Principles of Chemistry, Chapt. 2: Atomic
Structure and The Elements
I.
The Structure of Atoms
protons, neutrons, and electrons
II. Atomic Structure and Properties—the
Elements
atomic mass, atomic number, isotopes
III. The Mole Concept: 6.02 x 1023
IV. The Periodic Table
Homework: Chapt. 2 Problems 26, 29, 37, 43, 75
1
Atomic Theory in a single Slide
~ 10-10 meters = 1 angstrom (Å)
Smeared out electron
charge cloud
_
Most of the mass is
here
10-14 m
+
Most of the Chemistry
is here
Protons and
neutrons
+
+
+
+
+
+
+
2
STM Image: Oxygen atoms at the surface of
Al2O3/Ni3Al(111)
Electronic
charge
cloud
surrounding
the nucleus
S. Addepalli, et al. Surf. Sci. 442 (1999) 3464
3
What’s inside the nucleus:
Particle
Mass (amu)
Charge
Proton (p+)
Neutron (n0)
1.007 amu
1.009 amu
+1
0
What’s outside the nucleus:
Electron (e-)
.00055 amu
-1
Note: mass ratio of electron/proton (Mp+/Me-) = 1836
For any atom: # of electrons = # of protons: Why?
4
Atomic Theory: Late 19th Century
Atomic theory—everything is made of atoms—generally accepted
(thanks to Ludwig Boltzman, and others).
Mendeleev/periodic table—accepted, but the basis for periodic behavior
not understood
What are atoms made of?
How are they held together?
5
Electrical behavior: “+” attracts “-” but like charges repel
Radioactive material
β-particles (“–”)
Beam of
, ,
and 
Gamma ray (γ)
No charge,
no
deflection
Electrically
charged plates
α-particle (“+” )
Heavier, deflected less
than β
Atoms must contain smaller sub-units.
Alpha particle  2 n0 + 2p+
Beta particel  electron (e-)
Gamma photon
6
•Thomson
(1897) discovered the e-:
“Cathode rays”
– high voltage +
fluorescent
screen
cathode ray
• Travel from cathode (-) to anode (+).
• Negative charge (e−).
• Emitted by cathode metal atoms.
Electric and magnetic fields deflect the beam.
• Gives mass/charge of e- = −5.60 x 10-9 g/C
• Coulomb (C) = SI unit of charge
7
Essence of the Thompson Experiement (and old fashioned TV’s)
y
x
+
Electric field exerts Force
+ plate repels +charged particles
- Plate repels – charged particles
+
d
_
_
Phosphor screen
F = Eq = ma
d = displacement = ½ at2 = Eq/m (t = L/Vx)
Therefore, the greater the displacement, the lower the
mass of the particle
8
•
Millikan (1911) studied electrically-charged oil drops.
9
Charge on each droplet was:
n (−1.60 x 10-19 C) with n = 1, 2, 3,…
n (e- charge)
Modern value = −1.60217653 x 10-19 C.
= −1 “atomic units”.
•These
experiments give: me = charge x
= (-1.60 x 10-19 C)(-5.60 x 10-9 g/C)
mass
charge
= 8.96 x 10-28 g
Modern value = 9.1093826 x 10-28 g
10
•
Atoms gain a positive charge when e- are lost.
Implies a positive fundamental particle.
Hydrogen ions had the lowest mass.
• Hydrogen nuclei assumed to have “unit mass”
• Called protons.
Modern science: mp = 1.67262129 x 10-24 g
mp ≈ 1800 x me.
Charge = -1 x (e- charge).
= +1.602176462 x 10-19 C = +1 atomic units
11

