The Mole & Formulas
Dr. Ron Rusay
Spring 2008
© Copyright 2008 R.J. Rusay
Stoichiometry
Mole - Mass Relationships
Chemical Reactions
The Mole
•% Composition: Determining the Formula of
an Unknown Compound
•Writing and Balancing Chemical Equations
•Calculating the amounts of Reactant and
Product
•Limiting Reactant
The Mole
• The number of carbon atoms in exactly 12
grams of pure 12C. The number equals
6.02  10 23

1 mole of anything = 6.02  10 23 units
• 6.02  10 23 “units” of anything: atoms,
people, stars, $s, etc., etc. = 1 mole
Avogadro’s Number
Avogadro’s number equals 1 mole
….which equals
6.022 
23
10
“units”
Counting by Weighing
12 red marbles @ 7g each = 84g
12 yellow marbles @4g each=48g
55.85g Fe = 6.022 x 1023 atoms Fe
32.07g S = 6.022 x 1023 atoms S
Relative Masses of 1 Mole
CaCO3
100.09 g
Oxygen
32.00 g
Copper
63.55 g
Water
18.02 g
Atomic and Molecular Weights
Mass Measurements
• 1H weighs 1.6735 x 10-24 g and 16O
2.6560 x 10-23 g.
•DEFINITION: mass of 12C = exactly
12 amu.
•Using atomic mass units:
• 1 amu = 1.66054 x 10-24 g
• 1 g = 6.02214 x 1023 amu
QUESTION
What is the mass of one atom of copper in
grams?
1) 63.5 g
2) 52.0 g
3) 58.9 g
4) 65. 4 g
–22
5) 1.06  10 g
ANSWER
5)
–22
1.06  10
g
Section 3.3 The Mole (p. 82)
A mole of copper atoms has a mass of 63.55 g.
The mass of 1 copper atoms is
23
–22
63.55 g/(6.022  10 ) = 1.06  10 g.
Atomic and Molecular Weights
• Formula Weight a.k.a. Molecular Weight
• Formula weights (FW): sum of Atomic
Weights (AW) for atoms in formula.
• FW (H2SO4 ) = 2 AW(H) + AW(S) + 4 AW(O)
• = 2(1.0 amu) + (32.0 amu) + 4(16.0)
• = 98.0 amu
Atomic and Molecular Weights
• Molecular weight (MW) is the weight of
the molecular formula in amu.
• MW of sugar (C6H12O6 ) = ?
• MW = 6(12.0 amu) + 12(1.0 amu) + 6(16.0
amu)
• = 180 amu
Molar Mass
• A substance’s molar mass (equal to the
formula weight: atomic or molecular weight
in grams) is the mass in grams of one mole
of the element or compound.
C
= 12.01 grams per mole (g/mol)
 CO2

= ??
44.01 grams per mole (g/mol)
12.01 + 2(16.00) = 44.01
QUESTION
What is the molar mass of ethanol (C2H5OH)?
1) 45.07
2) 38.90
3) 46.07
4) 34.17
5) 62.07
ANSWER
3)
46.07
Section 3.4 Molar Mass (p. 86)
The molar mass is the sum of masses of all the
atoms in the molecule.
2  12.01 + 6  1.008 + 1  16.00 = 46.07
QUESTION
For which compound does 0.256 mole weigh
12.8 g?
1) C2H4O (MM = 44 g/mol)
2) CO2 (MM = ?)
3) CH3Cl (MM = 50 g/mol)
4) C2H6 (MM = 30 g/mol)
5) None of these
ANSWER
3)
CH3Cl
Section 3.4 Molar Mass (p. 86)
The molar mass has units of g/mol.
(12.8 g/0.256 mol) = 50.0 g/mol. The molecule
with the closest molar mass is CH3Cl.
QUESTION
How many grams are in a 6.94-mol sample of
sodium hydroxide?
1) 40.0 g
2) 278 g
3) 169 g
4) 131 g
5) 34.2 g
ANSWER
2)
278 g
Section 3.4 Molar Mass (p. 86)
The molar mass of sodium hydroxide, NaOH, is
22.99 g/mol + 16.00 g/mol + 1.008 g/mol = 40.00
g/mol. Convert to grams: 6.94 mol  (40.00 g/mol)
= 278 g.
The Mole & Stoichiometry
Stoichiometry - Mole - Mass
Relationships in Chemical Reactions
Stoichiometry:
•Writing and Balancing Chemical Equations
•Calculating the amounts of Reactant and Product
Chemical Reactions
Atoms, Mass & Balance: eg. Zn + S -->
QuickTime™ and a
Cinepak decompressor
are needed to see this picture.
