Chapter 3
Atoms: The Building
Blocks of Matter
1
Objectives
•Define the terms atomic mass and molar mass.
•Define the terms mole and Avogadro’s number.
•Write the name for common elements, given the symbol,
or the symbol, given the name.
•Calculate the molar mass of an element or compound,
given its formula.
•Calculate the mass of an element or compound given the
number of moles, or the number of moles of a given
mass of an element or compound.
•Calculate the number of atoms or molecules of an
element or compound given the number of moles, or the
number of moles given the number of atoms or
molecules.
2
Chapter 3
Section 1
The Atom
3
The modern definition of an element is a
substance that cannot be further broken
down by ordinary chemical means – H, C, O
Elements also combine to form compounds
that have different physical and chemical
properties than those of the elements that
form them – H2O.
The transformation of a substance into one
or more new substances is a chemical
reaction.
4
Relative Atomic Masses
Masses of atoms expressed in grams are
very small and not useful. (O: 2.66 x 10-23 g)
It is more convenient to use relative atomic
masses.
Therefore the standard used to govern units
of atomic mass is the carbon-12 atom.
It has been assigned a mass of exactly
12 atomic mass units, or 12 amu.
5
The atomic mass of any other atom is
determined by comparing it with the mass
of the carbon-12 atom.
Examples:
The hydrogen atom has an atomic mass of
1/12 that of the carbon-12 atom or 1 amu.
Oxygen has an atomic mass of 16/12 the
mass of a carbon-12 atom or 16 amu.
6
Relating Mass to Numbers of
Atoms
Introduction of three very important concepts:
1) The mole
2) Avogadro’s number
3) Molar mass
7
The Mole
Mole – the amount of carbon atoms that
are in exactly 12 g of carbon.
One mole of carbon weighs 12 grams.
The mole is the SI unit for amount of
substance. It is a counting unit.
Mole is related to the counting term dozen.
8
What is a counting unit?
You’re already familiar with one counting unit…a
“dozen”
A dozen = 12
“Dozen”
9
12
A dozen doughnuts
12 doughnuts
A dozen books
12 books
A dozen cars
12 cars
A dozen people
12 people
A Mole of Particles
Contains 6.02 x 1023 particles
Avogadro’s Number
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 NaCl molecules
10
Avogadro’s Number – 6.02 x 1023
is the number of particles in exactly one
mole of a pure substance.
1 mole of gold = 6.02 x 1023 particles
1 mole of uranium = 6.02 x 1023 particles
1 mole of water = 6.02 x 1023 particles
11
How big is a mole?
Enough soft drink cans to cover the surface
of the earth to a depth of over 200 miles.
If we were able to count atoms at the rate of
10 million per second, it would take about 2
billion years to count the atoms in one
mole.
12
Molar Mass
Molar Mass – The mass (in grams) of one
mole of a pure substance.
Molar masses are written in units g/mol.
The molar mass of an element is equal to
the atomic mass of the element.
Look on the periodic table for the atomic
masses.
13
Other terms commonly used for
the same meaning of molar mass:
Molecular Weight
Molecular Mass
Formula Weight
Formula Mass
14
Molar Mass
A molar mass of an element contains one
mole of atoms.
4.00 g helium = 1mole = 6.02 x 1023 atoms.
6.94 g lithium = 1mole = 6.02 x 1023 atoms.
200.6 g mercury = 1mole = 6.02 x 1023
atoms.
15
Molar Mass
Examples:
Molar mass of oxygen (O) = 15.99 g/mol
Molar mass of iron (Fe) = 55.85 g/mol
Molar mass of mercury (Hg) = 200.6 g/mol
16
One mole of carbon (12 grams) and
one mole of copper (63.5 grams)
Both contain 6.02 x 1023 atoms
17
Molar Mass
A molar mass of a compound is the sum
of the molar masses of the elements.
Example: Water, H2O:
2 H = 2 x 1g/mole = 2g/mole
1 O = 1 x 16g/mole = 16g/mole
molar mass of H20 =18g/mol
18
Molar Mass
A molar mass of a compound is the sum
of the molar masses of the elements.
Example: methane, CH4:
4 H = 4 x 1 g = 4 g/mole
1 C = 1 x 12 g = 12 g/mole
molar mass of CH4 =16 g/mol
19
Molar Mass for Compounds
The molar mass for a compound = the
sum of the molar masses of all the
atoms in the compound.
20
Example: Molar Mass &
Parenthesis
Be sure to distribute the subscript outside the
parenthesis to each element inside the parenthesis.
Example:
Find the
molar
mass for
Sr(NO3)2
21
Example
Example:
Find the
molar
mass for
Al(OH)3
22
Homework
Worksheet C.5 – molar masses of
compounds
Due:
23
Gram/Mole Conversions
How many roses are in 3 ½ dozen
roses?
