Chem 107 - Hughbanks
Exam 1 - Spring 2014
Name (Print)
Key
UIN #
Section 504
Exam 1, Version # B
On the last page of this exam, you’ve been given a periodic table and some physical constants. You’ll
probably want to tear that page off to use during the exam – you don’t need to turn it in with the rest of
the exam.
The exam contains 9 problems, with 4 numbered pages. You have the full 50 minutes to complete the
exam. Please show ALL your work as clearly as possible – this will help us award you partial credit
if appropriate. Even correct answers without supporting work may not receive credit. You may use an
approved calculator for the exam, one without extensive programmable capabilities or the ability to store
alphanumeric information. Print your name above, provide your UIN number, and sign the honor code
statement below.
On my honor as an Aggie, I will neither give nor receive unauthorized assistance on this exam.
SIGNATURE:
Name (Print)
Version A
(1) (12 points, 3 pts each) For each of the following, enter the answer in the line
provided. (No difficult calculations are necessary for these, so think first!)
(a) How many molecules are present in two moles of ozone (O3)?
Ans. (a) 1.2044 × 1024
(b) How many moles of fluorine atoms are present in four moles of CF4?
Ans. (b)
16 _
(c) How many moles of ammonium ions (NH4+) are present 4 L of 1.0 M
aqueous solution of (NH4)2SO4?
Ans. (c)
8 _
(d) What is the approximate mass (to the nearest gram) of 3 × 1023 F atoms?
Ans. (d)
9 (10 is acceptable) _
(For grading)
Scores
1
2
3
4
5
6
7
8
9
Tot.
/12
/12
/10
/8
/10
/10
/15
/12
/12
/100
(2) (12 pts, 4 pts each) Imagine that we had a balance that could compare the masses
of small numbers of atoms of different elements. The figures below show the
(hypothetical) results of 2 such experiments. The letters beside each balance pan identify the
elements being weighed. (These are NOT actual elements! Your answer should be based ONLY on
these pictures.)
D
C
B
A
Please circle your choice for each question below:
(a) Of elements A and B, which has more atoms per gram?
element A
element B
neither, they’re equal
(b) Of elements C and D, which has the greater molar mass?
element C
element D
neither, they’re equal
(c) Of elements C and D, which has more atoms per mole?
element C
element D
neither, they’re equal
1
Name (Print)
Version A
(3) (10 points) In the first row of blanks beneath each of the following compounds, classify them as
follows: weak acid (wa), strong acid (sa), weak base (wb), strong base (sb), or as a salt (s) that is
neither acid or basic. In the second row of blanks, classify the same compounds as a weak
electrolyte (we), strong electrolyte (se), or nonelectrolyte (ne).
RbOH
CH3COOH
HClO4
KCl
NH3
sb
wa
sa
s
wb
se
we
se
se
we
(4) (8 points) The following full sets of potential compounds are made up of common cations and
anions. Which full sets are most likely to actually exist? (All 3 compounds in a set must be sensible
to qualify.) There may be more than one correct choice; enter ALL the correct letters, in
alphabetical order.
(a) (NH4)2PO4, K2CO3, Zn3(PO4)2
(d) Ca3(PO4)2, Ag2SO4, NH4NO3
Ans. 4
(b) (NH4)2PO4, NaCO3, KNO3
(c) K2SO4, KClO4, Ca(OH)2
(e) Rb(SO4)2, CaCO3, ZnClO4
c, d _
(5) (10 points) Calcium hydroxide can be dissolved in phosphoric acid to form calcium phosphate and
water.
(i) (5 pts) Balance the equation below by filling in the blanks with coefficients:
_ 3 _ Ca(OH)2 + __ 2 __ H3PO4 → _1_ Ca3(PO4)2 + _6_ H2O
(ii) (5 pts) What volume (in mL) of 1.44 M H3PO4 is needed to react with 12.0 g of Ca(OH)2?
(Write the letter of the correct answer below.)
(a) 0.169 mL
(b) 0.075 mL
(c) 112.5 mL
(d) 168.8 mL
(e) 75.0 mL
Ans. 5 (ii) __ e__
2
Name (Print)
Version A
Questions 6 - 9 – show all work (48 pts)
(6)
(10 pts) Adipic acid is used in the production of nylon, so it is manufactured in large quantities.