Chemical Equation
•
Representation of a chemical reaction:
_ C2H5OH + _ O2  _ CO2 + _ H2O
reactants
products
• C=2; H =5+1=6; O=2+1
•
C=1; H=2; O=2+1
1 C2H5OH + 3 O2  2 CO2 + 3 H2O
Chemical Equation
• C2H5OH + 3 O2  2 CO2 + 3 H2O
The equation is balanced and the reaction can be
completely stated as:
1
mole of ethanol reacts with 3 moles of
oxygen
 to produce 2 moles of carbon dioxide and 3
moles of water.
QUESTION
Agriculturally, the following equation is important because it is used
to make millions of tons of urea. When the equation is balanced,
how many hydrogen atoms will be present on both sides of the
equation? Also, how many moles of NH3 would be needed to react
completely with 0.5 moles of CO2?
NH3 + CO2  (NH2)2CO + H2O
1.
2.
3.
4.
Three; two
Three; one
Six; one
Six ; two
ANSWER
Choice 3. Six; one 3 answers both stoichiometry questions
correctly. By placing a coefficient of 2 in front of NH3 six
hydrogen atoms would be represented on the left side of the
equation. This equals the six on the right side.
2 NH3 + CO2  (NH2)2CO + H2O
Next, the balanced equation shows a 2:1 relationship between NH3
and CO2. So, if only 0.50 mole of CO2 were present then one mole
of NH3 would be needed to maintain the 2:1 reacting ratio.
0.50 mole CO2  2 mol NH3/1mol CO2 = 1.0 mol NH3
Section 3.7: Balancing Chemical Equations
Section 3.8: Stoichiometric Calculations: Amounts of Reactants
and Products
The Chemical Equation:
Mole & Masses
• C2H5OH + 3 O2  2 CO2 + 3 H2O
 46g
(1 mole) of ethanol reacts with 3 moles
of oxygen to produce 2 moles of carbon
dioxide and 3 moles of water.
 How
many grams of carbon dioxide and
water are respectively produced?
The Chemical Equation:
Moles & Masses
• C2H5OH + 3 O2  2 CO2 + 3 H2O
 46g
(1 mole) of ethanol requires 3 moles of
oxygen to produce 2 moles of carbon
dioxide and 3 moles of water.
 How
many grams of oxygen are needed to
react with all of the 46g (1 mole) of ethanol ?
 How
many grams of oxygen are needed to
react with 15.3g of ethanol in a 12oz. beer ?
Chemical Stoichiometry
 Stoichiometry
is the study of chemicals and
their quantities that are consumed and
produced in chemical reactions.
 It
quantitatively relates the behavior of atoms
and molecules to observable chemical change
and measurable mass effects.
QUESTION
The fuel in small portable lighters is butane (C4H10). Suppose after
using such a lighter for a few minutes (perhaps to encourage your
favorite concert performer to play one more encore) you had used 1.0
gram of fuel. How many moles of butane would this be?
1.
2.
3.
4.
58 moles
0.077 moles
1.7  10–24 moles
0.017 moles
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
ANSWER
4) 0.017 mol is the answer for a proper grams to mole conversion.
You need to know the molar mass of butane: 4  12g/mol = 48 for
carbon + 10  1g/mol = 10.0 for hydrogen. Total = 48.0 + 10.0 =
58.0 g/mol. Next; 1.0 gram of butane  1 mol/58.0 g = 0.017 mol
Section 3.3: Molar Mass
Mass Calculations:
Products
Reactants
Chemically Relate:
Something (S)
Another Thing (AT)
Mass (S)
Mass (AT)
grams (S)
grams (AT)
© Copyright 1995-2002 R.J. Rusay
Mass Calculations:
Products
Reactants
grams (S)
grams (S)
1 mol (S)
grams (AT)
Avogadro's Number
Atoms
Molecules
Stoichiometry
grams (AT)
(Molecular
Weight)
?
grams (AT)
grams (S)
(Molecular
Weight)
© Copyright 1995-2008 R.J. Rusay
?
"Gatekeeper"
1 mol (AT)
QUESTION
The fuel in small portable lighters is butane (C4H10). Suppose after
using such a lighter for a few minutes (perhaps to encourage your
favorite concert performer to play one more encore) you had used 1.0
gram of fuel. . How many moles of butane would this be?
0.017 moles
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
How many grams of carbon dioxide would this produce?
1.) 750 mg
2.) 6.0 g
3) 1.5 g
4.) 3.0 g
ANSWER
Choice 4.) 3.0 g is the answer for the correct grams to moles to grams
conversion.
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
C4H10
1.0 g
grams (S)
1.0 g C4H10
?g CO2
grams (S)
grams (AT)
mol
C4H10
1 mol (S)
Avogadro's Number
Atoms
Molecules
44 g CO2
8Stoichiometry
mol
grams (AT)
CO2
(Molecular
Weight)
?
3.0 g CO2
?g CO2
grams (AT)
grams (S)
(Molecular
Weight)
58 g C4H10
2 mol
?
C4H10
"Gatekeeper"
1 mol (AT)
mol CO2