Atomic Structure and Subatomic Particles
John W. Moore
Conrad L. Stanitski
Peter C. Jurs
http://academic.cengage.com/chemistry/moore
Chapter 2
Atoms and Elements
Stephen C. Foster • Mississippi State University
Atoms are composed of subatomic particles:
• electron (e-), proton (p+) and neutron (n0).
Key discoveries:
Radioactivity
• Becquerel (1896)
• U ore emits rays that “fog” a photographic plate.
Marie and Pierre Curie (1898)
• Isolated new elements (Po & Ra) that did the same.
• Marie Curie called the phenomenon radioactivity.
radioactivity
Radioactivity
Electrons
Electrical behavior: “+” attracts “-”; like charges repel.
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
Atoms must contain smaller sub-units.
Electrons
Millikan (1911) studied electrically-charged oil drops.
• Coulomb (C) = SI unit of charge
Electrons
Charge on each droplet was:
n (−1.60 x 10-19 C) with n = 1, 2, 3,…
n (e- charge)
Modern value = −1.602176487 x 10-19 C.
= −1 “atomic units”.
These experiments give:
me = charge x
mass
charge
= (-1.60 x 10-19 C)(-5.60 x 10-9 g/C) = 8.96 x 10-28 g
Modern value = 9.10938215 x 10-28 g
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Protons
The Nuclear Atom
e- are
Atoms gain a positive charge when
Implies a positive fundamental particle.
lost.
Hydrogen ions had the lowest mass.
• Hydrogen nuclei assumed to have “unit mass”.
• Called protons.
protons
Modern science: mp = 1.672621637 x 10-24 g
mp ≈ 1800 x me
Charge = -1 x (e- charge).
= +1.602176487 x 10-19 C = +1 atomic units.
The Nuclear Atom
But … some had large deflections.
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 fired α-particles at thin metal foils.
Expected them to pass through with minor
deflections.
The Nucleus
Most of the mass and all “+” charge is concentrated
in a small core, the nucleus:
nucleus
• ≈10,000 times smaller diameter than the entire atom.
• e- occupy the remaining space.
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.”
Neutrons
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,
neutrons were ejected:
mn ≈ mp (0.1% larger).
mn = 1.674927211 x 10-24 g.
Present in all atoms (except normal H).
α particles
The Nuclear Atom
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+.
2
The Sizes of Atoms and Units
Atoms are very small.
• 1 teaspoon of water contains 3x as many atoms as
there are teaspoons of water in the Atlantic Ocean!
It’s impractical to use pounds and inches...
Need a universal unit system:
• The metric system.
• The SI system (Systeme International) - derived
from the metric system.
The Sizes of Atoms and Units
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
1.90 cm x 1 x 10 m x 1 pm-12
= 1.90 x 1010 pm
1 x 10 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
Metric Units
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
106
103
10-1
10-2
10-3
10-6
10-9
10-12
10-15
Examples
1 kilometer = 1 km = 1 x 103 meter
1 microgram = 1 μg = 1 x 10-6 gram
Some Common Unit Equalities
Length
1 kilometer
= 0.621 mile
1 inch
= 2.54 cm (exactly)
1 angstrom (Å) = 1 x 10-10 m
Volume 1 liter (L) = 1 dm3 = 1000 cm3 = 1000 mL
= 1.06 quarts
1 gallon = 4 quarts = 8 pints
Mass
= 1.661 x 10-24 g
= 453.6 g = 16 oz
= 1000 kg
= 2000 pounds
1 amu
1 pound
1 ton (metric)
1 ton (US)
Some Common Unit Equalities
Some Common Unit Equalities
What is the mass of a 5.0 lb bag of sugar in kilograms?
A patient’s blood cholesterol level measured 165
mg/dL. Express this value in g/L
1 lb = 453.6 g
1 mg = 1 x 10-3 g
;
1 dL = 1 x 10-1 L
453.6 g
= 2265 g
5.0 lb x
1 lb
= 2.3 x 103 g
-3
= 1.65 g/L
165 mg x 1 x10 g x 1 dL
dL
1 mg
1 x10-1 L
= 2.3 kg
3
Uncertainty and Significant Figures
Uncertainty and Significant Figures
All measurements involve some uncertainty.
