# Mass relations in chemistry and stoichiometry
## The Mole Concept

To begin with chemical relations to mass, perhaps the most important idea to understand is the mole concept. We begin by looking at an example from the periodic table:
* The symbols are simple enough:
* the 6 located at the top-center portion of the box is the atomic number for the corresponding element. As a basic rule for the periodic table, atomic numbers are always represented as whole numbers.
* the C situated in the center is the element's own specific symbol; in this case, Carbon.
* the number 12.0107 is what is known as the element's atomic mass (also commonly known as the molecular or molar mass).
The atomic mass is our primary concern here. By definition, a mole (symbolized as mol) is the quantitative amount of any given substance that contains as many atoms, molecules, or formula units as the number of atoms in exactly 12 g of carbon-12. This number is known as Avogadro's number (usually symbolized as N$_A$), and has recently been given the value of 6.0221367 × 10$^{23}$.
In this sense then, we might equate the term mole to dozen. With previous experience, we know that in a carton of a dozen eggs, you get twelve individual eggs. The same is said with the mole: for every 1 mole of a given substance, you have approximately 6.022 × 10$^{23}$ atoms/formula units/molecules in the sample amount. As you will learn later, the mole is very important in many calculations involving concentrations of substances, osmotic pressure, chemical kinetics, and equilibrium.
The main idea that must be understood is that for 1 mole of any substance, we have Avogadro's number of molecules, atoms, or formula units. Consider the chemical formula for water, H$_2$O. If we have one mole of H$_2$O, then we have approximately 6.022 × 10$^{23}$ different H$_2$O molecules. To continue further, we notice that there are two hydrogen atoms for every one oxygen atom in one formula unit of water, therefore we can say that there are approximately 1 × 6.022 × 10$^{23}$ oxygen atoms and 2 × 6.022 × 10$^{23}$ (1.204 × 10$^{24}$) hydrogen atoms present in one mole of water.
So how do we calculate the molar mass of H$_2$O? Simple. This is where the atomic mass is most important. On the periodic table we retrieve the atomic masses of hydrogen and oxygen, which are 1.00794 amu and 15.9994 amu respectively. (Note that "amu" stands for atomic mass units and has the units of $^g$/$_{mol}$). The atomic mass for a given substance is equal to one mole, and therefore contains roughly Avogadro's number of atoms/molecules/formula units. With this in mind, we add the following:
   2(Atomic Mass of Hydrogen) + Atomic Mass of Oxygen = Molecular Mass of H$_2$O 2(1.00794) + 15.9994 = 18.0152 $^g$/$_{mol}$
If you understood that, congratulations - you've completed your first stoichiometric calculation.

## Stoichiometry

Chemistry is a wonderful thing - and one of its least wonderful yet most important concepts is stoichiometry. Stoichiometry is the calculation of quantities of reactants and products involved in a chemical reaction. Too wordy? Let's make peanut butter and jelly sandwiches.
Think of how you make a PB&J sandwich - do you just slop the stuff together and hope it turns out right? Of course not! You take two pieces of bread, a tablespoon of peanut butter, and a table spoon of jelly, and there you have it. This is what stoichiometry is about: it's the recipe that counts, too.
So if we had an equation for a peanut butter and jelly sandwich, it might look like this: Pb + J + 2Bread → PbJBread$_2$
However, let's look at a real balanced chemical equation:
2C$_3$H$_8$ + 10O$_2$ → 6CO$_2$ + 8H$_2$O
What are those numbers in front of the compounds? In chemistry, we call these numbers coefficients which serve to balance out the amount of each individual type of atom on both sides (for example, there must be the same number of carbon atoms on the left side as there are on the right side of the equation).
So back to our equation:
2C$_3$H$_8$ + 10O$_2$ → 6CO$_2$ + 8H$_2$O
The easiest way to read it would be: "When you have two propane molecules (C$_3$H$_8$) and ten molecules of diatomic oxygen (O$_2$), you get six molecules of carbon dioxide (CO$_2$) and eight water molecules (H$_2$O)". Just like a recipe. Remember that in this scenario, molecules could possibly even be replaced with moles as well.
Now for an example. Let's take the same reaction with propane and oxygen: if you have 10 g of propane, how much carbon dioxide do you produce?
Believe it or not, this problem is very easy, and with some practice, you could become a professional at this. First, we look at the ratio of propane to carbon dioxide - which is 2 propanes for every 6 carbon dioxides, or in fraction: $^2$/$_6$ (do not reduce).
Let's start the calculation:
| 10 g C$_3$H$_8$ | × | 1 mol C$_3$H$_8$ | × | 6 mol CO$_2$ | × | 44.01 g | = | 29.99 g of CO$_2$ |
| 1 | × | 44.03 g | × | 2 mol C$_3$H$_8$ | × | 1 mol CO$_2$ | = | 29.99 g of CO$_2$ |
Notice the ratio of moles of propane to moles of carbon dioxide. This is called the key step. One might consider this part of the calculation a sort of "stepping stone" from one compound to the next.
Let's see how much H$_2$O we can get from 10 g of propane:
| 10 g C$_3$H$_8$ | × | 1 mol C$_3$H$_8$ | × | 8 mol H$_2$O | × | 18.02 g | = | 16.37 g of H$_2$O |
| 1 | × | 44.03 g | × | 2 mol C$_3$H$_8$ | × | 1 mol H$_2$O | = | 16.37 g of H$_2$O |
Now suppose we have 15 g of H$_2$O - how many grams of oxygen are there? Even though H$_2$O is a product, we can still work the problem the same way, but in reverse:
| 15 g H$_2$O | × | 1 mol H$_2$O | × | 10 mol O$_2$ | × | 32.00 g | = | 33.30 g of O$_2$ |
| 1 | × | 18.02 g | × | 8 mol H$_2$O | × | 1 mol O$_2$ | = | 33.30 g of O$_2$ |
   Note: you may have noticed that the molecular mass for O$_2$ is 32.00 rather than 15.999. There's a reason for this: gaseous oxygen is diatomic, meaning that in it's gaseous state, each oxygen molecule is comprised of two oxygen atoms that have been covalently bonded together.
   There are several different diatomic elements, and a simple way of memorizing them is through the acronym "Br I N C H O F" (pronounced "brink-hoff"), which represents the seven diatomic molecules.
   * Br$_2$ - Bromine
   * I$_2$ - Iodine
   * N$_2$ - Nitrogen
   * Cl$_2$ - Chlorine
   * H$_2$ - Hydrogen
   * O$_2$ - Oxygen
   * F$_2$ - Fluorine
From here, stoichiometry only gets more complicated (but not necessarily more difficult). As said before - with practice, these calculations can become second nature.