Transcript 4_3 How Atoms Differx

Section 4.3
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Standard: 1a, 11c, 3b
Mastering Concept: 112(47-52)
Terms: 98
Practice Problems: 99(11-13), 101(14), 104(15-17)
homework:
Cornell Notes: 4.3
Section Assessment: 104(18-20)
Mastering Problems: 113 (59-68)
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Section 4.3 How Atoms Differ
Section 4-3
•Explain the role of atomic number in determining
the identity of an atom.
•Define an isotope.
•Explain why atomic masses are not whole numbers.
•Calculate the number of electrons, protons, and
neutrons in an atom given its mass number and
atomic number.
Section 4.3 How Atoms Differ (cont.)
Section 4-3
periodic table: a chart that organizes all known elements into a
grid of horizontal rows (periods) and vertical columns (groups or
families) arranged by increasing atomic number
atomic number
isotopes
mass number
atomic mass unit
(amu)
atomic mass
The number of protons and the mass
number define the type of atom.
Atomic Number
Section 4-3
• Each element contains a unique positive charge in
their nucleus.
• The number of protons in the nucleus of an atom
identifies the element and is known as the
element’s atomic number.
Isotopes and Mass Number
Section 4-3
• All atoms of a particular element have
the same number of protons and
electrons but the number of neutrons in
the nucleus can differ.
• Atoms with the same number of protons but
different numbers of neutrons are called
isotopes.
Isotopes and Mass Number (cont.)
Section 4-3
• The relative abundance of each isotope is usually
constant.
• Isotopes containing more neutrons have a
greater mass.
• Isotopes have the same chemical behavior.
• The mass number is the sum of the protons
and neutrons in the nucleus.
Isotopes and Mass Number (cont.)
Section 4-3
Mass of Atoms
Section 4-3
• One atomic mass unit (amu) is defined as 1/12th
the mass of a carbon-12 atom.
• One amu is nearly, but not exactly, equal to
one proton and one neutron.
Mass of Atoms (cont.)
Section 4-3
atomic mass of an element is the
• The _____________
weighted average mass of the isotopes of
that element.
Mastering Concept: 112(47-52)
47. Does the existence of isotopes
contradict part of Dalton’s original atomic
theory? Explain. (4.3)
Yes; not all atoms of an element are
identical in mass.
Mastering Concept: 112(47-52)
48. How do isotopes of a given element
differ? How are they similar? (4.3)
Different number of neutrons, masses;
same: chemical properties, number of
protons and electrons
Mastering Concept: 112(47-52)
49. How is the mass number related to the
number of protons and neutrons an atom
has? (4.3)
Mass number = number of p+ + number of
n0
Mastering Concept: 112(47-52)
50. What do the superscript and subscript in
the notation 40
K represent? (4.3)
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The superscript represents the mass
number (40) and the subscript represents
the atomic number (19).
Mastering Concept: 112(47-52)
51. Explain how to determine the number of
neutrons an atom contains if you know its
mass number and its atomic number. (4.3)
Number of n0 = mass number – atomic
number
Mastering Concept: 112(47-52)
52. Define the atomic mass unit. What were
the benefits of developing the atomic mass
unit as a standard unit of mass? (4.3)
amu = 1/12 of the mass of a C-12 atom;
Scientists defined the atomic mass unit as a
relative standard that was closer in size to
atomic and subatomic masses.
Calculating Average Atomic Mass
isotope1 :  relative abundance(%) x isotopemass(amu)
isotope2 :  relative abundance(%) x isotopemass(amu)
isotope3 :  relative abundance(%) x isotopemass(amu)
• Sum of the products = average atomic mass
• Example:
– The isotopes of element “Bob” are found below:
– Bob-18, 25% 0.25 x18  0.60 x19  0.15 x 20
– Bob-19, 60%
– Bob-20, 15%
– What is the average atomic mass of naturally occurring
Bob?
1amu = 1.66x10-27kg
18.90 amu
A review…
Isotopes, Ions, and Allotropes (Oh my)
• Isotopes
– atoms of the same element with different number of
neutrons.
