Download Unit 3 Chap. 3 Atoms: The Building Blocks of Matter

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Chemistry Lecture ’93 B. Rife CHS
Text: Modern Chemistry; Holt, Rinehart & Winston 1993
Atoms: The Building Blocks of Matter Chapter 3
Homework:
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Section Reviews (pg 74,82,92)
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Reviewing Concepts (all) (pg 93-94)
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Problems (all) (pg 94-95)
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Chapter/Section Review (Handout)
page 1/6
Due Date
Exam Date _
3.1 The Atome: From Philosophical Idea to Scientific Theory
1000 BC 400 BC 0-1500
1627-1691
1660-1734
1733-1804
1743-1795
metallurgy and embalming
Greeks four fundamental substances: fire, earth, water, and air;
Demokritos - “atomos”
alchemy - discovered some elements and mineral acids
Robert Boyle - relationship btw. P & V
George Stahl - “phlogiston” from combustion
Joseph Priestley - discovered oxygen
Antoine Lavoisier - discovered combustion and respiration
3.1A Summarize the five essential points of Dalton’s atomic theory
(3)
Dalton's Atomic Theory
A New System of Chemical Philosophy (1808)
1. Each element is made up of tiny particles called atoms.
2. The atoms of a given element are identical and different from atoms of other elements.
3. Atoms cannot be subdivided, created, or destroyed.
4. Chemical compounds are formed when atoms combine with each other.
5. Chemical reactions involve reorganization of the atoms.
3.1B Explain the relationship between Dalton’s atomic theory and
the laws of conservation of mass and definite composition.
(6)
If atoms are indivisible and atoms of different elements can combine in chemical reactions then it
must be that mass is conserved in chemical reactions.
Law of Conservation of Mass states that the mass of substances formed by a chemical reaction
is the same as the mass of substances entering into the reaction.
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1754-1826
Joseph Proust
Law of Definite Proportion (Constant Composition) states that all samples of a compound
have the same composition, for example, the same proportions by mass of the constituent
elements.
3.1C Explain the law of multiple proportions
A.
(6)
John Dalton
Law of Multiple Proportions states that if two elements form more than a single compound,
the masses of one element combined with a fixed mass of the second are in the ratio of small
whole numbers.
Examples:
CO
CO2
12 g of C and 16 g of O
12 g of C and 32 g of O
The ratio of atoms (and masses) of oxygen is 2:1
FeCl2 56 g of Fe and 71 g of Cl
FeCl3 56 g of Fe and 106.5 g of Cl
The ratio of atoms (and masses) of chlorine is 2:3
3.2 The Structure of the Atom
3.2A Summarize the observed properties of cathode rays that led to
the discovery of the electron(2)
1898-1903
J.J. Thomson - discovered electrons (e-s);
“Plum Pudding Model”
Cathode Ray are negatively charged particles (electrons) emitted at the negative electrode
(cathode) in the passage of electricity through gases at very low pressures. Cathode rays are
deflected toward a positive magnetic field electrode (anode).
3.2B Summarize the experiment conducted by Rutherford that led to the discovery of
the nucleus. (5)
1871-1937
Ernest Rutherford - discovered proton (1911)
Nobel Prize in Chemistry (1908) for his work with radioactivity
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1909
Rutherford’s Experiment - bombarded a very thin gold foil with
alpha particles from a radioactive source. A fluorescent zinc sulfide screen recorded the
scattering of the alpha particles. The following observations were made.
1. Most alpha particles passed straight through.
2. Some alpha particles were deflected through moderate angles
3. A few alpha particles were scattered backward.
“as though you fired a 15” shell at a piece of tissue paper and it had bounced back and hit you”
thus developed the Nuclear Atom Model 1912
He was able to carry out the alchemist’s dream of transmutation when he change nitrogen
atoms into
oxygen atoms via nuclear reactions
3.2C Describe the properties of protons, neutrons, and electrons
(12)
The nucleus is the positively charged, dense central portion of the atom that contains nearly all
of its mass but takes up only an insignificant fraction of its volume.
nucleus diameter
electron cloud diameter
10-13 cm
10-8 cm
size of a dime (analogy)
football field (100 yds)
Proton are fundamental particles carrying the basic unit of positive electric charge and found in
the nuclei of all atoms. mass p = 1.673 x 10-27 kg
1932
James Chadwick - discovered neutrons
Neutrons are electrically neutral fundamental particles of matter found in all atomic nuclei
except that of the
simple hydrogen atom mass n = 1.675 x 10-27 kg
Electron is a negatively charged particle that moves around the nucleus of an atom.