6.3492 g has five significant figures.
figures
Scientists typically report a number with one uncertain
digit.
To determine the number of significant figures:
• Read numbers from left to right.
• Count all digits; start with the 1st non-zero digit.
• All digits are significant except zeros used to
position a decimal point (“placeholders”).
Consider a reported mass of 6.3492 g
• Last digit (“2”) is uncertain
• Close to 2, but may be 4, 3, 1, 0 …
0.00024030
placeholders
significant
(2.4030 x 10-4)
significant
5 sig. figs.
Uncertainty and Significant Figures
Number
Sig. figs.
Comment on Zeros
2.12
3
4.500
4
Not placeholders. Significant.
0.002541
4
Placeholders (not significant).
0.00100
3
Only the last two are significant.
500
1, 2, 3 ? Ambiguous
Ambiguous. May be placeholders or
Significant Figures in Calculations
Addition and subtraction
Find the decimal places (dp) in each number.
answer dp = smallest input dp.
Add:
17.245
0.1001
17.3451
dp = 3
dp = 4
Rounds to: 17.345
(dp = 3)
+
may be significant.
500.
3
Add a decimal point to show they are
significant.
5.0 x 102
2
No ambiguity.
Significant Figures in Calculations
Subtract 6.72 x 10-1 from 5.00 x 101
Significant Figures in Calculations
Multiplication and Division
Answer sig. fig = smallest input sig. fig.
Use equal powers of 10:
5.00 x 101
– 0.0672 x 101
4.9328 x 101
Rounds to: 4.93 x 101
dp = 2
dp = 4
dp = 2
x
17.245
0.1001
1.7262245
Rounds to: 1.726
sig. fig. = 5
sig. fig. = 4
sig. fig. = 4
Multiply 2.346, 12.1 and 500.99 = 14,221.402734
Rounds to:
1.42 x 104 (3 sig. fig.)
4
Rules for Rounding
Rules for Rounding
Examine the 1st nonnon-significant digit.
digit If it:
• > 5, round up.
• < 5, round down.
• = 5, check the 2nd non-significant digit.
Round the following numbers to 3 sig. figs:
Number
 round up if absent or odd; round down if even.
Round 37.663147 to 3 significant figures.
2nd
nonsignificant
digit
last retained
digit
Rounds up to 37.7
1st nonsignificant digit
Rules for Rounding
dp = 5
dp = 3
1st non-sig.
digit
2nd non-sig.
digit
Rounded
Number
2.123
2.123
3
-
2.12
51.372
51.37
72
51.372
2
51.4
131.5
131.5
5
-
132.
24.752
24.75
52
24.752
2
24.7
24.751
24.75
51
24.751
1
24.8
0.06744
0.06744
4
-
0.0674
Rules for Rounding
Answer dp = 3.
92.803 is the significant result.
(5 sig. figs).
To avoid rounding errors
• Carry additional digits through a calculation.
• Use the correct number of places in the final
answer.
92.80344
99.12444 – 6.321
=
= 3.37153195571
27.5256
27.5256
6 sig. figs.
Significant figures?
Note
Exact conversion factors:
(100 cm / 1 m) or
(2H / 1 H2O)
Have an infinite number of sig. figs.
= 3.3715 (5 sig. figs.)
Atomic Numbers & Mass Numbers
Same element: same number of p+
Atomic Numbers & Mass Numbers
The mass number (A
A) = number of p+ + number of n0
Atomic number (Z) = number of p+
Atomic mass unit (amu) =
contains 6 p+ and 6 n0.
1
(mass
12
of C atom) that
For element X, write:
1 amu = 1.66054 x 10-24 g
Particle
e−
p+
n0
Mass
(g)
9.10938215 x 10-28
1.672621637 x 10-24
1.674927211 x 10-24
Mass
(amu)
0.000548580
1.00728
1.00866
A ≈ mass (in amu) of an atom
Charge
(atomic units)
−1
+1
0
or
or
A
X
Z
A
X
X-A
e.g.
e.g.
12
C
6
12
C
e.g. carbon-12
(Z is constant for a given element)
5
Atomic Numbers & Mass Numbers
How many p+, n0 and e- are in the following elements:
29 p+ = 29 e- (neutral atom: e- = p+)
63
29 Cu
25
63−29 = 34 n0
12 p+ = 12 e- (periodic table; neutral)
Mg
25−12 = 13 n0
Isotopes and Atomic Weight
Isotopes
Atoms of the same element with different A.
