Stoichiometry 1

Do you ever find that you can’t get joy from something you sometimes get joy from? And when you do get joy from it, you feel this immense need to share it, but you deny that need and then the joy seems to go away?

It seems that unless I share what I learn each day, I don’t experience joy. It seems there’s no point to learning about the beauty of certain math things, or chemistry, astronomy, the universe, etc. It seems there’s no point to spending one’s time learning about rules and history and laws of the universe if there’s no one to share that knowledge with. At least, that is how I feel.

When I’m reading my chemistry book at my house, I’m not excited. When my teacher does chemistry at school, trying to engage the class, and it starts clicking in people’s heads—that’s the only time I really get excited.

To me, the most beautiful ideas in life don’t matter if I’m the only one that can appreciate them.

And for that reason I’m here now, and I’m about to share with you what I’m reading about. (Yay!)

In school we’re learning about stoichiometry right now. This chapter is all about conversions, where math is used a lot, but very simple math, and all of the same type. Stoichiometry can’t be understood well if other things aren’t understood, like chemical equations, balancing them, substances and how to write them, the meaning of coefficients and subscripts. I won’t go back to talk about all those things because I honestly do not have the time.

Sorry.

So, many stoichiometry problems are ones where there is a given and an unknown. Given refers to a mass (typically in grams) or an amount (like in moles) of a substance which is used in your mathematical conversions to figure out how much mass/ amount there is of another substance. Molar ratios and molar masses are used. More on those later.

There’s a linear map you might find it helpful to use, which allows you to see where you are, where you are headed, and how you are supposed to get there. It’s really simple and looks like this:

g –> mol –> mol –> g

This means that if your given is a substance in grams and you need to find the mass of another substance, you need to go from g to g, so you need to do 3 conversions. That’s how it works. Let me show you some examples which will clarify things, since examples are really good models. Simply putting it in words may make it seem more complicated than it is.

Example 1: How many moles of water are needed to react with 2.2 moles of Li2O?

Li2O + H2O –> 2LiOH

The equation is balanced, meaning there is the same number of atoms of each element on both sides. You have to use a balanced equation. To balance an equation, you must play around and change the coefficients (the regular-sized numbers in front of substances), NOT the subscripts (the tiny numbers following some elements). Changing coefficients changes the amounts of substances. H2O, for example, is a molecule; 2H2O means there are two H2O molecules. Change the coefficient, you change the number of molecules. The molecule itself has the unchanging structure of H2O, and changing that subscript will change the molecule, which you can’t do when balancing chemical equations. By the way, chemical equations represent chemical reactions.

Anyway, my known is the amount (in moles) of Li2OH. That “known” (switching to quotation marks here) is 2.2 moles of it. Don’t get confused by the fact this doesn’t match up with the number of mols there are of Li2OH in the chemical equation. We have the chemical equation in front of us simply to derive a molar ratio from it later.

Starting with 2.2 mols of Li2OH, we want to get to mols of water. This mols –> mols problem means there’s only one conversion step.

(2.2 mols Li2OH / 1) x (1 mol H2O / 1 mol Li2OH)

^That’s how I set it up. Mols Li2OH in the numerator of the fraction on the left cancel out with mols Li2OH in the denominator of the fraction on the right. The second fraction contains one of the molar ratios shown in the correctly balanced chemical equation. The molar ratio is shown in the coefficients. The fraction says that there is 1 mol of H2O alongside 1 mol of Li2OH, which is just as true as saying there is 1 mol of H2O alongside 2 mols of LiOH. Now you can see that, had the equation not been balanced, the molar ratio would not be correct.

(2.2 mols Li2OH / 1) x (1 mol H2O / 1 mol Li2OH)

We’re left with 2.2 mol H2O. This means that 2.2 mol H2O are needed to react with 2.2 mol Li2OH.

Now for a g –> g example problem.

Example 2: How many grams of CO2 will be produced if 39.6 grams of C2H5OH react with oxygen?

C2H5OH + 3O2 –> 2CO2 + 3H2O

(39.6 g C2H5OH / 1) x (1 mol C2H5OH / x g C2H5OH)

I make sure that the units (g in this case) of the given will cross out with the denominator of the fraction that the given is being multiplied by. In that way, we will convert, which is the purpose of these calculations. To get to mol of C2H5OH as opposed to g of C2H5OH, I had to multiply g of C2H5OH by a fraction with g C2H5OH in its denominator, so that they would cancel out and we no longer would be operating with g C2H5OH. So that is why I set up the multiplication as I did. However, instead of putting a number in the denominator of the second fraction, I put “x”. I did that because I’m going to show you how one figures out how many grams of a substance there are in a mol of that substance. In this case, that substance is C2H5OH. Well in a mole of C2H5OH, there is a mole of C2H5OH molecules. Each molecule has 2 C atoms, so there are 2 x 1 mol of C atoms. There are 2 mols of C atoms. There are 6 mols of H atoms. There is 1 mol of O atoms.

How do we find the mass of 1 mol of these molecules? We take the mass of 1 mol of C atoms and multiply it by 2 to get the mass of all the C atoms in 1 mol of C2H5OH.

(molar mass C x 2) + (molar mass H x 6) + (molar mass O x 1) = molar mass C2H5OH

In the periodic table of elements (that’s probably capitalized, my bad), C’s molar mass is rounded to 12.01. H’s molar mass is 1.01, and O’s molar mass is 16.00, so you write it like:

(12.01 x 2) + (1.01 x 6) + (16.00 x 1) = molar mass C2H5OH

24.02 + 6.06 + 16.00 = molar mass C2H5OH

46.08 = molar mass C2H5OH

Now we can change the “x” and perform this calculation:

(39.6 g C2H5OH / 1) x (1 mol C2H5OH / 46.08 g C2H5OH) =

(39.6 g C2H5OH / 1) x (1 mol C2H5OH / 46.08 g C2H5OH) =

.859375 mol C2H5OH

The reason that the number was not rounded is that my calculations are not done yet, and rounding it now would result in a less-accurate result than if I didn’t round it now, and rounded only at the end. We’re not done with converting, anyway—that was just step 1.

(.859375 mol C2H5OH / 1) x ( 2 mol CO2 / 1 mol C2H5OH) =

(.859375 mol C2H5OH / 1) x ( 2 mol CO2 / 1 mol C2H5OH) =

1.71875 mol CO2

Last step:

(1.71875 mol CO2 / 1) x ( 44.01 g CO2 / 1 mol CO2) =

(1.71875 mol CO2 / 1) x ( 44.01 g CO2 / 1 mol CO2) =

75.6 g CO2 is the theoretical yield; the amount of CO2 that should be produced when 39.6 g of C2H5OH react in the chemical reaction.

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