# High School Chemistry/Chemical Quantities & the Mole

A mole (moL) is a standard SI unit for the amount of a substance. This "mole" unit is used to measure mass, volume, and number of particles. It represents very large numbers. The mole is represented by Avogrado's number, which is 6.02 x 10$^{23}$. This was named after Italian scientist Amedeo Avogadro for his helpful clarifications on the differences between atoms and molecules.
Any mole of a substance contains Avogrado's number. For example, the number of eggs in one dozen, as we know it, is 12 eggs. Now, how many eggs are in one mole? How many planes are in one mole? How many books are in one mole? Surprisingly enough, they all have the same answer: 6.02 x 10$^{23}$. If you were to be asked: How many eggs are in 2 moles? -- the answer is 1.2 x 10$^{24}$ (6.02 x 10$^{23}$ times 2).

## Mole Hill

Not much explanation is needed here, but, the base-line:
* GOING UP = Divide
* GOING DOWN = Multiply
To calculate the grams (or mass) of a compound:
* Co$_2$O$_3$
Get the atomic mass for both cobalt and oxygen. NOTE: BRINClHOF elements, when alone, live in pairs (they are diatomic), so when calculating oxygen gass (O$_2$), you must multiply oxygen's atomic mass by two, which gives you 31.9988.
** 2 • 58.933195 = 117.86639
** 3 • 15.9994 = 47.9982
Add the atomic masses, and now you've got your mass in grams (remember sig-figs):
** 165.8646 grams

## Chemical Quantities
### Percent Composition
#### Empirical Formula → %

* Determine the molar mass.
MgSO$_4$ = (1 x 24.3050) + (1 x 32.065) + (4 x 15.9994) = 120.37 g/moL
* Divide the mass of each element (mass IN THE COMPOUND - not atomic mass) by the overall mass of the compound.
${\tfrac {24.3050}{120.37}}$ = 0.20192 ${\tfrac {32.065}{120.37}}$ = 0.26635 ${\tfrac {63.9976}{120.37}}$ = 0.53167
* Times the numbers by 100, and round to the hundreths place.
0.20192 x 100 = 20.19% 0.26635 x 100 = 26.64% 0.53167 x 100 = 53.17%
* Add all to 100 to see if these percentages are correct (also, use common sense).
20.19 + 26.64 + 53.17 = 100
* Finished! MgSO$_4$ is made out of 20.19% magnesium, 26.64% sulfur, and 53.17% oxygen.

#### % → Empirical Formula
##### 4 steps route

* Change percents straight to grams (57.61% → 57.61 grams).
* Divide the grams by the atomic mass of the element to make them into moles.
* Divide by the smallest number to make the numbers whole.
* If this does not work, then multiply the numbers by either 2 or 3. You must do this to all the other numbers too.

##### Example

Convert the percent composition to empirical formula: 69.55% O 30.45% N (O$_?$N$_?$)
* Convert the percents straight to grams.
69.55g 30.45g
* Divide the numbers by the atomic mass of the element.
69.55g/15.9994amu = 4.347moL 30.45g/14.0067amu = 2.174moL
* Divide the numbers by the lowest number, which is 2.174moL.
2.174/2.174 = 1 4.347/2.174 = 2
* Although "4.347/2.174 = 1.999", we can round it to 2 -- when numbers get really close to a whole number, like in this case, it is fine to round it.
O$_2$N

## Empirical Formula & Molecular Formula

The empirical formula is the basic ratio (lowest whole number ratio) of atoms in a chemical compound. For example, NO$_2$ is the empirical formula, as it is to the simplest terms (lowest whole number ratio). For ionic compounds, the formula unit is, most of the time, usually the empirical formula, while for molecular compounds, the formula does not always indicate the actual number of atoms in each molecule.
The molecular formula is the formula that is any whole number ratio of atoms in a chemical compound (example: N$_2$O$_4$).

### Molecular Formula → Empirical Formula

Simply simplify.

### Empirical Formula → Molecular Formula

In order to complete this process, you must know the molar mass of the molecular formula (remember, the empirical formula is the lowest whole number ratio while the molecular formula is any whole number ratio).
* Calculate the molar mass of the empirical formula.
* Divide M.F./E.F (molar masses).
* Multiply the empirical formula by the number you get from dividing the M.F. by the E.F., and your molecular formula is done.
EX1: What is the molecular formula of the following molecule given the empirical formula is CH$_2$O and the molar mass of the molecular formula is 90g/moL?
Calculate the molar mass of CH$_2$O.
* 1 x 12.0107 = 12.0107
* 2 x 1.00794 = 2.01588
* 1 x 15.9994 = 15.9994
** (with rounding): 30g/moL
Divide the M.F. by the E.F.
* 90/30 = 3
The M.F. is three times bigger than the E.F., so this is your final answer:
* C$_3$H$_6$O$_3$

### Hydrates

A hydrate is an ionic compound (salt) with water molecules loosely attached to its crystal structure. The formula for a hydrate shows the ratio of water molecules to formula units.
If you are asked how to get the percentage of water from a hydrate, simply:
* Calculate the molecular mass of the water, so (example) "10H$_2$O" --> "18.012 x 10 = 180.12"
* Divide the molecular mass of the water by the OVERALL molecular mass of the hydrate, so "180.12/(180.12 + [OVERALL molecular mass of hydrate])".
* Multiply the number by a 100 to get your percent.
When you are approached with a question along the lines of: "A 5.0 sample of a hydrate of MgCl$_2$ was heated and only 4.2g of anhydrous salt remained", you must subtract the overall mass of the hydrate by the grams of the anhydrous salt that has remained, so: 5.0 - 4.2, which will give you .8g. This .8g represents the water (H$_2$O).
When you are asked to find the number of water molecules in a hydrate, such as "When 5.00grams of FeCl$_3$ * xH$_2$O are heated, 3.00 grams of the salt remained. Find the chemical formula.", you must:
* Subtract the grams of salt that has remained by the overall mass before the heating.
5.00 - 3.00 = 2.00g of H$_2$O
* Divide 2.00grams of H$_2$O by the molar mass of water.
2.00g/18.02g = 0.111moL
* Now, you must find the molar mass of the salt. So, calculate the molar mass of FeCl$_3$ and divide 3.00g from 162.206g (molar mass of FeCl$_3$) to get your mole value.
3.00g/162.206g = 0.0185moL
* Divide these two numbers by the smallest number, which is 0.0185moL.
0.0185/0.0185 = 1 0.111/0.0185 = 6
* You have gotten your answer (note that 0.111moL belongs to water). Your full formula is FeCl$_3$ * 6H$_2$O.
FeCl$_3$ * 6H$_2$O