CHEM 167 FINAL REVIEW
Part 1
ELECTROLYTES VS. NONELECTROLYTES
Eletrolyte: Can break apart into ions
Examples: salts – NaCl, acids – HF, bases – NaOH
Types: Strong – complete dissociation into ions
Weak – not complete dissociation
 Nonelectrolyte: does not break apart into ions
Examples: ethanol, sugars (sucrose, fructose)
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MASS SPECTROMETER
VIADD
Vaporization: heat, laser, needs to be in gas phase
Ionization: electron shower/magnetic plate
Acceleration: ions are sped through spectrometer
Deflection: Magnetic field sorts larger/smaller ions
Detection: how many of which ion are present
 Gives a Mass:Charge ratio
 Data displayed on graph with peaks
 Helps to find average mass of an element
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FRACTIONAL ABUNDANCE OF AN ISOTOPE
Multiply fraction by the mass of the isotope
Example: 24Mg 23.99 amu – 78.99%
25Mg 24.99 amu – 10.00%
26Mg 25.99 amu – 11.01%
Work: (.7899x23.99) + (.1x24.99) + (.1101x25.99)
= 24.31 amu which is value on periodic table
 Can also be asked to find the fractional
abundance
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FIND THE FRACTIONAL ABUNDANCE
Problem: Bromine has two naturally occurring
isotopes 79Br 78.918 amu and 81Br 80.916 amu. The
average mass for bromine is 79.904 amu. What is
the fractional abundance of 79Br?
FINDING FRACTIONAL ABUNDANCE
Work: X + Y = 1 Y = 1-X  X= 79Br,Y=81Br
78.918(X) + 80.916(1-X) = 79.904
78.918X + 80.916 – 80.916X = 79.904
78.918X-80.916X = 79.904-80.916
-1.998X = -1.012
X = 0.51 = 79Br
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PERCENT BY MASS/EMPIRICAL FORMULA
Assume a 100g sample
 %mass  mass of element moles mole ratio
Problem: A compound is 54.05% Ca, 43.24% O, and
2.71 % H. What is its empirical formula?
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ANSWER
54.05%54.05g Ca/40.1g/mol1.348 mol Ca
43.24%43.24g O/16g/mol2.703 mol O
2.71%2.71g H/1.01g/mol2.683 mol H
Ca has the smallest number of moles
1.348mol/1.3481Ca
2.703mol/1.3482O
2.683mol/1.3482H
Answer: CaO2H2Ca(OH)2
LINE DRAWINGS AND THEIR FORMULAS
If no element is written assume is C at every
joint and end. Fill in H’s until number of bonds is
met
Examples: drawn on board/paper
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NAMING COMPOUNDS
Know group charges and specific element
charges, such as B, Al, Zn, and Ag
 Know polyatomic ions
 Carbon-carbon bonds: single –ane, double –ene,
triple –yne
 Transition metals need oxidation state in
parentheses
 Covalent bonding needs prefixes
Examples: a) Fe2O3  iron (III) oxide
b) Cl2O7  dichlorine heptoxide
c) (NH4)2S  ammonium sulfide
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SOLUBILITY VS. INSOLUBILITY
Review or write down solubility rules
Examples: are the following salts soluble or
insoluble?
a) KBr - soluble
b) AgCl - insoluble
c) Al(OH)3 - insoluble
d) BaSO4 - soluble
e) Ca3(PO4)2 - insoluble
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BREAKDOWN OF AN ELEMENT
An element contains a specific number of protons,
neutrons, and electrons
 If the element is neutral then the number of
electrons and protons is equal
 The number of protons and neutrons is equal to
the mass of the element (A)
 The number of protons is equal to the atomic
number of the element (Z)
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TYPES OF BONDING
Ionic bonding: between a metal and nonmetal; is
a transfer of electrons
 Covalent bonding: between nonmetals; is a
sharing of electrons
 Metallic bonding: between metals; electrons are
free flowing and shared between many nuclei
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BALANCING EQUATIONS
Matter cannot be created or destroyed, need the
same amount of each element on both sides of the
arrow.
Balance these equations:
a) C4H10 + O2  CO2 + H2O

b) NaClO3  NaCl + O2
c) CaCl2 + NaOH  NaCl + Ca(OH)2
d) C + SO2  CS2 + CO
ANSWERS
a) 2,13,8,10
b) 2,2,3
c) 1,2,2,1
d) 5,2,1,4
MOLES, MOLARITY, AND DILUTION
Moles = mass/molar mass; this is true for
elements and compounds
 Molarity = moles/liters
 Dilution = M1V1 = M2V2
Example Dilution problem:
What volume (mL) of 6M NaOH must be diluted to
create a solution of 200mL and 1.5M?

