 How
many atoms is 41 grams of Carbon
Dioxide?
 How
many grams is 12.04x10^23
molecules of water?
A
dozen?
• 12
A
century?
• 100
A
mole?
• 6.02 x 1023
(602,000,000,000,000,000,000,000)
If I want 2 dozen baseballs, I can count 24 baseballs
Or I can mass 16 kg of baseballs.
( 2 dozen)
6.02 x 1023 Tennis Balls
•Since we can’t count a mole of atoms (because it
would take too long), 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
 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!
 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!

Mole- is a # equal to the number of atoms in
12 g of C-12. We just use this as our base.



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
 Conveniently, the
atomic mass on your periodic table
is the mass of a mole of atoms of that element. That
has been figured out for you already!
 What
is the mass of a mole of copper atoms? Simple!
Look at the periodic table.
• 63.55 g
 So, to
count 6.02 x 1023 copper atoms, we mass out
63.55 g on the scale (instead of counting atoms,
because that would take forever!!!).
Mole and Mass Relationships
Substance Pieces in 1 mole
Weight of 1 mole
hydrogen
6.022 x 1023 atoms
1.008 g
carbon
6.022 x 1023 atoms
12.01 g
oxygen
6.022 x 1023 atoms
16.00 g
sulfur
6.022 x 1023 atoms
32.06 g
calcium
6.022 x 1023 atoms
40.08 g
chlorine
6.022 x 1023 atoms
35.45 g
copper
6.022 x 1023 atoms
63.55 g
1 mole
Sulfur
32.06 g
1 mole
Carbon
12.01 g
 What
 1)
 2)
are the 2 things 1 mole is equal to?
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
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
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
 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 (this is
what you did on your webquest!)
Hydrogen
gas
• H2
• = 2.02 g
Elemental
•H
• = 1.01 g
hydrogen
Ammonium
• NH4+
phosphate
(NH4)3PO4
PO43 =42.0 +12.12 +31.0 +64.0
 =149.12g
Carbon
• CO2
dioxide
 = 12.0 + 32.0
 = 44.0 g
 How
many atoms are there in 40 grams of
NaHCO3 …
To
convert between moles and
particles (atoms), simply multiply or
divide by Avogadro’s number.
• 2 mol x (6.02 x 1023 particles/mol) = 1.20 x 1024
particles
• 3.1 x 1024 particles x (1 mol/ 6.02 x 1023 particles) = 5
mol
 Converting
between moles and mass
requires the molar mass of the substance
from the periodic table.
 Element: Ag = 107.9g/mol
 Ionic compound: CaCl2 = 111.1 g/mol
 Covalent compound: NO2 = 46.0 g/mol
 Always keep at least one decimal place
on all values taken from the periodic
table.
 To
convert from moles to grams, multiply
by molar mass:
 0.50 mol H2O x (18.0g/mol) = 9.00g H2O
 To
convert from grams to moles, divide
by molar mass:
 54g H2O x (1mol/18.0g) = 3.0 mol H2O
 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°C
 Standard Pressure = 1 atmosphere = 760
mm Hg (barometric) = 100 kPa.
 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
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


Percentage of each element in a compound
•
By mass
Can be determined from
1.
2.
3.

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
1.
Formula of the compound
* H2O
2.
Mass of the compound
* 2 (1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
3.
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
 the
mass percent tells you the mass of a
constituent element in 100 g of the
compound
• the fact that NaCl is 39% Na by mass means that
100 g of NaCl contains 39 g Na
 this
can be used as a conversion factor
• 100 g NaCl  39 g Na
39 g Na
g NaCl 
 g Na
100 g NaCl
100 g NaCl
g Na 
 g NaCl
39 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.
1 mol C
80.0g C 
 6.67 mol C
12.0 g C
1 mol H
20.0g H 
 19.8 mol H
1.01 g H
6.67 mol
C
1
6.67 mol
19.8 mol
H
 2.96  3
6.67 mol
CH3
Example:
 A laboratory analysis of aspirin determined the
following mass percent composition. Find the
empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
Name
Molecular
Formula
glyceraldehyde
C3H6O3
Empirical
Formula
CH2O
erythrose
C4H8O4
CH2O
arabinose
C5H10O5
CH2O
glucose
C6H12O6
CH2O
Name
glyceraldehyde
Molecular
Formula
C3H6O3
Empirical
Formula
CH2O
Molar
Mass, g
90
erythrose
C4H8O3
CH2O
120
arabinose
C5H10O5
CH2O
150
glucose
C6H12O6
CH2O
180
 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
CH3= 15.03 g/mol
30.06 g/mol
ratio 
2
15.03 g/mol
Molecular formula = 2 x the empirical formula
C 2 H6