The Mole: Avogadro’s number
How much is:






A dozen?
12
A century?
100
A mole?
6.02 x 1023
(602,000,000,000,000,000,000,000)
Can you count a mole of
pennies?



If you could count 5 per second, it
would take you
6.02 x 10 23 ÷ 5 pennies/second ÷ 60
sec/min ÷ 60 min/hr ÷ 24hrs/day ÷
365 days/yr =
3,800,000,000,000,000 years!
Can you spend a mole of
dollars?



If you could spend $1,000,000 every
second it would take you
6.02 x 10 23 ÷ $1,000,000/sec ÷ 60
sec/min ÷ 60 min/hr ÷ 24hrs/day ÷
365 days/yr =
19,000,000,000 years!
The Mole Map
Gas
Volume
@ STP

x
#
Particles
x

Mass

Molar
Mole
Mass
x
When measuring amounts, you
can count or you can mass them.
If I want 2 dozen baseballs, I can count 24 baseballs
Or I can mass 16 kg of baseballs.
How many tennis balls are in
6 kg?
( 2 dozen)
How many tennis balls are in a
mole?
Since we can’t count a mole of
atoms, we MUST mass chemicals
to measure moles
6.02 x 10 23 atoms of sulfur
32.07 grams of sulfur
6.02 x 10 23 atoms of carbon
12.01 grams of carbon
How do we measure moles?

mole = number of particles equal to the
number of atoms in 12 g of C-12



1 atom of C-12 weighs exactly 12 amu
1 mole of C-12 weighs exactly 12 g
The number of particles in 1 mole is called
Avogadro’s Number = 6.0221421 x
1023

1 mole of C atoms weighs 12.01 g and has
6.022 x 1023 atoms

the average mass of a C atom is 12.01 amu
How do we measure moles?




The atomic mass on your periodic table
is the mass of a mole of atoms of that
element.
What is the mass of a mole of copper
atoms?
63.55 g
So, to count 6.02 x 1023 copper atoms,
we mass out 63.55 g on the scale.
Mole and Mass Relationships
1 mole
Sulfur
32.06 g
Substance
Pieces in 1 mole
Weight of 1 mole
hydrogen
6.022 x 10 23 atoms
1.008 g
carbon
6.022 x 10 23 atoms
12.01 g
oxygen
6.022 x 10 23 atoms
16.00 g
sulfur
6.022 x 10 23 atoms
32.06 g
calcium
6.022 x 10 23 atoms
40.08 g
chlorine
6.022 x 10 23 atoms
35.45 g
copper
6.022 x 10 23 atoms
63.55 g
1 mole
Carbon
12.01 g
Find the mass of:








A mole of silicon atoms
28.09 g
6.02 x 1023 atoms of nitrogen
14.01 g
6.02 x 1023 atoms of sodium
22.99g
2 moles of sodium atoms
45.98 g
How many atoms are in:








A mole of silicon
6.02 x 1023
14.01 g of nitrogen
6.02 x 1023
2 moles of sodium
12.04 x 1023
45.98 g of sodium
12.04 x 1023
How many things are in:








A mole of footballs
6.02 x 1023 footballs
A mole of water
6.02 x 1023 molecules
2 moles of pencils
12.04 x 1023
½ mole of lead
3.01 x 1023 atoms
Molar mass is the mass of one
mole of a substance


To find the molar mass of an element,
look on the periodic table.
To find the molar mass of a compound,
add all the masses of its elements
Chemical Formulas as
Conversion Factors



1 spider  8 legs
1 chair  4 legs
1 H2O molecule  2 H atoms  1 O
atom
Molar Mass of Compounds
• the relative weights of molecules can be
calculated from atomic weights
Formula Mass = 1 molecule of H2O
= 2(1.0 amu H) + 16.0 amu O = 18.0 amu
• since 1 mole of H2O contains 2 moles of H
and 1 mole of O
Molar Mass = 1 mole H2O
= 2(1.01 g H) + 16.00 g O = 18.02 g
Find the molar mass of:









Ammonium phosphate
NH4+
PO43(NH4)3PO4
=42.03 +12.12 +30.97 +64.00
=149.12g/mol
Carbon dioxide
CO2
= 12.01 + 32.00
= 44.01 g/mol
Find the molar mass of:

Hydrogen gas



H2
= 2.02 g/mol
Elemental hydrogen


H
= 1.01 g/mol
Find the molar mass of:

Iron



Fe
= 55.85 g/mol
Iron (III) hydroxide


Fe3+
Fe(OH)3


OH-
= 55.85 + 48.00 + 3.03
=106.88 g/mol
Converting to and from moles.





