Atoms in a Formula
• Each atom contained is listed by it’s symbol
– Na, K, Mn, O, S, ….
• Subscripts give atom quantities in formula
– NaOH (no need for “1” if only one atom)
– N2, O2, H2 (diatomic molecules have 2 atoms)
• Empirical formula gives only atom ratios
– C2H6O represents Ethyl Alcohol
– 3 elements and 9 atoms in the formula
• Structural formulas provide behavior clues
– C2H5OH is also Ethyl Alcohol (“alcohol” group is -OH)
– CH3OH is Methyl Alcohol (same –OH group)
Mass of Molecules
• Molecule mass is sum of element masses
– C2H6O represents Ethyl Alcohol (empirical)
• Carbon: 2 atoms * 12.01 amu = 24.02
• Hydrogen: 6 atoms * 1.008 amu = 6.048
• Oxygen: 1 atom * 16.00 amu = 16.00
– Formula or Molecule mass is Sig-Fig sum
• Round down result to least precise input
• 2 Sig-Fig after decimal in this case
• 24.02 + 6.05+16.00 = 46.07 amu (grams/mole)
Carbon Dioxide
Origin of Moles
• Periodic chart fundamentally based on particles
– Proton counting defines element number
– Proton + Neutron counting defines element mass
– How to assign laboratory weights to elements?
• Could define a reference number of atoms
– Use 10^10 or other arbitrary number?
– How to count the atoms ?
– Requires more conversions, calculations
• We define formula mass = total AMU in grams
– Simplifies calculations via periodic chart
– Links periodic chart to arbitrary mass unit
• 1 kilogram = 1000 cm^3 water = 1 liter water
Origin of Moles
• Mole comes from MOLEcular weight
– Reference laboratory mass (grams) set equal
to the atomic mass number (pure number).
– This is arbitrary (grams are based on water)
• We could have used pounds, grains, or stones
– Weights of elements in grams will be
proportional to the atomic mass numbers
– The weight in grams equal to the molecular
mass is defined as the MOLE
– Most precise definition has mole equal to
number of atoms in 12 grams of Carbon-12
Basis of Avogadro’s number
• Consider Carbon
– Carbon 6 protons, ≈ 6 neutrons  12.01 AMU
• Mass of particles
– Proton total 1.672622E-24g * 6 = 10.0356E-24
– Neutron total 1.6749E-24g * 6 = 10.0494E-24
– Sum of particle mass = 20.085E-24
• Assume 1 mole C weighs 12.00 grams
– Particles = 12 / 20.085E-24 = 5.974E23
– More exact value is 6.022E23
Avogadro’s Number
• Gram is an arbitrary value
– Originally based on 1 cubic centimeter of water, 1 liter=1kilogram
– Meter based on ¼ circumference of earth
• AMU is dimensionless, a counting method
– Protons = Z, the atomic number
– Mass = A = Z + n, including neutrons
• Atomic mass is an isotope average
– Not a whole number, a weighted average of all natural isotopes
• 1 “mole” defined as material equal to formula weight
– How many atoms is that ?