How were p+ and e- arranged?
Thompson:
• Ball of uniform positive charge, with small negative
dots (e-) stuck in it.
• The “plum-pudding” model.
1910 Rutherford (former Thompson graduate
student) fired α-particles at thin metal
foils.
Expected them to pass through with minor
deflections.
12
Rutherford Scattering Experiment
13
Different Models of the Atom: different scattering results
α particles
“Plum pudding model”
•+ and – charges evenly distrubted
•low, uniform density of matter
•No back scattering
Rutherford’s explanation of results:
α particles
Small regions of very high density
+ charge in the dense regions
- Charges in region around it
From wikipedia
14
Some Large Deflections were osbserved
α particles
Rutherford
“It was about as credible as if you had fired a 15-inch shell
at a piece of paper and it came back and hit you.”
15
Most of the mass and all “+” charge is
concentrated in a small core, the nucleus.
≈10,000 times smaller diameter than the entire atom.
e- occupy the remaining space.
α particles
16
Charge cloud
Diameter ~ 1 Å
Mass ~ 10-30 kg
Nucleus
diameter~ 10-4 Å = 10-14 meters
Mass ~ 10-27 Kg
17
Most Chemistry involves rearrangement of outermost electrons, not
nuclei
Example: H  1p+ , 1 e-
H + H  H2
+
18
19
7 Å Epitaxial Al2O3(111) film on Ni3Al(111) (Kelber group):
•Grown in UHV
Start with ordered films growth studies
•Uniform
Proceed to amorphous films on Si(100)
•No Charging
STM
Surface
terminated by
hexagonal array
of O anions
S. Addepalli, et al. Surf. Sci. 442 (1999) 3464
20
•Atomic
mass > mass of all p+ and e- in an atom.
•Rutherford proposed a neutral particle.
Chadwick (1932) fired -particles at Be atoms.
Neutral particles, neutrons, were ejected:
mn ≈ mp (0.1% larger).
mn = 1.67492728 x 10-24 g.
Present in all atoms (except ‘normal’ H).
21
~ 10-10 meters = 1 angstrom (Å)
Smeared out electron
charge cloud
_
10-14 m
+
Protons and
neutrons
+
+
+
+
+
+
+
22
Nucleus
• Contains p+ and n0
• Most of the atomic mass.
• Small (~10,000x smaller diameter than the atom).
• Positive (each p+ has +1 charge).
Electrons
• Small light particles surrounding the nucleus.
• Occupy most of the volume.
• Charge = -1.
Atoms are neutral. Number of e− = Number of p+
23
A neutron can decay into a proton and electron:
n0  p+ + eThis can cause decay of a radioactive element, e.g.,
# of p+ + n0
14
Elemental symbol (carbon)
C
6
Atomic No.
(# e- = # p+
Carbon with 6 protons and 8 neutrons is unstable (radioactive)
Carbon with 6 protons and 6 neutrons is stable (non-radioactive
14
12
6
6
C
radioctive
C
stable
24
An atom of 14C can undergo decay to N as a neutron turns into a
proton + an emitted electron
14
14
6
7
C
N
+ e-
1 p+  1 n0 + an electron (emitted)
25
Same
element - same number of p+
Atomic number (Z) = number of p+
Atomic mass unit (amu) =
contains 6 p+ and 6 n0.
1
12 (mass
of C atom) that
1 amu = 1.66054 x 10-24 g
Particle
e−
p+
n0
Mass
(g)
9.1093826 x 10-28
1.67262129 x 10-24
1.67492728 x 10-24
Mass
(amu)
0.000548579
1.00728
1.00866
Charge
(atomic units)
−1
+1
0
26


