Chemistry 100(02) Fall 2011
Instructor: Dr. Upali Siriwardane
e-mail: upali@chem.latech.edu
Office: CTH 311 Phone 257-4941
Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m Tu,Th,F 9:00 10:00 a.m.
Test Dates: March 25, April 26, and May 18; Comprehensive Fina
Exam: 9:30-10:45 am, CTH 328.
October 3,
October 26,
November 16,
November 17,
2011 (Test 1): Chapter 1 & 2
2011 (Test 3): Chapter 3 & 4
2011 (Chapter 5 & 6)
2011 (Make-up test) comprehensive: Chapters 1-6
9:30-10:45:15 AM, CTH 328
CHEM 100, FALL 2011 LA TECH
2-1
Chapter 2. Elements & Atoms
2.1 Atomic Structure and Subatomic Particles
2.2 The Nuclear Atom
2.3 The Size of Atoms and Units Used to represent them
2.4 Uncertainty and Significant Figures
2.5 Atomic Numbers and Mass Numbers
2.6 Isotopes and Atomic Weights
2.7 Amount of Substances-The mole
2.8 Molar Mass and Problem Solving
2.9 The Periodic Table
CHEM 100, FALL 2011 LA TECH
2-2
Chapter 2. KEY CONCEPTS
Elements & Atoms
Radioactivity
Subatomic Particles
Electrons
Electronic Charge
Nuclear atom Protons
Neutrons
Atomic number (Z)
Size of Atoms
SI Units
Unit Conversions
Mass Numbers
CHEM 100, FALL 2011 LA TECH
Isotopes
Isotopic symbols
Atomic Mass Units
Mass Spectrometer
isotope masses and %
composition?
Average atomic weights
Periodic Table
Abundance of Elements
Earth's Atmosphere
2-3
Atomic Structure
Early experiments showed the atom was
composed of three subatomic particles:
• Electron, proton and neutron.
The key discoveries:
Radioactivity
• Becquerel (1896)
– Uranium ore emits rays that “fog” a photographic plate.
•
Marie and Pierre Curie (1898)
– Isolated 2 new elements (Po and Ra) that did the same.
– Marie Curie called the phenomenon radioactivity.
CHEM 100, FALL 2011 LA TECH
2-4
Radioactivity
Types of Radiation
Alpha ray
Beta ray
Gamma ray γ
α
(positive charge)
β
(negative charge)
(no charge)
Electrical behavior:
attract)
+ attracted to - (opposites
(like charges
repel)
Radioactive material
β
+
−
Beam of α, β,
and γ
α
γ
Electrically
Charged plates
screen
CHEM 100, FALL 2011 LA TECH
2-5
Electrons
Thomson (1897) studied cathode rays and
discovered the electron:
fluorescent
screen
– high voltage +
•
•
•
cathode ray
Beam travels from the cathode (-) to the anode (+).
–
the beam flies through a ring anode and hits a fluorescent screen.
The cathode rays come from the cathode metal.
−
They are negative particles – electrons (e ).
CHEM 100, FALL 2011 LA TECH
2-6
Electrons
Thomson showed that electric and magnetic
fields deflect the beam.
+
– high voltage +
–
From the deflections,
Thomson calculated the mass/charge
ratio for an e :
-9
= −5.60 x 10 g/C
(Coulomb (C) = the SI unit of charge)
CHEM 100, FALL 2011 LA TECH
2-7
Electronic Charge
Robert Millikan (1911) studied electricallycharged oil drops.
•
•
For a single charged drop, he measured:
– the time to fall a fixed distance, and
– to rise the same distance in an electric field.
He showed that each drop
-19 had a charge that was an integer
multiple of −1.60 x 10
C.
(The charge of an electron. )
•
-19
The modern value is −1.602176462 x 10
C.
(Often written in “atomic units” as charge = −1).
CHEM 100, FALL 2011 LA TECH
2-8
Millikan’s Experiment
CHEM 100, FALL 2011 LA TECH
2-9
Mass of an Electron
The experiments by Thomson and Millikan
gave the mass/charge ratio and charge of
an e−.
me
= charge
x
mass
charge
-19
-9
= (−1.60 x 10
C)(−5.60 x 10 g/C)
-28
= 8.96 x 10
g
The modern value is:
me = 9.10938188 x 10-28 g
CHEM 100, FALL 2011 LA TECH
2-10
Protons
Atoms become positively charged when e- are
lost.