Converting between moles and mass
requires the molar mass of the substance
from the periodic table.
Element: Ag = 107.97g/mol
Ionic compound: CaCl2 = 110.98 g/mol
Covalent compound: NO2 = 46.01 g/mol
Always keep at least two decimal places on
all values taken from the periodic table.
Converting to and from moles.




To convert from moles to grams, multiply by
molar mass:
0.500 mol H2O x (18.02g/mol) = 9.01g H2O
To convert from grams to moles, divide by
molar mass:
54g H2O x (1mol/18.02g) = 3.0 mol H2O
Converting to and from moles.




For gases, use the fact that at STP, 1 mol
of any gas has a volume of 22.4 Liters.
What is STP? Standard Temperature and
Pressure
Standard Temperature = 273K or 0.0°C
Standard Pressure = 1 atmosphere = 760
mm Hg (barometric) = 101.325 kPa.
Converting to and from moles.




To go from moles to volume, multiply
by 22.4L.
3.00 mol x (22.4L/mol) = 67.2L of gas
To go from volume to moles, divide by
22.4L
44.8L x (1mol/22.4L) = 2.00 moles of
gas
Converting to and from moles.



To convert between moles and
particles, simply multiply or divide by
Avogadro’s number.
2.0 mol x (6.02 x 1023 particles/mol) = 1.2 x 1024
particles
3.1 x 1024 particles x (1 mol/ 6.02 x 1023 particles) =
5.0 mol
Remember unit
factors?
Converting to and from moles.


A convenient tool for making these
conversions is called a “mole map.”
With the mole at the center, we can put
all of the aforementioned calculations
together into one simple picture.
The Mole Map
Gas
Volume
@ STP

x
#
Particles
x

Mass

Molar
Mole
Mass
x
Percent Composition


1.
2.
3.

Percentage of each element in a compound

By mass
Can be determined from
the formula of the compound
the experimental mass analysis of the compound
the total mass of each element
The percentages may not always total to 100% due to
rounding
part
Percentage =
´100%
whole
What percentage of water is
Oxygen?
1.
Formula of the compound
* H2O
2.
Mass of the compound
* 2 (1.01
3.
g/mol)
+ 16.00
g/mol
= 18.02
Mass of each element
* H is 1.01
g/mol,
O is 16.00
g/mol
16.00 g O
´ 100% = 88.79% O
18.02 g H2O
g/mol
Mass Percent as a
Conversion Factor

the mass percent tells you the mass of a
constituent element in 100.0 g of the
compound


the fact that NaCl is 39.0% Na by mass
means that 100.0g of NaCl contains 39.0g Na
this can be used as a conversion factor

100.0 g NaCl  39.0 g Na
39.0 g Na
g NaCl ´
= g Na
100.0 g NaCl
100.0 g NaCl
g Na ´
= g NaCl
39.0 g Na
Empirical Formulas
• The simplest, whole-number ratio of atoms
in a molecule is called the Empirical
Formula
– can be determined from percent composition or
combining masses
• The Molecular Formula is a multiple of the
Empirical Formula
100g
%A
mass A (g)
100g
%B
mass B (g)
MMA
MMB
moles A
moles A
moles B
moles B
Empirical Formulas
Hydrogen Peroxide
Molecular Formula = H2O2
Empirical Formula = HO
Benzene
Molecular Formula = C6H6
Empirical Formula = CH
Glucose
Molecular Formula = C6H12O6
Empirical Formula = CH2O
Finding an Empirical Formula
1) convert the percentages to grams
a) skip if already grams
2) convert grams to moles
a) use molar mass of each element
3) divide all by smallest number of moles
4) round or multiply all mole ratios by
number to make all whole numbers
a) if ratio ?.5, multiply all by 2; if ratio ?.33 or
?.67, multiply all by 3, etc.
b) skip if already whole numbers

