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10/18/11
Electrons
Charge- negative
Mass- 0
Location- outside the nucleus
Electron Configuration- a way to describe the arrangement of the electrons around the
nucleus of an atom.
-arrangement
-location
-laid out
Hund’s Rule- Each orbital is occupied before pairing begins.
(orbital- probable location of each electron
- at least 2 electrons per orbital
- no electrons=no orbital)
Quantum Numbers- a number that describes the properties of electrons and consists of 4
numbers
(Quantum- a certain number)
Pauli’s Exclusion Principle- only 2 electrons can occupy a single orbital
10/19/11
We can describe an electron in 2 ways.
-Particle with a very small mass
-Wave
Electrons- determine chemical properties
-“Chemistry”
Nucleus- nuclear chemistry
-tremendous energy (ex: the sun)
Chemists describe electrons with:
- Wave-particle duality
- Quantum numbers
- Periodic table
Position of electron is energy which means orbitals are energy.
10/20/11
Orbitals- space where electrons could be found from Quantum numbers
Sublevels- S- 1- up to 2 electrons
P- 3- up to 6 electrons
D- 5- up to 10 electrons
F- 7- up to 14 electrons
Aufbau Principle
Lowest 
Energy
Level
Highest
Energy
Level
10/21/11
Electron Configuration PracticeBoron- 5 electrons
1s22s22p1 (2+2+1=5)
Hydrogen- 1 electron
1s1
Be- 4 electrons
1s22s2 (2+2=4)
Nitrogen- 7 electrons
1s22s22p3 (2+2+3=7)
Silicon- 14 electrons
1s22s22p63s23p2 (2+2+6+2+2=14)
Calcium- 20 electrons
1s22s22p63s23p64s2 (2+2+6+2+6+2=20)
10/24/11
valence electrons- electrons found in the outermost shell (energy level) of an atom
- also determines the atom’s properties
- outer shell electrons participate in chemicals reactions
There are 7 shells (energy levels)
Examples:
Mg- 12 electrons
valence electrons

2 6 2
1s22s 2p 3s

outermost shell
(2 valence electrons)
Ar- 18 electrons
2
valence electrons


2 6 2 6
1s 2s 2p 3s 3p
 
outermost shell
(8 valence electrons)
S- 16 electrons
valence electrons


2 2 6 2 4
1s 2s 2p 3s 3p
 
outermost shell
(6 valence electrons)
Electron Dots
Mg: (2 valence electrons)
..
·S: (6 valence electrons)
˙
..
:Ar: (8 valence electrons) (octet rule- there can be, at most, 8 valence electrons)
˙˙
Ca: (2 valence electrons)
10/25/11
Quiz review
Chapter 3.3- Electron Configuration
What’s involved? Periodic table, electron, atomic number
Electrons are arranged in orbitals around the nucleus
Things to know:
-Hund’s Rule, Aufbau Principle, Pauli’s Exclusion Principle
-Electron Dot- shows how many valence electrons it has.
-SPDF (orbitals)
S- 1- up to 2 electrons
P- 3- up to 6 electrons
D- 5- up to 10 electrons
F- 7- up to 14 electrons
- Valence electrons are found in the outer shell
- Atoms are mostly space
Examples:
Ca- 20 electrons
1s22s22p63s23p64s2
Ar- 18 electrons
1s22s22p63s23p6
Cl- 17 electrons
1s22s22p63s23p5
(2 valence electrons) Ca:
..
(8 valence electrons) :Ar:
˙˙
..
(7 valence electrons) :Cl:
˙
Chapter 4
Periodic Law- the properties of the elements are a periodic function of their atomic
number.
(repeats itself)
-Elements in each column of the periodic table have the same number of
valence electrons in their outer shell.
Group or Column- vertical- all have the same properties
Period or Row- horizontal (energy levels)- same energy levels
10/26/11
Main Group Elements (Representative Elements)
S-Block (groups 1-2)
P-Block (groups 13-18)
Group1-Alkali metals
-1 valence electron
- Li, Na, K, Rb, Cs, Fr
Group2- Alkaline Earth Metals
- 2 valence electrons
- Be, Mg, Ca, Sr, Ba, Ra
Group17- Halogens
- Highly reactive (really close to having a full outer shell)
- F, Cl, Br, I, At
Group18- Noble gases
- full outer shell, unreactive (8 valence electrons)
- Ne, Ar, Kr, Xe, Rn
10/27/11
Metals
- conduct electricity (conductors)
- malleable (can be shaped)
- ductile (drawn out into wire)
- left side of “staircase” on periodic table
Transition Metals
- D-block elements (groups 3-12)
- Lanthanides
- Actinides (radioactive) F- block
Nonmetals
- right of “staircase”
- not conductors
Metaloids
- touching “staircase”
- semi conductors
Big Bang Theory- Elements were created from Hydrogen being condensed in stars.
11/2/11
Periodic Trends
1) Atomic Radius- distance from the nucleus to the edge of the electron cloud


increases decreases
Electron cloud
2)
3)
4)
5)
6)

-Bond Radius – half the distance between bonding atoms
(nucleii)
Electron Affinity- energy change when a neutral atom gains an electron
Electronegativity- the measure of the ability of an atom in a compound to attract
electrons


decreases increases
Ionic Size- the increase or decrease in the size of an atom when it gains or loses
electrons (ion- an atom with a charge)
Ionization Energy- the energy required to remove an electron from an atom or ion
BP/MP(boiling point/melting point)- changes in state due to energy changes
11/9/11
Chemists measure moles
Atomic Mass
- Atomic Mass Unit
P+= 1AMU
N0= 1AMUmass of nucleus
e- = 0AMU
The Mole
1 mole = 6.02x1023
Mole - defined in terms of carbon
- the number of atoms in exactly 12 grams of the carbon isotope-12
- 1 mole = atomic mass on periodic table
Molar Mass- the mass, in grams, of 1 mole of an element
- g/mol
- the mass of 1 mole of an element = the element’s atomic mass (from P.T.)
Avogadro’s Number- the number of particles in 1 mole of a substance
grams  moles
#particles  moles
Examples:
Given: x grams oxygen
? moles oxygen
x grams oxygen 1 mol oxygen
16 grams oxygen
Given: x moles calcium
? grams calcium
x moles calcium 40 g calcium
1 mol calcium
= x/16 =
= x40 =
a mol oxygen
a g calcium
HW Review:
1) Given: 1 mol U
? mass U
1 mol U 238 g U = 1238 = 238gU
1 mol U
2) Given: .0050 mol U
? mass U
.0050 mol U
238 g U = .0050238 = 1.19gU
1 mol U
3) Given: .850gH
? moles H
.850 gH 1molH = .8501 = .850molH
1gH
Given: .850molH
? mass in gH
.850 molH 1gH = .8501 = .850gH
1molH
4) Given: 2.3456molPb
? mass in gPb
2,3456molPb 207.2gPb = 2.3456207.2 = .486gPb
1molPb
Given: 2.3456gPb
? molPb
2,3456gPb
1molPb = 2.3456/207.2 = .01132molPb
207.2gPb
11/15/11
23
1mol=6.02x10
Example:
Given: 0.30molF
? # atoms F
0.30molF 6.02x1023atomsF = .306.02x1023 = 1.8x1023atomsF
1molF
Given: 3.01x1023SO2
? mol SO2
3.01x1023 atomsSO2 1molSO2
= 3.01x1023/6.02x1023 = .5molSO2
23
6.02x10 atomsSO2
11/18/11
Practice:
Given: 3.01x1023SO2
? mol SO2
3.01x1023 atomsSO2 1molSO2
= 3.01x1023/6.02x1023 = .5molSO2
23
6.02x10 atomsSO2
Given: 25gF
? molF
25gF
1molF
19gF
= 25/19= 1.32molF
11/28/11
Find the mass, in grams, of 2.44x1024atomsC (particlesmass)
2.44x1024atomsC
1 molC
12gC = 2.44x102412 = 48gC
23
6.02x10 atomsC 1molC
6.02x1023
11/28/11
Molar Mass (7.2)
mass of each element from P.T.
