Memorial University of Newfoundland
Midterm Exam 1
St. John’s, Newfoundland and Labrador
Chemistry 1010 Intersession 2015
May 29th, 2015
Name:_________________________
Time: 50 minutes
MUN #:_________________________
READ THE FOLLOWING CAREFULLY!
1.
This exam has 7 pages with 2 sections. SECTION A is comprised of multiple
choice questions and shorter answer questions. SECTION B is comprised of
longer-answer questions. Ensure that this examination paper is complete, i.e. that
all pages are present.
2.
Failure to submit this paper in its entirety at the end of the examination may result
in disqualification.
3.
A Periodic Table and physical constants are provided. These are on page 7 of the
examination and may be detached for use during the examination.
4.
Answer each question in the space provided. Should you require more space, use
the back of the page and indicate clearly where this has been done.
5.
When answering questions show all relevant equations, all calculations, and
justify all simplifying assumptions.
6.
Correct units should be maintained in all steps of a calculation, though you may
show cancellation of units in early steps to eliminate them in later steps, if
possible.
7.
Numerical answers should be reported with the correct significant digits.
Do not write in the enclosed area below.
Question
Value
A1
10
A2
6
A3
4
B1
4
B2
8
Total
32
Mark
Page 1 of 7
Chemistry 1010 Midterm Exam 1
Section A
Shorter Answer Questions
(20 marks)
Complete ALL of the following questions in the space provided. Show all work as part
marks will be given. Unreadable work will NOT be marked, so be legible.
You should allow at least 25 minutes for this section.
A1.
Ten multiple choice questions (1 mark each)
i)
Dalton's atomic theory states:
a)
b)
c)
d)
that all elements have several isotopes.
that matter is composed of small indestructible particles.
that the properties of matter are determined by the properties of atoms.
that an atom is predominantly empty space.
ANSWER: ____B____
ii)
a)
b)
c)
d)
Which of the following represents a physical property?
Sodium metal is extremely reactive with chlorine gas.
Mercury is a silvery liquid at room temperature.
Aluminum has a tendency to "rust."
Argon has an unreactive nature.
ANSWER: ____B____
iii)
Which of the following contains BOTH ionic and covalent bonds?
a) CaI2
b) COS
c) CaSO4
d) SF6
ANSWER: ____C____
iv)
What species is represented by the following information?
protons = 47 neutrons = 62
electrons = 46
a) Ag+
b) Sm
c) Pd
d) Ag
ANSWER: ____A____
v)
a)
b)
c)
d)
Groups, or families, on the periodic table are
Elements which all occur naturally in the same state.
Horizontal rows of elements with increasing atomic numbers.
Extremely reactive with each other.
Vertical columns of elements with similar properties.
ANSWER: ____D____
Page 2 of 7
Chemistry 1010 Midterm Exam 1
vi)
Which of the following is a possible molecular formula for C4H4O?
a) C8H8O2
b) C12H12O2
c) C2H2O
d) C8H8O
ANSWER: ____A____
vii)
a)
b)
c)
d)
e)
How many mL are in 2.54 L?
2.54 × 10-3 mL
2.54 × 101 mL
2.54 × 103 mL
2.54 × 10-1 mL
2.54 × 102 mL
ANSWER: ____C____
viii)
Read the length of the metal bar with the correct number of significant figures.
a) 20 cm
b) 15 cm
c) 15.0 cm
d) 15.00 cm
e) 15.000 cm
ANSWER: ____D____
ix)
What answer should be reported, with the correct number of significant figures,
for the following calculation?
(16.84−15.93)
9.6318
𝑥 12.3
a) 1.16
b) 1
c) 1.162
d) 1.1621
e) 1.2
ANSWER: ____E____
x)
a)
b)
c)
d)
e)
Which of the following lists gives the symbol of a transition metal, a metal
and a non-metal in that order?
Fe, Cl, Ca
Na, Fe, Cl
Fe, In, Ar
Na, In, Cl
Ca, Fe, Ar
ANSWER: ____C____
Page 3 of 7
Chemistry 1010 Midterm Exam 1
[6]
[4]
A2.
