# Moles_notes

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```					Notes: Moles
Atoms are very, very small.
 1 atom of hydrogen weighs approximately 1.67 x 10-27 kg.

As a result, it’s not very practical to do chemical reactions by counting
out the number of atoms or molecules that will be reacting, because
we’ll be counting for a very long time!

You’ve seen this before, because when working with a large number
of objects, it’s frequently handy to use units that are easier to work
with.
   2 shoes = 1 pair
   12 eggs = 1 dozen
   144 pencils = 1 gross
   500 sheets of paper = 1 ream
   6.02 x 1023 atoms or molecules = 1 mole

The idea behind moles is the same as the idea behind “dozens”,
except that the number is much bigger.

Definition:
1 mole = 6.02 x 1023 of anything.
 6.02 x 1023 is referred to as “Avogadro’s number” in honor of
the dude who first worked with it.

If moles is such a handy number, why haven’t you used it before
now?
 1 mole of most objects that you work with on a daily basis is
very, very large. For example, 1 mole of M&M’s would cover
the continental United States to a depth of 125 km.
 Although we could use moles to describe numbers of things
that we work with everyday, it’s not really very practical.
Finding Molar Mass

Molar mass (also called “molecular weight” or “molecular mass”):
The weight of one mole of a chemical compound. The unit is
“g/mol”.
 For elements, the mass of one mole of atoms is called the
“atomic mass” and is found on the periodic table.

How to calculate the molar mass of a compound:
 For elements, the molar mass is the same thing as the atomic
mass.
 For chemical compounds, it’s the sum of the masses of all of
the atoms in the molecule.
Example: NaCl
Na:   23 grams x 1 atom =    23 grams/mol
Cl:   35 grams x 1 atom =    35 grams/mol
Total:           58 grams/mol

 More examples:
For these last two examples, tell them
o MgCl2             94 g/mol      that the molar mass for compounds
o Fe(OH)2           90 g/mol      like this is found by multiplying
everything in the parentheses by the
o Be3(PO4)2         217 g/mol     number outside the parentheses.

Mole calculations:

How do you count out a mole of atoms?
 You don’t. Even if it were possible to count out individual atoms
in a reasonable period of time, the equipment we have only
measures “grams.”
 As a result, we need to be able to convert between
atoms/molecules, moles, and grams.
Use the diagram below with the t-chart method of doing calculations

molar                             6.20 x 1023       molecules
grams      mass           moles
or atoms

 Handy hint: In conversion factors, always write “1” in front of
“moles”!

Quick recap of the t-chart method:
1. Make a T
2. Put what you’ve got in the top left
3. Put the units of what you’ve got in the bottom right
4. Put the units of what you want in the top right
5. Put in the conversion factors
6. [If needed, add another step to get to where you’re going]
7. Multiply the stuff on the top together and divide by the stuff on the bottom.

Examples: (Go over the T-chart with the first few):
 How many grams are in 2.1 moles of Be? (18.9)
 How many molecules are in 6.3 moles of CH4? (3.79 x 1024
molecules)
 How many molecules are there in 11.1 grams of carbon
dioxide? (1.51 x 1023 molecules)
 How many grams are in 4.1 x 1023 molecules of H2O? (18.7 g).

 Mole conversion practice worksheet

 Moles of chalk lab (General worksheet, honors DIY)

 HWK like this
Notes: Percent composition:

 Explain how to find percent composition.

o Make sure to mention that it’s a mass percent.

 Give them examples and have them solve:

o Al(OH)3     Al: 34.6%, O: 61.5%, H: 3.8%
o K2S         K: 70.9%, S: 29.1%

 Lab activity: The hydrate lab

 Homework on percent composition
Notes: Empirical and Molecular Formulas
Once you have a percent composition of a chemical compound, it’s
possible to figure out the molecular formula. Here’s how, using the
example of a compound with a molar mass of 28 grams/mol in which
the percent composition of the components are as follows:
 C: 85.7%
 H: 14.3%

1)   Assume you have 100 grams of the chemical. This serves
to convert the percentages into grams.
In our example, we now have 85.7 grams of carbon and 14.3
grams of hydrogen.

2)   Figure out how many moles of each element you have in
the compound.
Using the T-chart method we talked about before, we can
determine that there are 7.14 moles of carbon and 14.3 moles
of hydrogen.

3)   Find the empirical formula of the compound by finding the
ratio of the number of moles of elements.
 Review: Empirical formulas are reduced versions of the
molecular formula. For example, a compound that has an
empirical formula of BH may have a formula of B2H2, B3H3,
etc.
 In a practical sense, this means that we should divide the
number of moles of each element by the smallest answer for
number of moles. Since the smallest number is “7.14
moles”, we’ll divide both values by 7.14.
carbon: 7.14 / 7.14 = 1
hydrogen: 14.3 / 7.14 = 2
 Empirical formula: C1H2
 Handy hint: If the problem gives you the empirical formula,
you can start at the next step!

Continued on next page 
4)   Divide the experimentally determined molar mass by the
molar mass of the empirical formula.
 For our compound, the molar mass of the compound is 28
grams/mol (this was given in the problem).
 For the formula CH2, the molar mass is 12 + 2 = 14
grams/mol.
 28 / 14 = 2

5)   Multiply the coefficients in the empirical formula by the
number you found in step 4 to find the molecular formula.
 CH2 x 2 = C2H4, which is our answer!

More examples:
 Find the molecular formula of a compound with a molar mass of
142 g/mol and a percent composition of 43.7% P and 56.3% O.
P2O5
 Find the molecular formula of a compound with an empirical
formula of HO and a molar mass of 34 g/mol.
H2O2
 Find the molecular formula of a compound with a molar mass of 92
g/mol and a percent composition of 30.4% N and 69.6% O.
N2O4

Chemical formulas homework sheet

Hydrates Notes:

 Explain what hydrates are and what their structures look like:

o Ionic compounds sometimes have water molecules that
adhere to the metal ion in the compound. Such
compounds are called hydrates.
o These water molecules aren’t tightly bonded but rather
loosely associated with them. As a result, the water
molecules can be easily removed and replaced.
 Dehydration: When you remove the water
molecules from a hydrate (usually by heating). A
hydrate from which the water has been removed
is called an anhydrate.
 Hydration: When you add water to an
anhydrate to reform the hydrate.

 Unusual properties of hydrates:

o They may appear to have lower boiling points than
they really do:
 They may appear to “bubble” when the water is
removed, while in other cases the water may just
go away quietly.

o They may change colors when hydrated/dehydrated.
 This is used in many places to detect water
molecules.

 Formulas of hydrates:
o Hydrates have the formula “[ionic compound] . x H2O”.
This means that there are x water molecules stuck to the
ionic compound (the dot does not mean multiply, it means
“plus” in this case).
o They are named the same way as the ionic compound
except they have “[something]hydrate” added at the end.
 Prefixes: mono-, di-, tri-, tetra-, etc.
o Examples:
 CaCl2.2H2O = calcium chloride dihydrate
 CoCl2.6H2O = cobalt (II) chloride hexahydrate

Hydrate lab
 General: Do the “Percent composition Lab”.
 Honors: Give them magnesium sulfate heptahydrate and tell
them to figure out the empirical formula in the lab.
o Hand them the “General guidelines for lab reports”
worksheet.

Review

Quiz on Chapter 11
Hydrate lab:


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