Determine the empirical formula of a
compound containing 80.0 grams of
carbon and 20.0 grams hydrogen.
Grams to moles
1 mol C
80.0g C ´
= 6.67 mol C
12.01 g C
1 mol H
20.0g H ´
= 19.8 mol H
1.01 g H
Divide by smallest
6.67 mol
C=
= 1.00
6.67 mol
19.8 mol
H=
= 2.96 » 3.00
6.67 mol
Write Empirical Formula
CH3
All these molecules have the same
Empirical Formula. How are the
molecules different?
Name
Molecular
Formula
glyceraldehyde
C3H6O3
Empirical
Formula
CH2O
erythrose
C4H8O4
CH2O
arabinose
C5H10O5
CH2O
glucose
C6H12O6
CH2O
All these molecules have the same
Empirical Formula. How are the
molecules different?
Name
glyceraldehyde
Molecular
Formula
C3H6O3
Empirical
Formula
CH2O
Molar
Mass, g
90.09
erythrose
C4H8O4
CH2O
120.12
arabinose
C5H10O5
CH2O
150.15
glucose
C6H12O6
CH2O
180.18
Molecular Formulas


The molecular formula is a multiple of
the empirical formula
To determine the molecular formula you
need to know the empirical formula and
the molar mass of the compound
Molar Massreal formula = factor used to multiply subscripts
Molar Massempirical formula
What is the molecular formula for ethane if it
has a molar mass of 30.06 g/mol?
CH3= 15.04
g/mol
30.06 g/mol
ratio =
= 1.99867 = 2.000
15.04 g/mol
Molecular formula = 2 x the empirical formula
C2H6
Determine the Molecular Formula of
Cadinene if it has a molar mass of
204 g and an empirical formula of C5H8
Solutes and Solvents



Solution: a homogenous mixture
Solute: thing that dissolves
Solvent: thing that does the dissolving
(found in the largest amounts)

If the solvent is water, then it is called an
aqueous solution
Solubility




Why does sugar “disappear” in your
iced tea?
How do fish breathe underwater?
Why does soda go flat faster when left
out than when it is refrigerated?
It is all based on solubility!
Solubility

Example: iced tea


Solute
sugar
tea
Solvent
water
States and Solutions

Solutions can be any state of matter





Solid-solid: alloys (gold jewelry, brass, etc.)
Solid-liquid: salt water, sugar water
Liquid-liquid: vinegar, peroxide, rubbing
alcohol
Liquid-gas: soda, champagne, O2 in H2O
Gas-gas: air, air tanks (scuba)
How Things Dissolve

Need to find/ create “holes” in water
for the dissolving substance to move
into


Need to over come hydrogen bonding
between water (or solvent) molecules
Get interactions between water
molecules and molecules of the solute


Ion-dipole interactions
Dipole-dipole (and H bonding)
Why some coffees “Put hair on
your chest.”

“Strong” coffee has more
coffee dissolved in a given
amount (say 1 pot) than
“weak” coffee.



Strong coffee = concentrated
Weak coffee = dilute
Concentration: the amount
of solute in a given amount
of solvent (or solution).
Molarity (M)



Most common way to express concentration
Molarity is the number of moles of solute
dissolved in each liter of solution
Formula

M = moles of solute
liters of solution


Dependent on temperature
The higher the molarity the stronger the
concentration
Practice Problems
1. What is the molarity when 6.0 moles of
glucose is dissolved in water to make 3.0
L of solution.M = 6.0 moles = 2.0 M
3.0 L
2. How many moles of sodium chloride are
there in 500 mL of 4.0 M solution?
4.0 moles ö
æ
(0.5 L NaCl )ç
÷ = 2 moles NaCl
è 1L ø
3. What is the volume of 3.0 M solution that
contains 15 moles of glucose?
1L ö
æ
(15 moles )ç
÷ = 5.0 L
è 3.0 moles ø
How does something so strong
become so weak?




The answer is dilution.
The more dilute something is, the lower
the concentration (it’s weaker).
To accomplish this, add more solvent
How do we know how much to add?


M 1V1 = M 2V2
Typically start with a highly concentrated
solution and dilute down to what you need
Figure 15.8: Process of making 500 mL of a 1.00 M acetic
acid solution.
Mole Day is October 23!