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Here's a video discussion of moles (10 min): Khan Academy: The Mole and Avogadro's Number, on YouTube

This video discussion of formulas (empirical and molecular) and formula weights may also be useful (15 min): Khan Academy: Molecular and Empirical Formulas, on YouTube

Here's another video discussion of chemical equations and mole calculations (13 min): CrashCourse Chemistry, Stoichiometry: Chemistry for Massive Creatures on YouTube

Moles

Moles are a convenient unit used in chemistry to convert between amounts of a substance in grams and numbers of atoms or molecules. This is useful because we usually measure how much of a molecule is used or produced in a reaction by massing it, but as a chemical equation shows, the reaction will happen between atoms or molecules. For example, suppose we combine 1.0g of calcium oxide (CaO) with 1.0g of water (H2O). The product we get is Ca(OH)2. Here's the equation:
CaO(s) + H2O(l) → Ca(OH)2(s or aq)
This is balanced. Thus every molecule of water reacts with one CaO formula unit (it's not called a molecule because it's an ionic solid, and each Ca2+ ion is surrounded with oxide ions that it interacts with equally). How much calcium hydroxide is produced by this reaction? Once all of one reactant has been used, whatever is left of the other will stop reacting, because of the law of definite proportions: we won't change the ratio of O:H:Ca in the product. So will we get solid calcium hydroxide with calcium oxide left over, or will we have water left over, and thus get Ca(OH)2(aq)? To answer this question, we can convert both masses (1g of each) to the number of molecules (of water) or formula units (CaO, because this is ionic, it is not made of molecules), but this would be inconvenient because the number would be very very big!

Instead, we use moles. A mole (abbreviation: mol) is like a pair or a dozen in that it is a counted unit. You can have a pair of people, a pair of apples, whatever. A mole is 6.02 x 1023 of something. This is a convenient quantity because it converts amu (atomic mass units) to grams. The atomic weight of carbon is (on average) 12.011 amu/atom. The molar mass of carbon is also 12.011g/mol. In other words, 1g = 6.02 x 1023 amu. Usually, a mol of a substance is a useful, practical amount, somewhere between a few grams and a few kg. So the way to answer the question above is to convert both quantities to moles. The maximum amount of product that can be formed is the smaller number of moles.

First, we need to find the molar mass of the molecule or formula, which we will use as a conversion factor. To do this we sum the atomic masses for all the atoms in the molecule or formula. For water we get roughly 16 + 2*1 = 18 g/mol of water. Do the same for CaO, and we can find the moles of each. Note that in 1 mol of water, there are 1 mol of O atoms, and 2 mol of H atoms.


1.0 g CaO1 mol CaO = 17.8 mmol CaO
56.08 g CaO
1.0 g H2O1 mol H2O = 55.5 mmol H2O
18.01 g H2O
After we make 17.8 mmol (milimoles) of Ca(OH)2, we will use up all the CaO, so the reaction won't continue. The maximum amount of Ca(OH)2 possible to make is 17.8 mmol. If we wanted to know the theoretical yield (maximum mass of product) of Ca(OH)2, we could do it in a one-step calculation like this:
1.0 g CaO1 mol CaO1 mol Ca(OH)274.09 g Ca(OH)2 = 1.3 g Ca(OH)2
56.08 g CaO1 mol CaO1 mol Ca(OH)2
Here, we knew that the limiting reactant (or limiting reagent), which is the reactant that will run out first, is CaO because the masses are the same, the coefficients in the equation are the same, and the formula weight of CaO is bigger than the molecular weight of water. So we start with the limiting reactant mass, convert it to moles (using 1 mol = 56.08 g), then "convert" between moles of CaO and moles of Ca(OH)2 using the coefficients from the balanced equation (1 mol of CaO produces 1 mol of Ca(OH)2), then we convert to g of Ca(OH)2 (using 1 mol = 74.09 g). This is just an example of using dimensional analysis to convert units. We check to make sure we have always multiplied by 1 (because 1 mol CaO = 56g CaO, so (1 mol CaO/56 g CaO)=1), and that the units cancel out to leave the correct final units (g Ca(OH)2), and we can be pretty sure that we got it right.

Here's a slightly more complicated example. This time, we add 2.0g of water to 2.5g of Li2O. This will produce LiOH as the major product. What is the most LiOH (in g) that could be produced, also called the theoretical yield?

To answer, first we need to write and balance the chemical equation. It's going to look pretty similar to the previous one, because this is a similar reaction.
Li2O(s) + H2O(l) → 2LiOH(s or aq)
Which is the limiting reactant? The formula weights are 18.01 g and 29.88g. Water is still in excess, which means it will be left over. Here's the unit conversion:

2.5 g Li2O1 mol Li2O2 mol LiOH23.95 g LiOH = 4.0 g LiOH2
29.88 g Li2O1 mol Li2O1 mol LiOH
Notice that this time we used (1 mol Li2O produces 2 mol LiOH), using the coefficients from the balanced equation.

Avogadro's Number

The number of things in a mole is 6.02 x 1023, which is called Avogadro's number, and abbreviated NA. It is named after Avogadro, the scientist who proposed that a liter of any gas at the same temperature and pressure has the same number of molecules in it. To summarize:
1 mole of [thing] = NA things = 6.02 x 1023 things

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Last modified: Tue Mar 11 12:52:24 KST 2014