Lesson: Chapter - 7

The Mole

In the last chapter, we reviewed the process of balancing equations and based the rules for balancing equations on the principle that matter is neither created nor destroyed in the course of a chemical reaction. With this idea still in mind, let’s begin our discussion of moles and formula weights.

When you look at the periodic table, you see that one of the pieces of data given for each element is its atomic weight. But what exactly is the atomic weight of a substance? It is the mass of one mole of a substance. In turn, one mole of a substance is equal to 6.02 × 10²³ atoms or molecules of the substance (depending on what it is), and finally, the number 6.02 × 10²³ is known as Avogadro’s number. For example, carbon’s atomic weight is roughly 12 amu; this means that 6.02 × 10²³ carbon atoms, in a pile, weigh 12 grams.

In order to find the formula weight of a substance, you simply add up the atomic masses of all of the atoms in the molecular formula of a compound. But don’t forget to multiply the atomic mass of each element by the subscript behind that element. Formula weights have the units amu, or atomic mass units; for example, the formula weight of water, H₂O, is about 18 amu. (O = 16 amu plus 2 times H = 1 amu = 18 amu.) Similarly, the molar mass of a molecule is the mass (in grams) of 1 mol of a substance; so the molar mass of H₂O is also roughly 18.

Now try calculating some molar masses and formula weights on your own by filling in the following chart.

Example

Explanation

Three significant digits were used throughout, with the exception of molar masses, where two decimal places were used. But don’t stress over significant figures for this test: it’s multiple choice, and the answers will never be that precise. Here’s the table, filled in.

Now that you’ve had some practice figuring out molecular weights, let’s talk about how you’ll be expected to use them, and other stoichiometric tools, on the exam. For example, you will almost certainly be asked to find the percent composition of a compound, so let’s talk about that first.

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