• Ions
– atoms of the same element with different number of
electrons.
– Ions are easy to create; adding or removing electrons
can be done with electric current.
• Allotropes
– forms of the same element, bonded in different
structures.
– Diamond and pencil graphite are allotropes. They are
both pure carbon, but in different structures.
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Practice Problems: 99(11-13)
11. How many protons and electrons are in each of
the following atoms?
a. Boron
c. Platinum
b. Radon
d. Magnesium
12. An atom of an element contains 66 electrons.
What element is it?
13. An atom of an element contains 14 protons. What
element is it?
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Practice Problems: 99(11-13)
11. How many protons and electrons are in each of
the following atoms?
a. Boron
a. boron, 5
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Practice Problems: 99(11-13)
11. How many protons and electrons are in each of
the following atoms?
b. Radon
Radon, 86
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Practice Problems: 99(11-13)
11. How many protons and electrons are in each of
the following atoms?
c. Platinum
78
Pt
platinum, 78
Platinum
195.084
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Practice Problems: 99(11-13)
11. How many protons and electrons are in each of
the following atoms?
d. Magnesium
magnesium, 12
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Practice Problems: 99(11-13)
12. An atom of an element contains 66 electrons.
What element is it?
12. dysprosium
word=dysprosium
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Practice Problems: 99(11-13)
13. An atom of an element contains 14 protons. What
element is it?
13. silicon
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Practice Problems: 101 (14)
14. Determine the number of protons, electrons, and
neutrons, Name each isotope, and write its symbol.
Isotope Composition Data
Element
b. Calcium
c. Oxygen
d. Iron
e. Zinc
f. Mercury
Atomic
Mass
Number Number
20
8
26
30
80
46
17
57
64
204
P/
e
n
20
8
26
30
80
26
9
31
34
12
4
isotope
symbol
Calcium-46
Oxygen-17
Iron-57
Zinc-64
Mercury204
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Practice Problems: 104 (15-17)
15. Boron has two naturally occurring isotopes: boron-10 (abundance =
19.8%, mass = 10.013 amu), boron-11 (abundance = 80.2%, mass =
11.009 amu). Calculate the atomic mass of boron.
For isotope boron-10 :
Mass: 10.013 amu
abundance 19.8% = 0.198
Mass contribution = (mass)(percent abundance)
= 10.013 amu (0.198) = 1.98257
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Practice Problems: 104 (15-17)
15. Boron has two naturally occurring isotopes: boron-10 (abundance =
19.8%, mass = 10.013 amu), boron-11 (abundance = 80.2%, mass =
11.009 amu). Calculate the atomic mass of boron.
For isotope boron-11 :
Mass: 11.009 amu
abundance 80.2% = 0.802
Mass contribution = (mass)(percent abundance)
= 11.009 amu (0.802) = 8.829 amu
Sum the mass contributions to find the atomic mass.
1.98257 amu + 8.829 amu = 10.8116 amu
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Alternative
Isotope
%
Abunda
nce
Mass
Mass
contributio
n
boron-10 19.8%
0.198
10.013
1.98257
boron-11 80.2%
0.802
11.009
8.829
10.8116
Practice Problems: 104 (15-17)
16. Helium has two naturally occurring isotopes,
helium-3 and helium-4. the atomic mass of
helium is 4.003 amu. Which isotope is more
abundant in nature? Explain.
16. Helium-4 is more abundant in nature
because the atomic mass of naturally
occurring helium is closer to the mass of
helium-4 (~4 amu) than to the mass of
helium-3 (~3 amu).
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Practice Problems: 104 (15-17)
17.Calculate the atomic mass of magnesium. The
three magnesium isotopes have atomic masses
and relative abundances of 23.985 amu
(78.99%), 24.986 amu (10.00%), and 25.982 amu
(11.01%)
23.985 amu (0.7899) = 18.94575
24.986 amu (0.1000) = 2.4986
25.982 amu (0.1101) = 2.8606
24.31 amu
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