• 6.022 x 10^23 atoms/molecules per mole
– A consequence of the other definitions
– Can be demonstrated experimentally
• Oil on water 1 molecule thick (Chem 1A book example)
• Electrical generation of hydrogen, our future experiment
Los Alamos National Laboratory's Periodic Table
Group**
Period
1
IA
1A
2
3
1.008
3
4
H
Li
Be
6.941
9.012
11
12
Na Mg
22.99
4
5
8
9
10
3
4
5
6
7
11 12
------- VIII IIIB IVB VB VIB VIIB
IB IIB
-----3B
4B 5B 6B
7B
1B 2B
------- 8 ------
20
21
Ca
Sc
39.10
40.08
37
38
Rb
Sr
85.47
87.62
Cs
87
Fr
(223)
56
88
6
7
8
9
B
C
N
O
F
22
23
24
25
26
27
28
29
30
13
14
Al Si
32
Y
40
41
42
44
45
46
47
48
49
50
72
73
74
(98)
75
17
18
Cl
Ar
33
34
35
51
52
53
I
101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9
76
77
78
79
80
81
82
83
84
85
Pt Au Hg Tl Pb Bi Po At
138.9 178.5 180.9 183.9 186.2 190.2 190.2 195.1 197.0 200.5 204.4 207.2 209.0 (210) (210)
107
108
109
86
Rn
(222)
116
118
---
()
()
()
59
60
61
62
63
64
111
Xe
131.3
---
(257) (260) (263) (262) (265) (266)
110
54
114
58
106
83.80
---
Lanthanide
Series*
105
36
Kr
112
(227)
104
39.95
Ra Ac~ Rf Db Sg Bh Hs Mt --- --- --(226)
89
Ne
20.18
S
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
88.91 91.22 92.91 95.94
57
43
10
16
44.96 47.88 50.94 52.00 54.94 55.85 58.47 58.69 63.55 65.39 69.72 72.59 74.92 78.96 79.90
39
4.003
15
26.98 28.09 30.97 32.07 35.45
31
2
He
P
Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Ba La* Hf Ta W Re Os Ir
137.3
5
10.81 12.01 14.01 16.00 19.00
19
132.9
7
24.31
13
14
15 16
17
IIIA IVA VA VIA VIIA
3A 4A 5A 6A 7A
K
55
6
8A
2
IIA
2A
1
1
18
VIIIA
()
()
()
65
66
67
68
69
70
71
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
140.1 140.9 144.2 (147) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0
Mole Relationships
• 1 Mole defined as
– Element’s atomic mass # expressed in grams
• The “natural” weighted average of all isotopes
– Formula weight of atoms in a molecule
• Sum of atomic masses for elements involved
– 6.022 x 10^23 atoms or molecules
• Avogadro’s number is the SAME for all Moles
– 22.4 liters of gas (at STP)
• Gasses are compressible, liquids & solids are not
• “STP” ≡ Standard Temp (0oC) & Pressure (1 Atm)
Mole relationships
Mole Analogies
• Airport Hub & Spoke model
– Hub is stopping point to where you want to go
– Mole quantity is also path between units
• Google and Yahoo
– A means to go somewhere else
– 1st stopping point may not be the destination
• Mole is convenient path
– Mass of gas to volume
– Mass of input to product
Mass
• Formula mass or molecular mass
– Sum of atomic masses in the formula
– CH4 (methane)
• Carbon has mass of 12.01 (1 atom in formula)
• Hydrogen has mass of 1.008 (4 atoms in formula)
• 12.01 + (4*1.008) = 16.033 grams/mole
– Convenient to set up a M W spreadsheet
•
•
•
•
Do molecular weights only ONCE
Simple adaptations are easy
Refer to the chart visually and electronically
<< show spreadsheet example>>
FORMULA WEIGHT CALCULATOR
Generic Table of Values
gm/smpl
KAl(SO4)2.12H2O(solid)
Ag
Br
C
Cl
Cr
Cu
Fe
H
K
Mn
N
Na
O
P
S
Sn
Zn
Al
10.0000
atoms
0
0
0
0
0
0
0
24
1
0
0
0
20
0
2
0
0
gm/atom
107.8682
79.9040
12.0107
35.4527
51.9961
63.5460
55.8450
1.0079
39.0983
54.9380
14.0067
22.9898
15.9994
30.9738
32.0650
118.7100
65.3900
gm/mole