•
Implies a positively charged fundamental particle.
•
Hydrogen ions had the lowest mass.
– Hydrogen nuclei were assumed to have “unit mass” common to all
atoms and were called protons.
•
Modern science: mp = 1.67262158 x 10
-24
g
– mp is approximately 1800 times larger than me.
– I’ll use shorthand: p+ = proton.
– Charge equal, but of opposite sign, to the electron.
– Charge = +1.602176462 x 10-19 C (+1 in atomic units).
CHEM 100, FALL 2011 LA TECH
2-11
Nuclear Atom
•
Thompson thought it was a ball of uniform
positive
charge, with small negative dots (e ) stuck in it.
How were these particles (p+ and e-)
arranged?
•
The “plum-pudding” model.
CHEM 100, FALL 2011 LA TECH
2-12
The Nucleus
1910 Rutherford fired α-particles at thin metal
foils. He expected them to pass through the
foil.
But … a few α’s were deflected through large angles. Some
came almost straight back!
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.”
CHEM 100, FALL 2011 LA TECH
2-13
The Nucleus
Explanation?
• Most of the mass (and all the positive charge) is concentrated
in a tiny core – the nucleus.
•
– approximately 10,000 times smaller than entire atom.
The rest of the volume filled by the electrons.
α particles
CHEM 100, FALL 2011 LA TECH
2-14
•
•
•
•
Determination of nuclear charge
Rutherford estimated that the charge of the nucleus of an
atom was about one half of the atomic mass.
Moseley, while working for Rutherford, developed a more
accurate measurement.
While working with cathode rays on metal targets, he
measured the wavelength of the X-rays produced.
He found that a direct relationship exists between the
metal’s atomic number and the square root of the
frequency.
CHEM 100, FALL 2011 LA TECH
2-15
Discovery of Protons and Atomic
Number
Moseley, Henry & Gwyn Jeffreys
1887–1915, English physicist.
• studied the relations among x-ray spectra of different
elements.
• concluded that the atomic number is equal to the charge on
the nucleus based on the x-ray spectra emitted by the
element.
• explained discrepancies in Mendeleev’s Periodic Law.
CHEM 100, FALL 2011 LA TECH
2-16
Determination of nuclear charge
Atomic number
Moseley concluded that
the charge of the nucleus
was an integer.
Further, it was the same
as the number of electrical
units (electrons) but of
opposite charge.
1/2
X-Ray Frequency
CHEM 100, FALL 2011 LA TECH
2-17
Discovery of neutrons
• 1932 Chadwick observed that when beryllium-9 was exposed to
alpha particles, particles with the same mass as protons but no
charge were given off.
• These were called neutrons and are present in all atoms except
hydrogen-1.
• They contribute to the force that holds the nucleus together and
reduce the repulsive force between positively charged protons.
• Mass very similar to protons is 0.1% larger).
mn = 1.674928716 x 10-24 g.
CHEM 100, FALL 2011 LA TECH
2-18
Summary of Subatomic Particles
Particle
Proton
Charge
+1.6 x 10-19 C
Neutron
Electron
zero
-1. 6 x 10-19 C
Mass (g)
1.7 x 10-24 g
Mass (amu)
1.0073
1.7 x 10-24 g
1.0087
9.1 x 10-28 g
5. 5x 10-4
Remember: Atoms are usually electrically neutral, Indicating equal numbers
of protons and electrons!
CHEM 100, FALL 2011 LA TECH
2-19
Structure of the Atom
•
Nucleus
Atoms are composed of subatomic particles:
– Made of protons and neutrons.
– Contains most of the mass of the atom.
– Small (~10,000 times smaller than the entire atom).
– Positive (each p+ has unit positive charge).
– At the center of an atom.
•
Electrons
– Very small light particles that surround the nucleus.
– Occupy most of an atom’s volume.
– Each has unit negative charge.
CHEM 100, FALL 2011 LA TECH
−
Atoms are neutral. Number of e = Number of p
+
2-20
Sizes of Atoms and Units
Atoms are very small – we need to consider very
small numbers.
• A teaspoon of water contains 3x as many atoms as
there are teaspoons of water in the Atlantic Ocean!
•
It would be impractical to describe nanoscale objects in
pounds and inches.
– We need much smaller scales.
•
Need to use units in use around the world:
– The metric system.
– The SI system (Systeme International) - derived from the metric
system.