CaCO3 1Ca: 1x40= 40
1C: 1x12= 12
3O: 3x16= +48
100g/mol molar mass of CaCO3
CaSO3 1Ca: 1x40= 40
1S: 1x32= 32
3O: 3x16= +48
120g/mol molar mass of CaSO3
Al(NO3)3 1Al: 1x27= 27
3N: 3x14= 42
9O: 9x16= +144
213g/mol molar mass of Al(NO3)3
11/29/11
HW Review
7a) ? ions
3molNa2CO3
2molNa2+
6.02x1023ions = 6.02x102332= 3.61x1024ions
1molNa2CO3
1molNa2+
8a) ? moles
3.01x1023moleculesH2O
9a) ? mass
4.30x1016atomsHe
1molH2O
= 3.01x1023/6.02x1023= .5molH2O
23
6.02x10 moleculesH2O
1molHe
4gHe = 4.30x1016 4 = 2.86x10-7gHe
6.02x1023atomsHe 1molHe 6.02x1023
10a) ? ions
1gI- 1molI- 6.02x1023ionsI- = 6.02x1023/126.9= 4.74x1021ionsI126.9gI- 1molI11a) ? mass
6.02x1023moleculeIb
1molIb
206.31gIb = 6.02x1023206.31= 206.31gIb
6.02x1023moleculeIb 1molIb
6.02x1023
12a) ? mass
4.01x1023atomsCa
1molCa
40.08gCa = 64.01x102340.08 = 26.7gCa
23
6.02x10 atomsCa 1molCa
6.02x1023
13a) ? molecules
2.00molH2 6.02x1023moleculesH2 = 6.02x10232 = 1.204x1024moleculesH2
1molH2
Notes
KCl
2 atoms
1K: 1x40 = 40
1Cl: 1x35.5= +35.5
molar mass  75.5g/mol
Fe2O3
5 atoms
2Fe: 2x55.85 = 111.7
3O: 3x16
= +48
molar mass  159.7g/mol
C6H12O6
24 atoms
6C: 6x12 = 72
12H: 12x1 = 12
6O: 6x16 = +96
molar mass 180g/mol
11/30/11
Do Now: Find the Molar Mass
KClO4
1K: 1x39 = 39
1Cl: 1x35.5= 35.5
4O: 4x16 = +64
139.5g/mol
(NH4)2SO4
2N:
8H:
1S:
4O:
2x14 = 28
8x1 = 8
1x32 = 32
4x16 = +64
132g/mol
MUST KNOW THIS:
1mol
or
23
6.02x10
1mol
molar mass or atomic mass
or
6.02x1023
1mol
molar mass or atomic mass
1 mol
Practice
1) Calculate the grams in 1 molCaCl2
1Ca: 1x40 = 4
1molCaCl2 111gCaCl2 = 111gCaCl2
2Cl: 2x35.5 = +71
1molCaCl2
111g/mol
2) Calculate the grams in 3molCO2
1C: 1x12 = 12
3molCO2 44gCO2
= 132gCO2
2O: 2x16 = +32
1molCO2
44g/mol
3) 2) Calculate the number of moles in 32gCH4
1C: 1x12= 12
32gCH4 1molCH4 = 2molCH4
4H: 4x1 = + 4
16gCH4
16g/mol
4) 2) Determine the mass of 40 moles of Na2CO3
2Na: 2x23= 46
40molNaCO3 106gNaCO3 = 4240gNaCO3
1C: 1x12= 12
1molNaCO3
OR
3O: 3x16= +48
4.24x103gNaCO3
106g/mol
pg 239
1a)CsI 1Cs:1x133= 133
1I: 1x127= +127
260g/mol
b)CaHPO4 1Ca: 1x40 = 40
1H: 1x1 = 1
1P: 1x31 = 31
4O: 4x16 =+64
136g/mol
c) C12H22O11=342g/mol
d) I2=254g/mol
e) HC2H3O2=60g/mol
f) Mg3(PO4)2=262g/mol
3a) C6H5CH3=92g/mol
b) ?moles
7.51gToluene 1molToluene = .08molToluene
92gToluene
4a) PtCl2(NH3)2= 299g/mol
b) Find the mass of 4.115x1021 F.U. PtCl2(NH3)2
4.115x1021 FU 1mol PtCl2(NH3) 299g PtCl2(NH3)2 =2.04gPtCl2(NH3)2
6.02x1023FU 1mol PtCl2(NH3)2
12/2/11
Examples:
CuNH4Cl3
1) How many elements? 4 (Copper, Nitrogen, Hydrogen, Chlorine)
2) How many atoms? 9
3) How many H atoms in 1 mole CuNH4Cl3? 6.02x10234=2.408x1024atomsH
4) How many Cl atoms in 6 moles CuNH4Cl3? 6.02x102336=1.08x1025atomsCl
5) What is the mass in grams of 1 molecule CuNH4Cl3?
1CuNH4Cl3
1molCuNH4Cl3
299g PtCl2(NH3)2 =2.04gPtCl2(NH3)2
23
6.02x10 CuNH4Cl3 1molCuNH4Cl3
6) How many moles would be in 6.84gCuNH4Cl3?
6.84gCuNH4Cl3 1molCuNH4Cl3
=.0367CuNH4Cl3
186.55gCuNH4Cl3
12/5/11
7.2 & 7.3
Chemical Formulas
- Express Composition
ex: H2O Hydrogen and Oxygen
- Used to calculate molar mass
- Give ratios of Polyatomic Ions
(2 atoms) (has a charge)
ex: C6H12O6
Ca(OH)2
Percent Composition
-percent, by mass, of each element in a compound
molar mass of the part
molar mass of the whole
ex: XeF4
1Xe: 1x131= 131 %Xe: Xe = 131 100 = 63.3%
4F: 4x19 =+76
XeF4 207
207
%F: F4 = 76 100 = 36.7%
XeF4 207
ex: Fe2O3
2Fe: 2x56= 112 %Fe: Fe2 = 112 100 = 70%
3O: 3x16=+48
Fe2O3 160
160
%O: O3 = 48 100 = 30%
Fe2O3 160
Empirical Formula- A chemical formula that shows the composition of a compound in
terms of the relative numbers and kinds of atoms in the simplest ratio.
ex: C6H12O6= CH2O
ex: Ammonium Nitrate
Actual Formula: NH4NO2 (or N2H4O2)
Empirical Formula: NH2O (simplest ratio)
Start with % composition- chemical analysis
Ex: Sample problem G (from text)
Given: C: 60.0%
H: 13.4%
O: 26.6%
Empirical Formula?
1) Assume you are given 100g sample
Convert %’s to grams
C: 60.0%60gC
H: 13.4%13.4gH
O: 26.6%26.6gO
2) Convert gmoles
C: 60gC 1molC = 5molC
12gC
H: 13.4gH 1molH = 13.4molH
1gH
O: 26.6gO 1molO = 1.67molO
16gO
3) Divide each number of moles by the smallest mole.
5/1.67=3
13.4/1.67=8
new subscripts in chemical formula
1.67/1.67=1
(if these are not whole numbers or close to it, you
must multiply the equation by a number until they
are whole numbers)
4) Empirical Formula = C3H8O
12/6/11
Do Now: Determine the molar mass of C4H3FN2O2, Fluorouracil Cream.
What is its Empirical Formula?
4C: 4x12= 48
3H: 3x1 = 3
1F: 1x19= 19
2N: 2x14= 28
2O: 2x16= +32
130g/mol
Empirical Formula: C4H3FN2O2 (subscripts are already in the simplest ratio)
HW Review (pg 243 1-4)
1)Mn2O3
2)Cl= 38.77%38.77gCl
1molCl  1molCl
35.5gCl
2(ClO3.8) = Cl2O7
O= 61.23%61.23gO
3)Fe= 72.4%72.4gFe
1molO 3.8molO
16gO
1molFe 1.3molFe
55.8gFe
O= 27.6%27.6gO 1molO  1.7molO
16gO
1.3/1.3= 1
(divide by the
smallest mole)
1.7/1.3=1.3
3(FeO1.3)= Fe3O4
4) C= 18.0%18.0gC 1molC =1.5molC
12gC
1.5/1.5= 1
H= 2.26%2.26gH 1molH =2.26molH
1gH
2.26/1.51.5
Cl= 79.7%79.7gCl
2.28/1.51.5
1molCl =2.28molCl
35.5gCl
2(CH1.5Cl1.5)= C2H3Cl3
Empirical Formula- the simplest whole number ratio of atoms in a compound
-Sometimes empirical formulas and molecular formulas are the same.
ex: H2O Empirical and molecular formula
HNO3(nitric acid) Empirical and molecular formula
Molecular formulas are often whole number ratios of the empirical formula.
