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1
John W. MooreConrad L. StanitskiPeter C. Jurs
Stephen C. Foster • Mississippi State University
http://academic.cengage.com/chemistry/moore
Chapter 2Atoms and Elements
Atoms are composed of subatomic particles:
• electron (e-), proton (p+) and neutron (n0).
Key discoveries:
RadioactivityRadioactivity
• 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 radioactivityradioactivity.
Atomic Structure and Subatomic Particles
Radioactivity
Electrical behavior: “+” attracts “-”; like charges repel.
Atoms must contain smaller sub-units.
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
Electrons
Thomson (1897) discovered the e-:
“Cathode rays”
• Travel from cathode (-) to anode (+).
• Negative charge (e−).
• Emitted by cathode metal atoms.
fluorescentscreen
– high voltage + cathode ray
ElectronsMillikan (1911) studied electrically-charged oil drops. 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:
Modern value = 9.10938215 x 10-28 g
= (-1.60 x 10-19 C)(-5.60 x 10-9 g/C) = 8.96 x 10-28 g
me = charge xmass
charge
Electrons
2
ProtonsAtoms 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 protonsprotons.
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
How were p+ and e- arranged?
Thompson:
• Ball of uniform positive charge, with smallnegative dots (e-) stuck in it.
• The “plum-pudding” model.
19101910 Rutherford fired α-particles at thin metal foils.
Expected them to pass through with minordeflections.
ButBut …… some had large deflections.
α particles
Rutherford
“It was about as credible as if you had fired a 15-inchshell at a piece of paper and it came back and hit you.”
The Nuclear Atom
• ≈10,000 times smaller diameter than the entire atom.
• e- occupy the remaining space.
α particles
Most of the mass and all “+” charge is concentratedin a small core, the nucleusnucleus:
The Nucleus
Neutrons
Atomic mass > mass of all p+ and e- in an atom.
Rutherford proposed a neutral particle.
mn ≈ mp (0.1% larger).
mn = 1.674927211 x 10-24 g.
Present in all atoms (except normal H).
Chadwick (1932) fired -particles at Be atoms.Neutral particles, neutronsneutrons, were ejected:
NucleusNucleus
• Contains p+ and n0.
• Most of the atomic mass.
• Small (~10,000x smaller diameter than the atom).
• Positive (each p+ has +1 charge).
• Small light particles surrounding the nucleus.
• Occupy most of the volume.
• Charge = -1.
Atoms are neutral. Number of e− = Number of p+.
ElectronsElectrons
The Nuclear Atom
3
Atoms are very small.
• 1 teaspoon of water contains 3x as many atoms asthere 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) - derivedfrom the metric system.
The Sizes of Atoms and Units Metric Units
Prefix Factor
mega M 106
kilo k 103
deci d 10-1
centi c 10-2
milli m 10-3
micro μ 10-6
nano n 10-9
pico p 10-12
femto f 10-15
Prefixes multiply or divide a unit by multiples of ten.
ExamplesExamples1 kilometer = 1 km = 1 x 103 meter
1 microgram = 1 μg = 1 x 10-6 gram
1 pm = 1 x 10-12 m ; 1 cm = 1 x 10-2 m
How many copper atoms lie across the diameter of apenny? A penny has a diameter of 1.90 cm, and acopper atom has a diameter of 256 pm.
x 1 x 10-2 m1 cm
= 7.42 x 107 Cu atoms
1 pm1 x 10-12 m
x1.90 cm = 1.90 x 1010 pm
Number of atoms across the diameter:
1.90 x 1010 pm x 1 Cu atom256 pm
The Sizes of Atoms and Units
LengthLength 1 kilometer = 0.621 mile1 inch = 2.54 cm (exactly)1 angstrom (Å) = 1 x 10-10 m
VolumeVolume 1 liter (L) = 1 dm3 = 1000 cm3 = 1000 mL= 1.06 quarts
1 gallon = 4 quarts = 8 pints
MassMass 1 amu = 1.661 x 10-24 g1 pound = 453.6 g = 16 oz1 ton (metric) = 1000 kg1 ton (US) = 2000 pounds
Some Common Unit Equalities
5.0 lb
What is the mass of a 5.0 lb bag of sugar in kilograms?
