1. Chapter 2
The Components of Matter
2-1

2. Chapter 2: The Components of Matter
2.1 Elements, Compounds, and Mixtures:
An Atomic Overview
2.2 The Observations That Led to an Atomic View of Matter
2.3 Dalton’s Atomic Theory
2.4 The Observations That Led to the Nuclear Atom Model
2.5 The Atomic Theory Today
2-2

3. Chapter 2: The Components of Matter
2.6 Elements: A First Look at the Periodic Table
2-3

4. Atoms, Molecules and Ions
•
•
•
•
•
2-4
Dalton’s Atomic Theory
Atomic Structure
Atomic Mass
Chemical Formulas
Chapters 2.1-2.5, 3.1 (Silberberg)

5. Goals & Objectives
• The student will understand the
characteristics of the three types of
matter: elements, compounds, and
mixtures. (2.1)
• The student will understand the
significance of the three mass laws:
conservation of mass, definite
composition and multiple proportions.
(2.2)
2-5

6. Goals & Objectives
• The student will understand the
postulates of Dalton’s atomic theory
and how it explains the mass laws.
(2.3)
• The student will understand the major
contributions of experiments by
Thomson, Millikan, and Rutherford
concerning the atomic structure. (2.4)
2-6

7. Goals & Objectives
• The student will understand the
structure of the atom along with the
characteristics of the proton, neutron,
and electron and the importance of
isotopes. (2.5)
• The student will understand the
meaning and usefulness of the mole.
(3.1)
2-7

8. Goals & Objectives
• The student will understand the
relationship between the formula
mass and molar mass. (3.1)
• The student will understand the
relationship among the amount of
substance (in moles), mass (in
grams), and the number of chemical
atoms. (3.1)
2-8

9. Goals & Objectives
• The student will understand the
information in a chemical formula.
(3.1)
2-9

10. Master these Skills
• Distinguish between elements,
compounds and mixtures at the
atomic scale. (SP 2.1)
• Using the mass ratio of an element to
a compound to find the mass of an
element in a compound. (SP 2.2)
• Visualizing the mass laws. (SP 2.3)
2-10

11. Master these Skills
• Using atomic notation to express the
subatomic makeup of an isotope. (SP
2.4)
• Calculating an atomic mass from
isotopic composition. (SP 2.5)
• Calculating the molar mass of any
substance. (3.1; SP 3.3 and 3.4)
2-11

12. Master these Skills
• Converting between amount of
substance (in moles), mass (in
grams), and the number of atoms.
(SP 3.1-3.3)
2-12

13. Matter
• A scheme for the classification of
matter
2-13

14. Definitions for Components of Matter
Element - the simplest type of substance with unique physical and
chemical properties. An element consists of only one type of atom. It
cannot be broken down into any simpler substances by physical or
chemical means.
Molecule - a structure that consists of two or
more atoms that are chemically bound together
and thus behaves as an independent unit.
Figure 2.1
2-14

15. Definitions for Components of Matter
Compound - a substance
composed of two or more elements
which are chemically combined.
Figure 2.1
Mixture - a group of two or more
elements and/or compounds that
are physically intermingled.
2-15

16. Table 2.1
Some Properties of Sodium, Chlorine, and Sodium
Chloride.
Property
Sodium
Melting point
Chlorine
Sodium Chloride
97.8 C
-101 C
801 C
Boiling point
881.4 C
-34 C
1413 C
Color
Silvery
Yellow-green
Colorless (white)
Density
0.97 g/cm3
0.0032 g/cm3
2.16 g/cm3
Behavior in water
Reacts
Dissolves slightly
Dissolves freely
2-16
+

17. Figure 2.2 The law of mass conservation.
The total mass of substances does not change during a chemical reaction.
2-17

18. Law of Mass Conservation
The total mass of substances present does not change
during a chemical reaction.
reactant 1
+
product
reactant 2
total mass
calcium oxide + carbon dioxide
=
total mass
calcium carbonate
CaO
CO2
CaCO3
56.08 g
2-18
+
+
44.00 g
100.08 g

19. Law of Definite (or Constant) Composition
No matter the source, a particular compound is
composed of the same elements in the same parts
(fractions) by mass.
Figure 2.3
2-19

