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Chem 6A Michael J. Sailor, UC San Diego
AnnouncementsChem 6A Sept 27, 2011
• Course is being podcast: http://podcast.ucsd.edu/podcasts
• Course web link is:http://sailorgroup.ucsd.edu/Chem6A_sailor/
• Links should be available on courses.ucsd.edu and ted.ucsd.edu shortly
1
Chem 6A Michael J. Sailor, UC San Diego
Chapter 2: Matter
2
Chem 6A Michael J. Sailor, UC San Diego 3
Properties of sub-atomic particlesName Symbol Charge* Mass, g
Electron e-‐ -‐1 9.1 x 10-‐28
Proton p +1 1.673 x 10-‐24
Neutron n 0 1.675 x 10-‐24
*units are fundamental unit of charge, = 1.602 x 10-19 coulombs
Chem 6A Michael J. Sailor, UC San Diego
The Periodic Table of the Elements
4
1 18
1 2
H 2 13 14 15 16 17 He1.0079 4.0026
3 4 5 6 7 8 9 10
Li Be B C N O F Ne6.941 9.01218 10.811 12.011 14.0067 15.9994 18.9984 20.1797
11 12 13 14 15 16 17 18
Na Mg 3 4 5 6 7 8 9 10 11 12 Al Si P S Cl Ar22.9898 24.305 26.9815 28.0855 30.9738 32.066 35.4527 39.948
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr39.0983 40.078 44.9559 47.88 50.9415 51.9961 54.9381 55.847 58.9332 58.69 63.546 65.39 69.723 72.61 74.9216 78.96 79.904 83.8
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe85.4578 87.62 88.9059 91.224 92.9064 95.94 98.9063 101.07 102.906 106.42 107.868 112.411 114.82 118.71 121.75 127.6 126.905 131.29
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 Rn132.905 137.327 138.906 178.49 180.948 183.85 186.207 190.2 192.22 195.08 196.967 200.59 204.383 207.2 208.98 208.982 209.987 222.018
87 88 89 104 105 106 107 108 109
Fr Ra Ac Unq Unp Unh Uns Uno Une223.02 226.025 227.028 - - - - - -
58 59 60 61 62 63 64 65 66 67 68 69 70 71
Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu140.12 140.91 144.24 146.92 150.35 151.96 157.25 158.92 162.5 164.93 167.26 168.93 173.04 174.97
90 91 92 93 94 95 96 97 98 99 100 101 102 103
Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr232.038 231.04 238.03 237.05 239.05 241.06 247.07 249.08 251.08 254.09 257.1 258.1 255 262.1
Chem 6A Michael J. Sailor, UC San Diego 5
3 key experiments leading to the modern view of the atom
Person Experiment What it determined
J.J. Thomson1897
Effect of a charged plates on a cathode ray tube
mass/charge ratio of an electron, me/ze
Robert Millikan1909
Oil-drop experiment: suspending a charged drop of oil between two charged plates
charge on the electron, ze
Ernest Rutherford1910
Scattering of alpha-rays from metal foils
Atom consists of a very small, very dense nucleus
Chem 6A Michael J. Sailor, UC San Diego 6
Electrons
Cathode Ray (or Crookes) Tube measures the charge/mass ratio of the electron
Effect of magnetic and electric fields on the deflection of the beam can determine the mass to charge ratio, me/ze
Chem 6A Michael J. Sailor, UC San Diego 7
The CRT television display is a descendant of the Crookes tube
Chem 6A Michael J. Sailor, UC San Diego 8
Determination of the charge of the electron
Chem 6A Michael J. Sailor, UC San Diego 9
Problem: charge on the electron
Determine the fundamental charge on the electron, in esu
0
1 10-9
2 10-9
3 10-9
4 10-9
5 10-9
1 2 3 4 5 6 7 8 9 10
charge, esu
ch
arg
e,
esu
measurement
Charge on oil drop(esu)
9.60 x 10-10
1.92 x 10-9
2.40 x 10-9
2.88 x 10-9
4.80 x 10-9
Chem 6A Michael J. Sailor, UC San Diego 10
Solution: charge on the electron
(1) Sort by charge:
0
1 10-9
2 10-9
3 10-9
4 10-9
5 10-9
9 1 6 2 3 4 10 8 5 7
charge, esu
ch
arg
e,
esu
measurement
Chem 6A Michael J. Sailor, UC San Diego
0
1
2
3
