Honors Chemistry Topics & Final Exam Information
Dates: May 28(odd)/29(even)

This summative assessment will consist of multiple choice and short answer questions from Units 1-9.

You may use 2 sides of a 3 x5 note card, hand written, on the test.

all questions in blue are take from your unit assignments & are examples of some of the problems you
might encounter on your final…make sure you understand the concepts behind them too
Unit 2: Atomic Structure & Nuclear Chemistry
1.
Students know how to calculate the amount of a radioactive substance remaining after an integral number
of half-lives have passed.
A. The half-life of radon-222 is 3.824 days. After what time will one-fourth of a given amount of radon
remain?
(lots of ways to solve this...) example: if you started with 1 0 .5 0.25 so 2 half lives pass
7.648 days.
B. The half-life of cobalt-60 is 10.47 min. How many milligrams of cobalt-60 will remain after 104.7 min
if you start with 10.0 mg?
104.7/ 10.47 = 10 half lives
10 mg  5 …. (do this 10 times) = .009766 mg
2. Calculate the percent abundance of each isotope for a given element.
A. Copper is made up of two isotopes, Cu-63 (62.9296 amu) and Cu-65 (64.9278 amu). Given copper's
atomic weight of 63.546 amu, what is the percent abundance of each isotope?
62.9296x + 64.9278(1-x) = 63.546
62.9296x + 64.9278- 64.9278x = 63.546
-64.9278
-64.9278
2x = 1.3818
2
X=.6909
Cu-63 = 69.1%
& Cu-65 = 30.9%
B. 50.54% of the naturally occurring isotopes of bromine have an atomic mass of 78.92 amu while
49.46% of the naturally occurring isotopes of bromine have an atomic mass of 80.92 amu.
Calculate the average atomic mass of bromine.
(78.92)(.5054) = 39.886
(80.92)(.4946) = 40.023
79.9 amu
Unit 3: Periodic Trends
1. Related to electron configurations:
a.
Be able to write orbital notations for s, p, & d block elements.
A. Carbon
B. Ga
b. Be able to explain the concepts behind electron configurations.
B. Given that the electron configuration for oxygen is 1s22s22p4, answer the following
questions:


How many electrons are in each atom? 8
What is the atomic number of this element? 8

Write the orbital notation for oxygen’s electron configuration.