20. Calcium carbonate
Analysis by Mass
(grams/20.0 g)
8.0 g calcium
2.4 g carbon
9.6 g oxygen
20.0 g
2-20
Mass Fraction
(parts/1.00 part)
Percent by Mass
(parts/100 parts)
0.40 calcium
0.12 carbon
0.48 oxygen
40% calcium
12% carbon
48% oxygen
1.00 part by mass
100% by mass

21. Law of Multiple Proportions
If elements A and B react to form two compounds, the
different masses of B that combine with a fixed mass of A
can be expressed as a ratio of small whole numbers.
Example: Carbon Oxides A & B
Carbon Oxide I : 57.1% oxygen and 42.9% carbon
Carbon Oxide II : 72.7% oxygen and 27.3% carbon
2-21

22. Assume that you have 100 g of each compound.
In 100 g of each compound:
g O = 57.1 g for oxide I & 72.7 g for oxide
g C = 42.9 g for oxide I & 27.3 g for oxide II
For oxide I:
gO
57.1
=
gC
42.9
= 1.33
For oxide II:
gO
gC
= 2.66
=
72.7
27.3
2.66 g O/g C in II
1.33 g O/g C in I
2-22
=
2
1

23. Dalton’s Atomic Theory
Dalton postulated that:
1.
2.
3.
4.
2-23
All matter consists of atoms; tiny indivisible particles of
an element that cannot be created or destroyed.
Atoms of one element cannot be converted into atoms
of another element.
Atoms of an element are identical in mass and other
properties and are different from the atoms of any
other element.
Compounds result from the chemical combination of a
specific ratio of atoms of different elements.

24. Dalton’s Atomic Theory
explains the mass laws
Mass conservation
Atoms cannot be created or destroyed
postulate 1
or converted into other types of atoms.
postulate 2
Since every atom has a fixed mass,
postulate 3
during a chemical reaction the same atoms are
present but in different combinations; therefore
there is no mass change overall.
2-24

25. Dalton’s Atomic Theory
explains the mass laws
Definite composition
Atoms are combined in compounds in
specific ratios
postulate 3
and each atom has a specific mass.
postulate 4
Each element constitutes a fixed fraction of the total
mass in a compound.
2-25

26. Dalton’s Atomic Theory
explains the mass laws
Multiple proportions
Atoms of an element have the same mass postulate 3
and atoms are indivisible.
postulate 1
When different numbers of atoms of elements
combine, they must do so in ratios of small, whole
numbers.
2-26

27. The Structure of the Atom
•
•
•
•
•
2-27
Fundamental Particles in Atoms
Particle
Mass(amu)
Charge
electron
0.00054858 -1
proton
1.0073
+1
neutron
1.0087
0

28. The Structure of the Atom
• Electrons
2-28

29. Figure 2.4
Observations that established the properties of cathode rays.
Observation
Ray bends in magnetic field.
Conclusion
Ray consists of charged particles.
Ray bends toward positive plate Ray consists of negative particles.
in electric field.
Ray is identical for any cathode. These particles are found in ALL matter.
2-29

30. Figure 2.5
2-30
Millikan’s oil-drop experiment for measuring an
electron’s charge.

31. Millikan’s findings were used to calculate the mass on an
electron.
determined by J.J. Thomson
and others
mass of electron =
mass
charge
x charge
= (-5.686x10-12 kg/C) x (-1.602x10-19 C)
= 9.109x10-31 kg = 9.109x10-28 g
2-31

32. Figure 2.6
2-32
Rutherford’s a-scattering experiment and discovery
of the atomic nucleus.

33. Figure 2.7
General features of the atom.
The atom is an electrically neutral, spherical entity composed of a
positively charged central nucleus surrounded by one or more
negatively charged electrons.
The atomic nucleus consists of protons and neutrons.
2-33

34. Table 2.2
Properties of the Three Key Subatomic Particles
Charge
Name
Relative Absolute (C)*
(Symbol)
Mass
Relative
(amu)†
Absolute (g) Location in
Atom
Proton
(p+)
1+
+1.60218x10-19 1.00727
1.67262x10-24 Nucleus
Neutron
(n0)
0
0
1.67493x10-24 Nucleus
Electron
(e-)
1-
-1.60218x10-19 0.00054858 9.10939x10-24 Outside
nucleus
* The
†
2-34
1.00866
coulomb (C) is the SI unit of charge.
The atomic mass unit (amu) equals 1.66054x10-24 g.