Molar Mass of a compound is equal to the molar mass of the empirical formula times a
whole number.
12/7/11
HW Review (pg 248 1-5, 7)
1)a. Given the % composition, we need the atomic mass from the periodic table to find
the empirical formula.
b. Experimental molar mass
2) Molecular Formula- C8H18
Empirical Formula- C4H9 (lowest ratio)
3) You need the molar mass.
5) Experimental molar mass: 64g/mol
? molecular formula
(exp. Molar mass)64g/mol= 1(SO2)=SO2
(from 4b, we know the empirical formula is SO2) (emp. Molar mass)64g/mol
7) AgC2H3O2
1Ag:1x108= 108
Ag = 108 = 64.62%Ag
2C = 24 = 14.39%C
2C: 2x12 = 24
AgC2H3O2 167
AgC2H3O2 167
3H: 3x1 = 3
2O: 2x16 = +32
3H = 3 = 1.82%H
2O = 24 = 19.17%O
167g/mol AgC2H3O2 167
AgC2H3O2 167
Notes
n(Molar Mass of Emp. Formula)= Molar Mass of Molecular Formula
Sample H (from text pg 245)
Given: Empirical Formula- P2O5
Experimental Molar Mass- 284g/mol
? Molecular Formula
P2O5
2P: 2x31= 62
(exp molar mass)284g/mol = 2(P2O5)= P4O10
5O:5x16= +80
(emp molar mass)142g/mol
142g/mol
12/8/11
Do Now: Find the % composition of NaClO
Na = 23 = 30.89%Na
Na:1x23 = 23
NaClO 74.45
Cl: 1x35.45= 35.45
O: 1x16 = +16
Cl = 35.45 = 47.61%Cl
74.45g/mol
NaClO 74.45
O = 16 = 21.49%O
NaClO 74.45
HW Review(pg 245 1-3)
1) Given: Experimental Molar Mass- 78g/mol
Empirical Formula- CH
C: 1x12= 12
(exp molar mass)78g/mol = 6(CH) = C6H6
H: 1x1 =+1
(emp molar mass)13g/mol
13g/mol
2) Given: Experimental Molar Mass- 90g/mol
Empirical Formula- CH2O
C: 1x12= 12
(exp molar mass)90g/mol = 3(CH2O) = C3H6O3
H: 1x1 = 1
(emp molar mass)30g/mol
O: 1x16=+16
30g/mol
3) Given: Experimental Molar Mass- 46g/mol
Empirical Formula- NO2
N: 1x14= 14
(exp molar mass)46g/mol = 1(NO2) = NO2
O: 2x16=+32
(emp molar mass)46g/mol
46g/mol
Chapter 8 (8.1)
Chemical Equations
-chemical reactions rearrange atoms, but the number and kind of atoms remain the same.
2Na+2H2O2NaOH+1H2
2Na2Na
4H4H
2O2O
Due to the Law of Conservation of Mass, chemical equations must be balanced.
12/9/11
Do Now: Find the Empirical Formula for:
63.52%K63.52gK 1molK = .68molK
39gK
.68 =1
35.35%Cr35.35gCr 1molCr = .68molCr
52gCr
.68 =1
38.07%O38.07gO 1molO = 2.38molO
16gO
.68=3.5
Chemical Equations
Reactants yields Products
2(KCrO3.5)= K2Cr2O7
Solid-s
Liquid-l
Gas-g
Solid dissolved in water- aq (aqueous)
Writing Balanced Chemical Equations:
- Determine the reactants and products
- Assemble the equation
- Balance the equation
- # all reactant atoms = # all product atoms
When you balance chemical equations, you can only add numbers in front of the
elements or compounds.
12/12/11
Do Now- Determine the % composition of Fe(OH)2
Fe: 56
Fe = 56 = 62.2%Fe
2O = 32 = 35.5%O
2O: 32
Fe(OH)2 90
Fe(OH)2 90
2H: +2
90g/mol
2H = 2 = 2.2%H
Fe(OH)2 90
HW Review
4c) Sodium Hydrogen CarbonateSodium Carbonate + Carbon Dioxide + Water
2NaHCO3 Na2CO3+CO2+H2O
(we balance because of the Law of Conservation of Mass)
5c) 3 AgNO3 + AlCl3 3 AgCl + Al(NO3)3
7) Any compound with 2 molecules containing 2 atoms P each  4 atoms P
ex: 4Al(NO3)3
4Al
12N
36O
Notes
Balancing Chemical Equations
Coefficients- the number in front of the formula
- multiplies the number of each element in the formula that follows
ex: 2H2O2H2 + O2
ex: Ca(NO3)2
3Ca(NO3)2
1Ca
3Ca
2N
6N
6O
18O
- Never Change subscripts
- Polyatomic ions can be balanced as a unit
C3H8 + 5 O2  3 CO2 + 4 H2O
3C  3C
8H  8H
10 O  10 O
12/13/11
Do Now- Balance the Chemical Equations
2 Al + Fe3N2  2 AlN + 3 Fe
2 Na + Cl2  2 NaCl
2 H2O2  2 H2O + O2
HW Review (pg 285 #8 balance only)
8)a. Cl2 + 2NaBr  2NaCl + Br2
b. CaO + H2O  Ca(OH)2
c. Ca(ClO3)2  CaCl2 + 3O2
d. 2AgNO3 + K2SO4Ag2SO4 + 2KNO3
e. already balanced as written
f. 2C8H18 + 25O216CO2 + 18H2O
Types of Chemical Reactions
Single Displacement Rxn- one element displaces (or replaces) another in a compound
Ex: A + BC=B + AC
Ex: 2Al(s) + 3CuCl2(aq)3Cu(s) + 1AlCl3(aq)
Double Displacement Rxn- positive and negative portions of 2 compounds are
interchanged.
Ex: AB + CD  AD + CB
Ex: HCl + NaOH  HOH + NaCl
(or H2O)
Decomposition Rxn- substances break up into simpler substances when energy is applied.
Ex: ABA + B
Ex: 2H2O electricity 2H2 + O2
Synthesis Rxn- Two or more substances combine to form a new substance.
Ex: A + B  AB
Ex: C + O2 CO2
Combustion Rxn- Oxidation reaction of an organic compound in which energy is
released. Every combustion has O2, CO2, and H2O.
Ex:
+ O2 CO2 + H2O + energy
Ex: C3H8 + 5O23CO2 + 4H2O + energy
12/14/11
Do Now- Balance and name the type of reaction
2Na + Cl2  2NaCl Synthesis Rxn
C6H12O6 + 6O2  6H2O + 6CO2 Combustion Rxn
4(NH4)3PO4 + 3Pb(NO3)4  Pb3(PO4)4 + 12NH4NO3 Double Disp Rxn
2BF3 + 3Li2SO3  B2(SO3)3 + 6LiF Double Disp Rxn
8Ag2S16Ag + S8 Decomposition Rxn
Notes
Chemical Reactions are like recipes
-show the physical state
-reaction conditions
Ex: N2(g)+ 3H2(g) catalyst 2NH3(g)
Symbols:
(s)(l)(g)- physical state
(aq)- dissolved in water
heat or
add energy
catalyst
name or formula of catalyst
reversible reaction
12/15/11
Do Now
1) Determine the molar mass of (NH4)2SO4
2N: 2x14= 28
8H: 8x1 = 8
4O: 4x16 = 64
1S: 1x32 = +32
132g/mol
2)Find the empirical formula for:
Given: 50.1%S50.1gS 50.1gS 1molS = 1.565 =1 49.9gO 1molO = 3.12=2
SO2
49.9%O49.9gO
32gS
1.565
16gO 1.565
3)Given the experimental Molar Mass of 64g/mol, find the molecular formula.
2O: 2x16 = 32
1S: 1x32 = +32
64g/mol = (64g/mol) n n=1(SO2)=SO2
64g/mol
64g/mol 64g/mol
Balance:
2Ag(CH3COOH)+CaCl2  Ca(CH3COOH)2+2AgCl (double displacement reaction)
HW Review (pg 285 1-10)
1) 2 smaller compounds come together to form a larger one.
6) They are very reactive so they replace other elements.