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
24.1906
39.0983
0.0000
0.0000
0.0000
319.9880
0.0000
64.1300
0.0000
0.0000
% of FW
0.00%
0.00%
0.00%
0.00%
0.00%
0.00%
0.00%
5.10%
8.24%
0.00%
0.00%
0.00%
67.45%
0.00%
13.52%
0.00%
0.00%
mol/smpl
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0011
0.0017
0.0000
0.0000
0.0000
0.0142
0.0000
0.0028
0.0000
0.0000
1
0
26.9815
26.9815
0.0000
5.69%
0.00%
0.0012
0.0000
474.3884
100.00%
0.0211
48
atoms
FW =>
gm/mole
moles
Percent Composition
• Formula does NOT give weight % directly
– Some atoms are heavier than others
– Hydrogen = 1
Oxygen = 16
• For H2O, Hydrogen not 2/3 (67%) molecule mass
– Uranium = 238, Fluorine = 19
• For UF6, Fluorine is not 6/7 (86%) of the total mass
• Formulas must be converted to mass
– For water
• H=2*1.008 = 2.016, O = 16.00, H2O = 18.02 gm/mol
• Hydrogen therefore 2/18 or 11.1% of water’s mass
– For UF6
• U = 238.0 , F = 6*19.00, so UF6 = 352.0 gm/mole
• Fluorine is (6*19)/((6*19)+238) = 114/352 = 32.4% of mass
Mass % example
• C2H6O Ethyl Alcohol (empirical)
•
•
•
•
Carbon: 2 atoms * 12.01 amu = 24.02
Hydrogen: 6 atoms * 1.008 amu = 6.048
Oxygen: 1 atom * 16.00 amu = 16.00
Total = 24.02 + 6.05+16.00 = 46.07 grams/mole
• Mass percentages readily determined
•
•
•
•
Carbon is 24.02 / 46.07 = 52.14% of total mass
Hydrogen is 6.05 / 46.07 = 13.13% of total mass
Oxygen is 16.00 / 46.07 = 34.73% of total mass
Total = 100%
Empirical Formulas
• Empirical Formulas based only on atom ratios
– Does NOT provide chemical structure
– Empirical Formula “CH” is simply a ratio, NOT material
• Could be C2H2 (Ethylene)
• Could be C6H6 (Benzene)
• Any material with Carbon:Hydrogen ratio 1:1 fits the model
• What use is an empirical formula?
– Some analysis tools give only atomic percentages
• X-ray florescence (light emission characteristic of elements)
– Some methods provide mass ratios
• Burning fuels yield measurable grams of H2O and CO2
– Grams of products yield moles, moles provide atomic ratios
– Establishes ratio of atoms in formula
• Additional step to turn empirical into molecular formula
EDAX (X-Ray Fluorescence)
Yields ratios of elements
Iron into Iron Oxide … or rust
Empirical Formula
(reverse of finding % mass)
• Typical “real-world” problem
– Starting materials and end result are known
– What is formula of the resulting material?
• Rusting of Iron (text example)
– Pure Iron 1.62 gram, after rusting = 2.31 gram
•
•
•
•
Oxygen must be difference 2.31-1.62 = 0.69 gram
Moles of Iron = 1.62/55.85 = 0.029 mol
Moles of Oxygen = 0.69/18.00 = 0.043 mol
Mole Ratio Oxy / Iron =0.043 / 0.029 = 1.5
– Atomic ratios must be whole numbers
– Multiply by 2 to clear fractions  2 mole Fe + 3 mole Oxy
• Can also divide by lowest mole amount
– 0.029 / 0.029 = 1 unit Iron, 0.043/0.029 = 1.5 units Oxy
– Resulting whole number multiples are 2 and 3
– Empirical formula becomes Fe2O3
Elemental Analysis
• A means to determine empirical formulas
– Observed grams of each product material
– Convert grams to moles of elements
• Grams / (gram/mole) = moles of CO2, H2O, etc.
• Moles of products  moles of constituent atoms
– Do “conservation of mass” reality check
• Is everything accounted for?
• Grams of elements add up to observed products?
