## High School Chemistry: Stoichiometry

Most chemistry problems I'm called on to assist with involve all the
information you can extract from a chemical equation and the atomic chart.
For instance, suppose we're told that we have a chemical reaction combining
500 grams each of
hydrogen and oxygen. We know the first half of the chemical equation:

H + O -> something

What is the something that it produces? HO, obviously, but in what proportions?
We check the periodic chart and find that hydrogen has a valence — meaning
a charge in chemical reactions, essentially — of +1 and oxygen
-2, so we need two hydrogens to balance the charge of oxygen. Thus we
get H_{2}O (water, but you probably knew that already). The equation
so far is thus:

H + O -> H_{2}O

Now there's a gotcha here. One of the facts of chemistry which just has to
be memorized is that some atoms are *diatomic*, which means they're so
reactive that they're more stable when they're combined with other atoms of
the same kind than when they're alone. The diatomic atoms
are hydrogen, oxygen, nitrogen, and all the atoms in the column starting
with fluorine: fluorine, chlorine, bromine, iodine, astatine.
(In extreme conditions, other atoms are also diatomic. Certain metals when
heated to vapor are in this category. But under more or less standard conditions
of temperature and pressure, we don't have to worry about those.)

Both hydrogen and oxygen are in the diatomic list so our equation becomes:

H_{2} + O_{2} -> H_{2}O

If a chemist looked at that equation, he'd notice something alarming: There
are two oxygen atoms on the left side, but only one on the right side. What
happened to the other oxygen atom? It is an article of faith to a chemist
that atoms don't just disappear like that. (In the 20th century the physicists
discovered that the situation isn't quite so simple, but we don't
need to worry about nuclear reactions here. Oxygens don't disappear in chemical
reactions.) If we start with two oxygen atoms, we have to finish with two.
That's easily achieved:

H_{2} + O_{2} -> 2H_{2}O

Unfortunately, that's just as bad. We started with two hydrogen atoms and
now we finish with four. So we make another adjustment:

2H_{2} + O_{2} -> 2H_{2}O

Now we're ok. We start and end with 4 hydrogens and 2 oxygens. We have
neither created nor destroyed atoms; we just moved them around. The process
of ensuring the equation matches that reality is called *balancing* the
chemical equation. Every equation in chemistry must be balanced to be able
to use it as we're about to here.

You remember that at the beginning, I specified that we have 500 grams of each of the hydrogen and oxygen. Does that mean we have equal numbers of hydrogen and oxygen atoms? Well, if I told you that I have 5000 pounds of horses and 5000 pounds of rabbits, would I have equal numbers? Certainly not; horses and rabbits don't weigh the same. If we knew that every horse weighs, to pick a convenient number, 1000 pounds, and every rabbit weighs 5 pounds, we could easily figure that our 5000 pounds of horse is 5 horses, and our 5000 pounds of rabbit is 1000 rabbits.

Part of the wealth of information on the periodic chart will give us the same information about atoms. It says that an atom of hydrogen weighs 1 unit and an atom of oxygen weighs 16 units. So we have one sixteenth as many oxygen atoms as hydrogen atoms in our 500 grams.

In real life, we don't deal with individual atoms. At our gross level of
existence, we have only much larger aggregates. The unit that chemists use
is the *mole*. The atomic weight in the periodic table is the number of
grams per mole for each atom. So a mole of hydrogen weighs one gram, and a
mole of oxygen weighs 16 grams. And the key to the usefulness of the mole is
that every mole of chemical compound X has the same number of molecules
as a mole of another chemical compound Y.
Thus our 500 grams of hydrogen is 500 moles,
and the 500 grams of oxygen is 31.25 (= 500/16) moles. And to refine this a
little further, since we don't have individual hydrogen atoms but molecules
of two hydrogen atoms, and the same for oxygen, we have 250 moles of diatomic
hydrogen and 15.625 moles of diatomic oxygen.

Now that's interesting. From our chemical equation

2H_{2} + O_{2} -> 2H_{2}O

we can see that for each molecule of O_{2} we need two molecules of
H_{2}. And the same is true of moles, since every mole has the same
number of atoms or molecules: For each mole of O_{2} we need two moles
of H_{2}.

Now to put that together with our calculation that we have 250 moles of
H_{2} and 15.625 moles of O_{2}. We need twice as much
H_{2} as O_{2}. Since we have 15.625 moles of O_{2}
we need 31.25 moles of H_{2}. We actually have 250 moles of
H_{2}. The excess, 218.75 (= 250 - 31.25) moles, cannot be used,
because we run out of oxygen to combine it with.