1 lb = 453.6 g
= 2265 gx 453.6 g1 lb
= 2.3 x 103 g
= 2.3 kg
Some Common Unit Equalities
165 mgdL
A patient’s blood cholesterol level measured 165mg/dL. Express this value in g/L
1 mg = 1 x 10-3 g ; 1 dL = 1 x 10-1 L
x 1 x10-3 g1 mg
= 1.65 g/Lx 1 dL1 x10-1 L
Some Common Unit Equalities
4
All measurements involve some uncertainty.
Scientists typically report a number with oneone uncertaindigit.
Uncertainty and Significant Figures
Consider a reported mass of 6.3492 g
• Last digit (“2”) is uncertain
• Close to 2, but may be 4, 3, 1, 0 …
To determine the number of significant figures:
• Read numbers from left to right.
• Count all digits; startstart with the 1st non-zero digit.
•• AllAll digits are significant exceptexcept zeros used toposition a decimal point (“placeholders”).
Uncertainty and Significant Figures
6.3492 g has five significant figuressignificant figures.
placeholders significant
significant0.00024030 (2.4030 x 10-4)
5 sig. figs.
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 ? AmbiguousAmbiguous. May be placeholders oror
may be significant.
500. 3 Add a decimal point to show they are
significant.
5.0 x 102 2 No ambiguity.
Uncertainty and Significant Figures
dp = 4dp = 3
Addition and subtractionAddition and subtraction
Find the decimal places (dp) in each number.
answer dp = smallest input dp.
Add:17.245
+ 0.1001
17.3451
Rounds to: 17.345 (dp = 33)
Significant Figures in Calculations
dp = 2dp = 4
Subtract 6.72 x 10-1 from 5.00 x 101
Use equal powers of 10:
5.00 x 101
– 0.0672 x 101
4.9328 x 101
Rounds to: 4.93 x 101 dp = 22
Significant Figures in Calculations
Multiplication and DivisionMultiplication and Division
Answer sig. fig = smallest input sig. fig.
17.245
x 0.1001
1.7262245
Rounds to: 1.726 sig. fig. = 44
Multiply 2.346, 12.1 and 500.99
Rounds to: 1.42 x 104 (3 sig. fig.)
= 14,221.402734
Significant Figures in Calculations
sig. fig. = 4
sig. fig. = 5
5
Round 37.663147 to 3 significant figures.
Examine the 11stst nonnon--significant digitsignificant 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.
last retaineddigit
1st non-significant digit
Rounds up to 37.72nd non-
significantdigit
Rules for Rounding
Round the following numbers to 3 sig. figs:
1st non-sig. 2nd non-sig. RoundedNumber digit digit Number
2.123 2.1233 - 2.12
51.372 51.3772 51.3722 51.4
131.5 131.55 - 132.
24.752 24.7552 24.7522 24.7
24.751 24.7551 24.7511 24.8
0.06744 0.067444 - 0.0674
Rules for Rounding
Significant figures?
99.12444 – 6.32127.5256
=92.8034427.5256
= 3.37153195571
= 3.3715 (5 sig. figs.)
dp = 5 dp = 3
Answer dp = 3.92.803 is the significant result.(5 sig. figs).
6 sig. figs.
Rules for Rounding
To avoid rounding errorsTo avoid rounding errors
• Carry additional digits through a calculation.
• Use the correct number of places in the finalanswer.
NoteNote
Exact conversion factors:
(100 cm / 1 m) or (2H / 1 H2O)
Have an infinite number of sig. figs.
Rules for Rounding
Same element: same number of p+
Atomic numberAtomic number (Z) = number of p+
1 amu = 1.66054 x 10-24 g
Particle Mass Mass Charge(g) (amu) (atomic units)
e− 9.10938215 x 10-28 0.000548580 −1
p+ 1.672621637 x 10-24 1.00728 +1
n0 1.674927211 x 10-24 1.00866 0
Atomic Numbers & Mass Numbers
Atomic mass unit (amu) = (mass of C atom) thatcontains 6 p+ and 6 n0.