– Missing grams may mean missing element(s)
Elemental Analysis
• Note the moles of all elements
– Will NOT be integers, just values
– Find lowest common denominator
– Divide ALL elemental moles by common denominator
• Result will often be simple multiples
• If results are integer values, you’re done!
– The empirical formula is those integers beneath the element
– If results contain non-integers, 1 more step required
• Trial multiplication using integers which clear the fractions
• Final result WILL be integer values to use in formulas.
Route to Empirical Formula
Burning is typical Empirical method
• Fuel is unknown hydrocarbon
• Burning in air yields known products
– Carbon Dioxide and Water
– Grams of each used to calculate moles
– Moles yield amounts of carbon and hydrogen
– Empirical formula gives ratio of C and H
– More data required to determine real material
• C4H9 multiple possibilities (e.g. Octane = C8H18)
• Apparatus to determine percentages of
carbon and hydrogen in a compound.
• Copper oxide helps to oxidize traces of
carbon and carbon monoxide to carbon
dioxide, and to oxidize hydrogen to water.
Empirical Formula Limitations
• Only the elemental ratios are correct
– Sometimes Empirical = Molecular
• Methane CH4 is the same either way
• Does not have divisible numbers
– Empirical is often NOT a real material
• Ethylene molecular formula is C2H2
• Benzene molecular formula is C6H6
• Both have CH empirical … which does NOT exist
• More data turns empirical into molecular
Molecular Formulas
• Deduce molecular formula from empirical formula
– Most common situation
– Homework problems
• Need to know quantities involved
– Assume we know molecular mass of starting material
• Molecular mass / empirical mass = small number ratio
• Small number ratio * empirical formula = molecular formula
– Benzene example
• Molecular Mass =78, determined by other means
• Empirical formula mass = 13 (based on C:H ratio of 1:1)
• Ratio is 78/13=6, therefore Molecular Formula is C6H6
Molecular Formulas
• Once we know molecular formula
– Can calculate grams of constituents from moles
• “Percent Composition” questions
– Can calculate volumes of gases from moles produced
• 22.4 liters per mole of gas
• Final step is geometrical structure
– Benzene ring not obvious from molecular formula
• Many ring structures exist (5 and 6 sided)
• Many rings have non-carbon in backbone (N & S common)
– Will get into molecular modeling later in semester
Empirical  Molecular Formula
• Also a typical “real-world” situation
– Initial analysis gives empirical result
• Only the ratios are known
– Calculate mass of empirical formula
• For CH, 12.01 + 1.008 = 13.02
– Compare to molecular weight by other means
• If 26.04 then empirical is ½, and material is C2H2
• If 78.11 then empirical is 1/6 and material is C6H6
Summary, Empirical vs Molecular
• Formulas
– Empirical Formula is only a ratio of atoms
• whole numbers
• Benzene lowest order ratio would be CH
• Empirical often not very useful alone
– Molecular Formula = correct ratios and quantity
• Benzene would be C6H6
• Molecular formula accurately reflects composition
Isotopes
• Not all element atoms weigh the same
– Some have more neutrons than others
•
•
•
•
Uranium-235 (chain reaction fission) is element 92
Uranium-238 (no chain reaction) also element 92
Difference is 3 neutrons, different mass isotopes
Show isotope chart
• Natural elements include all the isotopes
– Mass in nature is the AVERAGE weight
• Might include tens of isotopes
• Weighted average relies on relative quantities
• Same average value everywhere on earth (we believe)
– One atom of element must be single isotope
Diatomic
• Diatomic Molecules
