Mol's, Equations, and Popcorn....Oh My!

Before I get into the chemistry aspect of this post, let me begin with (quite possibly) the most important statement of this entire post: I love popcorn.  Over the past 2(?) weeks since school started back up, we've been talking about some pretty interesting stuff, more specifically, mol's and how they relate to the measurement of reactants and products. (In other words, its time for a hypothetical discussion)

602,000,000,000,000,000,000,000-And why it matters
*Warning: Hypothetical Discussion Follows*

Lets say you work for a big chemical company (ie: DOW) and you need to mix up a batch of Sodium Benzoate (Na C6 H5 CO2) the first problem that you will encounter is how to you ensure that you get the proper ratio of atoms in the mixture. In a normal recipe one would typically use a cup of this ingredient and a teaspoon of another to gain that ratio. In chemistry though the ratios are specific to the number of molecules needed to complete the reaction. One unit of Sodium Benzoate contains one Na atom, 6 Carbon atoms, 5 hydrogen, and one molecule of CO2. Sadly you can't just measure out on atom of sodium to throw into the mix, so a old dead scientist (Jean Perrin) through a series of complicated math, came up with Avagadro's (attributed to Amedeo Avagadro) number. Avagadros number is approximately 6.02*10^23 or 602,000,000,000,000,000,000,000. What this number ends up meaning is that there are 602,000,000,000,000,000,000,000 atoms of the element in every mol of that element. Now the weight of a mol for each element is different (due to density and other fancy factors) but easily found on any periodic table. The atomic weight of an element is the weight of one mol of that element. Therefore when you mix up a batch of Sodium Benzoate you will mix 1 mol of sodium (22.990g), 6 mol's of Carbon (12.001g per mol), 5 mol's of Hydrogen (1.0097g per mol) and 1 mol of CO2 (44.009g). When you mix all of this together, due to the Law of Conversation of Mass, the total weight of the final mixture should be 144.0535g. Thereby allowing you to correctly calculate the amount of product needed to produce "X" grams of the solution, solving the second problem you would encounter: How much product do I need to purchase to produce the final chemical. By solving that issue you allow the people in accounting to allocate the funds needed to buy the products and by doing this they can calculate the price that they will charge for the final product, which ultimately means....YOU GET PAID (If you don't get why thats important...I would like to take a moment to welcome you to Earth!) And thats why 602,000,000,000,000,000,000,000 (six-hundred and two million billion, for the record) is important.

Equations: Why They Exist, and Why I Hate them

Welcome to Chem 111

Unfortunately for the non-mathmatically inclined, Chemistry involves equations. Today, these equations will involve going from precent of composition, Mass to Mol's and writing chemical equations. Equations like these are rather useful in chemistry. They can help you determine basically everything you could ever want to know about a mixture. In the above hypothetical situation, we use an equation to determine the ratio of grams of elements. To do this you determine the atomic weight of every element of the equation, multiply those numbers by the number of mol's needed of each element. To determine the ratio of the elements you divide each of the numbers by the smallest. This creates a gram to gram (is that a real term?) ratio. Which eventually tells you that for every gram of element X that you have you need to add 5 grams of element Y for the correct reaction.  Another way that equations are used in chemistry is to determine what a solution is (by way of empirical formulas) based on the percent composition of elements. Take the percent composition of 10.04% Carbon, 0.84% Hydrogen, and 89.12% Chlorine. To determine the empirical formula we are first going to assume that we have a 100g sample of the substance. By assuming this it makes it very easy to determine the grams (10.04% of 100g = 10.04g.) Once you have the number of grams of each element you divide the grams by the molecular  weight of the element, which gives you the number of moles of each element. This division gives you .846 mol of Carbon, .832 mol of Hydrogen, and 2.51 mol of Chlorine. The next step is to divide each of the number of mols by the smallest number of mols (.832 mol's of H) which will give you the (approximate(error of margin is always an issue to those super picky chem teachers)) ratio of elements. In this case the ratio is 1C : 1H : 3Cl, which translates into a molecular formula of CHCl3, otherwise known as Chloromethane or Methyl Chloride. 


Originally I planned to include data from the recent popcorn experiment we did, but since we just finished another much more interesting experiment I will put a post about that sometime soon, especially since I have a video of it! So, until then, in the words of Spock, "Live Long & Prosper."