One method for the industrial synthesis of adipic acid is the reaction of cyclohexane with oxygen.
O
2
+ 5 O2
cyclohexane
2
OH
HO
adipic acid
+ 2 H 2O
O
(a) Give the molecular formulas for
(2 pts) cyclohexane: ____C6H12____
(3 pts) adipic acid: ___ C6H10O4___
(b) (5 pts) Balance the skeleton equation above by putting in the smallest possible whole number
coefficients in the four boxes provided.
(7) (14 points) Shown below is the molecular structure of nepetalactone, which is a cat attractant first
isolated by steam distillation from catnip and also present in the wood of tartarian honeysuckle,
shavings of which are often used in cat toys.
nepetalactone
(a) (5 pts) Give the molecular formula for nepetalactone.
C10H14O2
(b) (4 pts) Assuming your answer is correct; give the molar mass of nepetalactone.
166.21 g/mol
(c) (5 pts) How many carbon atoms are there in 3.0 g of nepetalactone?
(3 g/166.21 mol/g)(10 C atoms/mol) = 1.086 × 1023 C atoms/g
3
Name (Print)
(8)
Version A
(12 points) A mixture of methane (CH4) and propane (C3H8) has a total mass of 59.68 g. When a
mixture is burned completely in excess oxygen, the CO2 and H2O products have combined mass of
285.94 g. Calculate the mass of methane in the original mixture.
The key to the problem is to realize that it is the ratio of product mass and reactant mass that matters. (This was a study
problem.)
Let x = mole fraction of CH4 , then (1 – x) = mole fraction of C3H8
x CH4 + excess O2 → x CO2 + 2x H2O
(1 – x) C3H8 + excess O2 → 3(1 – x) CO2 + 4(1 – x) H2O
––––––––––––––––––––––––––––––––––––––––––––––––
x CH4 + (1 – x) C3H8 + excess O2 → (3 – 2x) CO2 + (4 – 2x) H2O
285.94 g
59.68 g
= 4.791 =
total product mass
methane + propane mass
=
( 44.01 g/mol)( 3 − 2x ) + (18.02 g/mol)( 4 − 2x )
(16.04 g/mol) x + ( 44.09 g/mol)(1 − x )
solving for x, x = 0.690.
Now you need to multiply the mass fraction of methane by the total original mass,
16.04 g/mol 0.69 mol
original methane mass =
59.68 g = 26.70 g
16.04 g/mol 0.69 mol + 44.09 g/mol 0.31 mol
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(12 points) There are many isotopes of uranium, but only two are present in greater than trace
quantities on earth: 235U and 238U. The data in the table below given can be used to calculate the
molar mass of uranium, which is 238.0289 g/mol. The molar mass of uranium on earth hasn’t
always had this value, however, because both of these isotopes have been very slowly decaying
since the earth was formed about 4.5 billion years ago. The % abundances of the two isotopes
have therefore changed since the earth was formed. Measurements of the rates of decay of the two
isotopes allows us to calculate the how much of each isotope has decayed: 50.25% of the 238U
and 98.81% of the 235U that were present 4.5 billion years have decayed.
Current
Abundance (%)
235U
235.043993
0.727
238U
238.050788
99.273
Compute the molar mass of uranium on earth 4.5 billion years ago.
Isotope
Isotopic Molar Mass (g/mol)
If today you have 0.727 mol of
(
235
U , 4.5 billion years ago you would have had
100%
⎛
⎞
0.727 mol ⎜
= 61.09 mol
⎝ 100% − 98.81% ⎟⎠
)
If today you have 99.273 mol of
238
U , 4.5 billion years ago you would have had
100%
⎛
⎞
99.273 mol ⎜
= 199.54 mol
⎝ 100% − 50.25% ⎟⎠
The abundances 4.5 billion years ago were
61.09
199.54
⎛
⎞
⎛
⎞
235
238
U: ⎜
= 0.2344 (23.44%)
U: ⎜
= 0.7656 (76.56%)
⎟
⎝ 61.09 + 199.54 ⎠
⎝ 61.09 + 199.54 ⎟⎠
Molar mass 4.5 billion years ago = 235.043993 0.2344 + 238.050788 0.7656 = 237.35 g/mol
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