– 7 elements exist as diatomic, or paired atoms
• Formation of “double atom” molecules via covalent bond
• H2, N2, O2, F2, Cl2, Br2, I2
• Each atom shares outer electrons, so both have 8
– Bonding can include 1, 2, or 3 pairs of electrons
• Diatomic elements have twice the mass
– Oxygen molecule is O2, from two atoms of Oxygen
– Weight of O2 is twice that of Oxygen on chart
• 32 grams per mole of O2 not 16
• Volumes of diatomic same as mono-atomic,
• 22.4 liter/mole for atomic and diatomic
– Diatomic nature only applies to elements
• Oxygen bonds one at a time inside molecules
Mass Nomenclature
• Atomic Mass
– What an atom weighs on atomic scale
• Sum of protons, electrons, electrons
– Single atom must be a particular isotope
• NOT the average 35.45 AMU for all chlorine
– C12 atomic mass = 12.0 / 6.02*10^23
• Gram atomic Mass
– Mass of one mole of atoms (not molecules)
– Monatomic (not diatomic pairs)
– Isotopic average atomic weight in grams
– 35.45 gm/mole average for Chlorine isotopes
Mass Nomenclature
• Molecule Mass (molecular mass)
– What a molecule weighs on molecular scale
– Sum of individual isotopic masses in molecule
– Exact answer means knowing the isotopes
• Na23+Cl35=NaCl58 or 58 grams per mole
• (58 g/mole) / (6.02*10^23 molec/mole) = 9.63E-23 g/molecule
• Molar Mass, or Gram Molar Mass
– What a mole of a material weighs in grams
• Involves 6.02*10^23 atoms or molecules
– Utilizes average isotope mass of elements
• What’s on the periodic chart, for everyday calculations
Now to the Experiment
• Types of reactions
• Balancing a reaction
• Reaction drivers
Types of reactions
• Single Replacement
– More active element replaces less active
• Zn + CuSO4  Cu + ZnSO4
• Zn + H2SO4  ZnSO4 + H2(g)
• Double Replacement
– Swap partners to make gas, precipitate, water
• HCl + NaOH  NaCl + HOH
• AgNO3 + NaCl  AgCl(s) + NaNO3
• Dance analogy
– Jack & Jill go to dance, so do Arthur and Amanda
– Bonding reaction occurs on dance floor
» Jack goes home with Amanda, leaves the party
» Jill left with Arthur … no reaction?
• Chemical Activity (pg249)
– Elements are different
• High to low activity
• Is hydrogen liberated?
• Found in nature?
– “Active” elements
• Liberate H2 with acid
• Replace less reactive
– Transition = hydrogen
• Reference Activity point
– “Less Active” elements
• No acid reactions
• Replace least reactive
– Unreactive elements
• Few natural compounds
• Gold, Platinum
Reaction Drivers
• Formation from elements
– 2H2 + O2  2H2O
– Fe + Cl2  FeCl2
• Decomposition
– NaN3  2Na + 3N2
– 2NI3  N2 + 3I2
(Auto Air Bag, McMurry p 353)
• Water Formation, neutralization
– HCl + NaOH  NaCl + H2O
Now to the Reactions
• Balancing a reaction
– Conservation of Mass
• No added or missing materials
• Same number of element atoms on each side
– Conservation of charge (applies to ions)
• Same number and sign of charge on each side
– Usual pecking order
•
•
•
•
Balance Cations (Na, Fe) or Carbon first
Hydrogen balance (next to last if other cations)
Oxygen balance last, to fit water or oxides formed
Multiply to get rid of diatomic gas fractions (O2)
Rusting Iron example
• Fe (iron) + O2  Fe2O3 (rust) - unbalanced
– 2 Fe on right requires 2 Fe on left
• 2Fe + O2  Fe2O3 - still unbalanced
– 3 oxygen on right requires 3 on left
• 2Fe + 3/2 O2  Fe2O3 - balance OK but fractions
– Clear fraction by multiplying by denominator
• 4Fe + 3 O2  2Fe2O3 - Balanced, no fractions
– Double check number of atoms both sides
• You’re done
Burning methane example
• CH4 + O2  CO2 + H2O - unbalanced
– carbon on = carbon on right, carbon OK
– 4 hydrogen on left requires 4 on right
• CH4 + O2  CO2 + 2H2O – unbalanced
– Total of 4 oxygen on right, requires 4 on left
• CH4 + 2O2  CO2 + 2H2O – balanced !