Relationship: 1 dozen roses = 12 roses
3.5 dozen x 12 roses =
1 dozen
24
42 roses
Gram/Mole Conversions
Molar masses can be used as a conversion
factor in chemical calculations.
Example: The molar mass of helium is 4.00
g/mol. To calculate how many grams of
helium are in 2 moles of helium:
amount of He in moles
amount of He in grams
4.00 g He
2.00 mol He x 1 mol He
25
= 8.00 g He
Gram/Mole Conversions
Molar masses can be used as a conversion
factor in chemical calculations.
Amount in moles x molar mass (g/mol) = mass in grams
Example: What is the mass in grams of
2.50 mol of oxygen gas?
2.50 mol O2 x
26
32.00 g O2
1 mol O2
= 80.0 g O2
Gram/Mole Conversions
A chemist produced 11.9 g of aluminum, Al.
How many moles of aluminum were
produced?
mass of Al in grams
1 mol Al
11.9 g Al x
27 g Al
27
amount of Al in moles
= 0.44 mol Al
Gram/Mole Conversions
Review sample problem – page 82
Practice problems – Top of page 83, 1-4
(worksheet)
28
Gram/Mole Conversions
Practice problems – page 83 (bottom), 1-2
(worksheet)
29
Classwork
Worksheet: C-7
Grams to mole conversions
Moles to gram conversions
30
Homework
Worksheet: C-13
(skip)
Grams to mole conversions
Moles to gram conversions
31
Atoms / Gram Conversions
32
Atoms/Molecules and Grams
Since 6.02 X 1023 molecules = 1 mole
AND
1 mole = molar mass (grams)
You can convert atoms/molecules to
moles and then moles to grams! (Two
step process)
You can’t go directly from atoms to
grams!!!! You MUST go thru MOLES.
33
Calculations
molar mass
Grams
Avogadro’s number
Moles
atoms
Everything must go through
Moles!!!
34
Atoms/Molecules and Grams
How many atoms of Cu are
present in 35.4 g of Cu?
35.4 g Cu
1 mol Cu
63.5 g Cu
6.02 X 1023 atoms Cu
1 mol Cu
= 3.4 X 1023 atoms Cu
35
Problem
How many atoms of K are present in
78.4 g of K?
36
Atoms/Molecules and Grams
How many atoms of K are present
in 78.4 g of K?
78.4 g K
1 mol K
39.0 g K
6.02 X 1023 atoms K
1 mol K
= 12.1 X 1023 atoms K
37
Problem
What is the mass (in grams) of 1.20 X
108 atoms of copper (Cu)?
38
Problem
What is the mass (in grams) of 1.20 X
108 atoms of copper (Cu)?
1.20 x 108 atoms
1 mole Cu
63.5 g Cu
6.02 x 1023 atoms 1 mole Cu
= 1.27 x 10-14 grams copper
39
Problem
What is the mass (in grams) of 1.20 X
1024 molecules of glucose (C6H12O6)?
40
Problem
What is the mass (in grams) of 1.20 X
1024 molecules of glucose (C6H12O6)?
1.20 x 1024 mol.
1 mole glucose
6.02 x 1023 mol.
= 359 grams glucose
41
180 g glucose
1 mole glucose
Classwork
Problems – page 88
Questions: 21, 24 (a, c, e and f),
28 (a-e)
42
Chapter 7
Chemical Formulas and
Chemical Compounds
43
Chapter 7
Section 3
Chemical Formulas
44
A chemical formula indicates which
elements are in a compound and
how many of each element.
Example:
Water – H2O
Subscripts indicate there are two
atoms of hydrogen and one atom of
oxygen in water.
45
Example:
Aluminum sulfate – Al2(SO4)3
Parenthesis are used to surround the
polyatomic group to identify it as a unit.
The subscript 3 refers to everything inside
the parenthesis.
Al – 2 atoms
S – 3 atoms
O – 12 atoms
46
Classwork
Practice problems – page 226
Questions 1 - 3
47
Percentage Composition
It is often useful to know the percentage by
mass of a particular element in a
chemical compound.
Percentage Composition – the percentage by
mass of each element in a compound.
Mass of element in a compound
molar mass of compound
48
x 100
=
% element
in cmpd
Example:
Find the percentage composition of each
element in copper (I) sulfide, Cu2S.
The molar mass of Cu2S = 159.2 g/mol
2 mole Cu x 63.51 g Cu = 127.1 g Cu
mole
1 mole S x
49
32.07 g S = 32.07 g S
mole
Example:
Find the percentage composition of each
element in copper (I) sulfide, Cu2S.