35. The Modern Reassessment of the Atomic Theory
1. All matter is composed of atoms. The atom is the smallest body that
retains the unique identity of the element.
2. Atoms of one element cannot be converted into atoms of another
element in a chemical reaction. Elements can only be converted
into other elements in nuclear reactions.
3. All atoms of an element have the same number of protons and
electrons, which determines the chemical behavior of the element.
Isotopes of an element differ in the number of neutrons, and thus
in mass number. A sample of the element is treated as though its
atoms have an average mass.
4. Compounds are formed by the chemical combination of two or more
elements in specific ratios.
2-35

36. Atomic Symbol, Number and Mass
Figure 2.8
X = Atomic symbol of the element
A = mass number; A = Z + N
Z = atomic number
(the number of protons in the nucleus)
N = number of neutrons in the nucleus
2-36

37. Isotopes
Isotopes are atoms of an element with the same number
of protons, but a different number of neutrons.
Isotopes have the same atomic number, but a different
mass number.
Figure 2.8
2-37

38. Sample Problem 2.4
Determining the Number of Subatomic
Particles in the Isotopes of an Element
PROBLEM: Silicon (Si) has three naturally occurring isotopes:
28Si, 29Si, and 30Si. Determine the number of protons,
neutrons, and electrons in each silicon isotope.
PLAN: The mass number (A) is given for each isotope and is equal to
the number of protons + neutrons. The atomic number Z,
found on the periodic table, equals the number of protons.
The number of neutrons = A – Z, and the number of electrons
equals the number of protons for a neutral atom.
SOLUTION:
The atomic number of silicon is 14; therefore
28Si
29Si
has 14p+, 14e- and 15n0 (29-14)
30Si
2-38
has 14p+, 14e- and 14n0 (28-14)
has 14p+, 14e- and 16n0 (30-14)

39. Tools of the Laboratory
Figure B2.1
2-39
Formation of a positively charged neon particle (Ne+).

40. Tools of the Laboratory
Figure B2.2
2-40
The mass spectrometer and its data.

41. Isotopes and Neutrons
• atoms of the same element having
different atomic mass numbers
2-41

42. The Complete Symbol
• Competency III-1
• MZX where
– M = atomic mass number(p + n)
– Z = atomic number (p)
– = electrons in a neutral atom
• Number of neutrons = M - Z
•
•
2-42
7
74
3Li
p =___ e =____ n =____
34Se
p = ____ e = ____ n = ___

43. Ions
• Charged atoms obtained by adding or
removing electrons to an atom.
• Cations-positively charged ions
• Anions-negatively charged ions
•
•
2-43
39
K+ p = ___ e = ___ n = ___
19
32 S-2 p = ___ e = ___ n = ___
16

44. Atomic Weights(Masses)
• a relative scale based upon 126C
having a mass of exactly 12 amu
(atomic mass units)
• The atomic weight of an element is
the weighted average of all the
isotopes of that element.
2-44

45. Sample Problem 2.5
Calculating the Atomic Mass of an Element
PROBLEM: Silver (Ag, Z = 47) has two naturally occurring isotopes,
107Ag and 109Ag. From the mass spectrometric data
provided, calculate the atomic mass of Ag.
Isotope
107Ag
109Ag
Mass (amu)
Abundance (%)
106.90509
108.90476
51.84
48.16
PLAN: Find the weighted average of
the isotopic masses.
mass (g) of each isotope
multiply by fractional
abundance of each isotope
portion of atomic mass
from each isotope
add isotopic portions
atomic mass
2-45

46. Sample Problem 2.5
SOLUTION:
mass portion from 107Ag =
106.90509 amu x 0.5184 = 55.42 amu
mass portion from 109Ag =
108.90476amu x 0.4816 = 52.45amu
atomic mass of Ag = 55.42amu + 52.45amu
= 107.87amu
2-46