•
•
Isotopes
Atoms of the same element with different A.
equal numbers of p+
different numbers of n0
Hydrogen isotopes:
deuterium (D)
tritium (T)
1
H
1
1 p+, 0 n0
2
H
1
3
H
1
1 p+, 1 n0
1 p+, 2 n0
27
ISOTOPES: SAME Element, Different numbers of neutrons
Carbon: atomic no. = 6  6 protons in the nucleus+ 6 electrons
12
C
14
C
Atomic mass = 12 amu (12 gr/mole)
Therefore , 6 protons + 6 neutrons
Atomic mass = 14 amu
Therefore, 6 protons+ 8 neutrons
28
Isotopes Display the same chemical reactivities (which depend
mainly on the outer arrangement of the electrons)
12C
+ O2  CO2
14C
+O2  CO2
Isotopes display different nuclear properties
12
C
14
C
stable
Radioactive: spontaneously
emits electrons. Half-life ~
5730 years
29
Isotopes and Moles (more on this later) and isotope abundance:
1 mole = 6.02 x 1023 of anything!
1 mole of atoms = 6.02 x 1023 atoms
Molar Mass (in grams) = average atomic mass (in amu)
1 mole of H atoms = 1.008 gr.
Why not 1.000 gr??  most atoms are 1H, but some are 2H (deuterium)
30
Average atomic mass of H = 1.008 amu
100 atoms have a mass of 100.8 amu
# of 2H atoms = n
# of 1H atoms = 100 –n (assume these are the only two isotopes that matter)
Mass of 100 atoms = n x 2.000 +(100-n) x 1.000 = 100.8 amu
2n + 100-n = 100.8
n = 0.8 So, out of every 100 atoms , have 0.8 2H atoms
Out of every 1000 atoms, have 8 2H atoms
Natural abundance of “heavy hydrogen (deuterium) is then 0.8%
31
Most
elements occur as a mixture of isotopes.
Magnesium is a mixture of:
24Mg
number of p+
number of n0
mass / amu
12
12
23.985
25Mg
12
13
24.986
26Mg
12
14
25.982
32
For
most elements, the percent abundance of its
isotopes are constant (everywhere on earth).
The
periodic table lists an average atomic weight.
Example
Boron occurs as a mixture of 2 isotopes, 10B and 11B.
The abundance of 10B is 19.91%. Calculate the
atomic weight of boron.
33
Boron occurs as a mixture of 2 isotopes, 10B and 11B. The abundance of 10B is
19.91%. Calculate the atomic weight of boron.
Atomic mass = Σ(fractional abundance)(isotope mass)
10B
19.91 (10.0129 amu) = 1.994 amu
100
% abundance of 11B = 100% - 19.91% = 80.09%
11B
80.09 (11.0093 amu) = 8.817 amu
100
Atomic weight for B = 1.994 + 8.817 amu
= 10.811 amu
34
Periodic table:
5
B
Atomic number (Z)
Symbol
Boron
Name
10.811
Atomic weight
35
A
counting unit – a familiar counting unit is a “dozen”:
1 dozen eggs
= 12 eggs
1 dozen donuts = 12 donuts
1 dozen apples = 12 apples
1 mole (mol) = Number of atoms in 12 g of 12C
• Latin for “heap” or “pile”
• 1 mol = 6.02214199 x 1023 “units”
• Avogadro’s number
36




A green pea has a ¼-inch diameter. 48
peas/foot.
(48)3 / ft3 ≈ 1 x 105 peas/ft3.
V of 1 mol ≈ (6.0 x 1023 peas)/(1x 105 peas/ft3)
≈ 6.0 x 1018 ft3
U.S. surface area = 3.0 x 106 mi2
= 8.4 x 1013 ft2
height = V / area, 1 mol would cover the U.S. to:
6.0 x 1018 ft3
8.4 x 1013 ft2
=7.1 x 104 ft = 14 miles !
37
1
mole of an atom = atomic weight in grams.
1 Xe atom has mass = 131.29 amu
1 mol of Xe atoms has mass = 131.29 g
1 He atom has mass = 4.0026 amu
1 mol of He has mass = 4.0026 g
There are 6.022 x 1023 atoms in 1 mol of He and 1
mol of Xe – but they have different masses.
… 1 dozen eggs is much heavier than 1 dozen peas!
38
Example
How many moles of copper are in a 320.0 g sample?
Cu-atom mass = 63.546 g/mol (periodic table)
Conversion factor: 1 mol Cu = 1
63.546 g
1 mol Cu
nCu = 320.0 g x
= 5.036 mol Cu
63.546 g
n = number of moles
39
Calculate
the number of atoms in a 1.000 g
sample of boron.
nB = (1.000 g) 1 mol B
10.81 g
= 0.092507 mol B
B atoms = (0.092507 mol B)(6.022  1023 atoms/mol)
= 5.571  1022 B atoms
40
Dimensional Analysis and Problem Solving
Special Homework Problem: Due Tues. Recitation
Density = mass/volume
Problem:
Assume that a hydrogen atom has a spherical diameter of 1 angstrom
Assume that the nucleus (1 proton) has a diameter of 10-4 angstrom
1. Calculate the densities of the nucleus, and of the electron charge
cloud in kg/m3
2. Calculate the ratio of the two densities: R = dnucleus/delectron cloud
Mass of proton = 1.67 x 10-27 kg
Mass of electron = 9 x 10-31 kg
41

•
•
•
•

•
Summarizes
Atomic numbers.
Atomic weights.
Physical state (solid/liquid/gas).
Type (metal/non-metal/metalloid).
Periodicity
Elements with similar properties are arranged
in vertical groups.
42
The Periodic Table
In the USA, “A” denotes a
main group element…
International system
uses 1 … 18.
…”B” indicates a
transition element.
43
The Periodic Table
Main group metal
Transition metal
Metalloid
Nonmetal
44
Period
number
A period is a horizontal row
45
Group 1A
Alkali metals (not H)
A group is a vertical column
Group 7A
Halogens
Group 8A
Noble gases
Group 2A
Alkaline
earth metals
46
Alkali
metals (group 1A; 1)
Alkaline earth metals (group 2A; 2)
• Grey … silvery white
colored.
• Highly reactive.
• Never found as native
metals.
• Form alkaline solutions.
47