CHEM 100, FALL 2011 LA TECH
2-21
Atomic number, Z
• The number of protons in the nucleus
• The number of electrons in a neutral
atom
• The integer on the periodic table for each
element
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2-22
Structure of the Atom
electrons
found in electron
cloud
relative charge of -1.602 1773 x 10-19C
relative mass of
0.00055 amu
CHEM 100, FALL 2011 LA TECH
protons
• found in nucleus
• relative charge of +1
• relative mass of
1.0073 amu
neutrons
• found in nucleus
• neutral charge
• relative mass of
1.0087 amu
2-23
Relative
size of
atom
and
atomic
nucleus
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2-24
Scanning Tunneling Microscope
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2-25
Ions
Charged single atom
Charged cluster of atoms
• Cations: positive ions
• Anions: negative ions
Ionic compounds: combination of cations and
anions with zero net charge
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2-26
Nuclear Notation
X = atomic symbol
A = mass number
Z = atomic number
C-12, carbon-12
XA
C12
A
X
Z
12
C
6
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2-27
Mass Number, A
integer representing the approximate mass of an
atom
equal to the sum of the number of protons and
neutrons in the nucleus
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2-28
The atomic symbol & isotopes
Isotopic symbol: atomic symbol showing
atomic number (Z) and mass number (A)
Determine the number of protons, neutrons
and electrons in each of the following.
31
15
P
CHEM 100, FALL 2011 LA TECH
138
56
Ba
2+
238
92 U
2-29
Isotopes
*Atoms of the same element but having
different masses.
*All isotopes of an element have same atomic
number
*Each isotope has a different number of
neutrons.
Isotopes of hydrogen
Isotopes of carbon
CHEM 100, FALL 2011 LA TECH
1
1
H
2
1
H
3
1
H
12
6
C
13
6
C
14
6
C
2-30
Isotopes
Most elements occur in nature as a mixture of
isotopes.
• Element
•
•
•
•
•
H
C
O
Fe
Sn
Number of stable isotopes
2
2
3
4
10
This is one reason why atomic masses are not
whole numbers. They are based on averages.
CHEM 100, FALL 2011 LA TECH
2-31
Isotopes
• Atoms are composed of _____, _____ and ______.
• Almost all of the mass of an atom comes from the
______ and ______.
• All atoms of the same element will have the same
number of ______.
• The number of _______ may vary in isotopes of
an element.
• Most elements exist as a mixture of _______.
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2-32
Measurement
•
•
•
•
•
•
Measurements or observations are made
using our physical senses or using scientific
instruments.
1) Qualitative measurements.
Changes that cannot be expressed in terms of a
number.
2) Quantitative measurements.
expressed in terms of a number and an unit.
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Scientific Measurement
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2-34
Units
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2-35
SI units
SI - System International
• Systematic subset of the metric system.
Only uses certain metric units.
•
•
•
•
•
Mass
Length
Time
Temperature
Amount
kilograms
meters
seconds
Kelvin
mole
Other SI units are derived from SI base
units.
CHEM 100, FALL 2011 LA TECH
2-36
Metric prefixes
Changing the prefix alters the size of a unit.
Prefix
Symbol
Factor
mega
kilo
hecto
deka
base
deci
centi
milli
M
10
micro
m
nano
n
CHEM 100, FALL 2011 LA TECH
k
h
da
d
c
m
6
3
10
2
10
1
10
0
10
-1
10
-2
10
-3
10
-6
10
-9
10
1 000 000
1 000
100
10
1
0.1
0.01
0.001
0.0000001
0.000000000001
2-37
Metric Prefixes
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2-38
Unit Conversions
Example
How many grams of sugar are there in a 5.0 lb bag of sugar?
Look up conversion factor:
1 lb = 453. g
Too many digits for the
final answer!
5.0 lb
x
453. g
1 lb
= 2265 g
3
= 2.3 x 10 g
= 2.3 kg
CHEM 100, FALL 2011 LA TECH
2-39
Metric Units: Metric System
A decimal system.
Prefixes multiply or divide a unit by multiples of ten.