Now we can figure out how much water we'll finish with. Again referring to
the chemical equation for this reaction, we see that for every 2 moles of
H_{2} used, we get 2 moles of water. (Equivalently, for every mole of
O_{2} we get 2 moles of water.) We'll use 31.25 moles of H_{2}
so we'll
get 31.25 moles of water. How much does that weigh? Simple: Just add the
weights of the constituent atoms, 2 hydrogens and an oxygen. That's 18,
meaning 18 grams per mole of water, so the total weight of water is 562.5
(= 18 * 31.25) grams.

We can get the same answer in another way. Since we neither create nor destroy
matter in chemical reactions, the weight of water produced must be the same as
the weight of hydrogen used plus the weight of oxygen used. 31.25 moles of
H_{2} weigh 62.5 grams, 15.625 moles of O_{2} weigh 500 grams,
for a total weight of 562.5 grams.

It just keeps going from here. Suppose you're asked the volume of the
constituents at the start and finish? You'll learn that a gas at standard
temperature and pressure (which is abbreviated *STP*) occupies 22.4 liters
per mole. Then the 250 moles
of H_{2} that we started with will occupy 5600 liters at STP, and the
15.625 moles of O_{2} will occupy 350 liters at STP. After the reaction
has completed (which, incidentally, will cause quite a large bang as the
hydrogen explodes) all the oxygen will be used but we calculated that we'll
have 218.75 moles of H_{2} left over, which will occupy 4900 liters
at STP.

How about the
water? We can't do the same calculation for water because it isn't a gas at
STP. (Standard temperature and pressure is basically room temperature and
atmospheric pressure, and of course water is liquid at room temperature and
atmospheric pressure.) For the volume of water we'll have to use the
*density*, which is defined to be the mass per volume. Water's easy:
It's one gram per cubic centimeter at a standard reference temperature, which
is the same as one gram per
milliliter. Since we have 562.5 grams, that's 562.5 milliliters of water at
the reference temperature.

You might object that the temperature of the gasses and the resulting water
is not likely to be at the standard temperature, so the actual volumes
are going to be somewhat different from our calculations. But you'd be sorry
if you made that
objection, because the next thing the teacher will do is to say "Ok wise guy,
calculate the volume given the temperature at the end is, say, 90 degrees
Celsius." For the gasses, you'd use
the *ideal gas law*, which is PV = nRT, where

P is the pressure

V is the volume

n is the number of moles of the substance

R is the ideal gas constant (.082 liter * atmosphere per mole per degree Kelvin)

T is the temperature in degrees Kelvin

You can go on to do the calculations if you like,
and you *will* go on to do them in your high-school chemistry
class. And there's a similar rule for finding the (small) change in volume
of the water. And not only that, but if you know the enthalpy change for the
reaction and the starting temperature of the gasses, you can calculate the final
temperature.

### Very theoretical

How about an application, then? With human-induced climate change all the rage (literally), let's calculate the amount of CO_{2}that a vehicle will produce in its lifetime. I'll take my motorcycle with 134,000 miles and fuel economy around 34mpg for my test case.

In 134,000 miles at 34mpg, I'll have used about 3941 gallons of gasoline. The makeup of gasoline varies considerably, with hydrocarbon molecules containing anywhere from 4 to 12 carbon atoms, but I'll assume the average is 8 carbons per molecule (which would be octane). Octane has the form of a chain of carbon atoms attached by single bonds. Since carbon has a valence of 4, the six inner carbons of the chain have two extra slots, each occupied by a hydrogen atom, and the two carbons on the ends of the chain have three extra slots, again occupied by hydrogens. The total is 18 hydrogens. The balanced equation will be

2C

_{8}H

_{18}+ 25O

_{2}-> 16CO

_{2}+ 18H

_{2}O

And from that we observe that for every mole of octane (with 8 carbons) we'll get 8 moles of CO

_{2}(with one carbon atom).

First we convert gallons to moles. A gallon of gasoline weighs about 6 pounds, so my 3941 gallons will be about 23647 pounds. A kilogram is about 2.2 pounds, so that's 10749 kilograms, or 10748663 grams. Octane has a molecular weight of 8*12 (for the 8 carbons) + 18*1 (18 hydrogens) = 114 grams per mole. So 10748663 grams is 94287 moles. And since we get 8 moles of CO

_{2}when we burn a mole of octane, we'll produce 754292 moles of CO

_{2}.

Now to go from moles to weight of CO

_{2}: The molecular weight of CO

_{2}is 44 grams per mole, so that's 33188854 grams, 33189 kilograms, 73015 pounds, about 37 tons of CO

_{2}.

Having obtained this result, we can scale it to fit other vehicles without doing all the intermediate arithmetic. For example, if your big pickup gets 11mpg rather than the 34 of my motorcycle, then in 134000 miles you'll produce 34/11 times as much CO

_{2}, about 113 tons. If you drive it 200,000 miles rather than my 134,000, you'll produce 200,000/134,000 times as much CO

_{2}, about 168 tons.