7) For a double displacement reaction, 2 new compounds must be produced.
10) a) 2HgO2Hg+O2 (decomposition)
b) 2C3H7OH+O26CO2+8H2O (combustion)
c) Zn+CuSO4ZnSO4+Cu (single replacement)
d) BaCl2+Na2SO4BaSO4+Na2Cl2 (double displacement)
e) Zn+F2ZnF2 (synthesis)
f) C5H10+O2C5O2+H10 (single displacement)
Notes-Stoichiometry
2C3H2OH + 9O2 6CO2+8H2O
-Balanced: massR=massP
-combustion reaction
12/16/11
Do Now
1) What is the % composition of CuCl2?
CuCl2=1Cu: 1x63.55= 63.55
Cu:63.55100=48%
2Cl: 2x35.45= +70.90
134.45
134.45g/mol
2) Balance: CaCo3(S)CaO(S)+CO2
3) What type of reaction is this? Decomposition
4) What do these mean? (s): solid
(l): liquid
(g): gas
(aq): aqueous
5) Why do we use an activity series? To predict reaction activity.
Cl2:70.90 100=52%
134.45
12/20/11
Do Now- Define the following:
Mole- 6.02x1023 particles, atoms, molecules, ions
Molar Mass- mass in grams of 1 mole of a substance
Chapter 9- Stoichiometry
1mole 5moles
3moles
4moles
C3H8 + 5 O2 3 CO2 + 4 H2O (combustion reaction)
atoms = atoms
mass = mass
energy = energy
MOLES = MOLES
- need a balanced chemical equation
- Coefficients give the amounts of reactants and products
- Mass-Mass Problems
12/21/11
Do Now- pg 304 #1
Calculate the amount requested if 1.34 mol H2O2 completely reacts according to the
following equation: 2H2O22H2O + O2
a)moles of oxygen formed?
Given: 1.34 mol H2O2
1.34molH2O2 1molO2 = .67molO2
Unknown: ?molO2
2molH2O2
b)moles of water formed?
Given: 1.34 mol H2O2
Unknown: ?molH2O
1.34molH2O2 2molH2O = 1.34molH2O
2molH2O2
HW Review (pg 311 #1-7)
5) 2C2H2(g) + 5O2(g)4CO2(g) + 2H2O(g)
a) given:15.9L C2H2
15.9L C2H2 1molC2H2 4molCO2 = 1.4molCO2
? molCO2
22.4L C2H2 2molC2H2
(hint: 1molC2H2=22.4L C2H2)
4) Br2 + Cl2 2BrCl
a) 2.74molCl2 1molBr2 = 2.74molBr2
1molCl2
b) 239.7gCl2 1molCl2
35.4gCl2
c) 4.53x1025Cl2
2molBrCl
1molCl2
115.4gBrCl = 1562.8gBrCl
1molBrCl
1molCl2
1molBr2 159.8gBr2 = 12024.82gBr2
6.02x1023Cl2 1molCl2 1molBr2
12/22/11
Mole Ratio- uses coefficients
-Molesknown (or given)mole ratiomolesunknown
-Molesgiven Molesunknown Mole
Molesgiven
ratio
C6H12O6 + 6O26CO2 + 6H2O + energy
Coefficients are in green, tell us the number of moles
Do Now
Fe2O3 + 2Al2Fe + Al2O3
a) ?molesAl
MOLE RATIO
given: 3.3molFe2O3
3.3molFe2O3
2molAl = 6.6molAl
1molFe2O3
b) ?molFe
MOLE RATIO
given: 3.3molFe2O3
3.3molFe2O3
2molFe = 6.6molFe
1molFe2O3
MOLE RATIO
c) ?molAl2O3
3.3molFe2O3
1molAl2O3 = 3.3molAl2O3
1molFe2O3
12/23/11
Do Now:
Fe2O3 + 2Al2Fe + Al2O3
1) Given: 135gFe2O3
?gAl
135gFe2O3 1molFe2O3 2molAl
27gAl = 45.6gAl
160gFe2O3 1molFe2O3 1molAl
2) Given: 23.6gAl
?gAl2O3
23.6gAl 1molAl 2molAl2O3 102gAl2O3 = 44.5gAl2O3
27gAl 1molAl
1molAl2O3
3) Given: 475gFe
?gFe2O3
475gFe 1molFe 1molFe2O3 160gFe2O3= 679gFe2O3
56gFe 2molFe 1molFe2O3
1/3/12
Do Now
Write 23800 in scientific notation: 2.38x104
Balance : 4Fe(s) + 3O2(g)  2Fe2O3(s)
L#Atoms
#Moles State of
Molar Mass
4
3
2
55.8g/mol
32g/mol
159.6g/mol
Products
Reactants Mole Ratio (or flipped)
R
Fe
O2
Fe2O
4
6
4
6
2
S
G
S
Fe
O2
Fe2O3
4molFe/3molO2
3molO2/2molFe2O3
4molFe/2molFe2O3
3
Br2(l)+5F2(g)2BrF5(l) (synthesis reaction)
?moleculesBrF5
Given: 384gBr2
384gBr2 1molBr2 2molBrF5 6.02x1023moleculesBrF5 =
2.89x1024moleculesBrF5
160gBr2 1molBr2
1molBrF5
1/4/12
Do Now
- Where are p+ and no found? In the nucleus of an atom.
- Where are the e-‘s mostly to be found? In the electron cloud.
- Which subatomic particle does not carry a charge? Neutron
- Atomic number of an element = number of protons
- Mass # of an element with 12p+ and 13no is? 25
- What is the electron configuration for oxygen? 1s22s22p4
Br2(l)+5F2(g)2BrF5(l)
? moleculesBr2
Given: 1.11x1020moleculesF2
1.11x1020molecF2
1molF2
1molBr2 6.02x1023molecBr2 = 2.22x1019molecBr2
23
6.02x10 molecF2 5molF2 1molBr2
Ratios:
Molar Mass: 1mol/g OR g/1mol
Avogadro’s Number: 1mol/Avogadro’s Number OR Avogadro’s Number/1mol
Mole Ratio: GivenMol/UnknownMol OR UnknownMol/GivenMol
(from balanced chemical equation)
What mass of H2O is produced if 65.2gCaCO3 reacts with H3PO4 to form Ca3(PO4)2,H2O,
and CO2?
3CaCO3 + 2H3PO4 Ca3(PO4)2 + 3H2O + 3CO2?
?massH2O
65.2gCaCO3 1molCaCO3 3molH2O
18gH2O =
100gCaCO3 3molCaCO3 1molH2O
Do Now
.
Write Electron dot for Boron-B?
B
How many valence electrons are in Argon-Ar? 8
What is the total number of e-‘s in Li+? 2e-‘s
What are the 4 orbitals? S(sphere), P(dumbbells), D, F
How many shapes does each one have? 1, 3, 5, 7 (respectively)
How many e-‘s can each one hold? 2, 6, 10, 14 (respectively)
(reactants)
1/5/12
(products)
Notes- Stoichiometry
A+ B  AB
Limiting Reactant- the substance that controls the quantity of the product that can be
formed in a chemical reaction.
The limiting reactant forms the least amount of product. (runs out)
Excess Reactant- the substance that is not used up completely in a reaction.
HW (pg 319 #1,2,8)
1/6/12
Theoretical yield- the maximum quantity of product that a reaction could make if
everything works perfectly.
Actual yield-the mass of the product actually formed.
Why? -The actual yield is less than the theoretical yield.
- Many reactants do not completely use up limiting reactants.
- Purification
- Unwanted “side reactions”
Do Now
Who developed the PT? Mendeleev
How are the elements arranged on the PT? Increasing atomic number
Where are the: Alkali metals: group 1
Alkali Earth Metals: group 2
Metals: left side of staircase
Nonmetals: right side of staircase
Halogens: group 17
Noble Gases: group 18
Transition Metals: d block
S: from block 1-2
P: from block 13-18
D: from block 3-12
F: from Lanthanide-Actinide series
What are vertical columns called? Families/groups
What are horizontal rows called? Periods
1/10/12
Do Now
What is the most reactive nonmetal? Fluorine
Define: Atomic Radius- distance from the nucleus to the edge of the electron cloud
Electronegativity- the measure of the ability of an atom in a compound to attract
Electrons
Notes
moles(unknown)/moles(given)(mole ratio)
1mole/6.02x1023(Avogadro’s Number)
1mole/molar mass(PT)
1mole/22.4liters (molar volume)
PCl3+3H2OH3PO4+3Hcl
g
g
mol mol
g mol
% yield- actual yield
x100
theoretical yield
1/11/12
Do Now
1) 6.02x1023=1mole (Avogadro’s number)
2) How many moles of CO2 are present in 11.2L?
