The Mole
7.1 & 7.3
7.1 Chemical Measurement:
1. Counting units: pair = 2
dozen = 12
score = 20
gross = 144
ream = 500
mole = ?
2. Counting atoms, molecules and grams are
going to require much larger and much
smaller numbers.
3. Remember the mass of an atom is measured in
amu (atomic mass units) and the numbers on the
Periodic Table are based on 1/12 of a carbon-12
atom. So from the table the mass of 1 carbon
atom = 12.0 amu. This unit is very small, 1
atomic mass unit (amu) = 1.6605 x 10-24 g.
Determine how many carbon atoms equals 12.0 grams:

  1 C atom 
1
amu C
23

12.0 g C 

6.02
x
10
atoms


-24
 1.6605 x 10 g C   12 amu C 


Determine how many phosphorus atoms equal 31.0 g:

  1 P atom 
1
amu P
23

31.0 g P 

6.02
x
10
atoms


-24
 1.6605 x 10 g P   31 amu P 


4. This value is called Avogadro’s Number =
6.02 x 1023 = 1 mole
(602,000,000,000,000,000,000,000!)
5. A mole is the amount of a substance. It is
based on the number of atoms of an element
equal to the number of atoms in exactly 12.0g
of carbon-12.
Mole Analogies
• An Avogadro's number of standard soft drink cans would
cover the surface of the earth to a depth of over 200 miles.
• If you had Avogadro's number of unpopped popcorn kernels,
and spread them across the United States of America, the
country would be covered in popcorn to a depth of over 9
miles.
• If we were able to count atoms at the rate of 10 million per
second, it would take about 2 billion years
• 6.02 X 1023 Watermelon Seeds: Would be found inside a
melon slightly larger than the moon.
Mole Analogies
• 6.02 X 1023 Donut Holes: Would cover the earth
and be 5 miles (8 km) deep.
• 6.02 X 1023 Pennies: Would make at least 7 stacks
that would reach the moon.
• 6.02 X 1023 Grains of Sand: Would be more than
all of the sand on Miami Beach.
• 6.02 X 1023 Blood Cells: Would be more than the
total number of blood cells found in every human
on earth.
6. Atomic mass: mass of one atom of an element
measured in amu (atomic mass units).
Ex #1) H = 1.0 amu O = 16.0 amu C = 12.0 amu
7. Formula mass: mass of all the atoms in a
single molecule or formula unit of a compound.
Ex #2) H2O =
Ex #3) H2CO3 =
H: 1.0 x 2 = 2.0
H: 1.0 x 2 = 2.0
O: 16.0 x 1 =16.0 +
C: 12.0 x 1 = 12.0
18.0 amu
O: 16.0 x 3 = 48.0 +
62.0 amu
8. Molar Mass: mass of one mole of an element
The units used are g/mol. Round all
elements’ masses to the tenths place.
Ex #4) Elements:
Cu = 63.5 g/mol
1 mole Cu = 6.02 x 1023 atoms
Cl = 35.5 g/mol
1 mole Cl = 6.02 x 1023 atoms
Ex #5) Covalent Compounds:
H2O = 18.0 g/mol
1 mole H2O = 6.02 x 1023 molecules
H2CO3 = 62.0 g/mol
1 mole H2CO3 = 6.02 x 1023 molecules
Ex #6) Ionic Compounds:
NaCl = 58.5 g/mol
1 mole NaCl = 6.02 x 1023 formula units
MgO = 40.3 g/mol
1 mole MgO = 6.02 x 1023 formula units
9. Examples:
N2 = 28.0 g/mol = 6.02 x 1023 molecules
2N2 = 56.0 g/mol = 12.04 x 1023 molecules
10. Practice: Calculate the molar mass of sucrose,
C12H22O11.
C: 12.0 x 12 = 144.0
H: 1.0 x 22 = 22.0
O: 16.0 x 11 = 176.0 +
342.0 g/mol
Avogadro’s Number
11.
1 mole
= 6.02 x 1023 particles = molar mass
1 mole Ne = 6.02 x 1023 atoms Ne
= 20.2g
1 mole CO2 = 6.02 x 1023 molecules CO2 = 44.0g
1 mole CaF2= 6.02 x 1023 formula units CaF2 = 78.1g
12. The mole is the bridge between calculations
of # of particles, mass and volume of a gas.
13. Conversion Factors:
1 mole = 6.02 x 1023 particles (atoms, molecules or formula units)
1 mole = molar mass (grams) from the periodic table
May be used in a problem in one of two ways depending on the given units.
 6.02x1023   1 mole 

 or 
23 
1
mole
6.02x10




 X grams   1 mole 


 or 
 1 mole   X grams 
Ex #1) Given 11.2 g of NaCl, how many moles does
this represent?
 1 mole NaCl 
11.2 g NaCl 
  .191 mol NaCl *3 significant digits because 11.2 is 3
 58.5 g NaCl 


Ex #2) Given 2.50 moles of NaCl, how many
grams does this represent?
 58.5 g NaCl 
2.50 mol NaCl 
  146 g NaCl
 1 mol NaCl 
Ex #3) How many particles are in 2.00 moles of H2O?
 6.02 x 1023 molec H O 
2
2.00 mol H 2O 
  1.20 x1024 molec H 2 O


1 mol H 2O


How many atoms is this?
 3 atoms H 2 O 
24
1.20 x 10 molec H2 O 
=
3.60
x
10
atoms H2 O

 1 molec H 2O 
24
Ex #4) How many moles of CaCO3 are 8.74 x 1023 formula units?