112
A 12or X e.g. C
or X-A e.g. carbon-12
(Z is constant for a given element)
AZ
126
For element X, write: X e.g. C
A ≈ mass (in amu) of an atom
Atomic Numbers & Mass Numbers
The mass numbermass number (AA) = number of p+ + number of n0
6
How many p+, n0 and e- are in the following elements:
Cu6329
Mg25
29 p+ = 29 e- (neutral atom: e- = p+)
12 p+ = 12 e- (periodic table; neutral)
63−29 = 34 n0
25−12 = 13 n0
Al27 13 p+ = 13 e-
27−13 = 14 n0
Atomic Numbers & Mass Numbers
IsotopesIsotopes
Atoms of the same elementsame element with different A.
• equal numbers of p+
• different numbers of n0
deuterium (D)
tritium (T)
Hydrogen isotopes: H 1 p+, 0 n01
1
2
1 H 1 p+, 1 n0
3
1H 1 p+, 2 n0
Isotopes and Atomic Weight
Most elements occur as a mixture of isotopes.
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 Weight Isotopes and Atomic Weight
For most elements, the percent abundancepercent abundance of itsisotopes are constant (everywhere on earth).
The periodic table lists an average atomic weight.
ExampleExample
Boron is a mixture of 10B (10.0129 amu) and 11B(11.0093 amu). 10B is 19.91% abundant. Calculatethe atomic weight of boron.
Isotopes and Atomic Weights
Atomic weight for B = 1.994 + 8.817 amu= 10.811 amu
Atomic mass = Σ(fractional abundance)(isotope mass)
(11.0093 amu) = 8.817 amu11B 80.09100
(10.0129 amu)10B19.91100
= 1.994 amu
% abundance of 11B = 100% - 19.91% = 80.09%
10B = 10.0129 amu; 11B = 11.0093 amu (10B = 19.91%). Atomic weight of B?
B10.811
Boron
5 Atomic number (Z)
Symbol
Name
Atomic weightAtomic weight
Periodic table:
Isotopes and Atomic Weight
7
A counting unit – a familiar counting unit is a “dozen”:
1 dozen eggs = 12 eggs
1 dozen peas = 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 numberAvogadro’s number
Amounts of Substances: The Mole
1 mole eggs = 6.02 x 1023 eggs
1 mole peas = 6.02 x 1023 peas
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
height = V / area, 1 mol would cover the U.S. to:
U.S. surface area = 3.0 x 106 mi2
= 8.4 x 1013 ft2
6.0 x 1018 ft3
8.4 x 1013 ft2=7.1 x 104 ft = 14 miles !
Amounts of Substances: The Mole
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
There are 6.02 x 1023 atoms in 1 mol of He andand 1 molof Xe – but they have different masses.
1 He atom has mass = 4.0026 amu
1 mol of He has mass = 4.0026 g
… 1 dozen eggs is much heavier than 1 dozen peas!
Amounts of Substances: The Mole
ExampleExampleHow many moles of copper are in a 320.0 g sample?
Cu-atom mass = 63.546 g/mol (periodic table)
Conversion factor: 1 mol Cu63.546 g
= 1
nCu = 320.0 g x1 mol Cu63.546 g
= 5.036 mol Cu
n = number of moles
Molar Mass and Problem Solving
Calculate the number of atoms in a 1.000 g sample ofboron.
Molar Mass and Problem Solving
nB = (1.000 g) 1 mol B10.81 g
= 0.092507 mol B
B atoms = (0.092507 mol B)(6.022 1023 atoms/mol)
= 5.571 1022 B atoms
The Periodic Table
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.
8
In the USA, “A” denotes amain group element…
…”B” indicates atransition metal.
International systemuses 1 … 18.
The Periodic Table The Periodic Table
Main group metal
Transition metal
Metalloid
Nonmetal
The Periodic Table
A period is a horizontal row
Periodnumber
The Periodic Table
A group is a vertical columnGroup 7AHalogens
Group 8ANoble gasesGroup 1A
Alkali metals (not H)
Group 2AAlkaline
earth metals
Important Regions of the Periodic Table
•• AlkaliAlkali and alkaline earthalkaline earth metals are very reactive.
• Always found combined with other elements innature.
•• HalogensHalogens are highly reactive diatomic molecules.
• F2, Cl2, Br2, I2
•• Noble gasesNoble gases are the least reactive elements.
•• LanthanidesLanthanides and actinidesactinides are metals listedseparately at the bottom of the periodic table.