11.2LCO2 1molCO2 = .5molCO2
22.4LCO2
3) What is the %mass of Mg in MgO? Mg= 24 x100 = 60%Mg
MgO= 40
4) Define theoretical yield- the maximum quantity of product that a reaction could make
if everything works perfectly.
% yield- actual yield x100
theoretical yield
Notes
Determine the limiting reactant, theoretical yield and %yield if 14.0gN2 are mixed with
9gH2 and 16.1gNH3 form. N2+3H22NH3
14g 9g 16.1g (actual yield)
14gN2 1molN2 2molNH3 17gNH3 = 17gNH3 theoretical yield
28gN2 1molN2 1molNH3
(17 is smaller then 51 so N2 is the limiting reactant)
9gH2 1molH2 2molNH3 17gNH3= 51gNH3
2gH2
3molH2 1molNH3
%yield- actual yield = 16.1g x100 = 94.7%
theoretical yield
17g
1/12/12
Do Now- balance and classify
4Al+3O22Al2O3 (synthesis)
AlCl3+3NaOHAl(OH)3+3NaCl (double displacement)
2NaCl2Na+Cl2 (Decomposition)
Ba+2HClBaCl2+H2 (single replacement)
HW Review (pg 319 #7)
Determine the theoretical yield if 100gP4O10 are mixed with 200gH2O.
P4O10+6H2O4H3PO4
100gP4O10 1molP4O10 4molH3PO4 98gH3PO4 = 138gH3PO4 theoretical yield
284gP4O10 1molP4O10 1molH3PO4
200gH2O 1molH2O 4molH3PO4 98gH3PO4 = 544gH3PO4
18gH2O 6molH2O 1molH3PO4
%yield- 126.2gH3PO4 x100 = 91.4%
138gH3PO4
1/13/12
Notes
P4+10Cl24PCl5
What quantity, in moles Cl2, reacts with 4.0 moles of P4?
4molP4 10molCl2 = 40molCl2
1molP4
3NO2+H2ONO+2HNO3
How many moles of water are needed to react with 8.4 moles NO2?
8.4molNO2 1molH2O = 2.8 molH2O
3molNO2
P4+6F24PF3
How many grams of Fluorine are needed to react with 6.2g of phosphorous?
6.2gP 1molP 6molF2 38gF2 = 45.6gF2
31gP 1molP 1molF2
H2O+SO3H2SO4
If 36gH2O is used, how many moles of H2SO4 is made?
36gH2O 1molH2O 1molH2SO4 98gH2SO4 = 196gH2SO4
18gH2O 1molH2O 1molH2SO4
1/16/12
Do Now
4Al+3O22Al2O3
?gO2
8) given: 27gAl 1molAl 3molO2 32gO2 = 24gO2
27gAl 4molAl 1molO2
this graduated cylinder measures to 70.3mL
What is the last thing you do in a lab? Wash your hands.
Notes
Significant figures in measurements
-Significant figures in a measurement include all of the digits that are known,
plus one more digit that is estimated.
-Measurements must be reported to the correct number of significant figures.
Rules for counting significant figures
-Non-zeroes always count as significant figures
- 3456 had 4 significant figures
Zeros
- Leading zeroes do not count as significant figures.
- 0.0486 has 3 significant figures
- Captive zeroes always count as significant figures
- 16.07 has 4 significant figures
1/17/12
Do Now
A cube has a mass of 20 grams and a volume of 8.5mL. What is its density?
D=m/v=20g/8.5mL=2.35g/mL
What is the composition of Cl-35?
p+=17, e-=17, no=18
What is the electron configuration for N?
1s22s22p3
How many moles are present in 27gH20?
27gH2O 1molH2O = 1.5molH2O
18gH2O
Notes- Sig Figs
Zeros
- Trailing zeros are only significant if the number contains a written decimal point
- 9.300 has 4 significant figures
Two special situations have an unlimited number of significant figures
- Counted items (ex- 23 people or 425 thumbtacks)
- Exactly defined quantities (60 minutes = 1 hour)
Sig Fig Practice#1
1.0070m = 5 sig figs
17.10kg = 4 sig figs
100,890L = 5 sig figs
3.29x103s = 3 sig figs
0.0054cm = 2 sig figs
3,200,000mL = 2 sig figs
5 dogs = unlimited sig figs (counted value)
- In general, a calculated answer cannot be more precise than the least precise
measurement from which it was calculated.
- Ever heard that a chain is only as strong as the weakest link?
- Sometimes, calculated values need to be rounded off
Rounding Calculated Answers
- Decide how many significant figures are needed (more on this very soon)
- Round to that many digits, counting from the left.
- Is the next digit less than 5? Drop it
- Next digit 5 or greater? Increase by 1
1/18/12
Do Now
Define: Hypothesis- a reasonable and testable explanation for observations
Mass- the quantity of matter in an object (grams)
Kinetic energy- The energy of an object that is due to the objects motion.
Energy- the capacity to do work
How many grams is 3.75kg? 3750g
HW Review(pg 63 1-11)
6a) 0.8102mx3.44m = 2.79m2
b) 94.20g/3.16722mL = 29.74g/mL
7) 63J/8.0g(340k-314k) = .30J/gk
8a)129g/29.2mL = 4.42g/mL
11) 11.50g/9.03mL = 3 sig figs
Notes- Sig Figs
Addition and Subtraction- The answer should be rounded to the same number of decimal
places as the least number of decimal places in the problem.
Multiplication and Division
- round your answer to the same number of significant figures
as the least number of significant figures in the problem.
- # sig figs in the result equals the number in the least precise measurement
used in calculation.
- 6.38x2.0 = 12.76 = 13 (2 sig figs)
Sig Fig Practice#2
Calculation
Calculator Says
Answer
3.24m x 7.0m
22.68m2
23m2
100.0g/23.7cm3
4.219409g/cm3 4.22g/cm3
0.02cm x 2.371cm 0.04742cm2
0.05cm2
710m/3.0s
236.66666m/s 240m/s
1/19/12
Do Now
Standard metric system units: KHD-G-bDCM
Mass- grams (g)
Temperature- Celcius ©, Kelvin (K) = C+273
Volume- Liter (L or mL)
Distance- Meter (m)
Temperature (C)
Dependant Variable
y
100
0
-8
-20
ENERGY ADDED
Independent Variable

x
heat added
energy
(Joules)
1/20/12
Do Now
1) What is the most reactive nonmetal? F
2) Atomic Models: Chocolate Chip Cookie Model= Thompson’s Models
3) In order to occupy the same orbital, two electrons must have: opposite spin
4) How many moles of CO2 are present in 11.2L? (1mol=22.4L)
11.2LCO2 1molCO2 = .5molCO2
22.4LCO2
5) State of matter symbols: Solid-s, liquid-l, gas-g, aqueous-aq
1/23/12
Do Now
Balance: CaCO3+2HClCaCl2+CO2+H2O
How many moles of CaCl2 are produced when 50gCaCO3 react?
50gCaCO3 1molCaCO3 1molCaCl2 = .5molCaCl2
100gCaCO3 1molCaCO3
Balance: 2NO2N2+2O2
What type of reaction is this? Decomposition
How many moles of NO2 are used? 2
1/30/12
Chapter 5- Chemical Names and Formulas
Sec 5.1- Naming Ions
Objectives:
-Identify the charges on monatomic ions by using the periodic table, and name the
ions.
- Define a polyatomic ion and write the names and formulas of the most common
polyatomic ions.
-Identify the two common endings for the names of most polyatomic ions.
Atoms and Ions
-Atoms are electrically neutral
- Because there is the same numner of protons (+) and electrons (-).
-Ions are atoms, or groups of atoms, with a charge (positive or negative)
- They have different numbers of protons and electons.
- Only electrons can move, and gaining or losing electrons makes ions.
