Transcript WU_Chem101_F10_Ch3 - Chemistry at Winthrop University

INVESTIGATING CHEMISTRY
A FORENSIC SCIENCE PERSPECTIVE
second
edition
CHAPTER 3:
ATOMIC CLUES
This presentation authored by Kenneth A. French, PhD, Blinn College
Matthew E. Johll
Copyright © 2009 by W. H. Freeman and Company
CASE STUDY: TO BURN OR NOT TO BURN
A gruesome scene in Georgia
• In Georgia, in 2002, 334
bodies were found on a
grisly landscape.
• Who could be
responsible for all this?
• Families paid for
cremations and did get
urns with ashes.
• But did those urns all
contain the remains of
loved ones? Or what?
• Aristotle - believed that matter could be divided infinitely
without changing its basic nature.
• Lavoisier founded modern chemistry by laying the
foundation for the law of conservation of matter - reactant
and product masses equal
• Joseph Louis Proust proposed the law of definite
proportions – combine in set ratios
DALTON’S ATOMIC THEORY
1. Matter is composed of tiny, indivisible particles
called atoms.
2. Atoms can’t be created, destroyed, or transformed
into other atoms in a chemical reaction.
3. All the atoms of a given element are identical.
(Still true except for isotopes, which had not yet been
discovered.)
4. Atoms combine in simple whole- number ratios to
form compounds.
This reaction drives some fuel cells, but burning H2 also gives
NOx when N2 and O2 in the air unite at high temperature.
2H2 + O2  2H2O
Atoms combine in simple whole- number ratios to form
compounds.
THE LAW OF MULTIPLE PROPORTIONS
• Dalton’s theory led to the law of multiple proportions:
– Any time two or more elements combine in different
ratios, different compounds are formed.
– An example is: H2O and H2O2 (water and hydrogen
peroxide)
NO NO2
N2O
N2O3
N2O4
N2O5
3.4 ATOMIC STRUCTURE: SUBATOMIC
PARTICLES
• Atoms are composed of protons, neutrons, and electrons.
• Electrons were discovered in the 1850’s as cathode rays in
a cathode ray tube at low pressure, high voltage.
CATHODE RAY TUBE:
No Deflection
Upward Deflection
Downward
Deflection
In 1897, J. J. Thomson measured the charge to mass ratio of
the electron.
3.4 ATOMIC STRUCTURE:
SUBATOMIC PARTICLES
• In 1909, Ernest Rutherford bombarded thin gold foil with
alpha particles and was very surprised to see them
scattered or deflected; should not happen.
• In science a flawed theory is overturned when a new theory
is found to be more correct.
• Rutherford reasoned that the very massive alpha particles,
which have a positive charge, had encountered an even
more massive, positively charged entity within the atom.
• He suggested that atoms have a nucleus where all the
positive charge and most of the mass reside.
• Rutherford - The nuclear model of the atom.
A source of alpha particles (He nuclei) was allowed to impact
gold foil. Instead of always passing right through as expected,
a few particles were scattered at odd angles, and some even
came backwards!
Notes
•
•
•
•
•
•
•
•
Dalton’s atomic theory – 4 postulates
Law of Conservation of Matter (LCM) Lavoisier
Law of Definite Proportions (LDP)
Law of Multiple Proportions (LMP)
Thompson – mass to charge,
Rutherford – mostly empty space, dense nucleus
Millikan – oil drop experiment, electronic charge
Isotopes – radioactivity (spontaneous emission of
particles)
SUBATOMIC PARTICLES
• Electrons and protons have the same size charges but
very different masses.
• Neutrons and protons reside in the nucleus together.
• Neutrons have no charge but are comparable in mass
to the protons.
3.5 ISOTOPES
Hydrogen is unique in having different names for its isotopes:
Protium: Ordinary hydrogen, It has 1 proton, 1 electron, 0 neutrons.
Deuterium: Heavy hydrogen, 1 proton, 1 electron, 1 neutrons
Tritium: 1 proton, 1 electron, 2 neutrons
Same #
Carbon-14 is represented 146C. Subtracting Z from A gives the
number of neutrons. In this case, 14 – 6 = 8 n.
Figure 3.9, pg. 66
Investigating Chemistry, 2nd
Edition
© 2009 W.H. Freeman & Company
ISOTOPES: A SAMPLE PROBLEM
• How many neutrons are there in barium-136?
• Solution: Barium, Ba, is element 56, that is, it
contains 56 protons. Its atomic number is therefore
56.
• Its mass number is 136 (the sum of its protons and its
neutrons in this isotope).
• Answer: Subtracting 56 from 136 we get 80 neutrons.
