Chemistry 101
Homework – The Last Key
Instructions: This homework will be due on July 19th at the start of class. Answer each
of the following questions to the best of your ability. Make sure to use the proper number
of significant figures wherever appropriate. Note this homework is cumulative, like the
final exam will be. Please be sure to give it extra time.
1. Give the electron configuration for each of the following elements / ions:
a.
b.
c.
d.
e.
Vanadium = [Ar]4s23d3
Phosphorous = [Ne]3s23p3
Strontium = [Kr]5s2
Sb+ = [Kr]5s24d105p2
O2- = 1s22s22p6
2. Plot the trend in ionization energy for elements 3 – 18. Make sure to show any
deviations from a general trend. Give a brief written explanation for any
deviations.
Energy (kJ/mole)
Ionization Energy
2500
2000
1500
1000
500
0
0
5
10
15
20
2nd and 3rd Period Elements
The trend is generally increasing. There are a couple of deviations at B, O, Al,
and S, where removing an electron leaves an ion with either a half-full or
completely full orbital or orbital set. Removal of an electron from O and S leaves
a set of half-full ‘p’ orbitals. Removal of an electron from B or Al leaves a
completely full ‘s’ orbital.
3. Consider the following two molecules
C2H5OH and CH3CH2COCH3
If 1.5 moles of the each is used in its own combustion reaction, which one will
yield the least water? How much water will that be in grams?
C2 H 5OH  3O2  2CO2  3H 2O
2CH 3CH 2COCH 3  11O2  8CO2  8H 2O
The C2H5OH would yield the least amount of water.
1.5mole C2 H 5OH

3mole H 2O
18.0g

 81g
1mole C2 H 5OH 1mole H 2O
4. When benzene(C6H6) reacts with bromine(Br2), bromobenzene is obtained:
C6H6 + Br2  C6H5Br + HBr
a. What is the theoretical yield of bromobenzene (C6H5Br) when 30.0g of
benzene reacts with 65.0g of bromine?
30.0 g 1mole C6 H 6 1mole C6 H 5 Br
156.9g



 60.4 g
78 g
1 mole C6 H 6 1 mole C6 H 5 Br
65.0 g 1mole HBr 1mole C6 H 5 Br
156.9g



 126 g
80.9 g
1 mole HBr 1 mole C6 H 5 Br
The theoretical yield is 65.4g of bromobenzene.
b. If the actual yield of bromobenzene was 56.7g, what was the percentage
yield?
56.7 g
100  93.9%
60.4 g
c. Using the amount in part (a) how much of which reactant will be left over
in excess?
30.0 g 1mole C6 H 6 1mole Br2
159.8g



 61.5 g
78 g
1 mole C6 H 6 1 mole Br2
65.0 g  61.5 g  3.5 g
5. Answer each of the following:
a. Give one element, and two ions that have the electron configuration:
1s22s22p63s23p6.
Element = Ar
Ions = Cl- and K+
b. How many protons, neutrons, electrons, and valence electrons are in 28Si.
Protons = 14
Neutrons = 14
Electrons = 14
c. Give the Lewis structure and the molecular geometry for the following
molecules CCl2H2 and ICl4-. (The central atom is underlined)
H
Tetrahedral
C
Cl
H
Cl
Cl
Cl
Square Planar
I
Cl
Cl
6. Determine the empirical and molecular formulas for each of the following.
a. Ibuprofen, a headache remedy that contains 74.22%C, 9.28%H, and 16.50%O
by mass; molecular mass is about 194 g/mol.
74.22 gC 1moleC

 6.19moleC
12.0 g
9.28 gH 1moleH

 9.28moleH
1.0 g
16.50 gO 1moleO

 1.03moleO
16.0 g
C 6.19

6
O 1.03
H 9.28

9
O 1.03
O 1.03

1
O 1.03
C6H9O = 97 = Empirical
C12H18O2 = 194 = Molecular
b. Epinephrine (adrenaline) a hormone secreted at times of danger and stress contains
59.0%C, 7.1%H, 26.2%O, and 7.7%N by mass; molecular mass about 183 g/mol.
59.0 gC 1moleC

 4.916moleC
12.0 g
7.1gH 1moleH

 7.1moleH
1.0 g
26.2 gO 1moleO

 1.64moleO
16.0 g
7.7 gN 1moleN

 0.55moleN
14.0 g
C 4.916

9
O 0.55
H
7.1

 13
O 0.55
O 1.64

3
O 0.55
N 0.55

1
N 0.55
C9 H13 NO3  183  empirical  molecular