An Anion is…
-A negative ion
-Has gained electrons
-Nonmetals can gain electrons
-Charge is written as a superscript on the right
F1- has gained one electron (-ide is the new ending = fluoride)
O2- has gained two electrons (oxide)
A Cation is…
-A positive ion
-Formed by losing electrons
-More protons than electrons
-Metals can lose electrons
K1+ Has lost one electron (no name change for positive ions)
Ca2+ has lost two electrons
Predicting Ionic Charges
- Group 1A: Lose 1 electron to form 1+ ions (H1+, Li1+, Na1+, K1+, Rb1+, Cs1+, Fr1+)
- Group 2A: Loses 2 electrons to form 2+ ions (Be2+, Mg2+, Ca2+, Sr2+, Ba2+, Ra2+)
- Group 3A: Loses 3 electrons to forms 3+ ions (B3+, Al3+, Ga3+, In3+, Tl3+)
- Group 4A: Do they lose 4 electrons or gain 4 electrons? Neither! Group 4A elements
rarely form ions, the share)
- Group 5A: Gains 3 electrons to form 3- ions (N3-, P3-, As3-, Sb3-, Bi3-)
- Group 6A: Gains 2 electrons to form 2- ions (O2-, S2-, Se2-, Te2-, Po2-)
- Group 7A: Gains 1 electron to form 1- ions (F1-, Cl1-, Br1-, I1-, At1-)
- Group 8A: Stable noble gases no not form ions!
Group B elements: Many transition elements have more than one possible oxidation
state.
Iron (II) = Fe2+, Iron (III) = Fe3+ Note the use of Roman Numerals to show charges
Naming Cations
- Two methods can clarify when more than one charge is possible:
- Stock system- uses roman numerals in parenthesis to indicate the numerical
value
- Classical method- uses root word and suffixes (-ous, -ic)
- We will is the Stock System
- Cation- if the charge is always the same (like in the Group A metals) just write the name
of the metal.
- Transition metals can have more than one type of charge.
- Indicate the charge by roman numerals
Some of the post-transition metals elements also have more than one possible oxidation
State. Tin (II) = Sn2+, Tin (IV) = Sn4+
Exceptions:
-
Some of the transition metals have only one ionic charge
Do not need to use roman numerals for these
Silver is always 1+ (Ag1+)
Cadmium and Zinc are always 2+ (Cd2+ and Zn2+)
1/31/12
Do Now- What ions do the following form:
Ca- Ca2+ cation
Na- Na1+ cation
Cl- Cl1- anion
Al- Al3+ cation
Kr- none
O- O2- anion
CATIONS
Practice by naming these:
Na1+: Sodium Ion
Ca2+: Calcium Ion
Al3+: Aluminum Ion
Fe3+: Iron(III) Ion
Fe2+: Iron(II) Ion
Pb2+: Lead(II) Ion
Li1+: Lithium Ion
Write symbols for these:
Potassium Ion: K1+
Magnesium Ion: Mg2+
Copper(II) Ion: Cu2+
Chromium(VI) Ion: Cr6+
Barium Ion: Ba2+
Mercury(II) Ion: Hg2+
ANIONS
Naming Anions:
- Anions are always the same charge
- Change the monatomic element ending to – ide
- F1- a Fluorine atoms will become Fluoride ion
Practice by naming these:
Cl1- : Chloride
NaCl is Sodium Chloride
N3- : Nitride
Br1- : Bromide
O2- : Oxide
Ga3+ : Gallium Ion
Write symbols for these:
Sulfide Ion: S2Iodide Ion: I1Phosphide Ion: P3Strontium Ion: Sr2+
2/1/12
Do Now- Name
O2 : Oxide
Mg2+ : Magnesium Ion
Cl1- : Chloride
Al3+ : Aluminum Ion
Fe2+ : Iron(II) Ion
Pb4+ : Lead(IV) Ion
Write Formula
Potassium Ion: K1+
Sulfide Ion: S2Strontium Ion: Sr2+
Nitride Ion: N3Copper(II) Ion: Cu2+
Chromium(VI) Ion: Cr6+
Roman Numeral Review:
1-I
2-II
3-III
4-IV
5-V
6-VI
Polyatomic Ions are…
-Groups of atoms that stay together and have an overall charge, and one name.
-Usually end in –ide or –ite
-Acetate: C2H3O21-Nitrate: NO31-Nitrite: NO21-Permanganate: MnO41-Hydroxide: OH1- and Cyanide: CH1-Sulfate: SO42-Sulfite: SO32-Carbonate: CO32-Chromate: CrO42-Dichromate: Cr2O72-
-Phosphate: PO43-Phosphite: PO33-Ammonium: NH41+ (one of the few positive polyatomic ions)
If the polyatomic ion begins with H, then combine the word hydrogen with the other
polyatomic ion present:
H1+ + CO32-  HCO31Hydrogen + Carbonate Hydrogen Carbonate Ion
Naming and Writing Formulas for Ionic Compounds
-Objectives:
-Apply the rules for naming and writing formulas for compounds containing
polyatomic ions
Example: Barium nitrate (note the 2 word name)
Ba2+ + NO31-  Ba(NO3)2
1) Write the formulas for the cations and anions, including CHARGES!
2) Check to see if charges are balanced.
3) Balance charges, if necessary, using subscripts. Use parentheses if you need more than
one of a polyatomic ion. Use the criss-cross method to balance subscripts.
2/2/12
Do Now
Name: NO31- : Nitrate
Hg1+ : Mercury(I) Ion
SO32- : Sulfite
PO43- : Phosphate
Pb4+ : Lead(IV) Ion
Write Formula:
Carbonate: CO32Nitride: N3Chlorate: ClO31Nitrite: NO21Copper(II): Cu2+
Sulfate: SO42Notes
Example: Ammonium sulfate (note the two word name)
NH1+ + SO4 2-  (NH4)2SO4
1) Write the formulas for the cations and anions, including CHARGES!
2) Check to see if charges are balanced.
3) Balance charges, if necessary, using subscripts. Use parentheses if you need more than
one of a polyatomic ion. Use the criss-cross method to balance subscripts.
Example: Iron(III) Chloride
Fe3+ + Cl1- FeCl3
Example: Aluminum sulfide
Al3+ + S2-  Al2S3
Example: Magnesium carbonate
Mg2+ + CO32-  MgCO3
Example: Zinc hydroxide
Zn2+ + OH1-  Zn(OH)2
Example: Aluminum phosphate
Al3+ + PO43-  AlPO4
Naming Ionic Compounds
1) Name the cation first, then anion
2) Monatomic cation=name of the element
-Ca2+ = calcium ion
3) Monatomic anion = root + -ide
- Cl1- = chloride
- CaCl2 = calcium chloride
2/3/12
Write Formula:
Ammonium chloride- NH4Cl
Aluminum(III) oxide- Al3+O2-Al2O3
Potassium bromide- K1+Br1-KBr
Potassium sulfide- K1+S2-K2S
Tin(II) iodide- Sn2+I1-SnI2
Write name:
Ba(OH)2 : Barium hydroxide
Cr2(SO4)3 : Chromium(III) sulfate
Au(ClO3)3 : Gold(III) chlorate
NaClO : Sodium hypochlorite
SiC : Silicone carbide
HW Review- (Read 5.3, #1-8 pg 180)
5d) Copper(I) acetate
6b) Magnesium phosphide- Mg3P2
c) Silver(I) sulfide- Ag2S
d) Niobium(V) chloride- NbCl5
7a) Rb2O- Rubidium oxide
b) FeF2- Iron(II) flouride
c) K3N- Potassium nitride
8b) Lithium thiosulfate- Li2S2O3
c) Ammonium phosphate- (NH4)3PO4
d) Potassium permanganate- KMnO4
2/6/12
Do Now
Write Name:
Fe(C2H3O2)3- Iron(III) acetate
Cs2CO3- Cesium carbonate
MnSO3- Manganese(II) sulfite
Hg2Cl2- Mercury(I) chloride
Rb2CrO4- Rubidium chromate
Write Formula:
Radium nitrate- Ra(NO3)2
Barium hydroxide- Ba(OH)2
Aluminum(III) acetate- Al(C2H3O2)3
Magnesium nitrite- Mg(NO2)2
Gold(III) chlorate- Au(ClO3)3
Naming Ionic Compounds
- some metals can form more than one change (usually the transition metals)
- use a Roman numeral in their name:
PbCl2- use the anion to find the charge on the cation (chloride is always 1-)
Pb2+ is the Lead(II) cation
PbCl2 = Lead(II) chloride
Things to look for:
1) If cations have (), the number in the parenthesis is their charge.
