Chapter 7
The Mole
Collection Terms
1 trio
=
3 singers
1 six-pack Cola
=
6 cans Cola drink
1 dozen donuts
=
12 donuts
1 gross of pencils
=
144 pencils
A Moles of Particles
Contains 6.02 x 1023 particles
1 mole C
= 6.02 x 1023 C atoms
1 mole H2O = 6.02 x 1023 H2O molecules
1 mole NaCl = 6.02 x 1023 Na+ ions and
6.02 x 1023 Cl– ions
Examples of Moles
Moles of elements
1 mole Mg = 6.02 x 1023 Mg atoms
1 mole Au = 6.02 x 1023 Au atoms
Moles of compounds
1 mole NH3= 6.02 x 1023 NH3molecules
1 mole C9H8O4 = 6.02 x 1023 aspirin molecules
Avogadro’s Number
is 6.02 x 1023 or, 1 mole
6.02 x 1023 particles
1 mole
or
1 mole
6.02 x 1023 particles
Using Isotopic Abundances
to determine the Average
Atomic Mass of an Element
Atomic Mass Determination



What is this number 35.453 for Cl??
The atomic mass - is an average based on the
percents and masses of each of the isotopes
of chlorine
Chlorine consists of 75.77% chlorine-35 (mm =
34.968853 amu) and 24.23% chlorine-37
(36.965903 amu)
Atomic Mass Determination

Chlorine consists of 75.77% chlorine-35
(mm = 34.968853 amu) and 24.23% chlorine-37
(36.965903 amu)
1: convert the percentage to a decimal:
0.7577 chlorine-35
0.2423 chlorine-37
2: 0.7577 x 34.968853 amu = 26.4958992 amu
0.2423 x 36.965903 amu = 8.9651 amu
35.4527 amu
Atomic Mass Determination

Nitrogen consists of two naturally
occurring isotopes



99.63% nitrogen-14 with a mass of
14.003 amu
0.37% nitrogen-15 with a mass of 15.000
amu
What is the atomic mass of nitrogen?
Finding the Molar Masses
of Elements and
Compounds
Molar Mass of Elements
remember these are atoms


Number of grams in 1 mole of atoms
Equal to the numerical value of the
atomic mass from the periodic table
1 mole of C atoms
=
12.0 g
1 mole of Mg atoms
=
24.3 g
1 mole of Cu atoms
=
63.5 g
Learning Check
Give the molar mass to 1.0 mole
A. 1 mole of Br atoms
________
=
B. 1 mole of Sn atoms
________
=
Solution
Give the molar mass to 0.1 g
A. 1 mole of Br atoms =
79.9 g/mole
B. 1 mole of Sn atoms =
118.7 g/mole
Molar Mass of Compounds
Both molecules and Ionic Compounds
Mass in grams of 1 mole is equal to the sum of the
atomic masses of all the atoms present
1 mole of CaCl2 = 111.1 g/mole
(1 mole Ca x 40.1 g/mole) + (2 moles Cl x 35.5 g/mole)
1 mole of N2O4 = 92.0 g/mole
(2 moles N x 14.0 g/mole) + (4 moles O x 16.0 g/mole)
Learning Check
A. 1 mole of K2O = ______g
B. 1 mole of antacid Al(OH)3 = ______g
Solution
A. 1 mole of K2O
(2 K x 39.1 g/mole) + (1 O x 16.0 g/mole) = 94.2 g
B. 1 mole of antacid Al(OH)3
(1 Al x 27.0 g/mol) +(3 O x 16.0 g/mol) + (3 H x 1.0) = 78.0 g
Learning Check
Prozac, C17H18F3NO, is a widely used
antidepressant that inhibits the uptake of
serotonin by the brain. It has a molar mass of
1) 40.0 g/mole
2) 262 g/mole
3) 309 g/mole
Solution
Prozac, C17H18F3NO, is a widely used
antidepressant that inhibits the uptake of
serotonin by the brain. It has a molar mass of
17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0)
3) 309 g/mole
Determine the% composition of
H2SO4


