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Chapter 2

The Components of Matter

2-1
Chapter 2: The Components of Matter
2.1 Elements, Compounds, and Mixtures:
An Atomic Overview
2.2 The Observations That Le...
Chapter 2: The Components of Matter
2.6 Elements: A First Look at the Periodic Table

2-3
Atoms, Molecules and Ions
•
•
•
•
•

2-4

Dalton’s Atomic Theory
Atomic Structure
Atomic Mass
Chemical Formulas
Chapters 2...
Goals & Objectives
• The student will understand the
characteristics of the three types of
matter: elements, compounds, an...
Goals & Objectives
• The student will understand the
postulates of Dalton’s atomic theory
and how it explains the mass law...
Goals & Objectives
• The student will understand the
structure of the atom along with the
characteristics of the proton, n...
Goals & Objectives
• The student will understand the
relationship between the formula
mass and molar mass. (3.1)
• The stu...
Goals & Objectives
• The student will understand the
information in a chemical formula.
(3.1)

2-9
Master these Skills
• Distinguish between elements,
compounds and mixtures at the
atomic scale. (SP 2.1)
• Using the mass ...
Master these Skills
• Using atomic notation to express the
subatomic makeup of an isotope. (SP
2.4)
• Calculating an atomi...
Master these Skills
• Converting between amount of
substance (in moles), mass (in
grams), and the number of atoms.
(SP 3.1...
Matter
• A scheme for the classification of
matter

2-13
Definitions for Components of Matter
Element - the simplest type of substance with unique physical and
chemical properties...
Definitions for Components of Matter
Compound - a substance
composed of two or more elements
which are chemically combined...
Table 2.1

Some Properties of Sodium, Chlorine, and Sodium
Chloride.

Property

Sodium

Melting point

Chlorine

Sodium Ch...
Figure 2.2 The law of mass conservation.

The total mass of substances does not change during a chemical reaction.

2-17
Law of Mass Conservation
The total mass of substances present does not change
during a chemical reaction.
reactant 1

+

p...
Law of Definite (or Constant) Composition
No matter the source, a particular compound is
composed of the same elements in ...
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
(par...
Law of Multiple Proportions
If elements A and B react to form two compounds, the
different masses of B that combine with a...
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 = ...
Dalton’s Atomic Theory
Dalton postulated that:

1.
2.

3.

4.

2-23

All matter consists of atoms; tiny indivisible partic...
Dalton’s Atomic Theory
explains the mass laws

Mass conservation

Atoms cannot be created or destroyed

postulate 1

or co...
Dalton’s Atomic Theory
explains the mass laws

Definite composition

Atoms are combined in compounds in
specific ratios

p...
Dalton’s Atomic Theory
explains the mass laws

Multiple proportions

Atoms of an element have the same mass postulate 3
an...
The Structure of the Atom
•
•
•
•
•

2-27

Fundamental Particles in Atoms
Particle
Mass(amu)
Charge
electron
0.00054858 -1...
The Structure of the Atom
• Electrons

2-28
Figure 2.4
Observations that established the properties of cathode rays.

Observation
Ray bends in magnetic field.

Conclu...
Figure 2.5

2-30

Millikan’s oil-drop experiment for measuring an
electron’s charge.
Millikan’s findings were used to calculate the mass on an
electron.
determined by J.J. Thomson
and others
mass of electron...
Figure 2.6

2-32

Rutherford’s a-scattering experiment and discovery
of the atomic nucleus.
Figure 2.7

General features of the atom.

The atom is an electrically neutral, spherical entity composed of a
positively ...
Table 2.2

Properties of the Three Key Subatomic Particles
Charge

Name
Relative Absolute (C)*
(Symbol)

Mass
Relative
(am...
The Modern Reassessment of the Atomic Theory
1. All matter is composed of atoms. The atom is the smallest body that
retain...
Atomic Symbol, Number and Mass
Figure 2.8

X = Atomic symbol of the element

A = mass number; A = Z + N
Z = atomic number
...
Isotopes
Isotopes are atoms of an element with the same number
of protons, but a different number of neutrons.

Isotopes h...
Sample Problem 2.4

Determining the Number of Subatomic
Particles in the Isotopes of an Element

PROBLEM: Silicon (Si) has...
Tools of the Laboratory
Figure B2.1

2-39

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

2-40

The mass spectrometer and its data.
Isotopes and Neutrons
• atoms of the same element having
different atomic mass numbers

2-41
The Complete Symbol
• Competency III-1
• MZX where
– M = atomic mass number(p + n)
– Z = atomic number (p)
– = electrons i...
Ions
• Charged atoms obtained by adding or
removing electrons to an atom.
• Cations-positively charged ions
• Anions-negat...
Atomic Weights(Masses)
• a relative scale based upon 126C
having a mass of exactly 12 amu
(atomic mass units)
• The atomic...
Sample Problem 2.5

Calculating the Atomic Mass of an Element

PROBLEM: Silver (Ag, Z = 47) has two naturally occurring is...
Sample Problem 2.5
SOLUTION:
mass portion from 107Ag =
106.90509 amu x 0.5184 = 55.42 amu
mass portion from 109Ag =
108.90...
Atomic Weights(Masses)
• Naturally occurring copper consists of
two isotopes. It is 69.1% 6329Cu
which has a mass of 62.9 ...
2-48
2-49
Atomic Weights(Masses)

2-50

• Naturally occurring chromium
consists of four isotopes. It is
4.31% chromium-50 ,mass =
49...
2-51
Atomic Masses (Weights)
• Element
•
H2
•
N2
•
Mg

2-52

Atomic Weight
2.0158 amu
28.0134 amu
24.3050 amu
The Mole
The mole (mol) is the amount of a substance that
contains the same number of entities as there are
atoms in exact...
The Mole
• One mole is defined as 6.022 x 1023
objects. Also called Avagadro’s
number.
• The mass of one mole of an elemen...
Figure 3.1

2-55

One mole (6.022x1023 entities) of some familiar
substances.
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

6.0...
Figure
3.1

Counting objects of fixed relative mass.