2) If anions end in –ide they are probably off the periodic table (monatomic)
3) If anion ends in –ate or –ite, then it is polyatomic.
Practice:
Iron(II) phosphate- Fe3(PO4)2
Potassium sulfide- K2S
Ammonium chromate- (NH4)2CrO4
MgSO4- Magnesium sulfate
FeCl3- Iron(III) chloride
2/7/12
Naming and Writing Formulas for Molecular Compounds
-Objectives:
-Interpret the prefixes in the names of molecular compounds in terms of their
chemical formulas.
-Apply rules for naming and writing formulas for binary molecular compounds.
Molecular Compounds are…
- made of just nonmetals
- smallest piece is a molecule
- can’t be held together by opposite charge attraction
- can’t use charges to figure out how many charges of each atom (there are no
charges present)
Molecular compounds are easier!
-Ionic compounds use charges to determine how many of each.
-you have to figure out charges
-may need to criss-cross numbers
-Molecular compounds: the name tells you the number of atoms
-uses prefixes to tell you the exact number of each element present
2/8/12
1- mono
2- di
3- tri
4- tetra
5- penta
6- hexa
7- hepta
8- octa
9- nona
10- deca
- to write the name, write two words: prefix and name –ide
- one exception is we don’t write mono if there is only one of the first element.
- normally, we do not have double vowels when writing names (oa oo)
Practice by naming these:
-N2O = dinitrogen monoxide (also called nitrous oxide or laughing gas)
-NO2 = nitrogen dioxide
-Cl2O7 = dichlorine heptoxide
-CBr4 = carbon tetrabromide
-CO2 = carbon dioxide
-BaCl2 = barium chloride
Write formulas for these:
-diphosphorus pentoxide = P2O5
-tetraiodine nonoxide = I4O9
-sulfur hexafluoride = SF6
-nitrogen trioxide = NO3
-carbon tetrahydride = CH4
-phosphorus trifluoride = PF3
-aluminum chloride = AlCl3
Write the correct name for:
-As4O10 = tetrarsenic decoxide
-BrO3 = bromine trioxide
-BN = boron mononitride
-N2O3 = dinitrogen trioxide
-NI3 = nitrogen triodide
-SF6 = sulfur hexafluoride
-XeF4 = xenon tetrafluoride
-PCl3 = phosphorus trichloride
-CO = carbon monoxide
-PCl5 = phosphorus pentachloride
2/9/12
Do Now
Name:
NO- Nitrogen monoxide
CO- Carbon monoxide
H2O- Dihydrogen monoxide
P2S5- Diphosphorus pentasulfide
SbCl5- Antimony pentachloride
Write formula:
Sulfur trioxide- SO3
Nitrogen dioxide-NO2
Dihydrogen monoxide- H2O
HW Review (pg209 9&10)
9a) tin tetraiodide
b) dinitrogen trioxide
c) phosphorus trichloride
d) carbon diselenide
10a) PBr5
b) P2O3
c) AsBr3
d) CCl4
Naming and Writing Formulas for acids and bases
Objectives:
-Apply three rules for naming acids
2/14/12
Acids are…
-Compounds that give off hydrogen ions (H1+)when dissolved in water (the Arrhenius definition)
-Will start the formula with H
-There will always be some Hydrogen next to an anion
-The anion determines the name
Rules for naming acids: Name it as a normal compound first
1) If the anion attached to hydrogen ends in –ide, out the prefix hydro- and change –ide to
–ic acid
-HCl- hydrogen ion and chloride ion = hydrochloric acid
-H2S- hydrogen ion and sulfide ion = hydrosulfuric acid
-If the anion has oxygen in it, then it ends in –ate or –ite
2) change the suffix –ate to –ic acid (use no prefix)
-Example: HNO3 Hydrogen and nitrate ions = Nitric acid
3) change the suffix –ite to –ous acid (uses no prefix)
- Example: HNO2 Hydrogen and nitrite ions = Nitrous acid
Naming Acids
Normal Ending Acid name is….
-ide
hydro-ic acid
-ate
-ic acid
-ite
-ous acid
2 additional rules (not mentioned in book)
4) If the acid has 1 more oxygen than the –ic acid, add the prefix per- HClO3- (Hydrogen Chlorate) is chloric acid
- HClO4- would be perchloric acid
5)If there is 1 less oxygen than the –ous acid, add the prefix hypo- HClO2- (Hydrogen Chlorite) is chlorous acid then HClO would be hypochlorous
acid
2/15/12
Perchloric acid- HClO4
Chloric acid- HClO3
Chlorous acid- HClO2
Hypochlorous acid- HClO
Practice:
HF- Hydroflouric acid
H3P- Hyrdophosphoric acid
H2SO4- Sulfuric acid
H2SO3- Sulfurous acid
H2CrO4- Chromic acid
Writing Acid Formulas- in reverse!
- Hydrogen will be listed first
- the name will tell you the anion
- be sure the charges cancel out
- starts with prefix hydro?- there is no oxygen, -ide ending for anion
- no prefix hydro?
1) –ate anion comes form –ic endings
2) –ite anion comes from –ous ending
Examples:
Hydroiodic acid- HI
Acedic acid- HC2H3O2
Carbonic acid- H2CO3
Phosphorus acid- H3PO3
Hydrobromic acid- HBr
2/16/12
PRACTICE:
HNO3- Nitric acid
HClO- Hypochlorous acid
2/27/12
Properties of Substances with metallic, ionic, and covalent bonds
- metallic- electrons move throughout atoms
- ionic- transfer of electrons
- covalent- share electrons
Lewis Electron-Dot Structures – show valence electrons
- As you go from element to element across the periodic table, you add a dot to each side
of the element’s symbol.
- You do not begin to pair dots until all four sides of the element’s symbol have a dot
- An element with an octet of valence electrons has a stable configuration
- The tendency of bonded atoms to have octets of valence electrons is called the octet
rule.
- When two chlorine atoms form a covalent bond, each atom contributes one electron to a
shared pair.
- An unshared pair, or a lone pair, is a nonbonding pair of electrons in the valence shell of
an atom
- A single bond is a covalent bond in which two atoms share one pair of electrons
- The electrons can pair in any order. However, any unpaired electrons are usually filled
in to show how they will form a covalent bond.
H
Draw a Lewis structure for CH3I.
.. ..
C H H H :I:
H: C :I:
˙
˙˙
 ˙˙
H
Multiple Bonds
- For O2 to make an octet, each atom needs two more electrons. The two atoms share four
electrons.
:O::O: or :O=O:

- For N2 to make an octet, each atom needs three more electrons. The two atoms share six
electrons.
:N N: or :NN:
2/28/12
Draw the Lewis structure for formaldehyde CH2O
.
..
C H H O:


Objectives:
- Predict the shape of a molecule using VSEPR theory
- Associate the polarity of molecules with the shapes of molecules, and relate the polarity
and shape of molecules to the properties of a substance.
Determine Molecular Shapes
- The three-dimensional shape of a molecule is important in determining the molecule’s
physical and chemical properties
A Lewis Structure can help predict molecular shape
- You can predict the shape of a molecule by examining the Lewis structure of the
molecule
- The valence shell electron pair repulsion (VSEPR) theory is a theory that predicts some
molecular shapes based on the idea that pairs of valence electrons surrounding an atom
repel each other.
- According to the VSEPR theory, the shape of a molecule is determined by the valence
electrons surrounding the central atom
- Electron pairs are negative, so they repel each other
- Therefore, the shared pairs that form different bonds repel each other and remain as far
apart as possible.
- For CO2, the two double bonds around the central carbon atom repels each other and
remain far apart
CH4
but three dimensional
2/29/12
HW Review
2) They are linear because there are only 2
3) Lewis Dot helps us predict the shape of a molecule
6) Br2 is linear because there are only 2 shapes. HBr is linear because there are only
2 shapes. HBr is more polar.
3/1/12
The four shared pairs of electrons in CH4 are farthest when each pair is positioned at the
corners of a tetrahedron.
Draw the Lewis structure for H2O. Water has
2 shared and 2 unshared
pairs of electrons
Find the shape that allows the shared and unshared pairs of electrons to be as far apart as
possible. The water molecule will have a bent shape.