1 – Calc the # grams in a mole of the
formula
2 – Divide each of the elements total
masses by the molar mass

3 – Mult by 100 to make a percent
Determine the percent
composition of H2SO4

H 2 x 1.0 g

S 1 x 32.1 g

O 4 x 16.0 g
= 2.0 g x 100 =
98.1
= 32.1 g x 100 =
98.1
= 64.0 g x 100 =
98.1 g/mol
What is the percent composition of
C5H8NO4 MSG, a compound used to
flavor foods and tenderize meats?
What is the percent
composition of Ca(NO3)2
Molar Math
Converting moles
particle, and
Converting moles
mass
Convert 0.187 mol Na+ to ions.
(
) = 1.13 x 10
0.187 mol Na+ 6.02 x 1023 ions Na
1
23
ions Na
Convert 5.66 x 1023 atoms Xe to moles.
5.66 x 1023 atoms Xe
1 mole Xe
6.02 x 1023 atoms Xe
(
)=0.940 mol Xe
Moles = grams/molar mass
Determine the number of moles in 27.35 g of the element
sulfur.
Calculate the number of moles of SiO2 in a 18.95 g sample
Moles = grams/molar mass


Determine the number of grams in 0.0595 moles
of H2O ( mm = 18.0 g/mol)?
Determine the number of grams in 1.83 x 10-3
moles of SiO2 (SiO2 mm = 60.1 g/mol)
Learning Check
The artificial sweetener aspartame
(Nutri-Sweet) formula C14H18N2O5
is used to sweeten diet foods,
coffee and soft drinks. How many
moles of aspartame are present in
225 g of aspartame?
Solution
Molar mass of Aspartame C14H18N2O5
(14 x 12.0) + (18 x 1.01) + (2 x 14.0) + (5 x 16.0)
= 294 g/mole
Setup
moles = 225 g aspartame
= 0.765 mol
294 g aspartame/mol
= 0.765 mole aspartame
Moles and Grams
Aluminum is often used for the
structure of light-weight bicycle
frames. How many grams of Al are in
3.00 moles of Al?
3.00 moles Al
? g Al
Mole Mass Conversions
(grams = mol x molar mass)
Setup
grams = mol x molar mass
3.00 moles Al x 27.0 g Al
= 81.0 g Al
Formulas
Chemical Formulas
Include the symbols of the elements and the
subscripts, which indicate the number of
atoms, or ions, of each.
Example:
C6H12O6 – Glucose
Contains 6 atoms of Carbon, 12 atoms of
Hydrogen and 6 atoms of Oxygen.
Chemical Formulas
Any change to the symbols or suscripts
changes the formula. Changing the
formula changes the compound which
is represented.
Example:
C6H12O6 is Glucose!
C6H12O5 is Not Glucose!
Other Types of Formulas
The formulas for compounds can also be expressed
as: (1.) an empirical formula and as (2.) a molecular
(true) formula.
Empirical
Molecular (true)
Name
CH
C2H2
acetylene
CH
C6H6
benzene
CO2
CO2
carbon dioxide
CH2O
C5H10O5
ribose
Defining Empirical
and
Molecular Formula
Empirical formula: A formula that shows
the simplest whole-number ratio of atoms
in the formula.
Molecular formula: A formula that is a
whole-number multiple of the empirical
formula.
Learning Check
A. What is the empirical formula for C4H8?
1) C2H4
2) CH2
3) CH
1) C4H7
2) C6H12
3) C8H14
B. What is the empirical formula for C8H14?
C. What is a molecular formula for CH2O?
1) CH2O
2) C2H4O2
3) C3H6O3
Solution
A. What is the empirical formula of C4H8?
2) CH2
B. What is the empirical formula of C8H14?
1) C4H7
C. What is a molecular formula of CH2O?
1) CH2O
2) C2H4O2
3) C3H6O3
Determining Empirical Formulas
1. You need grams of each element to
calculate the moles of each element. Assume
you have a 100 g sample!
 2. Calculate the number of moles of
each element.
 3. Find the smallest whole number ratio by
dividing each mole value by the smallest mole
values:
 4. Write the simplest or empirical formula