Prefix
mega
kilo
deci
centi
Factor
M
k
d
c
10
10
10
10
milli
m
10
micro
μ
10
nano
pico
femto
n
p
f
CHEM 100, FALL 2011 LA TECH
10
10
10
6
3
-1
-2
-3
-6
-9
-12
-15
Example
6
1 x 10 tons
3
1 kilometer (km) = 1 x 10 meter (m)
-1
1 deciliter (dL) = 1 x 10 liter (L)
-2
1 centimeter (cm) = 1 x 10 m
-3
1 milligram (mg) = 1 x 10 gram (g)
-6
1 micrometer (μm) = 1 x 10 m
-9
1 nanogram (ng) = 1 x 10 g
-12
1 picometer (pm) = 1 x 10
m
-15
1 femtogram (fg) = 1 x 10
g
1 megaton =
2-40
Example. Metric Conversion
How many milligrams are in a kilogram?
1 kg
1g
=
=
1 kg x 1000
g
1000 g
1000 mg
x 1000
mg
kg
g
= 1 000 000 mg
CHEM 100, FALL 2011 LA TECH
2-41
Metric Units
Example
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.
Metric conversion factors:
1 pm
1 cm
-12
= 1 x 10-2 m
= 1 x 10 m
CHEM 100, FALL 2011 LA TECH
2-42
Metric Units
Convert the diameter into the same units as a Cu atom:
-2
10
1 pm-12
1
x
10
m
1.90
x
10
pm
1.90 cm
x
=
x 1 x 10
m
1 cm
Calculate the number of atoms across the diameter:
1.90 x 10
10
pm
CHEM 100, FALL 2011 LA TECH
x
1 Cu atom
256 pm
7
= 7.42 x 10 Cu atoms
2-43
Common Unit Equalities
Length1 kilometer = 1000 m = 0.62137 mile
1 inch
= 2.54 cm (exactly)
-10
1 angstrom (Å)
= 1 x 10
m
Volume
1 liter (L)
1 gallon
Mass 1 amu
-3
3
= 1 x 10 m
3
= 1000 cm = 1000 mL
= 1.056710 quarts
= 4 quarts = 8 pints
-24
= 1.6606 x 10
g
1 pound
= 453.59237 g = 16 ounces
1 ton (metric) = 1000 kg
1 ton (US) = 2000 pounds
CHEM 100, FALL 2011 LA TECH
2-44
Unit Conversions
A patient’s blood cholesterol level measured 165 mg/dL. Express
this value in g/L
Determine the conversion
relationships:
-3
1 mg = 1 x 10-1 g
1 dL = 1 x 10 L
165
mg
dL
-3
1
x10
g
x
1 mg
CHEM 100, FALL 2011 LA TECH
x
1 dL-1
1 x10 L
= 1.65 g/L
2-45
Uncertainty and Significant
Figures
All measurements involve some uncertainty.
Scientists write down all the digits that have
no uncertainty plus one additional uncertain
digit.
If an object is reported to have a mass =
6.3492 g, the last digit (“2”) is uncertain ( it
is probably close to 2, but may be 4, 1 …).
There are five significant figures in this
number. All the digits are meaningful.
CHEM 100, FALL 2011 LA TECH
2-46
Uncertainty and Significant Figures
To find the number of significant figures:
Read a number from left to right and count all
digits, starting with the first non-zero digit.
All digits are significant except those zeros that
are used to position a decimal point
(“placeholders”).
0.00034050
placeholders
significant
significant
5 sig. figs.
Scientific Notation (3.4050 x 10-4)
CHEM 100, FALL 2011 LA TECH
2-47
Uncertainty and Significant Figures
Examples
Number
2.12
4.500
Sig. figs.
Comment
3
4
The zeros are not placeholders. They are
significant.
0.002541
4
0.00100
3
500
1, 2, 3 ?
The zeros are placeholders (not significant).
500.
3
Adding a decimal point is one way to show that
the zeros are significant.
5.0 x 102
2
No ambiguity.
CHEM 100, FALL 2011 LA TECH
Only the last two zeros are significant.
Ambiguous. If a number lacks a decimal point
the zeros may be placeholders or may be
significant.
2-48
Significant Figures
The answer you report in a problem should only
include significant digits.
Addition and subtraction
Find the number of digits after the decimal point
(adp) in each number.
answer adp = smallest input adp.
Example
Add:
17.245
adp = 3
+ 0.1001
adp = 4
17.3451
Rounds to: 17.345
CHEM 100, FALL 2011 LA TECH
(adp = 3)
2-49
Significant Figures
Example
Subtract 6.72 x 10-1 from 5.00 x 101
Write the numbers down with the same power of 10:
5.00 x 101
– 0.0672 x 101
4.9328 x 101
Rounds to:4.93 x 101
CHEM 100, FALL 2011 LA TECH
adp = 2
adp = 4
adp = 2
2-50
Significant Figures
Multiplication and Division
Find the number of significant figures (sig. fig.) in each
number.