1909
Robert Millikan - mass of e- = 9.11 x 10-31 kg
Nuclear forces - (Strong & Weak) Atomic forces that overcome the electrical repelling forces
in proton-proton and thus hold the nuclear particles together.
Electrons are held in the electron cloud by electrical attraction.
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3.2D Define “atom” and “isotope” (3)
Atom - the smallest particle of an element which possesses the properties of that element.
Isotopes of an element are atoms with different numbers of neutrons in their nuclei. That is
,isotopes of an element have the same atomic numbers but different mass numbers.
3.2E Describe the atomic structures of the isotopes of hydrogen
Protium 99.985%
Deuterium - 0.015%
Tritium 0.001%
1p
1p & 1n
1p & 2n
(3)
most common
fuel for fusion
evidence of nuclear reaction
3.3 Weighing and Counting Atoms
3.3A Define “atomic number” and “mass number” and describe how they apply to isotopes
and nuclides. (5)
Atomic Number (Z) is the number of protons in the nucleus of an atom. It is also the number
of electrons outside the nucleus of an electrically neutral atom.
Mass Number (A) is the total of the number of protons and neutrons in the nucleus of an atom.
3.3B Determine the number of protons, neutrons, and electrons in a nuclide, given the
identity of the nuclide.
(5)
Nuclide - the name of each different variety of atom as determined by the number of protons
and neutrons.
Nuclide symbol - for an atom, X in which X is the symbol for an element, Z is its atomic
number, and A is its mass number.
Nuclear Symbol represents the composition of a nucleus
#p + #n A
#p
Z
±? #p - #e
X
3.3C Distinguish between relative atomic mass and average atomic mass.
(2)
Atomic Mass - The mass of an atom relative to the mass of a carbon-12 atom expressed
in atomic mass units (amu)
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Atomic Mass Unit - amu - µ - one twelfth of the mass of an atom of the carbon-12 iostope;
1.66 x 10-27 kg
Average Atomic Mass - Atomic Mass (Weight) of an element is the average of the isotopic
masses weighted according to the naturally occurring abundances of the isotopes of the
element.
The atomic mass of a nuclide is the relative atomic mass of atoms of the nuclide.
The atomic mass of an element is the average atomic mass of the naturally occurring mixture of
isotopes of the element.
3.3D Calculate the average atomic mass of an element given the relative abundances of each
isotope of the element
(2)
C-12
C-13
0.9890 (12.0)
+ 0.0110 (13.003355) = 12.011µ
3.3E Define a mole in terms of Avogadro’s number; define molar mass
(4)
Attempts to measure and work with the masses of individual atoms showed to be very difficult.
Thus a unit for describing large numbers of atoms is desirable.
The number of atoms in the atomic mass of a nuclide in grams is an important unit of measure in
chemistry.
This number is a constant and is called the Avogadro number or mole.
Mole is an amount of substance containing 6.02214 x 1023 (the Avogadro constant)
atoms, formula units, or molecules.
Avogadro’s Number (Constant NA) has a value of 6.02214 x 1023 mol-1.
It is the number of elementary units in one mole.
Molar Mass is the mass of one mole of atoms, formula units, or molecules.
The molar mass, M, in grams per mole, is numerically equal to the formula mass.
3.3F Solve problems involving mass, mole, and the number of atoms of an element.
(5)
Sample Exercise
Acetylsalicylic acid, C9H8O4, is the principal ingredient of aspirin.
What is the mass in grams of 0.287 mol of acetylsalicylic acid?
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Strategy: find the molar mass of C9H8O4 and use it to obtain the conversion factor
to convert mol to grams.
Solution:
M = 180.15 g/mol
0.287 mol C9H8O4 180.15 g C9H8O4 = 51.7 g C9H8O4
1 mol C9H8O4