47. Atomic Weights(Masses)
• Naturally occurring copper consists of
two isotopes. It is 69.1% 6329Cu
which has a mass of 62.9 amu and
30.9% 6529Cu which has a mass of
64.9 amu. Determine the atomic
weight of copper.
2-47

48. 2-48

49. 2-49

50. Atomic Weights(Masses)
2-50
• Naturally occurring chromium
consists of four isotopes. It is
4.31% chromium-50 ,mass =
49.946amu, 83.76% chromium-52,
mass = 51.941amu, 9.55%
chromium 53, mass = 52.941amu
2.38% chromium 54, mass =
53.939amu.
Determine the atomic mass of
chromium.

51. 2-51

52. Atomic Masses (Weights)
• Element
•
H2
•
N2
•
Mg
2-52
Atomic Weight
2.0158 amu
28.0134 amu
24.3050 amu

53. The Mole
The mole (mol) is the amount of a substance that
contains the same number of entities as there are
atoms in exactly 12 g of carbon-12.
The term “entities” refers to atoms, ions, molecules,
formula units, or electrons – in fact, any type of particle.
One mole (1 mol) contains 6.022x1023 entities (to
four significant figures).
This number is called Avogadro’s number and is
abbreviated as N.
2-53

54. The Mole
• One mole is defined as 6.022 x 1023
objects. Also called Avagadro’s
number.
• The mass of one mole of an element
will be its atomic mass expressed in
grams.
2-54

55. Figure 3.1
2-55
One mole (6.022x1023 entities) of some familiar
substances.

56. The Mole
• Element Mass
Contains Atoms
• Pt
6.022 x 1023
• Au
197.0 g
6.022 x 1023
• Pb
2-56
195.1 g
207.2 g
6.022 x 1023

57. Figure
3.1
Counting objects of fixed relative mass.
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
2-57

58. Figure
3.2
Oxygen
32.00 g
One mole of
common
substances.
CaCO3
100.09 g
Water
18.02 g
Copper
63.55 g
2-58

59. Table 3.1 Information Contained in the Chemical Formula of
Glucose C6H12O6 ( M = 180.16 g/mol)
Carbon (C)
Atoms/molecule of
compound
Hydrogen (H)
Oxygen (O)
6 atoms
12 atoms
6 atoms
12 mol of atoms
6 mol of atoms
Moles of atoms/mole 6 mol of atoms
of compound
Atoms/mole of
compound
6(6.022x1023) atoms 12(6.022x1023) atoms
6(6.022x1023) atoms
Mass/molecule of
compound
6(12.01 amu)
= 72.06 amu
12(1.008 amu)
= 12.10 amu
6(16.00 amu) =
96.00 amu
Mass/mole of
compound
72.06 g
12.10 g
96.00 g
2-59

60. Interconverting Moles, Mass, and
Number of Chemical Entities
Mass (g) = no. of moles x
no. of grams
g
1 mol
1 mol
No. of moles = mass (g) x
M
no. of grams
No. of entities = no. of moles x
6.022x1023 entities
1 mol
No. of moles = no. of entities x
2-60
1 mol
6.022x1023 entities

61. Figure 3.2
2-61
Mass-mole-number relationships for elements.

62. Sample Problem 3.1
PROBLEM:
Calculating the Mass of a Given Amount of
an Element
Silver (Ag) is used in jewelry and tableware but no
longer in U.S. coins. How many grams of Ag are in
0.0342 mol of Ag?
PLAN: To convert mol of Ag to mass of Ag in g we need the molar
mass of Ag.
amount (mol) of Ag
multiply by M of Ag (107.9 g/mol)
mass (g) of Ag
SOLUTION:
0.0342 mol Ag x
2-62
107.9 g Ag
1 mol Ag
= 3.69 g Ag

63. Sample Problem 3.2
Calculating the Number of Entities in a
Given Amount of an Element
PROBLEM: Gallium (Ga) is a key element in solar panels, calculators and
other light-sensitive electronic devices. How many Ga atoms
are in 2.85 x 10-3 mol of gallium?
PLAN: To convert mol of Ga to number of Ga atoms we need to use
Avogadro’s number.
mol of Ga
multiply by 6.022x1023 atoms/mol
atoms of Ga
2-63