Answer has sig. fig = smallest input sig. fig.
Example
Multiply 17.425 and 0.1001
sig. fig. = 5
17.245
sig. fig. = 4
x 0.1001
1.7262245
Rounds to: 1.726
sig. fig. = 4
Example
Multiply 2.346 x 12.1 x 500.99
Rounds to:
CHEM 100, FALL 2011 LA TECH
= 14,221.402734
1.42 x 104 (3 sig. fig.)
2-51
Rounding
Look at the 1st non-significant digit (the digit
after the last one retained). If it:
is > 5, round the last retained digit up by 1.
is < 5, make no change.
nd
equals 5, and the 2 non-significant digit is:
absent, round the last retained digit up by 1.
odd, round the last retained digit up by 1.
even, make no change.
Consider rounding 37.663147 to 3 significant figures.
2
last retained digit
1
CHEM 100, FALL 2011 LA TECH
st
nd
non-significant
digit
It rounds up to 37.7
non-significant digit
2-52
Rounding
Examples
Round the following numbers to 3 significant figures:
1
Number
2.123
51.372
131.5
24.752
24.751
0.06744
st
non-sig. 2
digit
2.123
51.372
131.5
24.752
24.751
0.06744
CHEM 100, FALL 2011 LA TECH
nd
non-sig.
digit
51.372
Rounded
Number
2.12
51.4
132.
24.752
24.751
-
24.7
24.8
0.0674
2-53
Significant Figures and Rounding
Example
Perform the following calculation and report the answer
with the correct number of significant figures:
adp = 5
92.803 is the significant result (adp = 3; 5
sig. figs.)
adp = 3
99.12444 – 6.321
27.5256
=
92.80344
27.5256
= 3.37153195571
6 sig. figs.
How many digits are significant?
= 3.3715 (5 sig. figs.)
CHEM 100, FALL 2011 LA TECH
2-54
Significant Figures and Rounding
To avoid rounding errors
• carry 1 additional digit through a calculation.
• round the final answer to the correct number of
places.
Remember
• Exact conversion factors like (100 cm / 1 m) or
(2H / 1 H2O) have an infinite number of
significant figures.
CHEM 100, FALL 2011 LA TECH
2-55
Problem Solving by
Factor Label Method
• State question in mathematical form
• Set equal to piece of data specific to the problem
• Use conversion factors to convert units of data specific
to problem to units sought in answer
• Other names used Unit Conversion Method
or dimensional (Unit) Analysis
CHEM 100, FALL 2011 LA TECH
2-56
Exact Numbers
conversion factors
should never limit the number of significant figures
reported in answer
12 inches = 1 foot
CHEM 100, FALL 2011 LA TECH
2-57
Calculation
8
-1
Speed of light is 3.00 x 10 m s . Convert the speed
of light to miles per year (1 mile = 1.61 km).
CHEM 100, FALL 2011 LA TECH
2-58
Temperature
Scales:
Fahrenheit
Celsius
Kelvin
• absolute scale using Celsius size degree
CHEM 100, FALL 2011 LA TECH
2-59
Three Temperature Scales
Fahrenheit
Celsius
Kelvin
212
100
373
98.6
37
310
32
0
273
- 273
0
Boiling Point of Water
Normal Body Temp.
Melting Point of Water
- 459
Absolute Zero
CHEM 100, FALL 2011 LA TECH
2-60
Temperature Conversions
oF
-- > oC ; C = 5/9 (F - 32)
oC -- > oF ; F =9/5 C + 32
oC -- > K ; K = C + 273.15
Human body temperature is 98.6
oF. Convert this temperature to
oC and K scale
oC
= 5/9 (98.6 - 32) = 5/9 (66.6) = 37.0
oC--> K = 37.0 oC +273.15 = 310.2 K
CHEM 100, FALL 2011 LA TECH
2-61
Measuring volume
Volume - the amount of space that an object
occupies.
The base metric unit is the liter (L).
The common unit used in the lab is the
milliliter (mL).
One milliliter is exactly equal to one cm3.
The derived SI unit for volume is the m3
which is too large for convenient use.