Finding the Empirical Formula
A compound is 71.66% Cl, 24.28% C,
and 4.06% H. The molar mass is known
to be 99.0 g/mol. What is the
empirical formula?
1. Determine the number of grams of
each element.
71.66 % Cl becomes 71.66 g Cl
24.28 % C becomes 24.28 g C
4.06 % H becomes 4.06 g H
2. Calculate the number of moles of
each element (divide by molar mass).
71.66 g Cl
= 2.02 mol Cl
35.5 g/mol Cl
24.28 g C
12.0 g/mol C
= 2.02 mol C
4.06 g H
1.0 g/mol H
= 4.06 mol H
Why moles?
Why do you need the number of
moles of each element in the
compound?
The subscripts in the chemical formula
will equal the mole to mole ratio of
the elements present.
3. Find the smallest whole
number ratio by dividing each mole
value by the smallest mole values:
Cl:
2.02
2.02
= 1 Cl
C:
2.02
2.02
= 1C
H:
4.06
2.02
= 2.005 H
4. Write the simplest or empirical
formula = CH2Cl
Determining a Molecular
Formula from an Empirical
Formula
In Order to Determine the
Molecular Formula from
the Empirical Formula you
must have the mass of
both the Molecular
Formula and the Empirical
Formula.
Empirical and Molecular Formulas
Molar Mass of the molecular formula
= n
Molar Mass of the empirical formula
or
Empirical formula x n = Molecular formula
5. Determine the molecular formula
from the Empirical form of
CH2Cl….remember,the formula mass
was 99.0 g/mol.
1(C) + 2(H) + 1(Cl) =
1(12.0) + 2(1.0) + 1(35.5) = 49.5g/mol
Determine multiplying factor = n
n =
mm molec form =
mm emp form
99.0 g/mol
49.5 g/mol
= 2
Solution: CH2Cl x 2 = C2H4Cl2
Learning Check
A compound has a formula mass of
176.0 and an empirical formula of
C3H4O3. What is the molecular formula?
Solution
A compound has a formula mass of 176.0
and an empirical formula of C3H4O3.
What is the molecular formula?
C3H4O3 = 88.0 g/mol
176.0 g
88.0 g
C6H8O6
= 2
Learning Check
Aspirin is 60.0% C, 4.50 % H
and 35.5% O. Calculate its
Empirical formula.
Solution
60.0 g C
=
______ mol C
Molar mass C
4.50 g H
=
______ mol H
=
______ mol O
Molar mass H
35.5 g O
Molar mass O
Solution
60.0 g C
= 5.00 mol C
12.0 g/mol C
4.50 g H
= 4.50 mol H
1.01 g/mol H
35.5 g O
16.0 g/mol O
= 2.22 mol O
Divide by the smallest # of moles.
5.00 mol C
2.22 mol O
=
2.25
4.50 mol H
2.22 mol O
=
2.00
2.22 mol O =
1.00
2.22 mol O
Are the results whole numbers?_____
Finding Subscripts
A fraction between 0.1 and 0.9 must not
be rounded if the given is in percents.
Multiply all results by an integer to give
whole numbers for subscripts.
(1/2) 0.500 x 2 =
1
(1/3)
0.333
x3 =
(1/4)
0.250
x4 =
(3/4) 0.750
x4 =
1
1
3
Multiply everything x 4
C: 2.25 mol C
x 4 = 9 mol C
H: 2.00 mol H
x 4 = 8 mol H
O: 1.00 mol O
x 4 = 4 mol O
Use the whole numbers of moles as the
subscripts in the simplest formula
C9H8O4
Practice
Finding the Molecular Formula
A compound is 71.65% Cl, 24.27% C, and
4.07% H. The molar mass is known to be
99.0 g/mol. What are the empirical and
molecular formulas?
Calculate the empirical formula
for the following compound
15.8% carbon and 84.2% sulfur
Calculate the empirical formula
for the following compound
28.7% K, 1.5% H, 22.8% P and 47.0% O
Calculate the empirical formula
for the following compound
43.6% phosphorus and 56.4% oxygen