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Introduction

This series of posts seeks to present the material covered in the first semester of a college level general chemistry course, in an easily digestible steemit blog post format. The series is intended to be read, and experienced in sequential order starting with Post 1. The material will build upon itself, and potential exercises included (problem sets), will pertain to the post they are contained in, or any previous post. Each post will pick up immediately where the previous in the series left off. Please check out the #chemistry-lesson tag for all posts in this series. I hope you find this series to be informative and beneficial toward your understanding of chemistry and science in general.

Immediate Preceding Post

Part 3: The Periodic Table and Nomenclature for Molecular and Ionic Compounds

Legend for This Section

As subscripts do not currently work on steemit, to symbolize when a number should be sub-scripted I will be writing it in the text as follows: /x/.

Avogadro’s Number and Molecular Mass

Avogadro’s Number

The mass of atoms is calculated in terms of atomic mass units (amu). One amu has been defined as 1/12 the mass of one carbon-12 atom (so 1 amu is equal to 1.660539*10^-24 g). The masses recorded on the periodic table are in amu, so we know precisely what values to use when calculating the masses of atoms of compounds.
When chemists do calculations with masses of compounds we usually refer to things in terms of number of Moles (mol is how the unit is written). Moles (no... not the animal) are the amount of a substance which contains the same number of units as are found in 12 g of carbon-12. The number of atoms found in 12 g of carbon-12 is 6.022 x 10²³. That number is referred to as Avogadro’s Number in honor of a scientist by the same surname. If I asked you how many atoms does 1 mol of Oxygen have, you would tell me 6.022 x 10²³, because it doesn’t matter what thing you are talking about, one mol is always 6.022 x 10²³ of something. If I asked you to get me 1 mol of TI-89 calculators, you would not be able to because that would be 2.185 x 10²³ Kg. Which is 3.6% of the mass of the earth, and I don’t think there are that many Ti-89 calculators. Now that we know and understand what a mole is I can tell you another fun fact, 1 gram is equal to 6.022 x 10²³ amu. Which means in addition to being the mass of each element in amu, the atomic masses on the periodic table are equal to the masses of 1 mole of each element in grams.

Molecular Mass

Since we know the masses of every element thanks to the periodic table, we can also easily calculate the masses of compounds right? Let’s begin with an example. What is the mass of a single molecule of Ethanol (C/2/H/5/OH)? Well we simply multiply the masses of each element by the number present in the compound and sum the masses up:

24.02 amu + 6.048 amu + 16 amu = 46.068 amu in one C/2/H/5/OH

If I then asked you what is the mass of 1 mol of C/2/H/5/OH you could easily tell me that its 46.068 g, because we know that one mol of something weighs the same amount in grams as one molecule does in amu!

Determining Percent Composition

Knowing the formula for a compound is essential to understanding what that compound is, and it tells us exactly what ratios the elements should be in that compound in an “ideal world”. However the lab is far from the ideal world, and the compounds we use are most often not 100% pure. So how can we go about determining what is present? We do this by determining percent composition for both the ideal compound (based on its molecular formula) and compare it with the composition we calculate from our own sample. Percent composition is the percent each element’s mass makes up in a compound. For an example of how to calculate this, let’s turn again to our old friend ethano (C/2/H/5/OH):

What we did in the example above was first multiplied the mass of carbon (12.01 g, from the periodic table) by the number of carbons in the molecule, then divided by the total mass of C/2/H/5/OH, then multiplied that fraction by 100%. The same process was repeated for the 6 hydrogens and one oxygen. Now we know the theoretical percent composition for our compound ethanol.

Say we were mining silver (because we are ballers like that) and we obtained a nice 500g hunk of silver ore (argentite, Ag/2/S) and we wanted to know how much silver we mined. Well we could use percent composition to calculate this! Let’s do it:

We know that our ore is 87% silver so (0.87 * 500 g) = 435 g Silver, not bad.

