Moles of Information!

See Mr. Ludwig?  I pay attention to your lecture slides!

The beginning of this quarter had a lot to do with moles.  Moles are the system used for measuring elements.  Each and every element on the periodic table has a different weight (in grams) to equal one mole.  One mole is 6.02x10^23, 6 billion trillion, Amadeo Avogadro's number.  That's how many molecules it takes an element to equal a mole.  In any other form but molecules that number is so large it would swamp us.  But when dealing with elements, it's perfect for making any measurement or equation we would need to solve a problems like how much of an element there is in a mixture and such.
But how can we tell how much of an element we need in order to have 1 mole of that element?  It's simple, the number is on the periodic table with the atom!  It's the decimal number by the abbreviation.  For example, the grams of Iron it takes to have 1 mole of iron is 55.845g.  For oxygen it's 15.999g.  Easy, and readily available through the work of other people! Yay!!

Mole Equations

Now we know what moles are and how to figure out how many g/mole are in a single element.  But what about compound elements?
Let's take a look at H2O.
That's 2 hydrogens for every oxygen.  So, to get the g/mole of H2O we simply add all of the elements together.
2 moles of hydrogen x 1.0079 + 1 mole of oxygen x 15.999 = 18.015 g/mole of H2O.
Alternately, you can find out how many moles are in, let's say, 262g of H2O.  Just take the H2O and calculate it to g/mole.  Then, we'll divide the 262 grams of H2O we have by that base number of 18.015.  Doing so gives us about 14.543 moles in 262 grams of H2O.  Easy!
Let's start with a number of moles now.  How  many grams of H2O are in 23 moles?  Again, you would add all the elements in H2O together to get a base number to multiply by.  Then we take that number and multiply it by the moles.
23 x 18.015 = 414.385 grams in 23 moles of H2O
I use H2O as my example because we used this compound a lot in our equations and calculations in class.

Believe it or not, this gets a little more complicated.


Not sure why I'd ever need this but through moles we can also find out how many atoms are in grams of an element.
Let's do this with 35.4g of Cu.
First, we've got to get out of the g part of the equation.
So we'll set it up in a way that we can do that.
35.4g Cu | 1 mole Cu | 6.02 x 10^23 atoms Cu
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               | 63.4g Cu   |        1 mole Cu
In English we're multiplying 35.4 by 1/63.4 by 6.02 x 10^23/1 to get 3.4 x 10^23 atoms Cu.  This way, we cancel out the unites of moles and the units of grams to end up with just the units of atoms.  And I use Mr. Ludwigs example for this equation because I always, without fail, mess up powers of 10 calculations, somehow.  I could probably do this exact same equation perfectly with x and pi instead of actual numbers but the basic algebra and math has been trained out of me by upper level mathematics.  But that's high school for you.  Pass until you fail.

There are three formulas for finding the atoms in a compound or in grams.  Someone kill me if I ever have to care about how many atoms are in 23g of vanadium.  Or any other element for that matter.  (Haha matter.)
The Empirical Formula calculates the smallest whole number ratio of atoms.
Ionic formulas are Empirical.
The Molecular Formula gives the number of each kind of atom in one molecule of a compound.
I think I did some of these in my notebook.  I'm looking at a bunch of scribbles now that look like they could be this.  Wanna see my test?  Well you can't because my cat ripped it to shreds.  With absolutely no encouragement from me.
I'm a reader, dudes, not a counter!!!