ANSWER
Use M1V1=M2V2
Assign M1=6M V1=? M2=1.5M V2=200mL
V1=(M2V2)/M1
V1=50mL
TOTAL IONIC, MOLECULAR, NET IONIC
Total ionic equation: write out all the ions on
both sides of arrow
 Molecular equation: combines ions and includes
states; aq, and l
 Net ionic equation: eliminate spectator ions from
the total ionic equation
Write the net ionic equation of:
AgNO3 + NaCl
*look for formed insoluble salts

ANSWER
AgNO3 + NaCl AgCl + NaNO3 AgCl is insoluble
Total ionic:
Ag+ + NO3- + Na+ + Cl-  AgCl + Na+ + NO3Spectator ions: Na+ and NO3Net ionic:
Ag+ + Cl-  AgCl
GRAPHS TO KNOW
Coulomb’s Law: repulsion on top half (same ions),
attraction on lower half (different ions). As radius
(r) increases force (F) decreases.
 Maxwell-Boltzmann distribution: explains how
the speed of molecules in a gas increases with
higher temperature, and the speed also increases
with smaller mass
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EXAM TWO MATERIAL
TYPES OF LIGHTS
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Incandescent bulbs: a wire inside is heated up by
electricity, excites the electrons in tungsten (W) which
then emits excited electrons to create light.
Fluorescent bulbs: an arc/plasma is sent through bulb,
excites the electrons of the gas in the bulb (gas varies),
then emitted excited electrons interact with
phosphorous coating to create light.
LEDs (light emitting diodes): promote electron from
ground to excited state, emits light to fall back to
ground state. Has a narrow wavelength and frequency
distribution, single color, high efficiency
Laser (light amplification by stimulated emission of
radiation): emits light same as LED. Need population
inversion. All emitted photons are in same phase
(coherent light). Single color with narrow wavelength
and frequency distribution
ATOMIC MODELS AND PRINCIPLES
Bohr’s: states that 1) electrons exist in well
defined orbits 2) each orbit has specific energy
associated with it and 3) energy is released or
absorbed when electrons change energy levels
 Heisenberg uncertainty principle: proves Bohr’s
model wrong because it states that it is
impossible to pin down where an electron is at
exactly.
 Another thing to prove Bohr’s model wrong is
that electrons diffract through slits so they can
also behave like waves
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ORBITALS
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Orbitals have a set of quantum numbers that describe
them: n, l, ml, and ms .
N: principle quantum level, gives number of energy
shell
L: secondary quantum number; determines shape of
orbital. Is equal to n-1or less. 0=s, 1=p, 2=d, 3=f
Ml: magnetic quantum number; tells how many
orbitals are in a subshell. Is equal to any number
from –l to +l
Ms: spin quantum number. Can only be +1/2 or -1/2.
Each orbital can only have 2 electrons and they spin
in opposite directions.
Pauli exclusion principle: each electron in an atom
has a unique set of quantum numbers
Hund’s rule: single electrons first fill empty orbitals
before they pair up in an occupied one.
SHAPE OF AN ORBITAL
S orbital: shaped like a sphere, groups 1&2
 P orbial: has two lobes, groups 3-8
 D orbital: has 4 lobes, transition metals
 F orbital: has 6 lobes, lanthanides &actinides
 Orbitals also have nodes (areas where no
electrons exist) the number of nodes is equal to n1. Nodes can be planar or spherical.
Examples:
Which orbital is this? n=2, l=1, ml=0, ms=+1/2
Which quantum numbers describe a 3p orbital?
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ANSWER
2p because n=2 and l=1p
 n= 3, l=1, ml=-1,0,1, ms= +1/2 or -1/2
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PHOTOELECTRIC EFFECT
Energy cost to remove one electron from a surface
E = h*v
 Energy cost to remove one electron using a
certain wavelength
E=(h*c)/l and c=lv
Example: What is the wavelength in meters of a
wave with a photon energy of 4.5*10^-28J?
 Know photoelectric graphs, there are 4
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ANSWER
Use E=(h*c)/l  l= (h*c)/E
h= Planck’s constant = 6.626*10^-34J*s
c= speed of light = 3*10^8m/s
l= 442m
ELECTRON CONFIGURATIONS
Fill in the orbitals across the periods of the
periodic table
 There are full electron configurations and
condensed configurations that use noble gases
Examples:
a) Write the full configuration of Ga.
b) Write the condensed state of Ga.
