Download Notes: Moles

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:
Cl:
23 grams x 1 atom =
35 grams x 1 atom =
Total:
23 grams/mol
35 grams/mol
58 grams/mol
 More examples:
o MgCl2
o Fe(OH)2
o Be3(PO4)2
For these last two examples, tell them
that the molar mass for compounds
like this is found by multiplying
everything in the parentheses by the
number outside the parentheses.
94 g/mol
90 g/mol
217 g/mol
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 to help you
convert between grams, moles, and molecules/atoms.
grams
molar
mass
moles
6.20 x 1023
molecules
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).
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
o K2S
Al: 34.6%, O: 61.5%, H: 3.8%
K: 70.9%, S: 29.1%
Notes: Empirical and Molecular Formulas
Honors Only!
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!
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
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