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Chem 103 MWF Exam 1 Date last revised 10/11/2022 Page 1 of 9 CHEMISTRY 103 MWF Name _________________________________ EXAM 1 October 5, 2022 Sec _________ TA ______________________ INSTRUCTIONS: 1. Do not separate the pages of the exam. 2. PRINT your name, section number, and TA’s name at the top of this page
NOW
. 3. The scantron form is machine-scored and must be filled out using a #2 pencil.
PRINT your last name and first name on the scantron form as it appears in Canvas, then fill in the circles corresponding to the letters in your name. Write your student ID number
under “Identification Number”, and fill in the circles. Write your lecture number un
der “Special Codes”. 1 = Zelewski 8:50 AM 2 = Zelewski 11:00 AM 3 = Zhou 1:20 PM 4 = Brunold 3:30 PM 4. Write your answers to the multiple choice questions in this test booklet so you have a record of what your answer is, then transfer your answers to the scantron form.
Only the scantron form will be graded. 5. Answer the remaining short answer questions directly in the space provided in this exam. Report all numerical answers with the correct number of significant figures and units. Show all relevant work to receive full credit.
6. Complete the following exam by yourself. You will be given 75 minutes. 7. When you are finished, turn in your notecard, reference sheet, scantron form and exam. 8. Read and sign the Honor Code statement. Honor Code: I affirm that I will not give or receive any unauthorized help on this exam, and that all work will be my own. Signature question 13 _________/10 question 14 _________/14 question 15 _________/8 question 16 _________/10 question 17 _________/10 _________/52 Short Answer Total KEY
Chem 103 MWF Exam 1 Date last revised 10/11/2022 Page 2 of 9 Enter your answers to the following questions on the scantron form. Each question is worth 4 points. There is ONE
best response to each question. 1. An ion has a charge of +2, and it has a total of 42 electrons. Which ion is it? A. Ca
2+
B. Mo
2+
C. Zr
2+
D. Pd
2+
E. Ru
2+
2. Molybdenum (Mo, element 42) is most likely to have similar chemical properties as …
A. Vanadium B. Chromium C. Manganese D. Niobium (Nb) E. Osmium (Os) 3. The mass spectrum of an unknown element (Q) is shown below. What is the average atomic mass of Q? A. 182 amu B. 183 amu C. 184 amu D. 185 amu E. 186 amu 4. Which of the following compounds is NOT an ionic compound? A. ZnO B. SrO C. TiO
2
D. SO
2
E. Ag
2
O 180 181 182 183 184 185 186 187 188 189 190
Relative Abundance
m/Z
Q
+
27.4%
22.5%
50.1%
Chem 103 MWF Exam 1 Date last revised 10/11/2022 Page 3 of 9 5. Which of the following statements is correct? A. An atom’s mass is primarily due to the mass of its electron cloud.
B. An atom’s size is primarily due to the size of its electron cloud. C. A proton and an electron have opposite charges but similar masses. D. A proton and a neutron have opposite charges but similar masses. E. Isotopes have the same number of protons but differing number of electrons. 6. Which of the following: H
2
SO
4
, Co(OH)
2
, LiCH
3
COO, CH
3
COOH are strong electrolytes when added to water? A. Only H
2
SO
4
and Co(OH)
2
B. Only LiCH
3
COO and CH
3
COOH C. Only Co(OH)
2
, LiCH
3
COO and CH
3
COOH D. Only H
2
SO
4
and LiCH
3
COO E. H
2
SO
4
, Co(OH)
2
, LiCH
3
COO and CH
3
COOH 7. When the following are combined, which will NOT result in a chemical change? (i) Zn(s) and CoCl
2
(aq) (ii) ZnSO
3
(s) and HCl(aq) (iii) CoCl
2
(aq) and Zn(NO
3
)
2
(aq) A. Only (i) B. Only (ii) C. Only (iii) D. Only (i) and (ii) E. Only (i) and (iii) 8. The net ionic
equation for the reaction between solid magnesium hydroxide and hydroiodic acid is A. Mg(OH)
2
(s) + 2 HI(aq) →
MgI
2
(aq) + O
2
(g) + 2 H
2
(g) B. MgOH(s) + HI(aq) →
MgI(aq) + H
2
O(
ℓ
) C. OH
–
(aq) + H
+
(aq) →
H
2
O(
ℓ
) D. MgOH(s) + H
+
(aq) →
Mg
+
(aq) + H
2
O(
ℓ
) E. Mg(OH)
2
(s) + 2 H
+
(aq) →
Mg
2+
(aq) + 2 H
2
O(
𝓵
) 9. Ammonia, NH
3
, is a A. weak electrolyte. It partially ionizes to form NH
4
+
and OH
–
ions in water. B. strong electrolyte. It ionizes completely to form N
3
–
and H
+
ions in water. C. weak electrolyte. It partially ionizes to form N
3
–
and H
+
and ions in water. D. nonelectrolyte. It is a molecular compound that does not form ions when added to water. E. strong electrolyte. It ionizes completely to form NH
4
+
and OH
–
ions in water.