64. Sample Problem 3.2
SOLUTION:
2.85 x 10-3 mol Ga atoms x 6.022x1023 Ga atoms
1 mol Ga atoms
= 1.72 x 1021 Ga atoms
2-64

65. Sample Problem 3.3
Calculating the Number of Entities in a
Given Mass of an Element
PROBLEM: Iron (Fe) is the main component of steel and is therefore the
most important metal in society; it is also essential in the body.
How many Fe atoms are in 95.8 g of Fe?
PLAN: The number of atoms cannot be calculated directly from the mass.
We must first determine the number of moles of Fe atoms in the
sample and then use Avogadro’s number.
mass (g) of Fe
divide by M of Fe (55.85 g/mol)
amount (mol) of Fe
multiply by 6.022x1023 atoms/mol
atoms of Fe
2-65

66. Sample Problem 3.3
SOLUTION:
95.8 g Fe x
1 mol Fe
= 1.72 mol Fe
55.85 g Fe
23
1.72 mol Fe x 6.022x10 atoms Fe
1 mol Fe
= 1.04 x 1024 atoms Fe
2-66

67. Figure 3.3
2-67
Amount-mass-number relationships for compounds.

68. Examples
• Competency I-2
• Determine the number of moles in
73.4g of Mg
• Calculate the number of atoms in
5.00 moles of magnesium.
• Determine the mass of a Mg atom in
grams to three significant figures.
2-68

69. 2-69

70. 2-70

71. Molar Masses
• The molar mass of a substance is the
sum of the atomic masses of the atoms
of each element in the formula.
• Determine the molar mass of propane
(C3H8).
• How many molecules are present in
44.1 g of propane?
• Determine the number of molecules of
C3H8 in 74.6 grams of propane.
2-71

72. 2-72

73. 2-73

74. 2-74

75. Sample Problem 3.4
PROBLEM:
PLAN:
Calculating the Number of Chemical Entities in
a Given Mass of a Compound I
Nitrogen dioxide is a component of urban smog that forms
from the gases in car exhausts. How many molecules are
in 8.92 g of nitrogen dioxide?
Write the formula for the compound and calculate its molar mass.
Use the given mass to calculate first the number of moles and then
the number of molecules.
mass (g) of NO2
divide by M
amount (mol) of NO2
multiply by 6.022 x 1023 formula units/mol
number of NO2 molecules
2-75

76. Sample Problem 3.4
SOLUTION: NO2 is the formula for nitrogen dioxide.
M = (1 x M of N) + (2 x M of O)
= 14.01 g/mol + 2(16.00 g/mol)
= 46.01 g/mol
8.92 g NO2 x
1 mol NO2
46.01 g NO2
= 0.194 mol NO2
23
0.194 mol NO2 x 6.022x10 molecules NO2
1 mol NO2
= 1.17 x 1023 molecules NO2
2-76

77. Sample Problem 3.5
PROBLEM:
Calculating the Number of Chemical Entities in
a Given Mass of a Compound II
Ammonium carbonate, a white solid that decomposes on
warming, is an component of baking powder.
a) How many formula units are in 41.6 g of ammonium
carbonate?
b) How many O atoms are in this sample?
PLAN:
Write the formula for the compound and calculate its molar mass. Use
the given mass to calculate first the number of moles and then the
number of formula units.
The number of O atoms can be determined using the formula and the
number of formula units.
2-77

78. Sample Problem 3.5
mass (g) of (NH4)2CO3
divide by M
amount (mol) of (NH4)2CO3
multiply by 6.022 x 1023 formula units/mol
number of (NH4)2CO3 formula units
3 O atoms per formula unit of (NH4)2CO3
number of O atoms
SOLUTION:
(NH4)2CO3 is the formula for ammonium carbonate.
M = (2 x M of N) + (8 x M of H) + (1 x M of C) + (3 x M of O)
= (2 x 14.01 g/mol) + (8 x 1.008 g/mol)
+ (12.01 g/mol) + (3 x 16.00 g/mol)
= 96.09 g/mol
2-78

