Chapter 3 pages 66 - 871 Modern Chemistry Chapter 3 Atoms: the building block of matter.
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Transcript of Chapter 3 pages 66 - 871 Modern Chemistry Chapter 3 Atoms: the building block of matter.
![Page 1: Chapter 3 pages 66 - 871 Modern Chemistry Chapter 3 Atoms: the building block of matter.](https://reader036.fdocuments.in/reader036/viewer/2022062304/56649f1a5503460f94c2feae/html5/thumbnails/1.jpg)
Chapter 3 pages 66 - 87 1
Modern ChemistryChapter 3Atoms:
the building block of matter
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Chapter 3 pages 66 - 87 2
Section 1Atoms:
From Philosophical Idea to
Scientific Theory
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Chapter 3 pages 66 - 87 3
Foundation of Chemical Atomic Theory
• Law of Conservation of Mass– Mass is neither created or destroyed
during ordinary chemical reactions or physical changes
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Chapter 3 pages 66 - 87 4
Law of Conservation of Mass Image
p.
69
*
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Chapter 3 pages 66 - 87 5
Law of Conservation of Mass Image
p.
69
*
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Chapter 3 pages 66 - 87 6
Foundation of Chemical Atomic Theory
• Law of Definite Proportions– A chemical compound contains the
same elements in exactly the same proportions by mass regardless of the size of the sample or the source of the compound.
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Chapter 3 pages 66 - 87 7
Foundation of Chemical Atomic Theory
• Law of Multiple Proportions– If two or more different compounds
are composed of the same two elements then the ratio of the masses of the second element combined with a certain mass of the first element is always a ratio of small whole numbers.
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Chapter 3 pages 66 - 87 8
Law of Multiple Proportions Image
p.
69
*
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Chapter 3 pages 66 - 87 9
Dalton’s Atomic Theory
1. All matter is composed
of extremely small
particles called atoms
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Chapter 3 pages 66 - 87 10
Dalton’s Atomic Theory
2. Atoms of a given
element are identical in
size, mass and other
properties; atoms of
different elements differ
in size, mass and other
properties.
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Chapter 3 pages 66 - 87 11
Dalton’s Atomic Theory
3. Atoms cannot be
subdivided, created or
destroyed.
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Chapter 3 pages 66 - 87 12
Dalton’s Atomic Theory
4. Atoms of different
elements combine in
simple whole-number
ratios to form chemical
compounds.
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Chapter 3 pages 66 - 87 13
Dalton’s Atomic Theory
5. In chemical reactions,
atoms are combined,
separated or
rearranged.
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Chapter 3 pages 66 - 87 14
Modern Atomic Theory
Leucippus
Democritus
Atomic Theory Tested by experiment and
modified with new discoveries and experiments
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Chapter 3 pages 66 - 87 15
Section 2
The Structure of the Atom
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Chapter 3 pages 66 - 87 16
Discovery of the Electron• Cathode Ray Tube Experiment -
Thompson• Observations
– Cathode Rays are deflected a magnetic field.
– Cathode rays are deflected from a negatively charged object.
– Charge to mass ratio is always the same for the cathode rays.
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Chapter 3 pages 66 - 87 17
Discovery of the Electron• Cathode Ray Tube
Experiment - Thompson• Conclusion
– Cathode rays are composed of negatively charged particles
– Named “electrons”
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Chapter 3 pages 66 - 87 18
Discovery of the Electron• Oil Drop Experiment - Millikan
– Measured the charge of the electron– Calculated the mass of an electron
•9.109 x 10-31 kg
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Chapter 3 pages 66 - 87 19
Discovery of the Electron• Inferences
–Atoms are neutral, so there must be a positive charge.
–Electrons are small, so there must be other particles.
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Chapter 3 pages 66 - 87 20
Discovery of the Electron• Plum Pudding Model
–Negative electrons were spread evenly throughout the positive charge.
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Chapter 3 pages 66 - 87 21
Discovery of the Electron
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Chapter 3 pages 66 - 87 22
Discovery of the Atomic Nucleus
• Gold Foil Experiment – Rutherford et. al– Hypothesis: Alpha particles would
pass through with slight deflection.– Observation: 1 in 8000 particles were
deflected back to the source.– Conclusion: The atom contains a
small densely packed bundle of matter with a positive charge
– Named the “nucleus”
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Chapter 3 pages 66 - 87 25
Gold Foil Experiment Imagep
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5
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Chapter 3 pages 66 - 87 26
Discovery of the Atomic Nucleus
Relative size of the nucleus
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Chapter 3 pages 66 - 87 27
Composition of The Atomic Nucleus
• Nuclei contain protons and neutrons• Neutral because number of protons
equal number of electrons• Each element has a different number of
protons in their nucleus – The number of protons determines
the atom’s identity• Nuclear forces hold protons & neutrons
together
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Chapter 3 pages 66 - 87 28
Properties of Subatomic Particles
p.
76
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Chapter 3 pages 66 - 87 29
While the number of protons in the nucleus defines an element's
identity, variations on the number of neutrons in the nucleus give
rise to different isotopes of the same element.