– Double check, same # atoms both sides?
– You’re done
Samples from exercise
• Now we will do a few examples on board
• Students choose one from each page
• Work it out, then class review
Los Alamos National Laboratory's Periodic Table
Group**
Period
1
IA
1A
2
3
1.008
3
4
H
Li
Be
6.941
9.012
11
12
Na Mg
22.99
4
5
8
9
10
3
4
5
6
7
11 12
------- VIII IIIB IVB VB VIB VIIB
IB IIB
-----3B
4B 5B 6B
7B
1B 2B
------- 8 ------
20
21
Ca
Sc
39.10
40.08
37
38
Rb
Sr
85.47
87.62
Cs
87
Fr
(223)
56
88
6
7
8
9
B
C
N
O
F
22
23
24
25
26
27
28
29
30
13
14
Al Si
32
Y
40
41
42
44
45
46
47
48
49
50
72
73
74
(98)
75
17
18
Cl
Ar
33
34
35
51
52
53
I
101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9
76
77
78
79
80
81
82
83
84
85
Pt Au Hg Tl Pb Bi Po At
138.9 178.5 180.9 183.9 186.2 190.2 190.2 195.1 197.0 200.5 204.4 207.2 209.0 (210) (210)
107
108
109
86
Rn
(222)
116
118
---
()
()
()
59
60
61
62
63
64
111
Xe
131.3
---
(257) (260) (263) (262) (265) (266)
110
54
114
58
106
83.80
---
Lanthanide
Series*
105
36
Kr
112
(227)
104
39.95
Ra Ac~ Rf Db Sg Bh Hs Mt --- --- --(226)
89
Ne
20.18
S
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
88.91 91.22 92.91 95.94
57
43
10
16
44.96 47.88 50.94 52.00 54.94 55.85 58.47 58.69 63.55 65.39 69.72 72.59 74.92 78.96 79.90
39
4.003
15
26.98 28.09 30.97 32.07 35.45
31
2
He
P
Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Ba La* Hf Ta W Re Os Ir
137.3
5
10.81 12.01 14.01 16.00 19.00
19
132.9
7
24.31
13
14
15 16
17
IIIA IVA VA VIA VIIA
3A 4A 5A 6A 7A
K
55
6
8A
2
IIA
2A
1
1
18
VIIIA
()
()
()
65
66
67
68
69
70
71
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
140.1 140.9 144.2 (147) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0
Formula Unit or Empirical Formula
• Formula Unit is a ratio
– Lowest whole number ratio of atoms
– Smallest representative unit of a substance
– Lowest reduced ratio of ions in a compound
– It’s a ratio, may (or may not) be real material
• Example is mass of formula unit
– Formula Unit of Sodium Chloride = NaCl
• Na=22.99 amu, Cl=35.45 amu, NaCl=58.44 amu
• 1 mole (6.02*10^23 atoms) of NaCl is 58.44 grams
Molecular Formulas
• Atomic percentage from molecular formulas
– Easy to do, need only formula and periodic chart
• Molecular (or empirical) formula gives atom ratios
– Ethanol example
•
•
•
•
•
Molecular formula is C2H5OH, empirical is C2H6O
Molecular Mass = sum of atomic masses = 45 grams/mole
Atomic Masses: C=12gm/mole, H=1gm/mole, O=16gm/mole
Percent mass: 2*12/45=52%C, 6*1/45=13%H, 1*16/45=35%O
100 grams alcohol contains: 52gm C + 13gm H + 16gm O
The index of refraction varies by frequency, causing monochromatic
light rays from the left to emerge from the prism at different angles.
