the division to find an answer. While this method is not wrong, the extra handling of the numbers has seen to produce many more errors.

This example starts at "mass given" and goes through the mol ratio to "mass find."

Notice by the chart above we may get the number of mols of material given if we change the mass by the formula weight, but in our continuous running math problem, we don't have to stop and calculate a number of mols. Students who insist on doing problems piecemeal tend to get more calculator errors.

The more traditional formula for converting mols to mass would be, where Fw is the formula weight, m is the mass, and n is the number of mols: n x Fw = m. You should be able to "see" these formula relationships on the roadmap.