Isotopes of hydrogen
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Chapter 3 pages 66 - 87 30
Gold Foil Experiment Photo
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Chapter 3 pages 66 - 87 31
Gold Foil Experiment Photo
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Chapter 3 pages 66 - 87 32
Thompson and Rutherford Photo
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Chapter 3 pages 66 - 87 33
Section 2 Homework
Ch 3 Sec 2 Review Page 76 #1-5
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Chapter 3 pages 66 - 87 34
Section 3
Counting Atoms
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Chapter 3 pages 66 - 87 35
Atomic Number
• The number of protons of each atom of that element
• Identifies the element
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Chapter 3 pages 66 - 87 36
Isotopes• Atoms of the same element that
have different masses equals the isotope
• Isotopes do not differ significantly in their chemical behavior
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Chapter 3 pages 66 - 87 37
Mass Numbers
• Mass numbers = # of p+ + # of n0
of a specific isotope
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Chapter 3 pages 66 - 87 38
Designating Isotopes• Hyphen notation
– name of element – mass number– Hydrogen – 3
• Nuclear symbol
mass number
atomic number
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Chapter 3 pages 66 - 87 39
Number of neutrons in an atom
neutrons = mass number – atomic number
(This is how to calculate isotopes…)
Nuclide – a general term for a specific isotope of an element
**Practice Problems page 87 #2-3
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Chapter 3 pages 66 - 87 40
Relative Atomic Mass• One atom, carbon-12, is set as a
standard• All masses are expressed in
relation to this standard• 1 atomic mass unit = 1/12 the
mass of a carbon-12 atom
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Chapter 3 pages 66 - 87 41
Relative Atomic Mass• Examples
– Hydrogen – 1 = 1.007825 amu– Oxygen – 16 = 15.994915 amu– Magnesium – 24 = 23.985042 amu
• p+ = 1.007276 amu, n0 = 1.008665 amu, e- = 0.0005486 amu
• Relative mass and mass number are close in value but not the same
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Chapter 3 pages 66 - 87 42
Average Atomic Mass• The weighted average of the
atomic masses of the naturally occurring isotopes of an element
• Example– Copper
Cu-63: .6915 x 62.93 amu = 43.52Cu-65: .3085 x 64.93 amu = 20.03 63.55 amu
percentrelative mass
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Chapter 3 pages 66 - 87 43
The Mole• An amount of a substance that
contains as many particles as there are atoms in exactly 12 g carbon-12.
• Similar to a dozen or a pair or a gross
• 6.022 x 1023 carbon-12 atoms = 12 grams of carbon-12
• Avogadro’s number = 6.022 x 1023 particles
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Chapter 3 pages 66 - 87 44
Molar mass• The mass of one mole of a pure
substance • Unit = g/mol• On the periodic table, use 4 sig.
figs.
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Chapter 3 pages 66 - 87 45
Gram-Mole Conversions• The conversion factor for gram-
mole conversion is molar mass.
• What is the mass, in grams, of 3.50 moles of Cu?– 222 grams Cu
ORg
molg
mol
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Chapter 3 pages 66 - 87 46
Practice Problems page 851. What is the mass in grams of 2.25
mol of the element iron?2. What is the mass in grams of 0.357
mol of the element potassium?3. What is the mass in grams of
0.0135 mol of the element sodium?4. What is the mass in grams of 16.3
mol of the element nickel?
p. 8
5
126 g Fe
14.7 g K
0.310 g Na
957 g Ni
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Chapter 3 pages 66 - 87 47
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Chapter 3 pages 66 - 87 48
Gram-Mole Conversions• The conversion factor for gram-
mole conversion is molar mass.
• A Chemist produced 11.9 g of Al. How many moles of Al were produced?– 0.411 moles Al
ORg
molg
mol
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Chapter 3 pages 66 - 87 49
Practice Problems page 851. How many moles of calcium are in
5.00 g of calcium?2. How many moles of gold are in 3.60
x 10-5 g of gold?3. How many moles of zinc are in
0.535 g of zinc?
p. 8
5
0.125 mol Ca
1.83 x 10-7 mol Au
8.18 x 10-3 mol Zn
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Chapter 3 pages 66 - 87 50
Conversions with Avogadro’s Number
• The conversion factor for particle-mole conversion is Avogadro’s number.
• How many moles of silver are in 3.01 x 1023 atoms of silver– 0.500 moles Ag
OR6.022x1023atoms
1 mol 6.022x1023atoms
1 mol
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Chapter 3 pages 66 - 87 51
Practice Problems page 861. How many moles of lead are 1.50 x
1012 atoms of lead?2. How many moles of tin are in 2500
atoms of tin?3. How many atoms of aluminum are
in 2.75 mol of aluminum?
p. x
x
2.49 x 10-12 mol Pb
4.2 x 10-21 mol Sn
1.66 x 1024 atoms Al
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Chapter 3 pages 66 - 87 52
Conversions with Avogadro’s Number
• The conversion factor for particle-mole conversion is Avogadro’s number.
• What is the mass, in grams, of 1.20x1018 atoms of Cu?– 1.27 x 10-4 g Cu
OR6.022x1023atoms
1 mol 6.022x1023atoms
1 mol
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Chapter 3 pages 66 - 87 53
Practice Problems page 871. What is the mass in grams of 7.5 x
1015 atoms of nickel?2. How many atoms of sulfur are in
4.00 g of sulfur?3. What mass of gold contains the
same number of atoms as 9.0 g of aluminum?
p. x
x
7.3 x 10-7 g Ni
7.51 x 1022 atoms S
66 g Au
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Chapter 3 pages 66 - 87 54
Con
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Chapter 3 pages 66 - 87 55
Section 1 Homework
Section Review Page 87 #1-7