Transcript Moles - Teacher Notes

Unit 4 – The Mole
Honors Chemistry
Part 1
WHAT IS A MOLE?

602214199000000000000000

6.02 x 1023
Mole Facts
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6.02 X 1023 Pennies: Would make at least 7
stacks that would reach the moon.
6.02 X 1023 Watermelon Seeds: Would be found
inside a melon slightly larger than the moon.
6.02 X 1023 Blood Cells: Would be more than
the total number of blood cells found in every
human on earth.
1 Liter bottle of Water contains 55.5 moles H20
Definition of Mole
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The amount of atoms in 12.0 grams of Carbon
12 (6.02 x 1023 atoms known as Avogadro’s
number).
A sample of any element with a mass equal to
that element's atomic weight (in grams) will
contain precisely one mole of atoms (6.02 x
1023 atoms).
Molecular Mass

The sum of the masses of all the atoms in a
molecule of a substance
CaCO3
1 atom of Ca = 40.08 amu
1 atom of C = 12.00 amu
3 atoms of O = 48.00 amu
100.08 amu
}
Add
these
Formula Mass (Molar Mass)
The mass of 1 mole (in grams)
 Equal to average atomic mass but the unit is
grams
1 mole of C atoms
=
12.01 g
1 mole of Na atoms
=
22.99 g
1 mole of Cu atoms
=
63.55 g

Example Problem
Find the mass of 1 mole of KAl(SO4)2 12H2O
●
1K
= 39.10
1 Al
= 26.98
2 (SO4)
2(32.06 + ((16.00 x 4))=192.12
12 H2O
12(2.02 + 16.00)
=216.24
Mass of 1 mole
= 474.44 g/mol
Try These:
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Find the molecular mass for these :
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HNO3
CO2
Find the molar mass for these compounds:
 C6H10O5
 H2SO4
The Mole
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1 mole of gas always contains 6.02 x 1023
molecules of that gas
1
1
1
1
mole
mole
mole
mole
Cl2 gas = 6.02 x 1023 molecules of Cl2
NO2 gas = 6.02 x 1023molecules of NO2
CO gas = 6.02 x 1023 molecules of CO
CO2 gas = 6.02 x 1023molecules of CO2
The Mole
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Also applies to other particles! (not only
molecules in a gas)
1
1
1
1
1
mole
mole
mole
mole
mole
C = 6.02 x 1023 C atoms
H2O = 6.02 x 1023 H2O molecules
NaCl = 6.02 x 1023 NaCl formula units
of Na+ = 6.02 x 1023 Na+ ions
of Cl- = 6.02 x 1023 Cl– ions
Avogadro’s Number