• equal numbers of p+
• different numbers of n0
Hydrogen isotopes:
deuterium (D)
27
13
Al
p+ =
13
e-
27−13 = 14 n0
tritium (T)
Isotopes and Atomic Weight
Most elements occur as a mixture of isotopes.
1
1H
1 p+, 0 n0
2
1H
1 p+, 1 n0
3
H
1
1 p+, 2 n0
Isotopes and Atomic Weight
For most elements, the percent abundance of its
isotopes are constant (everywhere on earth).
Magnesium is a mixture of:
24Mg
25Mg
26Mg
number of p+
12
12
12
number of n0
12
13
14
mass / amu
23.985
24.986
25.982
Isotopes and Atomic Weights
10B
= 10.0129 amu; 11B = 11.0093 amu (10B = 19.91%). Atomic weight of B?
The periodic table lists an average atomic weight.
Example
Boron is a mixture of 10B (10.0129 amu) and 11B
(11.0093 amu). 10B is 19.91% abundant. Calculate
the atomic weight of boron.
Isotopes and Atomic Weight
Periodic table:
Atomic mass = Σ(fractional abundance)(isotope mass)
5
19.91 (10.0129 amu) = 1.994 amu
100
B
10B
% abundance of 11B = 100% - 19.91% = 80.09%
11B
80.09 (11.0093 amu) = 8.817 amu
100
Atomic number (Z)
Symbol
Boron
Name
10.811
Atomic weight
Atomic weight for B = 1.994 + 8.817 amu
= 10.811 amu
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Amounts of Substances: The Mole
A counting unit – a familiar counting unit is a “dozen”:
1 dozen eggs
1 dozen peas
= 12 eggs
= 12 peas
1 mole (mol) = Number of atoms in 12 g of 12C
• Latin for “heap” or “pile”
• 1 mol = 6.02214179 x 1023 “units”
• Avogadro’s number
1 mole eggs = 6.02 x 1023 eggs
1 mole peas = 6.02 x 1023 peas
Amounts of Substances: The Mole
1 mole of an atom = atomic weight in grams.
Amounts of Substances: The Mole
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 !
Molar Mass and Problem Solving
Example
How many moles of copper are in a 320.0 g sample?
1 Xe atom has mass = 131.29 amu
1 mol of Xe atoms has mass = 131.29 g
Cu-atom mass = 63.546 g/mol (periodic table)
1 He atom has mass = 4.0026 amu
1 mol of He has mass = 4.0026 g
Conversion factor: 1 mol Cu = 1
63.546 g
There are 6.02 x 1023 atoms in 1 mol of He and 1 mol
of Xe – but they have different masses.
nCu = 320.0 g x
… 1 dozen eggs is much heavier than 1 dozen peas!
Molar Mass and Problem Solving
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
1 mol Cu
= 5.036 mol Cu
63.546 g
n = number of moles
The Periodic Table
Summarizes
•
•
•
•
Atomic numbers.
Atomic weights.
Physical state (solid/liquid/gas).
Type (metal/non-metal/metalloid).
Periodicity
B atoms = (0.092507 mol B)(6.022  1023 atoms/mol)
= 5.571  1022 B atoms
•
Elements with similar properties are arranged in
vertical groups.
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The Periodic Table
In the USA, “A” denotes a
main group element…
The Periodic Table
Main group metal
International system
uses 1 … 18.
Transition metal
…”B” indicates a
transition metal.
Metalloid
Nonmetal
The Periodic Table
The Periodic Table
Period
number
A period is a horizontal row
Group 1A
Alkali metals (not H)
A group is a vertical column
Group 7A
Halogens
Group 8A
Noble gases
Group 2A
Alkaline
earth metals
Important Regions of the Periodic Table
• Alkali and alkaline earth metals are very reactive.
• Always found combined with other elements in
nature.
• Halogens are highly reactive diatomic molecules.
• F2, Cl2, Br2, I2
• Noble gases are the least reactive elements.
• Lanthanides and actinides are metals listed
separately at the bottom of the periodic table.
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