Shape affects polarity
- One property that shape determines is the polarity of a molecule
- The polarity of a molecule that has more than two atoms depends on the polarity of each
bond and the way the bonds are arranged in space.
- If two dipoles are arranged in opposite directions, they will cancel each other.
- If two dipoles are arranged at an angle, they will not cancel each other out.
- Compare the molecules of nonpolar carbon dioxide, CO2, which has a linear shape, and
polar water, H2O, which has a bent shape.
3/2/12
linear
tetrahedral
trigonal planar
trigonal bipyramidal
9) If water is not a good conductor of electric current, why is it dangerous to handle an
electrical appliance when hands are wet or when you are standing on wet ground?
Because of all the minerals and
3/6/12
Review for 6.3 quiz
Lewis Dot structure- element and valence electrons
Covalent bond- a bond that shares electrons
Ionic bond- a bond that transfers electrons
Double bonds have 4 electrons
Electrons that exist outside the bond are unshared, unbonded, lone pairs of electrons.
Planar trigonal- 4 atoms
Tetrahedral- 5 atoms
bent
Angular or bent- 3 atoms
Linear can be 2 or 3 atoms
What does VSEPR stan d for? Valence shell electrons pair repulsion
Single, double, triple bonds
3/8/12
Heat -q- is the energy transferred between objects which are at different temperatures.
The unit for heat is joules. Energy is never created nor destroyed.
Temperature -T- a measure of the average kinetic energy of a substance. (kinetic energythe energy of motion)
3/12/12
Temperature
Heat
u
Density is intensive
Intensive Property- Extensive PropertyDoes not depend on the
amount of the substance
Does depend on the
amount of the sample
Enthalpy- the total energy of a sample
motion- KE
light- solar
electromagnetic energy
heat- thermal
chemical
nuclear
potential
sound
Thermodynamics- the branch of science concerned with the energy changes that
accompany chemical and physical changes.
Gases- Intro to gases
Properties- great distance between molecules.
- low densities
- highly compressible
- fill any container
- fluid
Kinetic Theory
- Gas molecules are in constant, rapid, random motion
- Gas molecules are very far apart relative to their size
-The pressure exerted by a gas is a result of collisions of the gas molecules against the
walls of the container
Pressureunits- KPa- kilopascals
- Atm- atmospheres
- mmHg- millimeters Mercury
- Torr- millimeters Mercury
- psi- pounds per square inch
1atm = 101.3kPa = 760mmHg = 14.7psi = 760 torr
3/14/12
Gas Laws
- Volume
- Pressure
- Temperature
3/15/12
Boyle’s Law- Pressure and volume of a gas are inversely related. Temperature remains
constant. V1P1=V2P2
Charles Law- Volume and temperature of a gas are directly related. Pressure remains
constant. V1/T1=V2/T2 They move in the same direction.
Gay-Lussac’s Law- Pressure and Temperature of a gas are directly proportional.
P1/T1=P2/T2
3/16/12
Do Now
Write the untis for:
Pressure: atm, psi, kPa, mmHg, Torr
Temperature: C, K, F
Volume: L, mL, cm3
3/19/12
Notes
STP- standard temperature and pressure, 0C and 1 atm
Combined Gas Laws- V1P1 = V2P2
T1
T2
3/23/12
Practice:
-What is the volume of a gas stored at 45C and 1376mmHg if the same gas occupies
148cm3 at -43C and 3608mmHg
V1=?
V1P1 = V2P2 V1(1376mmHg) = (148cm3)(3608mmHg)
T1 = 45C = 318K
T1
T2
318K
230K
P1 = 1376mmHg
V2 = 148cm3
(148cm3)(3608mmHg)(318K) = V1(1376mmHg)(230K)
T2 = -43C = 230K
V1 = 536.5cm3
P2 = 3608mmHg
-A 35,800L balloon is held at 15875mmHg and 120C. The balloon is then stored at
24,928mmHg and 40C. What is the new volume?
V1= 35,800L
V1P1 = V2P2 (35,800L)(15,875mmHg) = (V2)(24,928mmHg)
T1 = 120C = 393K
T1
T2
393K
313K
P1 = 15,875mmHg
V2 = ?
(35,800L)(15,875mmHg)(313K) = V2(24,928mmHg)(393K)
T2 = 40C = 313K
V2 = 18,158L
P2 = 24,928mmHg
Ideal Gas LawIdeal Gas Equation- PV=nRT
P= pressure
V= volume
n= moles = n/M = g/(g/mol)
M= molar mass
R= gas constant = 8.31(LkPa/molK) = 0.082(Latm/molK)
T= temperature
3/26/12
Example:
-A 240mL sample of gas at 100C ad 755mmHg is found to have a mass of 1.12g. What
is the Molar Mass of the gas?
R=8.31(LkPa/molK) = 0.0821(Latm/molK)
1atm = 760mmHg = 101.kPa
Given: P= 755mmHg = 100.63kPa
V=240mL = .24L
R= 8.31(LkPa/molK) = 0.0821(Latm/molK)
T= 100C = 373K
m= 1.12g
M= molar mass= ?
VP=nRT  VP=(m/M)RT  M=mRT  M=(1.12g) 8.31LkPa (373K) = 143g/mol
VP
(.24L)
molK (100.63kPa)
3/27/12
Notes:
Effusion- to move from a high concentration to low concentration through a hole
Diffusion- the movement from an area of high concentration to low concentration
Dalton’s Law of Partial Pressure
PT = P1+P2+P3+…+Pn
Patm = PN2+PO2+PH2O+PCO2+…Pn
1atm=101.3kPa=760mmHg=14.7psi
3/29/12
Do Now
Name: CaCl2 = Calcium Chloride
LiBr = Lithium Bromide
Write Formula: Strontium Phosphate = Sr +2 PO4 -3 = Sr3(PO4)2
Aluminum Sulfide = Al+3S-2 = Al2S3
NotesMixtures: Iced Tea, Lemonade, Meat, cereal, crunch yogurt, rock, salt water, pool water,
medicine
Heterogeneous Mixture- the particles can be seen with (at least) a microscope, has at least
2 phases (same properties)
3/30/12
Do Now
Write formula: Calcium phosphate Ca+2PO4-3 Ca3(PO4)2
Sodium sulfate Na+1SO4-2 Na2SO4
Write the name: AgCl2 Silver(II) chloride
H2O Dihydrogen monoxide
4/3/12
Do Now
Write the name: PF3- Phosphorous trifluoride
SCl4- Sulfur tetrachloride
BrCl4- Bromine tetrachloride
Write formula: Carbon dioxide- CO2
Sulfur trioxide- SO3
Krypton difluoride- KrF2
Notes
Heterogeneous Mixtures- 2 phases, depending on size
Homogeneous Mixtures- 1 phase (same phase)
- appear uniform,
- solutions-solvent- the substance that does the dissolving
-solute- the substance that is dissolved
Suspension- heterogeneous mixture in which different parts separate over time.
- largest size solutes
Colloid- a mixture consisting of particles that are sized between a suspension and a
solution.
- Does not separate over time.
- Stable
4/4/12
Write the name: H2SO4- Sulfuric Acid
HCl- Hydrochloric Acid
H3PO4- Phosphoric Acid
H2S- Hyrdosulfuric Acid
Write formula: Nitric Acid- HNO3
Hydrobromic Acid- HBr
Carbonic Acid- H2CO3
Solutions- Smallest particle size, molecules, atoms
Colloids- Middle size
Suspension- Largest size
Separating Mixtures- Distilling
4/5/12
Do Now
MgF2-(Ionic) Magnesium fluoride
H2O2- (Covalent) Dihydrogen dioxide
H3PO4- (acid) Phosphoric Acid
Sodium chloride- (salt) NaCl
Carbon tetrahydride- (methane) CH4
Acetic Acid- (vinegar) HC2H3O2
`
Separating Mixtures- Distilling (based on BP), Filtering, Evaporation, centrifuge,
chromatography
Liquids- fluid
- enough energy to allow molecules to move past one another
- Viscosity- the degree to which a fluid will flow
viscosity  flow
- high viscosity- molasses
- low viscosity- water
Cohesion- the attractive forces molecules have for each other
Ex- beads of water on a waxed car
Adhesion- the attractive forces molecules have for a solid
Ex- meniscus