Moving charged particles effected
by magnetic field as if in a “prism”
Mass Spectroscopy
• Interaction of charged particle and magnetic
field
– Deflection based on elect charge and magnetic field
• F = q (v * B)
– Deflection inversely proportional to mass of Ion
• Curvature of path, R = m*v /(e*B)
– Curved path varies with mass of particle
– Heavy ions less deflected (bigger Radius) than lighter ones
– This was original method for separating Uranium
• Used for WW2 nuclear (“Atomic”) bomb
Original Calutron
Calutron Details
• Feasibility shown by UC Berkeley Cyclotron
– Germany also working on Uranium enrichment
• 10 full scale Calutron separators in 1942
– Collected 4 grams of enriched Uranium per day
• “Enriched” Uranium went from 0.7% to 15% U235
– Ions traveled a 48 inch arc
• Isotope collectors were 0.6” apart
– Bigger “racetrack” models built to increase production
• Chamber was 122 feet long, 77 feet wide, 15 feet high
• Magnet wire =14,700 tons of silver borrowed from the mint
– Copper wire was in short supply due to wartime need
• Initial yield was 200 grams of 12% U235 (1 kg goal)
• Successive passes increased U235 percentage
– U235 from Calutrons and other technologies were
combined to make initial bombs that ended WW2
Stochiometry
• Formulas
– Empirical Formula is ratio of atoms (whole numbers)
• Benzene lowest order ratio would be CH (not too useful)
– Molecular Formula has correct ratios AND quantity
• Benzene would be C6H6 which accurately reflects composition
• Stoichiometry
– converting between grams & moles
• FW of NaCl=59.44 gm/mol, so 15 grams approx 0.25 mol
– Mind the units !
• Common error is inverting a constant
– Multiply by moles or Avagadro # … when should have divided
Chemical Equation Balancing
• Material Balance
– Conservation of mass: Same mass  different form
– Must have same amount of material before and after reaction
• Equation required to maintain same quantities of same atoms
• OK to have changes between phases (e.g. solids  Liquid  gas)
– We ignore tiny mass of electrons and energy conversions
• Ions would weight slightly more or less than elements
• Total matierial IN = total materials OUT
• Charge Balance
– Chemical reactions only relocate electrons
• No creation or distruction of charge
• Electrical neutrality of material must exist before and after reaction
• Net electrical charge must be same before and after
Chemical Equation Balancing
• Material Form, and reality requirement
– Avoid fractional molecules
• ½ O2, = O, but single “O” does not exist in nature
• O3 exists in upper atmosphere … not in usual practice
– What makes math sense may give wrong impression
• 1/3 C6H6 = CH empirical ratio, but structurally does not exist
– Starting and ending materials must make sense
• No impossible products to balance numbers (e.g. Na(OH)6)
Balancing Equations
• Same atoms before & after rearrangement
– Mass not created or destroyed in reactions
– Need to find correct multipliers (coefficients)
– Reactants and products must be realistic
• Trading atomic partners is common
– AgNO3 + KCl  AgCl + KNO3
– Underlined material indicates precipitate
– Precipitates takes reactants out of solution,
typically makes reaction irreversible
Balancing Equations
• Chemical reactions involve transfer of electrons
– Na0 + H2O  Na+ OH- + H20 (g)
• Evolution of a gas