CHEM 100, FALL 2011 LA TECH
2-62
Density
Density is an intensive property of a substance
based on two extensive properties.
Mass
Density =
Common units are g /
» g / cm3
Air
Water
Gold 19.3
CHEM 100, FALL 2011 LA TECH
Volume
cm3
or g / mL.
3
cm = mL
g / cm3
0.0013
Bone
1.7 - 2.0
1.0
Urine
1.01 - 1.03
Gasoline 0.66 - 0.69
2-63
Density Calculations
Equation method:
Density = mass ÷ volume; d = m/v
Factor Label method: 14.2 g -- > ? cm3
conversion factor?
2.70 g
1 cm3
-------- or -----1 cm3
2.70 g
14.2 g x 1 cm3
--------------------2.70 g
CHEM 100, FALL 2011 LA TECH
= 5.26 cm3
2-64
Specific gravity measurement
Hydrometer
Float height will
be based on
Specific Gravity =
density of substance
density of reference
Specific Gravity is unitless.
o
Reference is commonly water at 4 C.
CHEM 100, FALL 2011 LA TECH
2-65
Average atomic masses
• Most elements exits as a mixture of isotopes.
• Each isotope may be present in different
amounts.
• The masses listed in the periodic table reflect the
world-wide average for each isotope.
• One can calculate the average atomic weight
(AAM) of an element if the abundance of each
isotope for that element is known.
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2-66
Masses of Atoms
Carbon-12 Scale
Masses of the atoms are compared to the mass of
C-12 isotope having a mass of
12.0000 amu
Atomic mass units (amu)
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2-67
Mass Spectrometer
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2-68
Mass Spectrum of Neon
AAM(Ar) = [(90 x 20) + (10 x 22)] / 100 = 20.20
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2-69
Atomic Masses and
Isotopic Abundances
natural average atomic masses =
[(atomic mass of isotope)  (fractional isotopic abundance)]
CHEM 100, FALL 2011 LA TECH
2-70
How do you calculate average
Atomic Mass?
Ma x a + Mb x b
------------------------ = AAM
100
Ma
=
mass of isotope a
Mb
=
mass of isotope b
a
=
percent abundance of a
b
=
percent abundance of b
AAM =
Average atomic mass (Reported
on the Periodic Table)
CHEM 100, FALL 2011 LA TECH
2-71
How do you calculate average
Atomic Mass?
Ma x a + Mb x b = AAM
Ma = mass of isotope a
Mb
=
mass of isotope b
a
=
fractional abundance of a
b
=
fractional abundance of b
AAM =
Average atomic mass (Reported
on the Periodic Table)
CHEM 100, FALL 2011 LA TECH
2-72
Average atomic masses
Example.
• Silicon exists as a mixture of three isotopes. Determine
it’s average atomic mass based on the following data.
• Isotope
• 28Si
• 29Si
• 30Si
CHEM 100, FALL 2011 LA TECH
Mass (u)
27.976 9265
28.976 4947
29.973 7702
Abundance
92.23 %
4.67 %
3.10 %
2-73
Calculation
Gallium in nature consists of two isotopes, gallium-69,
with a mass of 69.926 amu and a fractional abundance
of 0.601; and gallium-71, with a mass of 70.925 amu
and a fractional abundance of 0.399.
Calculate the weighted average atomic mass of
gallium.
1) Ma x a + Mb x b = AAM
Ma x a(%) + Mb x b(%)
2) ----------------------------------- = AAM
100
CHEM 100, FALL 2011 LA TECH
2-74
AAM Calculation
Ma (69Ga ) =68.926 u,
a = percent abundance of 69Ga = 0.601 x 100
Mb (71Ga ) = 70.925 u,
b = percent abundance of 71Ga = 0.339 x 10
We can obtain an equation with one unknown, AAM.
AAM = 68.926x(0.601 x 100)+70.925 x(0.339x100)
100
AAM (Ga) = 4142.5 + 2829.9
100
AAM (Ga) = 6972.3 = 69.723
100
AAM (Ga) = 69.723 u (amu)
CHEM 100, FALL 2011 LA TECH
2-75
The Mole
a unit of measurement, quantity of matter present
Avogadro’s Number
6.022  1023 particles
Latin for “pile”
CHEM 100, FALL 2011 LA TECH
2-76
Molar Mass
Sum atomic masses (amu or g/mol) represented by
formula
atomic masses  gaw (g/mol)
molar mass  MM
CHEM 100, FALL 2011 LA TECH
2-77
Example
How many grams of Cu are there in 5.67 mol Cu?