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Chem 103 MWF Exam 1 Date last revised 10/11/2022 Page 4 of 9 10. Which of the statements about the balanced chemical reaction shown below is/are correct? 8 HNO
3
(aq) + 6 KI(aq) → 6 KNO
3
+ 3 I
2
(s) + 2 NO(g) + 4 H
2
O(
ℓ
) (i) NO
3
–
is the oxidizing agent. (ii) The oxidation number of nitrogen changes from +6 to +2. (iii) I
–
is reduced. A. Only (i) B. Only (i) and (ii) C. Only (ii) D. Only (i) and (iii) E. (i), (ii) and (iii) 11. The following diagram represents the collection of CO
2
and H
2
O molecules formed by the complete combustion of a hydrocarbon. What is the empirical
formula of the hydrocarbon? A. C
4
H
8
B. CH
4
C. C
4
H
16
D. CH
2
12. The combustion of an unknown hydrocarbon produces 0.234 mol H
2
O and 0.187 mol CO
2
. Which of the following could be the molecular
formula of this unknown compound? A. C
5
H
10
B. C
4
H
6
C. C
4
H
10
D. C
2
H
6
E. C
6
H
14
Make sure to enter your multiple choice answers on the scantron form!
Chem 103 MWF Exam 1 Date last revised 10/11/2022 Page 5 of 9 Answer the following questions directly in the space provided in this exam. Show all relevant work to receive full credit. Circle your final answer. 13. [10 points total] Maple syrup is made by boiling the sap of sugar maple trees at ~100 °C. Several sap solutions of varying sugar concentrations were analyzed for their density. Use the graph below to answer the following questions. A. [2 pts] A jar of maple syrup you purchase at the store has a sugar concentration of 88 g/100 mL. What is the density of this maple syrup? __
1.33 g/mL
__ Correct answer (2 pts); –
1 pt for missing units B. [3 pts] How many grams of sugar is in 2.0 tablespoons of this maple syrup? Show your work.
(1 tablespoon = 14.8 mL) (2.0 tbsp)(14.8 mL/tbsp)(0.88 g/mL) = 26 g Correct setup (2pts) Correct final answer with units (1pt) (0 pt if no work is shown) –
1 pt for >4 significant figures for final answer Some information that may help you answer the following two questions are listed in the table below. Boiling point (°C) Density (g/mL) Water (H
2
O) 100 0.96 Sugar (C
12
H
22
O
11
) >200 1.58 C. [2 pts] As the sap of the sugar maple tree is cooked at ~100 °C over time, you notice steam rising out of the solution. What is the major constituent of the steam? Circle your answer. H
2
O molecules
C
12
H
22
O
11
molecules H
2
and O
2
molecules CO
2
molecules No partial credit D. [3 pts] As the sap of the sugar maple tree is cooked at ~100 °C over time, you notice that the density of the solution increases. Explain why the density would increase. Sugar has higher density than water. (1pt)
As water is boiled away, the solution becomes more concentrated in sugar. (1pt)
Increasing concentration of the higher density component increases the overall density of the solution. (1pt)
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
30
40
50
60
70
80
90
100
110
120
Solution Density (g/mL)
Grams of Sugar in 100 mL Solution
Density of Maple Syrup at 100 °
C
_____ /10 pts ____/2 pts
____/3 pts
____/2 pts
____/3 pts
Chem 103 MWF Exam 1 Date last revised 10/11/2022 Page 6 of 9 14. [14 points total] Reaction Types and Chemical Logic A. [6 pts] Fill in the table by writing what you would observe
