MOLES AND CALCULATIONS USING THE MOLE CONCEPT
INTRODUCTORY TERMS
A. What is an amu ?
1.66 x 10-24 g
B. We need a conversion to the macroscopic world.
1. How many hydrogen atoms are in 1.00 g of hydrogen?
1 H atom
1.00 g H x
=
6.02 x 1023 H atoms
1.66 x 10-24 g H
Avogadro’s Number
2. Consider carbon-12 (the most abundant isotope of C)
What is the mass of one carbon-12 atom ?
amu
12
g
x 1.66 x 10-24
C atom
g
=
1.99 x 10-23
amu
C atom
3. What is the mass of Avogadro’s number of C atoms?
g
6.02 x 1023 C atoms x
1.99 x 10-23
=
C atom
12.00 g
C. Definition of a mole
1. A mole is the amount of any substance that contains as
many elementary entities as there are atoms in
exactly 1.00 g of hydrogen-1.
2. A mole is the amount ... in exactly 12.00 g of carbon-12.
3. 6.02 x 1023 of anything
4. It is important to state the entities involved:
atoms, molecules, ions, electrons, etc.
5. How large a number is this?
How Large a Quantity is Avogadro’s Number??
D. Definition of a molar mass (g/mol)
1. Molar mass of an element - The mass in grams that is
numerically equal to the atomic weight of the element.
C, 12.011 g/mol
Na, 22.99 g/mol
Xe, 131.30 g/mol
2. Molar mass of a molecular form of an element May be different
N2, 28.02 g/mol
Cl2, 70.906 g/mol (O2, I2, F2, Br2)
3. Molar mass of a compound - The mass in grams that
is numerically equal to the formula weight of the
compound.(The sum of the atomic weights involved)
H2O, 18.0152 g/mol
(1.01 + 1.01 + 15.99)
BaCl2, 208.23 g/mol
(137.33 + 35.45 + 35.45)
CO2, 44.01 g/mol
(12.01 + 15.99 + 15.99)
4. Sample Calculations of Molar Mass
a. Na2HPO4
Na2
H
P
O4
2 x 22.99 = 45.98
1 x 1.008 = 1.008
1 x 30.97 = 30.97
4 x 16.00 = 64.00
141.96 g/mol
b. Ca3(PO4)2
Ca3
(P )2
( O 4)2
3 x 40.08 = 120.24
2 x 30.97 =
61.94
8 x 16.00 = 128.00
310.18 g/mol
c. C15H22ClNO2
(Demerol)
C15
H22
Cl
N
O2
15 x 12.01 =
22 x 1.008 =
1 x 35.45 =
1 x 14.04 =
2 x 15.99 =
180.15
22.22
35.45
14.01
31.98
283.81 g/mol
THE ABOVE CONCEPTS LEAD TO
NEW "CONVERSION FACTORS"
1 mole
-------------------------6.02 x 1023 objects
or
6.02 x 1023 objects
-------------------------1 mole
How many atoms are represented by 3.00 moles of calcium (Ca) ?
3.00 mol Ca
X
6.02 x 1023 atoms
1.00 mol Ca
= 1.81 x 1024 atoms
of Ca
How many moles are represented by
1,000,000 molecules of carbon monoxide (CO) ?
1.0 x 106 molecules CO
1.00 mole CO
X
6.02 x 1023 molecules CO
=1.66 x 10-18 mol CO
MOLES
OF
SUBSTANCE
PARTICLES
OF
SUBSTANCE
Avogadro’s Number
ANOTHER PAIR OF
"CONVERSION FACTORS"
44.01 g of CO2
----------------------1 mole of CO 2
or
1 mole of CO 2
-------------------44.01 g of CO2
How many grams are contained in
4.25 moles of carbon dioxide (CO2) ?
4.25 moles CO 2 X
44.01 g of CO2
----------------------1 mole of CO 2
= 187 g CO2
3.77 g of CO2 is equal to how many moles of CO2 ?
3.77 g CO 2
X
1.00 mole of CO 2
-------------------- = .0856 mol CO 2
44.01 g of CO2
MOLES
OF
SUBSTANCE
GRAMS
OF
SUBSTANCE
Molar Mass
How many atoms are in 7.67 g of cobalt (Co) ?
1 mol of Co
7.67 g Co X -------------------58.93 g of Co
X
6.02 x 1023 atoms
1.00 mol Co
= 7.84 x 1022 atoms of Co
What is the mass of 1.77 x 1030 molecules of CO2 ?
1.77 x 1030 molecules of CO 2 X
X
44.01 g of CO2
----------------------1 mole of CO 2
1.00 mol CO 2
6.02 x 1023 molecules CO 2
= 1.29 x 108 g CO 2
GRAMS
OF
SUBSTANCE
MOLES
OF
SUBSTANCE
PARTICLES
OF
SUBSTANCE
Information Available from A Balanced Equation
Conversion Factors Relevant to Stoichiometry
Use Avogadro’s
number as a
conversion factor.