#g Cu = (5.67 mol)(63.546g/mol)
= 360. g
Atomic mass of Cu
CHEM 100, FALL 2011 LA TECH
2-78
Example
Calculate the number of boron atoms in 1.000g
sample of the element.
#B atoms = (1.000g)(1mol / 10.81g)
 (6.022  1023atoms/mol)
= 5.571  1022 B atoms
CHEM 100, FALL 2011 LA TECH
2-79
Example
How many moles of silicon, S, are in 30.5g of S?
#mol Si = (30.5g)(1 mol/32.06g)
= 0.951 mol Si
CHEM 100, FALL 2011 LA TECH
2-80
Example
What is the molar mass of methanol, CH3OH?
MM = 1(gaw)C + (3 + 1)(gaw)H + 1(gaw)O
= 1(12.011)C + 4(1.00794)H + 1(15.9994)O
= 22.042 g/mol
CHEM 100, FALL 2011 LA TECH
2-81
Example
How many moles of carbon dioxide molecules are
there in 6.45g of carbon dioxide?
MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol
#mol CO2 = (6.45g)(1 mol/44.01g)
= 0. 147 mol
CHEM 100, FALL 2011 LA TECH
2-82
Periodic Table
• Periodic table is an arrangement of
all known element according to their
atomic number and chemical
properties.
CHEM 100, FALL 2011 LA TECH
2-83
Development of Periodic Table
Newlands - English Scientist
1864 – Law of Octaves – every 8th element
has similar properties
CHEM 100, FALL 2011 LA TECH
2-84
Who is Dmitri Mendeleev?
Mendeleev, Dmitri (1834-1907): Russian chemist
Mendeleev is best known for
his work on the periodic table;
arranging the 63 known
elements into a Periodic
Table based on Atomic Mass
CHEM 100, FALL 2011 LA TECH
2-85
Mendeleev’s Periodic Table
the elements are arranged according to
increasing atomic weights
Missing elements: 44, 68, 72, & 100 amu
CHEM 100, FALL 2011 LA TECH
2-86
Dimitri Mendeleev created this, the original,
periodic table.
CHEM 100, FALL 2011 LA TECH
2-87
Predicted Properties of
Ekasilicon
Property
Ekasilicon Germanium
Atomic Weight 72
72.6
Color
gray
gray
Density, g/mL
5.5
5.36
Oxide
EsO2
GeO2
Chloride
EsCl4
GeCl4
CHEM 100, FALL 2011 LA TECH
2-88
Modern Periodic Table
the elements are arranged according to increasing
atomic numbers
CHEM 100, FALL 2011 LA TECH
2-89
Organization of Periodic Table
Period – horizontal row
Group – vertical column
CHEM 100, FALL 2011 LA TECH
2-90
Modern periodic table
1
2
I A II A
13 1 4
15
16
III A IV A V A VI A VIIA 0
H
1
2
17 18
Li
He
Be
3
3
4
5
III B IVB V
6
7 8
B VIB VIIB
9 10 11 12
VIII B
IB IIB
B
C
N
O
F
Ne
Al
Si
P
S
Cl
Ar
Na
Mg
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
Cs
Ba
*Lu
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
La
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
CHEM 100, FALL 2011 LA TECH Ac
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
4
5
6
7
+
Fr
Ra
Lr
*
+
2-91
Information that may be in the table
47
Silver
Atomic number
Name of the element
Elemental Symbol
Ag
107.87
CHEM 100, FALL 2011 LA TECH
Average Atomic mass
2-92
Vertical columns- groups,families
Horizontal columns- periods
Elements in a group have similar
chemical
properties
Group IA - alkali
metal: Li, Na, K Rb, Cs, Fr
Group IIA- alkaline earth metals:
Be, Mg, Ca, Sr, Ba, Ra
Group VIIA - Halogens: Cl, Br, I, At
Group 0 - Noble gases: He, Ne, Ar, Kr, Xe, Rn
CHEM 100, FALL 2011 LA TECH
2-93
Groups are assigned
Roman numerals
with an A or B
A group or family
I A II A
III A IV A V A VI A VIIA 0
H
He
Li
Be
Na
Mg
III B IVB V B VIB VIIB
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Cs
Ba
Lu
Hf
Ta
W
Re
Os
Ir
Pt
Fr
Ra
Lr
CHEM 100, FALL 2011 LA TECH
VIII
B
C
N
O
F