12 red marbles @ 7g each = 84g
12 yellow marbles @4g each = 48g 55.8...
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
Table 3.1 Information Contained in the Chemical Formula of
Glucose C6H12O6 ( M = 180.16 g/mol)
Carbon (C)
Atoms/molecule o...
Interconverting Moles, Mass, and
Number of Chemical Entities
Mass (g) = no. of moles x

no. of grams

g

1 mol
1 mol

No. ...
Figure 3.2

2-61

Mass-mole-number relationships for elements.
Sample Problem 3.1
PROBLEM:

Calculating the Mass of a Given Amount of
an Element

Silver (Ag) is used in jewelry and tabl...
Sample Problem 3.2

Calculating the Number of Entities in a
Given Amount of an Element

PROBLEM: Gallium (Ga) is a key ele...
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
Sample Problem 3.3

Calculating the Number of Entities in a
Given Mass of an Element

PROBLEM: Iron (Fe) is the main compo...
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
...
Figure 3.3

2-67

Amount-mass-number relationships for compounds.
Examples
• Competency I-2
• Determine the number of moles in
73.4g of Mg
• Calculate the number of atoms in
5.00 moles of ...
2-69
2-70
Molar Masses
• The molar mass of a substance is the
sum of the atomic masses of the atoms
of each element in the formula.
...
2-72
2-73
2-74
Sample Problem 3.4
PROBLEM:

PLAN:

Calculating the Number of Chemical Entities in
a Given Mass of a Compound I

Nitrogen ...
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...
Sample Problem 3.5

PROBLEM:

Calculating the Number of Chemical Entities in
a Given Mass of a Compound II

Ammonium carbo...
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/mo...
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 ...
Molar Mass
The molar mass (M) of a substance is the mass per
mole of its entites (atoms, molecules or formula units).

For...
For molecular elements and for compounds, the
formula is needed to determine the molar mass.
The molar mass of O2 = 2 x M ...
Molecular Masses from Chemical Formulas
Molecular mass = sum of atomic masses

For the H2O molecule:
molecular mass =
(2 x...
Sample Problem 2.15

Calculating the Molecular Mass of a
Compound

PROBLEM: Using the periodic table, calculate the molecu...
Sample Problem 2.15

Calculating the Molecular Mass of a
Compound

PROBLEM: Using the periodic table, calculate the molecu...
Molar Masses
• Determine the number of (a) moles,
(b) molecules, and (c) oxygen atoms
in 60.0 g of ozone, O3.
• Determine ...
2-86
2-87
Tools of the Laboratory

Basic Separation Techniques
Filtration: Separates components of a mixture based upon
differences ...
Tools of the Laboratory

Figure B2.3

2-89

Distillation
Tools of the Laboratory

Figure B2.4

2-90

Procedure for column chromatography
Tools of the Laboratory

Figure B2.5

2-91

Principle of gas-liquid chromatography (GLC).
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  1. 1. Chapter 2 The Components of Matter 2-1
  2. 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. 3. Chapter 2: The Components of Matter 2.6 Elements: A First Look at the Periodic Table 2-3
  4. 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. 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. 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. 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. 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. 9. Goals & Objectives • The student will understand the information in a chemical formula. (3.1) 2-9
  10. 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. 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. 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. 13. Matter • A scheme for the classification of matter 2-13
  14. 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. 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. 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. 17. Figure 2.2 The law of mass conservation. The total mass of substances does not change during a chemical reaction. 2-17
  18. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 28. The Structure of the Atom • Electrons 2-28
  29. 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. 30. Figure 2.5 2-30 Millikan’s oil-drop experiment for measuring an electron’s charge.
  31. 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. 32. Figure 2.6 2-32 Rutherford’s a-scattering experiment and discovery of the atomic nucleus.
  33. 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. 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. 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. 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. 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. 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. 39. Tools of the Laboratory Figure B2.1 2-39 Formation of a positively charged neon particle (Ne+).
  40. 40. Tools of the Laboratory Figure B2.2 2-40 The mass spectrometer and its data.
  41. 41. Isotopes and Neutrons • atoms of the same element having different atomic mass numbers 2-41
  42. 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. 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. 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. 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. 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. 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. 48. 2-48
  49. 49. 2-49
  50. 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. 51. 2-51
  52. 52. Atomic Masses (Weights) • Element • H2 • N2 • Mg 2-52 Atomic Weight 2.0158 amu 28.0134 amu 24.3050 amu
  53. 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. 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. 55. Figure 3.1 2-55 One mole (6.022x1023 entities) of some familiar substances.
  56. 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. 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. 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. 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. 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. 61. Figure 3.2 2-61 Mass-mole-number relationships for elements.
  62. 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. 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. 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. 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. 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. 67. Figure 3.3 2-67 Amount-mass-number relationships for compounds.
  68. 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. 69. 2-69
  70. 70. 2-70
  71. 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. 72. 2-72
  73. 73. 2-73
  74. 74. 2-74
  75. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 86. 2-86
  87. 87. 2-87
  88. 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. 89. Tools of the Laboratory Figure B2.3 2-89 Distillation
  90. 90. Tools of the Laboratory Figure B2.4 2-90 Procedure for column chromatography
  91. 91. Tools of the Laboratory Figure B2.5 2-91 Principle of gas-liquid chromatography (GLC).
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