4
5
6
9 1 6 2 3 4 10 8 5 7
Divide by LCD
ch
arg
e,
esu
/9.6
x1
0-1
0
measurement
11
Solution: charge on the electron
(1) Sort by charge:(2) Divide by
smallest charge:
Chem 6A Michael J. Sailor, UC San Diego
0
1
2
3
4
5
6
7
8
9
10
11
12
9 1 6 2 3 4 10 8 5 7
multiply by 2
ch
arg
e,
esu
/9.6
x1
0-1
0
measurement
12
Solution: charge on the electron
(1) Sort by charge:(2) Divide by
smallest charge:(3) Multiply by 2:
Chem 6A Michael J. Sailor, UC San Diego 13
Solution: charge on the electronEsu Divide by 9.6 x 10-10 Multiply by 2
9.60x10-10 1 21.92x10-9 2 42.40x10-9 2.5 52.88x10-9 3 6
4.80x10-9 5 10
So the fundamental charge on the electron is:(9.6 x 10-10)/2 = 4.8 x 10-10 esu
Chem 6A Michael J. Sailor, UC San Diego 14
Discovery of the nucleus
Chem 6A Michael J. Sailor, UC San Diego 15
3 key experiments leading to the modern view of the atom
Person Experiment What it determined
J.J. Thomson1897
Effect of a charged plates on a cathode ray tube
mass/charge ratio of an electron, me/ze
Robert Millikan1909
Oil-drop experiment: suspending a charged drop of oil between two charged plates
charge on the electron, ze
Ernest Rutherford1910
Scattering of alpha-rays from metal foils
Atom consists of a very small, very dense nucleus
Chem 6A Michael J. Sailor, UC San Diego 16
A mass spectrometer measures masses* of ions
*more correctly, the mass to charge ratio, m/z
Chem 6A Michael J. Sailor, UC San Diego 17
Definition of atom masses is based on 12C
• A carbon-12 atom is defined as our mass standard• it weighs exactly 12 atomic mass units (amu).
• 1 amu = 1.66054 x 10-24g
…so why does the periodic table list C as 12.011?
Chem 6A Michael J. Sailor, UC San Diego 18
Mass ratios, isotopes, and %abundance
12C13C
12C, 13C
More 12C: bigger spot
Less13C: smaller spot
mass
inte
nsity
12 13 1411
Chem 6A Michael J. Sailor, UC San Diego 19
Isotopes: mass ratios and percent abundance
Element Mass (position on detector)
Abundance (intensity on detector)12C 12.00 98.89
13C 13.003355 1.11Mass ratio: 1.0836129
Mass ratio:
€
13C12C
=13.003355
12=1.0836129
Chem 6A Michael J. Sailor, UC San Diego 20
Problem: percent abundance of isotopes
A student uses a mass spectrometer to determine the masses and percent abundances of all the naturally occurring isotopes of chlorine, the values of which are presented below. Calculate the average mass of chlorine, in amu.
Isotope % abundant Atomic Mass, amu 35Cl 75.77 % 34.97 37Cl 24.23 % 36.97
Chem 6A Michael J. Sailor, UC San Diego 21
Solution: percent abundance of the isotopes
This is just a weighted average:
Mass = (0.7577 x 34.97) + (0.2423 x 36.97) = 35.45 amu
The average of all naturally occurring isotopes is called the atomic mass (or atomic weight) of an element
Chem 6A Michael J. Sailor, UC San Diego 22
Detection of peptides in human blood by mass spectrometry
mass spectrum of blood serum from human
inte
nsity
Villanueva, J. et al., Analytical Chemistry 2004, 76, 1560-1570
white = peptide present
blue = peptide not present
Chem 6A Michael J. Sailor, UC San Diego 23
Application of mass spectrometry to detection of cancer
mass spectra of blood serum obtained from patients with and without brain tumors (Villanueva, J. et al., Analytical Chemistry 2004, 76, 1560)
Chem 6A Michael J. Sailor, UC San Diego
The Periodic Table of the Elements
24
1 18
1 2
H 2 13 14 15 16 17 He1.0079 4.0026
3 4 5 6 7 8 9 10
Li Be B C N O F Ne6.941 9.01218 10.811 12.011 14.0067 15.9994 18.9984 20.1797
11 12 13 14 15 16 17 18