Ne
Al
Si
P
S
Cl
Ar
Zn
Ga
Ge
As
Se
Br
Kr
Ag
Cd
In
Sn
Sb
Te
I
Xe
Au
Hg
Tl
Pb
Bi
Po
At
Rn
IB IIB
La
Ce
Pr
Nd
Ac
Th
Pa
U
Pm Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Np
Am Cm
Bk
Cf
Es
Fm
Md
No
Pu
2-94
A row or period
Periods are
assigned numbers
1
H
He
2
3
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
Cs
Ba
Lu
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
Fr
Ra
Lr
4
5
6
7
CHEM 100, FALL 2011 LA TECH
La
Ce
Pr
Nd
Ac
Th
Pa
U
Pm Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Np
Am Cm
Bk
Cf
Es
Fm
Md
No
Pu
2-95
Elemental states at
room temperature
Solid
Liquid
Gas
He
H
Li
Be
B
C
N
O
F
Na
Mg
Al
Si
P
S
Cl
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
Cs
Ba
* Lu
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
Ne
Ar
+
Fr
Ra
Lr
*
La
Ce
Pr
Nd
Pm Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
+
Ac
Th
Pa
U
Np
Am Cm
Bk
Cf
Es
Fm
Md
No
CHEM 100, FALL 2011 LA TECH
Pu
2-96
The known elements
118 elements are currently known
• 89 are metals
• 31 are radioactive
• 22 are synthetic (all radioactive)
• 11 occur as gases
• 2 occur as liquids
• Let’s take a look at them on the table.
CHEM 100, FALL 2011 LA TECH
2-97
Periodic Table of the Elements
IA
1
1
2
3
4
5
6
7
II A
III B
IV B
VB
VI B
VII B
VIII B
IB
II B
III A
IV A
VA
VI A
VII A
1
VIII A
2
H
H
He
1.008
1.008
3
4
5
6
7
8
9
4.0026
10
Li
Be
B
C
N
O
F
Ne
6.939
9.0122
10.811
12.011
14.007
15.999
18.998
20.183
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
22.99
24.312
26.982
28.086
30.974
32.064
35.453
39.948
19
20
31
32
33
34
35
36
21
22
23
24
25
26
27
28
29
30
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
39.102
40.08
44.956
47.89
50.942
51.996
54.938
55.847
58.932
58.71
63.54
65.37
69.72
72.59
74.922
78.96
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
Br
Kr
79.909
53
83.8
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.468
87.62
88.906
91.224
92.906
95.94
* 98
101.07
102.91
106.42
107.9
112.41
114.82
118.71
121.75
127.61
126.9
131.29
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
**La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.91
137.33
138.91
178.49
180.95
183.85
186.21
190.2
192.22
195.08
196.97
200.29
204.38
207.2
208.98
* 209
* 210
* 222
87
88
89
104
105
106
107
108
109
110
111
112
Fr
* 223
Ra ***Ac
226.03 227.03
Rf
Ha
Sg
Ns
Hs
Mt
* 261
* 262
* 263
* 262
* 265
* 268
* 269
* 272
* 277
58
59
60
61
62
63
64
65
* Designates that **Lanthanum
all isotopes are
Series
radioactive
*** Actinium
Series
114
Uun Uuu Unb
116
Uuq
118
Uuh
*285
*289
Based on symbols used by ACS
66
67
68
69
Uuo
*293
S.M.Condren 2001
70
71
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
140.12
140.91
144.24
* 145
150.36
151.96
157.25
158.93
162.51
164.93
167.26
168.93
173.04
174.97
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
232.04
231.04
238.03
237.05
* 244
* 243
* 247
* 247
* 251
* 252
* 257
* 258
* 259
* 260
CHEM 100, FALL 2011 LA TECH
2-98
Family Names
Group IA
alkali metals
Group IIA
alkaline earth metals
Group VIIA
halogens
Group VIIIA
noble gases
transition metals
inner transition metals
lanthanum series
rare earths
actinium series trans-uranium series
CHEM 100, FALL 2011 LA TECH
2-99
What are these?
• Transition Metals
• Actinides
• Lanthanides
• Semimetals or Metalloids
• Ionic Charges
• Poly atomic ions and their charges
CHEM 100, FALL 2011 LA TECH
2-