1 mol CaCO3
8.74 x 10 f.u. CaCO3 
 = 1.45 mol CaCO3
23
 6.02 x 10 f.u. CaCO3 
23
Multistep Problems:
Ex #5) How many particles of copper are in 56.3 g of copper?
 1mole Cu
56.3 g Cu 
 63.5 g Cu

  6.02 x 1023 atoms Cu 
23
=
5.34
x
10
atoms Cu



1mole Cu


We use atoms because copper is an element. There are 3 significant digits because 56.3 has 3.
Ex #7) How many sugar particles of sugar (C12H22O11) are in 250 g of sugar?
 1mole C12 H 22O11
250 g C12 H 22O11 
 342.0 g C12 H 22O11

  6.02 x 1023 molec C H O 
12 22 11

 = 4.4 x 1023 molec C12 H 22O11


1mole C12 H 22O11


Moles & Gases:
Molar Volume: 1 mole of any gas at Standard Temperature and Pressure
(STP = 0oC and 1 atm) has a
volume of 22.4 dm3 or 22.4 L
Ex #8) How many particles of CO2 gas are in a 1.0L flask at STP?
 1 mole CO2
1.0 L CO2 
 22.4 L CO2
  6.02 x 1023 molecules CO2 
22
=
2.7
x
10
molecules CO2


1 mole CO2


Ex #9) How many atoms of radon gas are contained in a 6500. dm3 basement
room at STP?
 1 mole Rn
6500. dm Rn 
 22.4 L Rn
3
  6.02 x 1023 atoms Rn 
26
=
1.747
x
10
atoms Rn


1 mole Rn


Ex #10) If a room has a volume of 4000. L, how many moles of air is this at
STP?
 1 mole air
4000. L air 
 22.4 L air

 = 178.6 mol air

7.3 % Composition, Empirical & Molecular Formulas:
% Composition: The mass of each element in a compound compared to the entire mass
of the compound x by 100%. This can be calculated experimentally from the
grams of each element in a compound, or the expected % can be calculated by
using the molar masses of the elements in the compound compared to the molar
mass of the compound.
Ex #11) What is the % hydrogen in water? Use the molar masses when not
provided with the lab data.
%H=
2.0 g H
x 100% = 11% H
18.0 g H20
%O=
16.0 g O
x 100%  88.9%
18.0 g H 2O
The numbers should add up to 100% unless significant digits prevent it.
Ex #12) Find the % composition of a compound that contains 2.30 g of sodium,
1.60 g of oxygen, and 0.100 g of hydrogen.
2.30 g Na
1.60 g O
0.100 g H
x 100% = 57.5% Na
x 100%  40.0%O
x 100% = 2.50% H
4.00 g
4.00 g
4.00 g
Ex #13) A sample of an unknown compound with a mass of 0.562 g has the
following % composition: 13.0 % carbon, 2.20 % hydrogen, and 84.5 %
fluorine. When this compound is decomposed into its elements, what mass of
each element should be recovered?
(0.562 g)(0.130 C) = 0.0731 g C
(0.562 g)(0.0220 H) = 0.0124 g H
(0.562 g)(0.845 F) = 0.475 g F
Empirical Formula: A formula that gives the simplest whole-number ratio of
the atoms of the elements in the compound.
Molecular Formula: Gives the actual number of atoms of each element in a
molecular compound.
The molecular formula may be the same as the empirical formula. H20
Ex #14) hydrogen peroxide = H2O2 molecular formula
= HO empirical formula (a 1: 1 ratio)
Ex #15) glucose
= C6H12O6
molecular formula
= CH2O empirical formula (a 1: 1 :1 ratio)
Ex #16) Find the empirical formula of the compound if you have 80.g of carbon
and 20.g of hydrogen.
 1 mol C 
6.7
80. g C 
=1
 = 6.7 mol C
6.7
 12.0 g C 
 1 mol H 
20.
20. g H 
=3
 = 20. mol H
6.7
 1.0 g H 
CH 3
Ex #17) An unknown compound is analyzed. It’s composition was determined
to be 51.85 % carbon, 8.64 % hydrogen and 39.51 % oxygen. Find the
empirical formula of the compound.
 1 mol C 
51.85 g C 
 = 4.321 mol C
 12.0 g C 
 1 mol H 
8.64 g H 
 = 8.64 mol H
 1.0 g H 
8.64
= 3.5 x 4 = 14
2.469
 1 mol O 
39.51 g O 
 = 2.469 mol O
 16.0 g O 
2.469
=1
2.469
4.321
= 1.75 x 4 = 7
2.469
x4= 4
C7 H14O4
Ex #18) Find the molecular formula of ribose (molar mass = 150.0 g/mol). It
has a chemical composition that is 40.0% carbon, 6.67% hydrogen and 53.3%
oxygen. Assume a 100 g sample. Hint: First find the empirical formula!
 1 mol C 
40.0 g C 
 = 3.33 mol C
 12.0 g C 


 1 mol H 
6.67 g H 
 = 6.67 mol H
 1.0 g H 


 1 mol O 
53.3 g O 
 = 3.33 mol O
 16.0 g O 


150.0 g/mol
Molecular Mass
=
= 5
Empirical Mass
30.0 g/mol
3.33
=1
3.33
6.67
=2
3.33
Empirical Formula =
CH 2O
3.33
= 1 Empirical Mass = 30.0 g/mol
3.33
Molecular Formula = C5H10 O5