79. Sample Problem 3.5
41.6 g (NH4)2CO3 x
1 mol (NH4)2CO3
96.09 g (NH4)2CO3
= 0.433 mol (NH4)2CO3
23
0.433 mol (NH4)2CO3 x 6.022x10 formula units (NH4)2CO3
1 mol (NH4)2CO3
= 2.61x1023 formula units (NH4)2CO3
2.61x1023 formula units (NH4)2CO3 x
3 O atoms
1 formula unit of (NH4)2CO3
= 7.83 x 1023 O atoms
2-79

80. Molar Mass
The molar mass (M) of a substance is the mass per
mole of its entites (atoms, molecules or formula units).
For monatomic elements, the molar mass is the
same as the atomic mass in grams per mole. The
atomic mass is simply read from the Periodic Table.
The molar mass of Ne = 20.18 g/mol.
2-80

81. For molecular elements and for compounds, the
formula is needed to determine the molar mass.
The molar mass of O2 = 2 x M of O
= 2 x 16.00
= 32.00 g/mol
The molar mass of SO2 = 1 x M of S + 2 x M of O
= 32.00 + 2(16.00)
= 64.00 g/mol
2-81

82. Molecular Masses from Chemical Formulas
Molecular mass = sum of atomic masses
For the H2O molecule:
molecular mass =
(2 x atomic mass of H) + (1 x atomic mass of O)
= (2 x 1.008 amu) + (1 x 16.00 amu)
= 18.02 amu
By convention, we read masses off the periodic
table to 4 significant figures.
For ionic compounds we refer to a formula mass
since ionic compounds do not consist of molecules.
2-82

83. Sample Problem 2.15
Calculating the Molecular Mass of a
Compound
PROBLEM: Using the periodic table, calculate the molecular (or
formula) mass of:
(a) tetraphosphorous trisulfide
(b) ammonium nitrate
PLAN: Write the formula and then multiply the number of atoms by the
respective atomic masses. Add the masses for each compound.
SOLUTION:
(a) P4S3
molecular mass = (4 x atomic mass of P) + (3 x atomic mass of S)
= (4 x 30.97 amu) + (3 x 32.07 amu) = 220.09 amu
(b) NH4NO3
formula mass = (2 x atomic mass of N) + (4 x atomic mass of H) +
(3 x atomic mass of O)
= (2 x 14.01 amu) + (4 x 1.008 amu) + (3 x 16.00 amu)
= 80.05 amu
2-83

84. Sample Problem 2.15
Calculating the Molecular Mass of a
Compound
PROBLEM: Using the periodic table, calculate the molecular (or
formula) mass of:
(a) tetraphosphorous trisulfide
(b) ammonium nitrate
PLAN: Write the formula and then multiply the number of atoms by the
respective atomic masses. Add the masses for each compound.
SOLUTION:
(a) P4S3
molecular mass = (4 x atomic mass of P) + (3 x atomic mass of S)
= (4 x 30.97 amu) + (3 x 32.07 amu) = 220.09 amu
(b) NH4NO3
formula mass = (2 x atomic mass of N) + (4 x atomic mass of H) +
(3 x atomic mass of O)
= (2 x 14.01 amu) + (4 x 1.008 amu) + (3 x 16.00 amu)
= 80.05 amu
2-84

85. Molar Masses
• Determine the number of (a) moles,
(b) molecules, and (c) oxygen atoms
in 60.0 g of ozone, O3.
• Determine the molar masses of
Ca(NO3)2.
2-85

86. 2-86

87. 2-87

88. Tools of the Laboratory
Basic Separation Techniques
Filtration: Separates components of a mixture based upon
differences in particle size. Filtration usually involves separating
a precipitate from solution.
Crystallization: Separation is based upon differences in solubility
of components in a mixture.
Distillation: separation is based upon differences in volatility.
Extraction: Separation is based upon differences in solubility in
different solvents (major material).
Chromatography: Separation is based upon differences in
solubility in a solvent versus a stationary phase.
2-88

89. Tools of the Laboratory
Figure B2.3
2-89
Distillation

90. Tools of the Laboratory
Figure B2.4
2-90
Procedure for column chromatography

91. Tools of the Laboratory
Figure B2.5
2-91
Principle of gas-liquid chromatography (GLC).