Notes
• 1. Fill in the blank spaces and write out all the
symbols in the left hand column in full, in the
form (i.e., include the appropriate values of Z
and A as well as the correct symbol X).
•
•
•
•
Symbol
…
Au
…
# protons
17
…
…
# neutrons # electrons
18
…
118
…
20
20
3.7 MATHEMATICS OF ISOTOPE
ABUNDANCE AND ATOMIC MASS
• Each element is composed of isotopes.
• There is a uniform ratio of these on Earth. It is called
the natural abundance.
• Use weighted average
– The atomic mass is the sum of the masses of each of the
isotopes multiplied by its fraction found in nature.
• For silver 51.8% of the atoms found have a mass no.
of 107, and 48.2% have a mass no of 109.
3.7 MATHEMATICS OF ISOTOPE
ABUNDANCE AND ATOMIC MASS
• Here is the equation:
– Atomic Mass = (mass of isotope A)(fraction of
isotope A) + (mass of isotope B)(fraction of
isotope B) + …
• For Silicon, atomic mass 28.09 it is:
– Atomic Mass of Si = (27.97693)x(0.9223) +
(28.976495)x(0.0467) = 28.09
• 92.23% is Si-28 and 4.67% is Si-29.
3.8 ATOMIC STRUCTURE: ELECTRONS & EMISSION SPECTRA
• Through a prism, white light produces all the colors of the rainbow, a
continuous spectrum. [ROYGBIV]
• Atoms excited in a flame or by high voltage produce what seems to be one
color, but when passed through a prism gives a set of bright lines called a
bright line or emission spectrum, not a continuous spectrum.
• At room temperature, the electrons in atoms are in their lowest energy
levels, the ground state. When atoms absorb energy, the electrons can enter
an excited state.
• When the electrons move from the excited state back to their ground state,
energy is released as a photon of light (one line).
• The energy states resemble not a ramp but are analogous to an irregular set
of stairs.
3.8 ATOMIC STRUCTURE: ELECTRONS
AND EMISSION SPECTRA
• Since several excited states exist, each atom produces
a unique set of bright lines, one line for each color of
photon.
• So each element can be identified by the color it
produces in the flame, but even better by the pattern
of lines in its spectrum.
• Not all atoms produce visible spectra. The photons
may correspond to ultraviolet light or other forms of
electromagnetic radiation.
- Passing the red-orange light from excited neon atoms through a
prism gives a bright line spectrum (top). Passing sunlight or other
white light through a prism gives instead a continuous spectrum of
all the colors.
- Since several excited states exist, each atom produces a unique set
of bright lines, one line for each color of photon.
- So each element can be identified by the color it produces in the
flame, but even better by the pattern of lines in its spectrum.
- Not all atoms produce visible spectra. The photons may
correspond to ultraviolet light or other forms of electromagnetic
radiation.
Emission spectra can be used to ID elements because each
atom has its own distinct line spectrum.
Sodium, Na, looks intensely yellow in a flame.
Potassium, K, is lilac or violet in a flame.
Cesium, Cs, is pale violet in a flame.
3.9 MATHEMATICS OF LIGHT
• Nu (ν) is the frequency in cycles per second (1/s) or
hertz (Hz). It corresponds to how many waves go
past a fixed point each second.
• The shorter the wavelength, λ, the more waves that go
by each second.
• c = wavelength times frequency
3.9 MATHEMATICS OF LIGHT
• Sample Problem: If the lead from gunshot residue is
detected by light emitted at 220.4 nm, what frequency
does this correlate with? 1 nm = 109 m
• Solution: c = lambda  nu (or λ x ν)
• So nu = c / lambda
• Frequency = (3.00  108 m/s) / 220.4 nm
• Answer: 1.36  1015 Hz (or s1) (cycles/second)
3.9 MATHEMATICS OF LIGHT
• The energy of a photon E = h  nu
• The h represents Planck’s constant , The Energy = 6.626 
1034 J . s  1.361  1015 (1/s)
• The dual nature of light (or any electromagnetic radiation)
allows us to determine the photon’s energy, E.
• Frequency units: Hz = 1/s or cps = s1.
• Sample Problem: Find the energy of a single photon from
a lead atom if the photon has a frequency of 1.361  1015
Hz.
• So E for this photon is 9.018  1019 J.
29. Boron has how many protons?
A) 2
B) 3
C) 4
D) 5
33. The isotope of cobalt with the symbol
59 Co has how many electrons?
27
A) 33
B) 32
C) 59
D) 27
How many electrons are in atoms of Co?
A)
B)
C)
D)
27
28
58
59