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
6.02 x 1023 particles
1 mole
or
1 mole
6.02 x 1023 particles
Learning Check
Suppose we invented a new collection unit
called a mep. One mep contains 8 objects.
A. How many paper clips in 1 mep?
1) 1
2) 4
3) 8
B. How many oranges in 2.0 meps?
1) 4
2) 8
3) 16
C. How many meps in 40 gummy bears?
1) 5
2) 10
3) 20
Solution
Suppose we invented a new collection unit
called a mep. One mep contains 8 objects.
A. How many paper clips in 1 mep?
3) 8
B. How many oranges in 2.0 meps?
3) 16
C. How many meps in 40 gummy bears?
1) 5
Learning Check
1. Number of atoms in 0.500 mole of Al
1) 500 Al atoms
2) 6.02 x 1023 Al atoms
3) 3.01 x 1023 Al atoms
Solution
1. Number of atoms in 0.500 mol of Al
0.500 mol Al x 6.02 x 1023 Al atoms = # Atoms
3) 3.01 x 1023 Al atoms
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?
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 percent
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
Calculations with Molar Mass
grams
= molar mass
moles
grams
= moles
molar mass
molar mass x moles = grams
Calculations with Molar Mass
grams = molar mass
moles
grams
=
moles
molar mass
molar mass x moles = grams
# Atoms etc
=
moles
(# atoms, ions, or molecules)
Avo #
moles x Avo # = # Atoms etc
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
moles
=
grams
molar mass
Setup
moles = 225 g aspartame
= 0.765 mol
294 g aspartame/mol
Solution
Molar mass = 146.0 g/mole
%=
total g C
x 100
total g compound
=
60.0 g C
146.0 g MSG
x 100 = 41.1% C
Molar Mass Factors
Methane CH4 known as natural gas is used
in gas cook tops and gas heaters. Express
the molar mass of methane in the form of
conversion factors.
Molar mass of CH4 =
16.0 g CH4
1 mole CH4
and
16.0 g
1 mole CH4
16.0 g CH4
Learning Check
Acetic acid CH3COOH is the acid in vinegar
. It has a molar mass of 60.0 g/mole.
1 mole of acetic acid
1 mole acetic acid
g acetic acid
=
____________
or
g acetic acid
1 mole acetic acid
Solution
Acetic acid CH3COOH is the acid in vinegar . It
has a molar mass of 60.0 g/mole.
1 mole of acetic acid
1 mole acetic acid
60.0 g acetic acid
=
or
60.0 g acetic acid
60.0 g acetic acid
1 mole acetic acid
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 types or number of
atoms, or ions, changes the compound
which is represented.
Example:
C6H12O6 is Glucose!
C6H12O5 is Not Glucose!
Other Types of Formulas
The formulas for compounds can be expressed as an
empirical formula and as a molecular(true) formula.
Empirical
Molecular (true)
Name
CH
C2H2
acetylene
CH
C6H6
benzene
CO2
CO2
carbon dioxide
CH2O
C5H10O5
ribose
Empirical Formulas
Define:
Empirical formula
Molecular 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 calc
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
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 =
2 =
n 88.0
Learning Check
Aspirin is 60.0% C, 4.50 % H
and 35.5% O. Calculate its
simplest formula.
Solution
60.0 g C
= ______ mol C
Molar mass C
4.50 g H
= ______ mol H
Molar mass H
35.5 g O
Molar mass O
= ______ mol 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
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?
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 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)
(1/3)
(1/4)
(3/4)
0.500
0.333
0.250
0.750
x2 =
1
x3 =
1
x4 =
1
x4 =
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 mols as the
subscripts in the simplest formula
C9H8O4
Learning Check
A compound is 27.4% S, 12.0% N and
60.6% Cl. If the compound has a
molar mass of 351 g/mol, what is the
molecular formula?
Change % to g and determine
moles of each element
27.4 g S x 1 mol C = 5.00 mol C
32.1 g C
12.0 g N x 1 mol H = 4.50 mol H
14.0 g H
60.6 g Cl x 1mol O = 2.22 mol O
35.5 g O
Divide by smaller # of moles
0.853 mol S /0.853 = 1 S
0.857 mol N /0.853 = 1 N
1.71 mol Cl /0.853
= 2 Cl
Empirical formula = SNCl2 =117.1 g/mol
Calculate the empirical formula for
the following compound
a. 15.8% carbon and 84.2% sulfur
Calculate the empirical formula
for the following compound

b. 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
Determine Molecular Formula
A compound has an empirical formula of
SNCl2, and a molar mass = 351.0 g/mol.
Determine it’s molecular formula.
Mol. Mass
=
=
Empirical mass
One More Time!
A compound is 40.0 % carbon, 53.3
% oxygen, and 6.66 % hydrogen.
What is its empirical formula? If
the compound has a molecular
weight equal to 60. g/mol, what is
the molecular formula?