The molar mass of Cu2S = 159.2 g/mol
% Cu: 127.1 g Cu
X 100 = 79.85% Cu
159.2 g Cu2S
% S:
50
32.07 g S
159.2 g Cu2S
X 100 = 20.15% S
Classwork
Practice problem – page 228
Questions 1 A and B
51
Chapter 7
Section 4
Determining
Chemical Formulas
52
Empirical Formula – consists of the
symbols for the elements in a compound,
with subscripts showing the smallest
whole-number mole ratio of the different
elements in the compound.
Example:
53
H2O not H4O2
Empirical formula does not necessarily
indicate the actual numbers of each
element in a molecule.
Example:
Diborane gas
Empirical formula is BH3
Molecular formula is B2H6
54
Calculating Empirical Formula
To determine a compound’s empirical
formula from its percentage composition,
begin by converting percentage
composition to a mass composition.
Example:
Assume you have a 100 g sample of
diborane, B2H6.
55
The percentage composition of B2H6 is:
The molar mass of B2H6 = 28 g/mol
% B:
22 g B
X 100 = 78% B
28 g B2H6
% H:
6gH
28 g B2H6
X 100 = 22% H
Therefore, 100 g of diborane contains
78 g of B and 22 g of H.
56
Next the mass composition of each
element is converted to composition in
moles:
1 mol B
78.1 g B x 10.81 g B = 7.22 mol B
21.9 g H x 1 mol H = 21.7 mol H
1.01 g H
57
These values give a mole ratio of 7.22 mol B
to 21.7 mol H.
Convert these numbers to the smallest
whole numbers:
7.22 mol B
7.22
21.7 mol H
7.22
58
= 1
= 3.01
Thus, diborane contains a mole ratio of
1 B : 3 H to give an empirical formula of:
BH3
Review Sample Problem 7-12 and 7-13,
Page 230 and 231
59
Calculating Molecular Formula
The molecular formula is the actual formula
for the compound.
To know the molecular formula you must
know the compound of interests molecular
weight.
60
Example:
Experimentation shows the molecular
weight for diborane is 27.67 g/mol.
The molecular weight for the empirical
formula BH3 is 13.84 g/mol.
Dividing the formula mass by the empirical
mass yields:
27.67 = 2
13.84
61
Take this number and times it by the
subscripts in the empirical formula to get
the molecular formula:
2 X BH3 = B2H6
62
Classwork
Do Practice Problems 1-3 , page 231
Page 233: Section Review
Questions 1- 3
63
Homework
Page 237: Chapter 7 Review Problems
Questions 34, 35, 36, 37
64
Chapter 11
Section 1
Volume-Mass
Relationships of
Gases
65
Objectives
 Calculate the volume of a gas at STP
(standard temperature and pressure), or
the number of moles of a given volume
of an element or compound.
66
Gases
 Monoatomic Gases – gases which are
composed of only one atom of the
element. Ex: He, Ar, Xe – Noble gases
 Diatomic Gases – gases which are
composed of only two atom of the
element. Ex: H2, N2, O2, F2, Cl2
67
Molar Volume of Gases
Recall that one mole of any substance
contains a number of molecules equal to
Avogadro's number – 6.02 x 1023
One mole of oxygen gas (O2), contains
6.02 x 1023 diatomic oxygen molecules and
has a mass of 32.0 g/mol.
68
One mole of hydrogen gas (H2), contains
6.02 x 1023 diatomic hydrogen molecules
and has a mass of 2.0 g/mol.
One mole of helium gas (He), contains
6.02 x 1023 monatomic helium atoms and
has a mass of 4.0 g/mol.
69
Avogadro’s Law – equal volumes of gases
at the same temperature and pressure
contain equal number of molecules.
According to Avogadro’s law, one mole of
any gas will occupy the same volume as
one mole of any other gas at the same
temperature and pressure, despite mass
differences.
One mole of H2, O2 and He occupy the
same volume at same temp. and pressure.
70
Standard Molar Volume of a Gas – the
volume occupied by one mole of a gas at
STP (standard temperature and pressure).
The volume = 22.4 L
Standard temperature = 273K = 0oC
Standard pressure is 1 atm.
71
Knowing the volume of a gas, you can use
1 mol/22.4 L as a conversion factor to find
the number of moles and mass of a given
volume of gas at STP.
1 mol of O2
volume = 22.4 L
mass = 32.0 g/mol
72
vs.
