Reading Notes for Chapter 3

These are Dr. Bodwin's reading notes for Chapter 3 of "Chemistry 2e" from OpenStax. I am using a local .pdf copy that was downloaded in May 2020.

Chapter Summary:

This chapter builds upon Chapter 2 by developing the tools that we will use to make macroscopic observations of the microscopic world of atoms, ions, and molecules. Keep on reviewing chemical formulas and know your polyatomic ions - if you don't have those right, formula masses and moles are going to be much more challenging.

Formula Mass:

In many ways, formula mass is just another way of expressing a chemical formula, and because of that it is essential to have a correctly balanced chemical formula if you want to have a correct formula mass.
There are a few terms that are commonly used to describe this same idea: "formula mass", "formula weight", "molecular weight", "molecular mass", "molar mass"... there are some subtle differences in exactly what these terms mean, but they pretty much refer to the same idea.
Don't make this harder than it is. If we were doing this with grocery store items instead of atoms and molecules, you'd be able to figure it out without a second thought. Each type of atom in a formula has a mass listed on the Periodic Table. Add up the masses of all the pieces in the chemical formula, and you have the formula mass. Carbon dioxide has 1 carbon and 2 oxygens. A carbon has an atomic mass of 12.011amu and each oxygen has an atomic mass of 15.999amu. Carbon dioxide has a formula mass of (12.011+15.999+15.999) = 44.009amu.

The Mole:

Another example of "don't make it harder than it is". A mole is a grouping unit. When we're counting atoms in a macroscopic sample, the number of atoms is HUGE, so to make it more manageable to discuss that amount, we use a great big number that we call a "mole". One of my favorite cookies come in boxes of 36 cookies. That box is a grouping unit. When I go to the store, I don't think "hey, I should pick up 36 cookies", I think "hey, I should pick up a bot of cookies". Moles work the same way. I don't say "Here's 602214179000000000000000 sodium atoms", I say "Here's 1 mole of sodium atoms"
Back to formula mass... I used "atomic mass units" or "amu" in the example above. But that's not a really useful macroscopic mass unit. Here comes the mole... when we're looking at an atom, amu is a convenient mass unit; when we're macroscopically looking at a mole of some substance, grams makes more sense. Because some scientists were smart about the way they defined the mole, it turns out that the "mass" listed on the Periodic Table has units of amu when you're looking at atoms, but has units of grams when you're looking at moles of atoms. An average sodium atom has a mass of 22.990amu... a mole of sodium atoms has a mass of 22.990g!
Because we will almost always be making macroscopic observations of the microscopic world, we're going to be using moles a LOT. Again, a mole is just a grouping unit, so we can talk about moles of anything... atoms, ions, molecules...
And just to try and get a better idea of just how big a mole is...think of something small, like a blueberry. If we say a blueberry is 0.5cm in diameter and we pile them in a simple cubic packing arrangement (http://www.ntci.on.ca/chem/sch4u/crystalpacking.pdf), we would have a cube of blueberries a bit over 260 miles wide. That's a lot of blueberries.

Empirical Formulas and Molecular Formulas:

Percent composition and finding empirical formulas are the same process in opposite directions. Start with percent composition:
Sodium thiocyanate (NaSCN) has a formula mass of (22.990+32.066+12.011+14.007) = 81.074g/mol. Its percent composition is:

%Na = (22.990g/mol / 81.074g/mol)*100 = 28.357% Na
%S = (32.066g/mol / 81.074g/mol)*100 = 39.552% S
%C = (12.011g/mol / 81.074g/mol)*100 = 14.815% C
%N = (14.007g/mol / 81.074g/mol)*100 = 17.277% N

{NOTE: Those don't add up to exactly 100% because of rounding.}
OK, let's go the other way... start with percents:
A substance is found to be 36.112% calcium and 63.888% chlorine. What is its formula?

The problems almost always start with "assume you have 100g of sample" because it makes the math easier, but I think students sometimes get confused because the math was made easier! Let's assume that we have some amount of substance... let's try 9.37g. From the percent composition data, we know that if we have9.37g of substance, then (9.37*0.36112) = 3.38g of the sample is calcium and (9.37*0.63888) = 5.99g of the sample is chlorine.That means there are (3.38g Ca / 40.078g/mol) = 0.0843mols of Ca in the sample and (5.99g Cl / 35.453g/mol) = 0.169mols of Cl in the sample. We can take a ratio of those moles (with the smaller number on the bottom...) and simplify that ratio to see that this samplecontains 2 moles of chlorine for every mole of calcium. The empirical formula is CaCl ₂. {Which is what we would predict based upon expected charges from the Periodic Table.}

An "empirical formula" represents the smallest whole-number ratio of elements. For larger molecules, there are often multiple empirical formula units, so the "molecular formula" might be a multiple of the empirical formula. Without additional information, we cannot determine if the molecular formula contains more than one empirical formula unit.

Solutions and Concentration Units:

Solutions are homogeneous mixtures of 2 or more substances.
The major component of a solution is called the "solvent".
The minor component(s) is called the "solute(s)".
We use "concentration units" to describe the amount of solute relative to the amount of solvent or solution.
The most important concentration unit for us to use in this class is Molarity, moles of solute per liter of solution, designated with an "M".
That capital "M" is important! If you use a lower case "m", it means something completely different.
The other concentration units mentioned in this chapter show up in a number of different places and contexts, but we won't worry about them at this point. We will revisit solutions at the beginning of Gen Chem II.
When diluting a solution, you can just about always use C₁V₁=C₂V₂ to determine the new concentration. This simple formula is your friend when you use it for dilutions!

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