when the two solutions are mixed together. Use the following abbreviations: NOC –
No Observed
Change ppt –
precipitate is formed gas –
gas is formed Solutions H
2
SO
4
Cu(NO
3
)
2
K
3
PO
4
Na
2
S H
2
SO
4
Cu(NO
3
)
2
NOC K
3
PO
4
NOC ppt Na
2
S
gas ppt NOC Correct observation (1 pt each) B. [4 pts] Write the chemical formulas of ALL precipitates that formed. Cu
3
(PO
4
)
2
CuS Correct formula for Cu
3
(PO
4
)
2
(2 pts); Correct formula for CuS (2 pts) –
1 pt for each additional incorrect answer C. [4 pts] Write the balanced overall chemical equation for ONE reaction that produced a gas. Include states: g, ℓ,
s, aq in your equation. Na
2
S(aq) + H
2
SO
4
(aq) → H
2
S(g) + Na
2
SO
4
(aq) Na
2
S and H
2
SO
4
in reactants (1 pt) Correct formula for Na
2
SO
4
(1 pt) and H
2
S (1 pt) in products Reaction is balanced and all correct states (1 pt) –
1 pt for each additional incorrect species in the equation _____ /14 pts ____/6 pts
____/4 pts
____/4 pts
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Chem 103 MWF Exam 1 Date last revised 10/11/2022 Page 7 of 9 15. [8 points total] Upon heating, a 3.060 g sample of an unknown metal hydroxide, M(OH)
2
(s), decomposes into the corresponding metal oxide, MO(s), and 0.6134 g of H
2
O(g) (18.015 g/mol). A [2 pts] Write the balanced equation for this reaction. M(OH)
2
(s) →
MO(s) + H
2
O(g)
Correct formula for reactant and products (1 pt) Reaction is balanced and all correct states (1 pt) B. [6 pts] Calculate the atomic weight of M and determine the identity of M. Show your work.
moles of M(OH)
2
: 0.6134 g H
2
O × (1 mol/18.015 g) × (1 mol M(OH)
2
/1 mol H
2
O) = 0.03405 mol M(OH)
2 →
molar mass of M(OH)
2
is (3.060 g M(OH)
2
/0.03405 mol M(OH)
2
) = 89.87 g/mol →
average atomic mass of M is 89.87 g/mol –
2 × (15.999 g/mol +1.008 g/mol) = 55.86 g/mol Correct moles of M(OH)
2
(2 pt) Correct molar mass of M(OH)
2
(2 pt) Correct average atomic mass of M (1 pt) Correct identity of M (1 pt) –
1 pt for <3 or >5 figures for atomic weight of M (0 pt if only answer is written and no work is shown) Alternative approach: moles of MO: 0.6134 g H
2
O × (1 mol/18.015 g) × (1 mol MO/1 mol H
2
O) = 0.03405 mol MO
→
molar mass of MO is (3.060 g M(OH)
2
- 0.6134 g H
2
O)/0.03405 mol MO = 71.85 g/mol →
average atomic mass of M is 71.85 g/mol –
15.999 g/mol = 55.86 g/mol Correct moles of MO (2 pt) Correct molar mass of MO (2 pt) Correct average atomic mass of M (1 pt) Correct identity of M (1 pt) –
1 pt for <3 or >5 figures for atomic weight of M (0 pt if only answer is written and no work is shown) Atomic weight of M: __
55.86
__ [no units] Identity of M: __
Fe
__ _____ /8 pts ____/2 pts
____/6 pts
Chem 103 MWF Exam 1 Date last revised 10/11/2022 Page 8 of 9 16. [10 points total] Consider the reaction of iron metal, Fe(s), with O
2
(g) to form Fe
2
O
3
(s). A. [2 pts] Balance the following equation for this reaction. 4 Fe(s) + 3 O
2
(g) →
2 Fe
2
O
3
(s) [or 2 Fe(s) + 3
/
2
O
2
(g) →
Fe
2
O
3
(s)] All correct coefficients (2 pt); no partial credit B. [4 pts] What is the theoretical yield (in grams) of Fe
2
O
3
(s) if you allow 10.0 mol of Fe(s) to react with 10.0 mol of O
2
(g)? Show your work.
