Download Chem 107 - Hughbanks Exam 1

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Chem 107 - Hughbanks
Exam 1 - Spring 2011
Name (Print)
SOLUTIONS
UIN #
Section 503
Exam 1, Version # B
Scores
1
2
3
4
5
6
7
8
9
10
11
Tot.
/7
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/7
/12
/10
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/12
/100
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 the to use during the exam – you don’t need to turn it in with the rest
of the exam.
The exam contains 11 problems, with 6 numbered pages. You have the full 75 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:
(1) (7 points) Which of the following is not a strong electrolyte?
(a) KBr
(b) CH3NH2
(c) H2SO4
(d) Ba(OH)2
(e) NaOH
Ans. 1____ b_____
(2)
(7 points) Consider the position of the elements rubidium (Rb), chlorine (Cl), and barium (Ba) in
the periodic table. Which one of the following full sets of compounds is the most likely to be easily
available (or exist at all)?
(a) Ba2Cl, RbCl, RbCl2
(d) Cl2, Ba2Cl, RbCl
(b) Cl2, BaCl, Rb2Cl
(e) Cl2, BaCl2, RbCl
(c) BaCl, RbCl, RbCl2
Ans. 2_____ e____
(3) (7 points) Calcium carbide, CaC2, is an important preliminary chemical for industries producing
synthetic fabrics and plastics. CaC2 may be produced by heating calcium oxide with coke (a source
of carbon). CaC2 is be reacted with nitrogen, N2 at 1000 – 1200 ˚C to produce another important
chemical precursor called calcium cyanamide, CaCN2:
CaC2(s) + N2(g) → CaCN2(s) + C(s)
What is the amount of CaCN2 that can be produced from the reaction of excess nitrogen gas and
12.8 g of calcium carbide? Assume that the percent yield in this process is 85%.
(a) 18.8 g
Ans. 3_____ b____
(b) 13.6 g
(c) 14.2 g
(d) 16.0 g
(e) 12.1 g
(4) (7 points) Which of the following choices correctly completes this statement: A cation with a
charge of +2 that has an atomic number of 12 and a mass number of 25,
(a) has the same number of electrons as a Si atom.
(b) has 10 electrons.
(c) has the same number of protons as an Ne atom.
(d) has 12 neutrons.
(e) is a Si2+ ion.
Ans. 4___ b ___
(5) (7 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(SO4)2, ZnCO3, CaCN
(b) NH4PO4, NH4Cl, KNO3
(d) NH4SO4, K2CO3, Ca(OH)2
(c) Zn3(PO4)2, MgSO4, KNO3
(e) (NH4)2SO4, K2CO3, Ca3(PO4)2
Ans. 5___ c, e___
(6)
(7 points) Which of the following quantities are conserved in all chemical reactions? There may be
more than one correct choice; enter ALL the correct letters, in alphabetical order.
(a) Number of atoms of each element
(b) Volume
(c) Total number of atoms
(d) Number of molecules
(e) Mass
Ans. 6__ a, c, e_
Questions 7-11 – show all work (58 pts)
(7)
(12 points) The combustion of heptane with oxygen is a well-known process.
(a) (5 points) Write a balanced equation for the reaction of heptane, C7H16, with oxygen, O2,
forming the products carbon dioxide, CO2, and water, H2O.
C7H16 + 11 O2 → 7 CO2 + 8 H2O
(b) (7 points) What mass of carbon dioxide will be produced by the complete reaction of 8.93
grams of heptane?
(8.93 g C7H16)/(100.198 g/mol) = 0.08913 mol (× 7 mol CO2/mol C7H16)
⇒ 0.62386 mol CO2 produced
(0.62386 mol CO2)(44.01 g/mol) = 27.45 g CO2 produced
(8)
(10 points) Boron has two naturally occurring isotopes, 10B and 11B. Use the data in the table
given to calculate the missing isotopic molar mass of 10B. Give an answer with 4-digit precision.
The molar mass of naturally occurring boron is 10.811 g/mol.
Isotope
10
11
B
B
Isotopic Molar Mass (g/mol)
11.00931
Abundance(%)
19.78
80.22
Let x = isotopic molar mass of 10B (in g/mol)
0.1978x + 0.8022(11.00931) = 10.811 ⇒ x = + 0.8022(11.00931) = (1/0.1978)(10.811 – 8.83167)
x = 10.01 g/mol
(9)
(12 points) The molecule shown below is one of many used in devices ranging from wristwatches
to displays for laptop computers. These molecules form highly ordered “liquid crystals”. Electric
fields are used to control the extent of ordering of the molecules, thus allowing segments of the
liquid to be switched between transparent and opaque forms.
(a) (7 points) Give the chemical
formula for this liquid crystal
molecule.
N
O
H
H
H3C
H
C
O
C
C
C
C
H
C
H
H
C
N
C
H
H
C
C
C
C
C
CH2
H2C
H
CH2
CH3
C18H21NO
(b) (5 points) How many carbon atoms are there in 1.0 g of this compound?
Total molar mass = 267.358 g ⇒ (1.0 g)/(267.358 g/mol) = 3.74 × 10–3 mol in 1 g
(18 C atoms/molecule)(3.74 × 10–3 mol)(6.022 × 1023/molecules/mol) = 4.054 × 1022 C atoms
(10) (12 points) Cryolite, Na3AlF6, is used in the commercial production of aluminum metal from
bauxite ore (Al2O3). Cryolite itself is produced by the following (unbalanced) reaction:
.
6
NaOH +
1
Al2O3 +
12
HF →
2
Na3AlF6 +
9
H2 O
(a) (5 points) Using the blanks provided, balance this equation with the smallest possible integer
coefficients (Hint: One of the reactants has a coefficient of 1.)
(b) (7 points) A mixture containing 800.0 kg of NaOH, 300.0 kg of Al2O3, and 600.0 kg of HF is
heated to 950 ˚C and reacts to completion. What is the maximum mass of Na3AlF6 formed?
ratio
(800.0 kg NaOH)/(0.040 kg/mol) = 20000 mol NaOH
6.797
(300.0 kg Al2O3)/(0.10196 kg/mol) = 2942.3 mol Al2O3
1
(600.0 kg HF)/(0.02001 kg/mol) = 29985 mol HF
10.19
Compared to the quantity of Al2O3, there is an excess of NaOH and there is too little HF.
Therefore, HF is the limiting reagent ⇒ only (2/12) × (29985 mol) Na3AlF6 can be formed.
max. # moles of Na3AlF6 is 4997.5 ⇒ (4997.5 mol) × (0.20995 kg/mol) = 1049 kg
(11) (12 points) One compound found in decaying flesh goes by the descriptive name cadaverene.
Analysis of cadaverene shows it to contain 58.77% carbon, 13.81% hydrogen, and 27.42%
nitrogen by mass. Reactions with strong acids show that cadaverene molecules each contain two
basic –NH2 groups (it is a diamine) that can react with HCl solutions and pick up protons (H+) that
convert these into two –NH3+ groups. Determine the molecular formula of cadaverene, assuming
the –NH2 groups contain all the nitrogen.
In a 100.0 g sample of cadaverene, we have 58.77g C, 13.81g H, and 27.42g N
ratio
2.5
(58.77g C)/(12.01 g/mol C) = 4.893 mol C
(13.81g H)/(1.008 g/mol H) = 13.70 mol H
⇒
7.0
(27.42g N)/(14.01 g/mol N) = 1.957 mol N
1.0
ratio
5
⇒
14
2
Other data shows there are two nitrogen atoms, so formula is C5H14N2
(The actual structure is H2N
NH2
.)