Jeremy and Steve at work


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 H2O (water, but you probably knew that already). The equation so far is thus:
H + O -> H2O

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:
H2 + O2 -> H2O

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:
H2 + O2 -> 2H2O
Unfortunately, that's just as bad. We started with two hydrogen atoms and now we finish with four. So we make another adjustment:
2H2 + O2 -> 2H2O
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
2H2 + O2 -> 2H2O
we can see that for each molecule of O2 we need two molecules of H2. And the same is true of moles, since every mole has the same number of atoms or molecules: For each mole of O2 we need two moles of H2.

Now to put that together with our calculation that we have 250 moles of H2 and 15.625 moles of O2. We need twice as much H2 as O2. Since we have 15.625 moles of O2 we need 31.25 moles of H2. We actually have 250 moles of H2. 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 H2 used, we get 2 moles of water. (Equivalently, for every mole of O2 we get 2 moles of water.) We'll use 31.25 moles of H2 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 H2 weigh 62.5 grams, 15.625 moles of O2 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 H2 that we started with will occupy 5600 liters at STP, and the 15.625 moles of O2 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 H2 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 CO2 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 equation will be
C8H18 + O2 -> CO2 + H2O
Now that equation isn't balanced, but all we need from it is to observe that for every mole of octane (with 8 carbons) we'll get 8 moles of CO2 (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 CO2 when we burn a mole of octane, we'll produce 754292 moles of CO2.

Now to go from moles to weight of CO2: The molecular weight of CO2 is 44 grams per mole, so that's 33188854 grams, 33189 kilograms, 73015 pounds, about 37 tons of CO2.

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 CO2, 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 CO2, about 168 tons.