10.0 mol Fe × (2 mol Fe
2
O
3
/4 mol Fe) = 5.00 mol Fe
2
O
3
→
limiting reagent
10.0 mol O
2
× (2 mol Fe
2
O
3
/3 mol O
2
) = 6.67 mol Fe
2
O
3
5.00 mol Fe
2
O
3
× (159.69 g Fe
2
O
3
/1 mol Fe
2
O
3
) = 798 g Alternative approach: Need 10.0 mol Fe × (3 mol O
2
/4 mol Fe) = 7.50 mol O
2
for complete reaction →
Fe =
limiting reagent
Correct identification of limiting reagent (2 pt) Correct mass of product Fe
2
O
3
(2 pt) –
1 pt for <2 or >4 figures for mass of Fe
2
O
3 (0 pt if only answer is written and no work is shown) C. [4 pts] How many moles of the excess reagent remain after the reaction has come to completion? Show your work.
used
: 5.00 mol Fe
2
O
3
× (3 mol O
2
/2 mol Fe
2
O
3
) = 7.50 mol O
2
remaining
: 10.0 mol O
2
–
7.50 mol O
2
= 2.5 mol O
2 Correct moles of O
2
used (2 pt) Correct moles of O
2
remaining (2 pt) –
1 pt for >3 figures for mol O
2
remaining (0 pt if only answer is written and no work is shown)
Alternative approach: remaining
: (6.67 –
5.00) mol Fe
2
O
3
× (3 mol O
2
/2 mol Fe
2
O
3
) 2.5 mol O
2 Correct moles of additional Fe
2
O
3
that could have been produced with O
2
(2 pt) Correct moles of O
2
remaining (2 pt) –
1 pt for >3 figures for mol O
2
remaining (0 pt if only answer is written and no work is shown) There is one more question on the back page. _____ /10 pts ____/2 pts
____/4 pts
____/4 pts
Chem 103 MWF Exam 1 Date last revised 10/11/2022 Page 9 of 9 17. [10 points total] When 0.4773 g of a compound that only contains carbon
, hydrogen
, and nitrogen
are burned completely in excess O
2
(g), 1.234 g of CO
2
(g) (44.010 g/mol) and 0.3788 g of H
2
O(g) (18.015 g/mol) are produced along with an unknown amount of N
2
(g). A. [7 pts] What is the empirical
formula of this compound? Show your work.
[Unbalanced equation: C
x
H
y
N
z
+ O
2
(g) →
CO
2
(g) + H
2
O(g) + N
2
(g)] 1.234 g CO
2
× (1 mol/44.01 g) × (1 mol C/1 mol CO
2
) = 0.02804 mol C 0.02804 mol C × (12.011 g/1 mol) = 0.3368 g C
0.3788 g H
2
O × (1 mol/18.015 g) × (2 mol H/1 mol H
2
O) = 0.04205 mol H 0.04205 mol H × (1.008 g/1 mol) = 0.04239 g H →
C
x
H
y
N
z
additionally contains 0.4773 g –
(0.3368 g + 0.04239 g) N = 0.09811 g N 0.09880 g N × (1 mol/14.007 g) = 0.007054 mol N →
C : H : N = 4 : 6 : 1 Correct mol C (1 pt) Correct mass of C (1 pt) Correct mol H (1 pt) Correct mass of H (1 pt); if mol H is off by factor of 2 then give credit for mass of H if conversion is correct Correct mass of N (1 pt) Correct mol of N (1 pt) Correct C:H:N ratio/ empirical formula (1 pt) (0 pt if only answer is written and no work is shown) Empirical formula: __
C
4
H
6
N
__
B. [3 pts] Draw the hard
mass spectrum for this compound. Write the relative intensities of the signals in the figure
. Assume that all elements have a single isotope, 12
C, 1
H, and 14
N. If you are unable to determine the empirical formula, then use C
7
H
10
N
3
. Correct masses and relative signal intensities (3 pt); no partial credit Answer must be consistent with empirical formula in part A. or C
7
H
10
N
3
if the student was unable to complete part A. 0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Relative Abundance
mass (amu)
_____ /10 pts ____/7 pts
____/3 pts
6
4
1
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Part A
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Combustion
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