1 mol of H2
volume = 22.4 L
mass = 2.0 g/mol
Problem:
A chemical reaction produces 0.068 mol of
oxygen gas. What volume in liters is
occupied by this gas at STP?
moles of O2
22.4
L
mol x
1 mol
73
volume of O2 in L
=
volume of O2 in L
moles of O2
mol x 22.4 L
1 mol
volume of O2 in L
=
22.4 L
0.068 mol x
1 mol
74
volume of O2 in L
= 1.52 L of O2
Problem:
A chemical reaction produced 98.0 mL of
sulfur dioxide gas, SO2, at STP. What is the
mass in grams of the gas produced?
liters of SO2
L x
75
moles of SO2
1 mol SO2
22.4 L
grams of SO2
g
SO
__
2
x
=
1 mol SO2
g SO2
liters of SO2
1 mol SO2
L x
22.4 L
0.098 L x 1 mol SO2
22.4 L
76
moles of SO2
grams of SO2
g SO2__
x
=
1 mol SO2
x 64.07g SO2__
1 mol SO2
g SO2
= 0.28 g SO2
We will use the concept of molar
volume of gases when we talk about
the gas laws later in the year!
77
Classwork
Page 336-337: Practice Problems
Questions 1-3
Page 337: Practice Problems
Questions 1-3
78
Chapter 13
Section 3
Concentrations
of
Solutions
79
Objectives
 Define a solution’s concentration in terms of
molarity.
 Calculate the concentration of a solution, given
the mass of solute and volume of solution, or
the mass of solute needed to prepare a
solution of a given concentration.
80
What is solubility?
• Imagine dissolving a
spoonful of salt in water
• What is the solute?
• What is the solvent?
81
81
Concentration of a solution – a measure of
the amount of solute in a given amount of
solvent.
Common solution terms:
Dilute – small amount of solute in a
solvent.
Concentrated – a large amount of solute in
a solvent
82
Molarity
Molarity (M) – the number of moles of
solute in one liter of solution.
To determine the molarity of a solution
you must know the molar mass
(molecular weight) of the solute.
83
For example, a one molar solution of
sodium hydroxide, (NaOH), contains one
mole of NaOH in one liter of solution.
1 M NaOH
= 1 mol NaOH
1L
Since 1 mole of NaOH = 40 g
40.0g NaOH
1M NaOH =
1L
84
The relationship between molarity, moles
and volume is the following:
Amount of solute (mol)
Molarity (M) =
Volume of solution (L)
85
Example problem:
What is the molarity of 20 g of NaOH in 1 L
of solution? (NaOH = 40 g/mol)
Molarity (M) =
Amount of solute (mol)
Volume of solution (L)
Molarity (M) =
0.50 mol NaOH
1.0 L
Molarity (M) = 0.50 M
86
Example problem:
What is the molarity of a 3.50 L solution
that contains 90.0 g of sodium chloride,
NaCl?
Molarity (M) =
Amount of solute (mol)
Volume of solution (L)
You first need to convert 90.0 g of NaCl to
moles of NaCl.
90.0 g___
Moles of NaCl = 58.5 g/mol = 1.54 mol
87
Molarity (M) =
Amount of solute (mol)
Volume of solution (L)
Molarity (M) =
1.54 mol NaCl
3.50 L
Molarity (M) = 0.440 M NaCl
88
You have 0.8 L of a 0.5 M HCl solution.
How many moles of HCl does this solution
contain? (HCl = 36.5 g/mol)
Molarity (M) =
0.5 M =
Amount of solute (mol)
Volume of solution (L)
moles of HCl
0.8 L
moles of HCl = (0.50 M)(0.8 L) = 0.4 mol
HCl
89
Example problem:
To produce 40.0 g of silver chromate, you
will need at least 23.4 g of potassium
chromate (K2CrO4) in solution as a
reactant. All you have in the stock room is
5 L of a 6.0 M K2CrO4 solution. What
volume of this solution is needed to give
you the 23.4 g of K2CrO4?
Molarity (M) =
90
Amount of solute (mol)
Volume of solution (L)
Given:
concentration of solution = 6.0 M of K2CrO4
mass of solute needed = 23.4 g of K2CrO4
maximum volume of solution = 5 L
(not used in the calculation of molarity)
Calculated:
Molar mass of K2CrO4 = 194.2 g/mol
Molarity (M) =
91
Amount of solute (mol)
Volume of solution (L)
Molarity (M) =
Amount of solute (mol)
Volume of solution (L)
You first need to convert 23.4 g of K2CrO4
to moles of K2CrO4.
23.4 g___
Moles of K2CrO4 = 194.2 g/mol = 0.12 mol
0.12 mol
6.0 M K2CrO4 =
L
0.12 mol
Liter (L) = 6.0M
= 0.020 L of K2CrO4
92
Classwork
Page 415:
Questions 1-3
93
Homework
Page 421: Chapter 13 Review Problems
Questions 15 (b), 16 (a-b), 17, 18 (a,c), 19
Period 8: Due November 6th
94
Gram/Mole Conversions
How many moles are present in 352 g of
iron(III) oxide, Fe2O3?
mass of Fe2O3 in grams
amount of Fe2O3 in moles
Calculate the molar mass of Fe2O3 = 160 g/mol
352 g Fe2O3 x 1 mol Fe2O3 = 2.2 mol Fe2O3
160 g
95