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ANSWER
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1s22s2 2p63s23p64s23d104p1

[Ar] 4s23d104p1
GROUND STATE AND EXCITED STATE
Electrons can either be in their ground state or
their excited state. This can be demonstrated in
electron configurations.
Examples:
1) Write the stable configuration of nitrogen (N)
2) Which is an excited state?
a) 1s22s22p63s23p64s1
b) 1s22s22p64s1
c) 1s22s22p63s1
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ANSWER
1s22s2 2p3
b
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PERIODIC PROPERTIES
Zeff: increases as move left to right because Zeff =
Z-S. S= shielding (core) and remains the same,
Z=atomic number and it increases.
 Radius, diameter, and volume: increase as move
left and down towards francium (Fr).
 Ionization energy: increases as move right and up
towards fluorine (F). Is a positive value
 Electron affinity: increases in magnitude as
move right and up towards fluorine (F). Is a
negative value. Noble gases are extremely
positive and do not follow trend.
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LATTICE ENERGY
The energy cost to break an ionic species into
cations and anions
 Larger energy is characterized by larger ionic
charges and smaller distance apart (smaller r)
Example:
a) Which has the smallest radius?
O2-, F-, Ne, Na+, Mg2+
b) Which will have the largest lattice energy?
NaCl, MgCl2, MgO, Na2O
 In general cations are smaller than anions
because of the missing electron(s) which occupy a
lot of space by repulsion.
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GAS LAWS
P1V1=P2V2, P1/T1=P2/T2, V1/T1=V2/T2
 Combines into ideal gas law PV=nRT where
R=0.08206 L*atm/mol*K
 Conditions where any gas behaves most ideally
are at high temperatures and low pressures
 Van der Waals equation helps account for not
ideal conditions (P+(n2/V2)a) (V-nb) = nRT
Example:
What is the V occupied by 2 moles of N2 at P=1atm
and T=273K?
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ANSWER
Use ideal gas law
V=(nRT)/P
V=(2*0.08206*273)/1=44.8L
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DALTON’S LAW AND MOLE FRACTION
Dalton’s Law: total pressure is equal to the sum
of the partial pressures.
 Mole fraction: equal to partial moles divided by
total moles is symbolized by Xi
 Partial pressure can be found by applying mole
fraction and Dalton’s law. Pi=Xi * PT
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CALCULATING THE EXTENT OF A REACTION
Need to make sure the equation is balanced and
then use the stoichiometry to decide if there is a
limiting reactant.
Example:
a) N2 + H2  NH3 *balance it
What mass of N2 is needed to fully react with 5.11g
of H2?
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b) O2 + 2H2  2H2O
What is the yield of H2O if 7.9g O2 and 2.9g H2
react to completion?
ANSWERS
N2 + 3H2 2NH3
5.11g H2/2.02  2.53 mol H2
H2 and N2 in 1:3 ratio; 2.53mol/30.843 mol N2
0.843mol * 28g/mol = 23.6g N2
 7.9g O2/32  0.247mol O2
2.9g H2/2.021.435mol H2
Limiting reactant is O2 because of mole ratio
0.247mol O20.494mol H2O from stoichiometry
0.494mol * 18g/mol = 8.89g H2O
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REACTIONS WITH SOLIDS AND GASES
Can use gaseous reactants in reactions to
calculate product yields by using the ideal gas
law.
Examples:
a) O2(g) + 2H2(g)  2H2O(g)
What is the V of H2 needed to make 36g H2O at
T=30C and P=0.995 atm?
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b) CaO(s) + CO2 (g) CaCO3 (s)
What V of CO2 can be reacted exactly with 112g
CaO at T=800C and P=1.25 atm?
ANSWERS
First find moles of H2O created
36g/18g/mol  2mol H2O2mol H2 by reaction
stoichiometry
Use number of moles in PV=nRTV=(nRT)/P
V=(2*0.08206*303)/0.995 = 50L
 First find moles of CaO used
112g/56.1g/mol2mol CaO2mol CO2
Use number of moles in V=(nRT)/P
V=(2*0.08206*1073)/1.25 = 141L
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PRESSURE MEASURING DEVICES
Capacitance manometer: uses a flexible
diaphragm and measures the distance change
(capacitance) between the diaphragm and a
metal plate.
 Ionization gauge: measures the current created
by ions, an increase in P makes the current
increase
 Thermocouple gauge: as pressure decreases the
measured temperature of the filament increases
 Mass spectrometer: measures partial pressures of
the sample
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