Na Mg 3 4 5 6 7 8 9 10 11 12 Al Si P S Cl Ar22.9898 24.305 26.9815 28.0855 30.9738 32.066 35.4527 39.948
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr39.0983 40.078 44.9559 47.88 50.9415 51.9961 54.9381 55.847 58.9332 58.69 63.546 65.39 69.723 72.61 74.9216 78.96 79.904 83.8
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe85.4578 87.62 88.9059 91.224 92.9064 95.94 98.9063 101.07 102.906 106.42 107.868 112.411 114.82 118.71 121.75 127.6 126.905 131.29
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 Rn132.905 137.327 138.906 178.49 180.948 183.85 186.207 190.2 192.22 195.08 196.967 200.59 204.383 207.2 208.98 208.982 209.987 222.018
87 88 89 104 105 106 107 108 109
Fr Ra Ac Unq Unp Unh Uns Uno Une223.02 226.025 227.028 - - - - - -
58 59 60 61 62 63 64 65 66 67 68 69 70 71
Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu140.12 140.91 144.24 146.92 150.35 151.96 157.25 158.92 162.5 164.93 167.26 168.93 173.04 174.97
90 91 92 93 94 95 96 97 98 99 100 101 102 103
Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr232.038 231.04 238.03 237.05 239.05 241.06 247.07 249.08 251.08 254.09 257.1 258.1 255 262.1
Chem 6A Michael J. Sailor, UC San Diego 25
Bonding in Compounds
covalent ionic
Covalent bond = neutral atoms held together by sharing a pair of electrons
Ionic bond = charged atoms (ions) held together by electrostatic forces
€
E =z1z2q
2
4πεor1−2
Coulomb’s law:charge
distance
Na+
Cl-H
H
O ClCl
water moleculechlorine molecule
Chem 6A Michael J. Sailor, UC San Diego 26
The rock salt lattice
Ionic solids are held together by ionic bonds
Chem 6A Michael J. Sailor, UC San Diego 27
Energy of an ionic bondIonic bond = charged atoms
(ions) held together by electrostatic forces
€
E =z1z2q
2
4πεor1−2
Coulomb’s law:charge
distanceNa+Cl-
Energy of attraction between 1 Na+ and 1 Cl- is
€
E =z1z2q
2
4πε or1−2=
(+1)(−1)(1.602 ×10−19)2
(4)(3.14)(8.85 ×10−12)(2.81×10−10)= 8.2 ×10−19J
0.281 nm
Chem 6A Michael J. Sailor, UC San Diego 28
Lattice enthalpies and ionic radius
650
700
750
800
850
900
950
3 3.2 3.4 3.6 3.8 4 4.2 4.4
Lattice Enthalpy vs 1/(Ionic radii)
Latt
ice
enth
alpy
, kJ/
mol
1/(Na-X) distance, Å- 1
NaF
NaCl
NaBr
NaI
Ion Radius (pm)F- 13.3
Cl- 18.1
Br- 19.6
I- 22.0
Lattice enthalpy = energy that holds the crystal together
The closer the ions, the stronger the lattice
Chem 6A Michael J. Sailor, UC San Diego
Silberberg, pg. 53: Common Monatomic Ions
Chem 6A Michael J. Sailor, UC San Diego 30
Covalent bonds
covalent
Covalent bond = neutral atoms held together by sharing a pair of electrons
•An assembly of atoms held together by covalent bonds is a molecule
•If it has a net charge, it is called a molecular ion
ClCl
water moleculechlorine molecule
H
H
O
Chem 6A Michael J. Sailor, UC San Diego 31
Problem: Naming polyatomic ions (Table 2.5)
Write the names of the following ions:Write the names of the following ions:Write the names of the following ions:1. CH3CO2
-
2. Cr2O72-
3. ClO3-
4. MnO4-
acetatedichromatechloratepermanganate
Memorize all the ions, names, formulas and charges in Table 2.3, 2.4, and 2.5
Chem 6A Michael J. Sailor, UC San Diego 32
Problem:
What is the MOLAR MASS of Aspirin (acetylsalicylic acid):
(C9H8O4)
O
O
CH3
O OH
H
H
H
H
Chem 6A Michael J. Sailor, UC San Diego 33
Solution:
MOLAR MASS of Aspirin (acetylsalicylic acid):
O
O
CH3
O OH
H
H
H
H
(C9H8O4) = 9(12.011) + 8(1.0079) + 4(15.9994) = 180.16 g/mol
Chem 6A Michael J. Sailor, UC San Diego 34
Problem:
What is the percent by mass of carbon in Aspirin?
(C9H8O4)
O
O
CH3
O OH
H
H
H
H