Moles of A
Particles of A
Use molar mass as a
conversion factor.
Moles of A
Grams of A
Use mole ratio as a
conversion factor.
Moles of A
Moles of B
Solution Map for Stoichiometry
Grams of
A
Grams of
B
Molar Mass
Molar Mass
Moles of
A
Coefficients
Moles of
B
Avogadro’s Number
Avogadro’s Number
Particles of
A
Particles of
B
Limiting Reactant, Theoretical Yield, and Percent Yield
Limiting Reactant
The reactant that is completely consumed in a chemical reaction
Theoretical Yield
The amount of product that can be made in a chemical reaction based on the amount of limiting
reactant
Actual Yield
The amount of product actually produced by a chemical reaction
Actual Yield
Percent Yield = --------------------------------- x 100%
Theoretical Yield
Consider the Reaction:
Ti (s) + 2 Cl2 (g) -------> TiCl4 (s)
If we begin the reaction with 1.8 mol of titanium and 3.2 mol of chlorine, what
is the limiting reactant and theoretical yield of TiCl4 in moles?
(theoretical yield)
If only 1.5 mol of TiCl4 is actually isolated from the reaction, what is the percent yield for the process?
Actual Yield
Percent Yield = --------------------------------Theoretical Yield
x
100%
=
1.5
-------1.6
x
100%
=
94%
Consider the Reaction:
2 Al (s) + 3 Cl2 (g) -------> 2AlCl3 (s)
If we begin the reaction with 0.552 mol of aluminum and 0.887 mol of
chlorine, what is the limiting reactant and theoretical yield of AlCl3 in moles?
(theoretical yield)
If only 0.448 mol of AlCl3 is actually isolated from the reaction, what is the percent yield for the process?
Actual Yield
Percent Yield = --------------------------------Theoretical Yield
x
100%
=
.448
-------.552
x
100%
=
81.1%
Consider the Reaction:
2 NO (g) + 5 H2 (g) -------> 2 NH3 (g) + 2 H2O (g)
What is the maximum amount of ammonia in grams that can be synthesized
from 45.8 g of NO and 12.4 g of H2?
(theoretical yield)
If only 16.4 g of NH3 is actually isolated from the reaction, what is the percent yield for the process?
Actual Yield
Percent Yield = --------------------------------Theoretical Yield
x
100%
=
16.4
-------26.0
x
100%
=
63.1%
Consider the Reaction:
2 Na (s) + Cl2 (g) -------> 2 NaCl (s)
If we begin the reaction with 53.2 g of sodium and 65.8 g of chlorine, what is the
limiting reactant and theoretical yield of NaCl in grams?
(theoretical yield)
If only 86.4 g of NaCl is actually isolated from the reaction, what is the percent yield for the process?
Percent Yield
=
Actual Yield
--------------------------------Theoretical Yield
x
100%
=
86.4
-------108
x
100%
= 80.0%
Consider the Reaction:
2 Fe (s) + 3 S (l) -------> Fe2S3 (s)
When 10.4 g of Fe are allowed to react with 11.8 g of S, 14.2 g of Fe2S3 are obtained. Find the
limiting reactant, theoretical yield and percent yield?
(theoretical yield)
Actual Yield
Percent Yield = --------------------------------Theoretical Yield
x
100%
=
14.2
-------- x 100% = 73.2%
19.4
Consider the Reaction:
Cu2O (s) + C (g) -------> 2 Cu (s) + CO (g)
When 11.5 g of C are allowed to react with 114.5 g of Cu2O, 87.4 g of Cu
are obtained. Find the limiting reactant, theoretical yield and percent yield?
(theoretical yield)
Percent Yield
=
Actual Yield
--------------------------------Theoretical Yield
x
100%
=
87.4
-------101.7
x
100%
=
85.9%
EMPIRICAL FORMULA
A chemical formula that indicates the relative
proportions of the elements in a molecule rather than
the actual number of atoms of the elements.
(An empirical formula may be obtained from percentage
composition of elements in a compound.)
MOLECULAR FORMULA
A chemical formula that indicates the actual number
of atoms of the elements in a molecule.
(Information in addition to percentage composition of
elements is needed to determine a molecular formula.)
Molecular Formulas
May Differ from Empirical Formulas
Benzene
Empirical Formula, CH
Molecular Formula, C 6H6
Acetylene
Empirical Formula, CH
Molecular Formula, C 2H2
Molecular Formulas
May Differ from Empirical Formulas
Glucose
Empirical Formula, CH2O
Molecular Formula, C 6H12O6
Fructose
Empirical Formula, CH2O
Molecular Formula, C 6H12O6
Solutions of Emprical Formula Problems