How many unpaired electrons does oxygen have? 2

What is the highest occupied energy level? 2nd main energy level

How many inner-shell electrons does the atom contain? 2

In which orbital(s) are these inner-shell electrons located? 1s
2. Related to Periodic Trends:
a.
Be able to explain & use the concept of atomic radii as related to ions
A. Is the radius of Mg2+ greater than, equal to, or smaller than the radius of Mg? Explain.
Mg2+ = 12 protons/ 10 electron (2 energy levels)
Mg= 12 protons/ 12 electrons (3 energy
levels)
Mg2+ is smaller than Mg because it has lost its 3s energy level when it lost the 2 electrons. Also,
the nuclear charge of 20 protons pulls very hard on the 18 electrons.
Unit 4: Bonding
1.
Related to Lewis Dot Structure:
a.
Be able to draw Lewis structures for polyatomic ions & compounds that show resonance.
A. Resonance Below:
Draw 1 of the resonance structure for Ozone (O3):
b. Draw the other
resonance structure
for O3 :
c. In reality, the
electrons are shared
equally, or
delocalized between
all three Oxygen
atoms. Draw the
Resonance Hybrid:
b. Be familiar with the exceptions to the octet rule for Lewis structures (electron deficiency &
expanded octets) & be able to draw them.
A. Boron & Beyllium are the most common examples of electron deficient atoms. What does it
mean for Be & B to be electron deficient?
Some stable molecules simply do not have enough electrons to achieve octets around all atoms.
This usually occurs in compounds containing Be or B.
Beryllium only has two valence atoms, and can only form electron pair bonds in two locations.
Boron has three valence electrons.
B. Phosphorus & Sulfur (and some atoms existing in Periods 3 and higher) are sometimes known to
have an expanded octet. What does this mean?
“Expanded octet” refers to the Lewis structures where the central atom ends up with more
than an
octet, such as in PCl5.
Third period elements occasionally exceed the octet rule by using their empty d orbitals to
accommodate additional electrons.
Although elements such as Si, P, S, Cl, Br, and I obey the octet rule in many cases, under other
circumstances they form more bonds than the rule allows.
For example, in the case of phosphorus, an octet would be 3s23p6. However, the 3d subshell is
also available, and some of the 3d orbitals may also be involved in bonding. This permits the
extra pair of electrons to occupy the valence shell of phosphorus in PF5.
2. Be able to draw and name VSEPR Shapes.
A. SeCl6
B. BF3
C. PCl5
Unit 5: Thermodynamics
1.
Students know how to identify solids and liquids held together by different intermolecular forces and
relate these forces to volatility and boiling/ melting point temperatures.
Substance #1
Strongest Force of
Attraction
London DF
Substance #2
a.
between
molecules of I2
b.
between
molecules of H2O
(liquid)
Hydrogen bonding
between molecules
of H2S
dipole-dipole
c.
between
molecules of NH3
Hydrogen bonding
between molecules
of CH4
London DF
d. between atoms
of Fe
Metallic bonds
between atoms of Kr
e.
Dipole-dipole
Between particles
of NaCl
London DF (noble
gases can have
LDF’s)
Ionic Bond
between
molecules of HCl
between molecules
of Cl2
Strongest Force of
Attraction
London DF
(metal & non-metal)
f.
between
molecules of
CH3F
Dipole-Dipole
between molecules
of CH3OH
Hydrogen Bonding
Substance with
Higher Boiling Point
I2
both are nonpolar
molecules so only
London dispersion
forces are present.
London dispersion
forces get stronger
as molar mass
increases.
H2O (l)
Hydrogen bonds are
stronger
NH3
Hydrogen bonds are
stronger
Fe
metallic bonds are
stronger
NaCl
Ionic Bonds are very
strong.
CH3OH
Hydrogen bonds are
stronger
Unit 6: Gases & Phases of Matter
1. Students know how to apply Dalton’s law of partial pressures to describe the composition of gases
A. A sample of gas is collected over water at a temperature of 35.0 oC when the barometric pressure
reading is 742.0 torr. What is the partial pressure of the dry gas?
Ptotal = Pgas + PH2O
Given: Ptotal = 742.0 torr PH2O @ 35oC = 42.2 mmHg (or 42.2 torr) use chart to look this up
Solve: 742 torr = Pgas + 42.2 mmHg
Pgas = 699.8 torr
2. Use Graham’s Law to predict diffusion of gases
Effusion is the process by which gas molecules under pressure pass through a tiny opening. Diffusion is the
mixing of gases by random molecular motion and there is not a constraint (small hole) preventing their
ability to spread out.
A. What is the ratio of the average velocity of hydrogen molecules to that of neon atoms at the same
temperature and pressure.
FYI: Velocity & rate of effusion are proportional to each other and have same formula. See pg. 35
velocity or rate H2
velocity or rate Ne
=
√20.18
√2.02
=3.16
B. At a certain temperature and pressure, chlorine molecules have an average velocity of 0.0380 m/s.
What is the average velocity of sulfur dioxide molecules under the same conditions?
rate of SO2
m
.0380 s
Cl2
=
√70.9
√64.06

X
m
.0380 s Cl2
= 1.052
X=.0400 m/s this is the rate of SO2
Unit 7: Chemical Reactions
1.
Empirical Formula, & molecular formula
a.
What is the molecular formula of the molecule that has an empirical formula of CH 2O and a molar
mass of 120.12 g/mol?
Step 1: find the multiplying factor between molecular & empirical
Molecular formula mass =
x
Empirical formula mass
Empirical mass = C + H2 + O = 30.02 (plug into equation)
120.12g = 4