– Typically removes material from reaction vessel
– Eliminates reversal, forces a 1-way conclusion
– Gas evolution indicated with a bracketed (g)
• Sometimes indicated by an “up” arrow
• Precipitates
– Converts soluble ions into insoluble compounds
– Forces a conclusion via removal of reactant(s)
– Precipitates indicated with a bracketed (s) for “solid”
• Sometimes indicated by an underlined material
Molecular Formulas
• Atomic percentage from molecular formulas
– Easy to do, need only formula and periodic chart
• Molecular (or empirical) formula gives atom ratios
– Ethanol example
•
•
•
•
•
Molecular formula is C2H5OH, empirical is C2H6O
Molecular Mass = sum of atomic masses = 45 grams/mole
Atomic Masses: C=12gm/mole, H=1gm/mole, O=16gm/mole
Percent mass: 2*12/45=52%C, 6*1/45=13%H, 1*16/45=35%O
100 grams alcohol contains: 52gm C + 13gm H + 16gm O
Today’s Lab
Moles and Molar Mass Exercises
• Use Prof. John Song’s sheet (or update)
– 14 questions, most with sub-sections
• Show your answers on separate sheets
– Must include calculations for full credit
– “Devine Inspiration” does not count …
(right final number, but nothing else)
• Usual report procedure
– Due 1 week after the classroom work
– Includes evaluation sheet – cover letter
More Rusty Iron
One Example from Handout
Professor Song's Question #14
Element
Smbl
weight %
Sample
Grams
100
Carbon
C
38.7%
38.7
Hydrogen
H
9.7%
9.7
Oxygen
O
51.6%
51.6
100.0%
100.00
Carbon
Hydrogen
Oxygen
grams/mole
Atom Mass
÷
÷
÷
moles of
elements
12.01
=
3.22
1.008
=
=
9.62
16.0
3.23
simplest
grams/mole
element
mole ratio Atomic Mass
grams
1.0
x
12.01
= 12.01
3.0
x
1.008
= 3.01
1.0
x
16.0
= 16.01
Formula Weight (grams/mole)-->
31.03
lowest
simplest
mole value mole ratio
÷
÷
÷
3.22
1.0
3.22
3.0
3.22
1.0
Empirical
Formula
C1 H 3O 1
Molecular
Formula
Given
Empirical
Compound molecule
gm/mole
ratio
62.0
2.0
C2 H 6O 2
Los Alamos National Laboratory's Periodic Table
Group**
Period
1
IA
1A
2
3
1.008
3
4
H
Li
Be
6.941
9.012
11
12
Na Mg
22.99
4
5
8
9
10
3
4
5
6
7
11 12
------- VIII IIIB IVB VB VIB VIIB
IB IIB
-----3B
4B 5B 6B
7B
1B 2B
------- 8 ------
20
21
Ca
Sc
39.10
40.08
37
38
Rb
Sr
85.47
87.62
Cs
87
Fr
(223)
56
88
6
7
8
9
B
C
N
O
F
22
23
24
25
26
27
28
29
30
13
14
Al Si
32
Y
40
41
42
44
45
46
47
48
49
50
72
73
74
(98)
75
17
18
Cl
Ar
33
34
35
51
52
53
I
101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9
76
77
78
79
80
81
82
83
84
85
Pt Au Hg Tl Pb Bi Po At
138.9 178.5 180.9 183.9 186.2 190.2 190.2 195.1 197.0 200.5 204.4 207.2 209.0 (210) (210)
107
108
109
86
Rn
(222)
116
118
---
()
()
()
59
60
61
62
63
64
111
Xe
131.3
---
(257) (260) (263) (262) (265) (266)
110
54
114
58
106
83.80
---
Lanthanide
Series*
105
36
Kr
112
(227)
104
39.95
Ra Ac~ Rf Db Sg Bh Hs Mt --- --- --(226)
89
Ne
20.18
S
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
88.91 91.22 92.91 95.94
57
43
10
16
44.96 47.88 50.94 52.00 54.94 55.85 58.47 58.69 63.55 65.39 69.72 72.59 74.92 78.96 79.90
39
4.003
15
26.98 28.09 30.97 32.07 35.45
31
2
He
P
Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Ba La* Hf Ta W Re Os Ir
137.3
5
10.81 12.01 14.01 16.00 19.00
19
132.9
7
24.31
13
14
15 16
17
IIIA IVA VA VIA VIIA
3A 4A 5A 6A 7A
K
55
6
8A
2
IIA
2A
1
1
18
VIIIA
()
()
()
65
66
67
68
69
70
71
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
140.1 140.9 144.2 (147) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0