Determining Empirical Formulas with Experimental Data

Remember how I said we would be seeing empirical Formulas again? Well, here is one of those times! We can utilize the percent compositions we just learned about to back calculate an empirical formula (empirical means verifiable by observation after all)! We have a compound it contains 40% C, 6.7% H and 53.3% O. What is the empirical formula for this compound? Well the easiest thing for us to do would be to assume we had 100 grams of this compound and figure out how many moles of each element are there:

We multiply each of the percent compositions by 100 g and then divide by the molecular mass of each element respectively to obtain the number of mols. We next divide all of the mole values we determined by the smallest one (in this case both C and O are 3.3 so we divide each number by 3.3) This gives us 1 C, 2 H and 1 O. These numbers constitute our empirical formula for this compound. CH/2/O. If we know the molar mass for the compound we can also convert this empirical formula into the actual molecular formula. This compound has a molar mass of 180.2 g/mol. Lets figure out its molecular formula.
We take the following steps, calculate the total mass of the empirical formula (1 x 12.01 (C) + 2 x 1.008 (H) + 1 x 16 (O)) which is 30.026, we then divide the molar mass by the empirical mass 180.2/30.026 = 6. Now we multiply the empirical mass by this ratio number so 6* CH/2/O = C/6/H12/O6. Which is the formula for glucose (and also what we were working with ).

Writing Out Chemical Reactions

So now let’s start to (FINALLY!!!) talk about chemical reactions! Chemical reactions are simply the processes by which one or more substances become one or more different substances. Let us begin with the following chemical reaction (combustion of butane C/4/H/10/):

A chemical reaction has two sides, on the left side of the arrow are the reactants, and on the right side of the arrow are the products. The plus sign means reacts with, so on the reactants side we have butane and oxygen reacting with each other. The arrow means yields, so the reaction of butane and oxygen (combustion) yields carbon dioxide gas and water vapor. This chemical reaction is also balanced, so that each atom on the reactants side is present on the products side. Balancing occurs by adding whole numbers in front of each molecule, as these are the pieces we have to work with in the reaction. Our balanced reaction indicates that to burn butane (as so many of us do whenever we use a lighter), 2 butane molecules are reacted with 13 oxygen molecules to yield 8 carbon dioxide molecules and 10 water vapor molecules. These are the lowest whole numbers that are possible for this reaction, so this is the balanced set.

Balancing a Chemical Reaction

When you do a chemical reaction in lab, you don’t usually immediately know the ratios of the things that reacted. It is also likely that you are going to have to do a significant amount of analysis to identify what the products are. After you have done that, you must write a balanced chemical reaction. I usually do this in a few steps:

1. Write out your reactants on the left side of your reaction arrow, and your products on the right side of the reaction arrow.
2. Begin assigning coefficients to any elements that occur in only one reactant and only one product, make the number of those elements equal.
3. Assign coefficients to elements that occur in more than one reactant and/or product.
   For this to make sense, it will be best to do an example. Let us look at the following reaction:

   

   We first wrote out our reactants and products for the reaction, next we balanced the sulfur there were 8 reactant sulfur atoms so there must be 8 in the product. We placed the coefficient of 8 in front of the SO/2/ product. Now oxygens were no longer balanced, there were 16 on the product side, so we placed the coefficient of 8 in front of the O2 on the reactant side to give us 16 on that side as well. Now the reaction is balanced. I will provide you a tougher one to try to work through in the problem set! Once we have established the correct coefficients (the numbers in front of the molecules) for the reaction, these ratios between the molecules of products and reactants are referred to as the stoichiometry of the reaction. I will go into more detail about this in the next installment!

   End of Calculating Atomic Mass and Writing Chemical Reactions Problem Set

   We’ve reached the end of this section, and as with the end for previous sections, I will provide you with a problem set so you could test yourself to see if you are fully understanding the material.

   Problem Set 3
   Problem Set 3 Answer Key

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   Future Posts

   Subsequent posts will cover: Chemical Reactions continued, Acids Bases and other Aqueous Reactions, Electronic Configuration of Atoms, Chemical Bonding, and Molecular Geometry, and more.
   
   

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   Reference Figure: Periodic Table

   

   Source

   Other References

   Constants and Conversions List
   Source for Additional Constants
   Some Common Ions

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