We can use Avogadro’s # as a conversion
factor:
1 mole
6.02 x 1023 particles
Or
6.02 x 1023 particles
1 mole
 Note that a particle could be an atom,
molecule, formula unit, or ion !
Example Problems
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How many molecules are in 3.5 moles of H2O?
How many moles are present in 4.65
molecules of NO2?
Mass and Mole Relationship
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1 mole of any substance = the molar mass
of that substance (in grams)
Find the number of moles present in 56.7 g
of HNO3.
56.7 g HNO3 1 mole HNO3
63.01 g HNO3
Example Problems
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Find the number of grams present in 4.5
moles of C6H10O5.
Find the number of moles present in 12.31 g
of H2SO4.
How many molecules are in 4.5 grams of
NaCl?
Gas Volumes and Molar Mass
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Avogadro’s Law
Equal volumes of gases under the same
conditions of temperature and pressure
contain equal numbers of molecules
1 mole of any gas at standard temperature
and pressure (STP) occupies 22.4 liters
Standard temperature: 0ºC or 273K
Standard pressure: 1 atm or 101.325 kPa
Gas Volumes and Molar Mass
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32.00 g O2 = 1 mole = 22.4 L
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2.02 g H2 = 1 mole = 22.4 L
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44.01 g CO2 = 1 mole = 22.4 L
Example Problems
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How many liters are present in 5.9 moles of
O2?
How many liters are present in 3.67 moles
of CO2?
How many atoms of O are present in 78.1 g
of O2?
Percent Composition
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Finding what percent of the total
weight of a compound is made up
of a particular element
Formula for calculating %
composition:
Total mass of the element in the
compound
X 100
Total formula mass
Example Problem:
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Calculate the % composition of calcium in
Ca(OH)2.
Example Problems
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Find the percentage composition of a
compound that contains 1.45 g of carbon,
4.23 g of sulfur, and 1.00 g of hydrogen in a
6.68 g sample.
21.7% carbon
63.3% sulfur
15.0% hydrogen
Example Problem
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A sample of an unknown compound with a
mass of 5.00 grams is made up of 75%
carbon and 25% hydrogen. What is the mass
of each element?
3.75 g of carbon
1.25 g of hydrogen
Formulas
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Empirical Formula - expresses the smallest
whole number ratio of atoms present
 E.g.
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CH2O
Ionic formulas are always empirical formulas
Molecular Formula - states the actual
number of each kind of atom found in one
molecule of the compound.
 E.g.
C6H12O6
Empirical Formula
1.
Determine mass in grams of each element
2.
Calculate the number of moles of each
3.
4.
Divide each by the smallest number of moles
to obtain the simplest whole number ratio
If whole numbers are not obtained in step 3,
multiply all by the smallest number that will
give whole numbers
Empirical Formula
Remember this:
 Percent to mass
 Mass to mole
 Divide by small
 Multiply ‘till whole
Example Problem
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Given that a compound is composed of
60.0% Mg and 40.0% O, find the
empirical formula.
Example Problem
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A compound is found to contain 68.5%
carbon, 8.63% hydrogen, and 22.8%
oxygen. The molecular weight of this
compound is known to be approximately
140.00 g/mol. Find the empirical and
molecular formulas.
Hydrates
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Ionic compounds
Water is bonded to the crystal structure
Ex: CuSO3 • 7H2O
The percentage of water in a hydrate can
easily be calculated using the formula:
% Water = Mass of water
Mass of hydrate
x 100
Example Problem
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What is the percentage of water in CuSO3 •
7H2O?
A 3.5 g sample of a hydrate is heated and only
1.7 g of the anhydrous salt remain. What is
the percentage of water?
Law of Definite Proportions
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Formulas give the numbers of atoms or moles
of each element
Always a whole number ratio
1 molecule NO2 : 2 atoms of O for every 1
atom of N
1 mole of NO2 : 2 moles of O atoms to every
1 mole of N atoms
Law of Multiple Proportions
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When any two elements, A and B, combine to
form more than one compound, the different
masses of B that unite with a fixed mass of A
have a small whole-number ratio
Example:
In H2O, the proportion of H:O = 2:16 or 1:8
In H2O2, H:O is 2:32 or 1:16
How Do We Determine Concentration?
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Molarity
Molality
How do we make solutions?
M1 = m1/V1
M2 = m2/V2
rearrange to M1V1 = m1
rearrange to M2V2 = m2
If m1=m2
then, M1V1 = M2V2
M1V1 = M2V2
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M1 = concentration of the first solution
V1 = volume of the first solution
M2 = concentration of the second solution
V2 = volume of the second solution
Let's consider a sample problem:
You have 1 L of a 0.125 M aqueous solution of
table sugar. You want to dilute the solution to
0.05 M. What do you do?
Dilution
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To solve the problem, you simply plug in the
three numbers you know:
(0.125 M) (1 L) = (0.05 M) V2
2.5 L = V2
Using the equation, you determine that the
volume of the diluted solution should be 2.5 L.
So we simply add enough water to the first
solution so that the solution's volume becomes
2.5 L.
What is Saturation?
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A solution is saturated if it contains as much solute as can possibly be
dissolved under the existing conditions of temperature and pressure
Unsaturated: Has less than maximum amount of solute that can be dissolved
Supersaturated: Contains more than maximum (How can this happen?)