Multiply all the subscripts by 4 -
4(CH2O) = C4H8O4
30.02g
b.
A sample compound with a formula mass of 34.00 amu is found to consist of .44 g H and 6.92 g O.
Find its molecular formula.
(Hint: even though it’s asking for the molecular formula, you will still need to find the empirical formula first. In
the easier problems, like #5, they give you the empirical formula, but here, you’ll have to determine it first)
Step 1: Find Empirical Forumla & then the mass
H= 0 .4g
1mol
1.01 g
= .396 / .375 = 1mol
O= 6g
1mol
16g
= .375 / .375 = 1mol
Empirical Formula: HO
Empirical Mass: 17.01g
Step 2: Determine multiplying factor & multiply by the empirical formula
34.00 = 2  2(HO) =
**
H2O2
**
17
Unit 8: Reaction Rates, Equilibrium, and Stoichiometry
1.
Be able to solve various enthalpy problems (including using Hess’s law)
A. Find the ΔH for the reaction below:
N2H4(l) + H2(g) → 2NH3(g)
Use the following information:
N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3H2 (g)
ΔH = -37 kJ
N2(g) + 3H2(g) → 2NH 3(g)
ΔH = -46 kJ
CH4O(l) → CH2O(g) + H 2(g)
ΔH = -65 kJ
Step 1:
N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3H2 (g)
N2(g) + 3H2(g)
→ 2NH 3(g)
CH2O(g) + H 2(g)
→ CH4O(l)
N2H4(l) + H 2(g)
→
2NH 3(g)
ΔH = -37 kJ
ΔH = -46 kJ
ΔH = +65 kJ (flip this one)
ΔH = -18 kJ
Be able to perform calculations involving limiting reactants & % yield at a higher level than your regular chem.
classmates (you will be asked to write formulas & then solve)
A. If 75.0 g of CO reacts to produce 68.4 g CH3OH, what is the percent yield of CH3OH?
CO(g) + 2H2(g)

% yield = actual/ theoretical

68.4g = actual yield

Use stoich to find theoretical yield
75g CO
CH3OH(l)
1 mol CO
1 mol CH3OH
32.04g CH3OH
28.01g CO
1 mol CO
1 mol
=85.79g CH3OH Theoretical yield

Plug into formula
68.4g
85.79g
=
79.7%
(Book says 79.8% but I can’t seem to get that…)
B. Metallic magnesium reacts with steam to produce magnesium hydroxide and hydrogen gas.

If 16.2 g of Mg are heated with 12.0 g of H2O, what is the limiting reactant?
16.2g Mg
1mol Mg
1mol H2
2.02g
24.31g
1mol Mg
1 mol
1mol H20
1mol H2
2.02g
18.02g
2mol H20
1 mol
= 1.35g H2
12g H20
= .673g H2
**H2O is my limiting reactant. I can only make 0.673g H2.**

How many moles of the excess reactant are left?
Mg is my excess reactant
I started with 0.666 moles Mg
2 ways I could solve this:
12g H2O  moles Mg used
12g H20
1mol H20
18g
1mol Mg
2mol H20
= .333 mole used
.666 starting Mg- .333 moles mg used =.333
moles remain
Other way to solve:
0.673 g H2  moles Mg
.673g H2
1mol H20
2.02g
1mol Mg
1mol H2
= .333 mole used
.666 starting Mg- .333 moles mg used =.333

moles remain
How many grams of each product are formed?
I already figured out grams of H2 above in part A
= .673g H2
(used the limiting reactant)
To find g Mg(OH)2
12g H2O
1mol
1mol Mg(OH)2
58.32g
18g H2O
2mol H2O
1 mol
= 19.4g Mg(OH)2
Unit 9: Solutions, Acids, & Bases
1.
Differentiate between strong & weak acids/bases & perform calculations related to pH and ion
concentrations.
A. If [H3O+]= 2.0x 10-5 M
 Find [OH] = [H3O+] [OH] = 1 x 10-14
[2.0x 10-5 M ] [OH] = 1 x 10-14
= 5 x 10-10
= -Log [H3O+]
 Find pH
= -Log[2.0x 10-5 M]
Ph= 4.69
B. If pH= 2.70
 Find [OH]
5.0 x 10-2
(I’d find the [H3O+] below first and then do the OH using: [H3O+] [OH] = 1 x 10-14
 Find [H3O+]=
2.70 = = -Log [H3O+]
Antilog of -2.7= [H3O+]
2.0 x 10-3