•
•
•
Transition Elements (groups 1B – 8B)
Also called transition metals.
Middle of table, periods 4 – 7.
Includes the lanthanides & actinides.
Lanthanides and Actinides
• Listed separately at the bottom.
• Chemically very similar.
• Relatively rare on earth.
 (old name: rare earth elements)
48

•
•
Groups 3A to 6A
Most abundant elements in the Earth’s crust
and atmosphere.
Most important elements for living organisms.
Halogens (group 7A; 17)
• Very reactive non metals.
• Form salts with metals.
• Colored elements.
Noble gases (8A; 18)
• Very low reactivity.
• Colorless, odorless gases.
49

•
Atoms are very small.
1 tsp of water contains 3x as many atoms as
there are tsp of water in the Atlantic Ocean!
Impractical to use pounds and inches...
Need a universal unit system
• The metric system.
• The SI system (Systeme International) - derived
from the metric system.
50
•
•
A decimal system.
Prefixes multiply or divide a unit by multiples of ten.
Prefix
mega
kilo
deci
centi
milli
micro
nano
pico
femto
M
k
d
c
m
μ
n
p
f
Factor
Example
106
1 megaton = 1 x 106 tons
103
1 kilometer (km) = 1 x 103 meter (m)
10-1 1 deciliter (dL) = 1 x 10-1 liter (L)
10-2 1 centimeter (cm) = 1 x 10-2 m
10-3 1 milligram (mg) = 1 x 10-3 gram (g)
10-6 1 micrometer (μm) = 1 x 10-6 m
10-9 1 nanogram (ng) = 1 x 10-9 g
10-12 1 picometer (pm) = 1 x 10-12 m
10-15 1 femtogram (fg) = 1 x 10-15 g
51
How many copper atoms lie across the diameter of a
penny? A penny has a diameter of 1.90 cm, and a
copper atom has a diameter of 256 pm.
1 pm = 1 x 10-12 m
;
1 cm = 1 x 10-2 m
-2 m
1 pm
1
x
10
1.90 x 1010 pm
=
1.90 cm x
x
1 x 10-12 m
1 cm
Number of atoms across the diameter:
1.90 x 1010 pm x 1 Cu atom = 7.42 x 107 Cu atoms
256 pm
52
Length
1 kilometer
= 0.62137 mile
1 inch
= 2.54 cm (exactly)
1 angstrom (Å) = 1 x 10-10 m
1 gallon
= 1000 cm3 = 1000 mL
= 1.056710 quarts
= 4 quarts = 8 pints
1 amu
1 pound
1 ton (metric)
1 ton (US)
= 1.66054 x 10-24 g
= 453.59237 g = 16 ounces
= 1000 kg
= 2000 pounds
Volume 1 liter (L)
Mass
53
Example: What is the volume of a 2 gallon container in Liters?
1 gallon x 4 quarts/gallon = 4 quarts
4 quarts x 1Liter/1.057 quarts = 3.784 Liters (L)
54
A patient’s blood cholesterol level measured 165
mg/dL. Express this value in g/L
1 mg = 1 x 10-3 g
;
1 dL = 1 x 10-1 L
-3 g
mg
1
x10
= 1.65 g/L
165
x 1 dL
x
1 x10-1 L
dL
1 mg
55
All
measurements involve some uncertainty.
Reported
numbers include one uncertain digit.
Consider a reported mass of 6.3492 g
• Last digit (“2”) is uncertain
• Close to 2, but may be 4, 1, 0 …
• Five significant figures in this number.
56
Read
numbers from left to right.
Count all digits, starting with the 1st non-zero
digit.
All
digits are significant except zeros used to
position a decimal point (“placeholders”).

0.00024030


-4)
(2.4030
x
10
placeholders significant
significant
5 sig. figs.
57
58
Examine
the 1st non-significant digit. If it:
• > 5, round up.
• < 5, round down.
• = 5, check the 2nd non-significant digit.
 round up if absent or odd; round down if even.
Round 37.663147 to 3 significant figures.
last retained
digit
2nd nonsignificant
digit
Rounds up to 37.7
1st nonsignificant digit
59