Complete the following table. Classify each species as an atomic element, a
molecular element, a molecular compound, or an ionic compound. For the row
where the Classification reads “(any) molecular element” give the name and
formula for a molecular element of your choosing.
Chemical Name
Chemical Formula
Classification
Chromium
Cr
Atomic element
Potassium Sulfide
K2S
Ionic compound
Lithium hydrogen sulfite
LiHSO3
Ionic compound
Your choice
Your choice
(any) molecular element
Triphosphorous
heptachloride
P3Cl7
Molecular compound
Cobalt (II) carbonate
hexahydrate
CoCO3∙ 6 H2O
Ionic compound
A3.
An element has two naturally occurring isotopes. Isotope 1 has a mass of
120.9038 amu and fractional abundance of 0.574, and isotope 2 has a mass of
122.9042 amu. Find the atomic mass of the element and identify it using the
periodic table.
The atomic mass of this element will be
Atomic mass =
(mass isotope 1 x natural abundance isotope 1)
+ (mass isotope 2 x natural abundance isotope 2)
Atomic mass =
(120.9038 amu x 0.574) + (122.9042 amu) x (1.000-0.574)
Note that since natural abundance 1 + natural abundance 2 must equal 1.0000, then
natural abundance 2 must be 1.000-0.574. That is, since there are only two isotopes,
and isotope 1 makes up 57.4 % of all the atoms of that element, the other isotope must
make up the remaining 42.6 %.
Atomic mass =
69.39 amu + 52.36 amu =
121.75 amu
Since the atomic mass is 121.8 amu (appropriate sigfigs) this element must be antimony
(Sb)
Page 4 of 7
Chemistry 1010 Midterm Exam 1
Section B
Longer Answer Question
(12 marks)
Complete the following questions in the space provided. Be sure to complete all the
questions. Show all work as part marks will be given. Be as legible as possible as
unreadable work will NOT be marked.
You should allow at least 25 minutes for these questions.
[4] B1.
Most fertilizers consist of nitrogen-containing compounds. An example includes
ammonium carbonate (NH4)2CO3. Calculate the mass % of N in this fertilizer.
The molar mass of ammonium carbonate is (and be careful here since there are TWO
ammonium ions in the formula!):
Molar mass (NH4)2CO3
= (1 x 2 x molar mass of N) + (4 x 2 x molar mass of H)
+ molar mass of C +(3 x molar mass of O)
Molar mass (NH4)2CO3
= (2 x 14.0067 g mol-1) + (8 x 1.0079 g mol-1)
+ 12.0107 g mol-1 + (3 x 15.9994 g mol-1)
Molar mass (NH4)2CO3
= 28.0134 g mol-1 + 8.0632 g mol-1 + 12.0107 g mol-1
+ 47.9982g mol-1
Molar mass (NH4)2CO3
= 96.0855 g mol-1
The mass of N atoms in one mole of ammonium carbonate would be (2 x molar mass of
N) = (2 x 14.0067 g mol-1) = 28.0134 g mol-1
Therefore the mass % of N atoms in ammonium carbonate is
Mass %
= mass N per mole / molar mass x 100 %
= 28.0134 g mol-1 / 96.0855 g mol-1 x 100%
= 0.291547 x 100 % = 29.1547 %
The sigfigs in your answer will depend on how many sigfigs you used in your atomic
molar masses. Each person’s sigfigs were graded on that basis…
Page 5 of 7
Chemistry 1010 Midterm Exam 1
[8] B2.
a)
The foul odor of rancid butter is due largely to butyric acid, a compound
containing only carbon, hydrogen and oxygen. Combustion analysis of a 4.300 g
sample of this compound produced 8.590 g CO2 and 3.520 g H2O.
Find the empirical formula for butyric acid.
Given molar masses: H2O = 18.0152 g mol-1 CO2 = 44.0095 g mol-1
The number of moles of carbon dioxide produced was
Moles CO2 = mass CO2/ molar mass CO2 = 8.590 g / 44.0095 g mol-1 = 0.19519 mol CO2
Since each CO2 contains one carbon atom, the original butyric acid sample contained
0.19519 moles of C. The mass of carbon atoms in the butyric acid sample was
Mass C = moles C x molar mass C = 0.19519 mol x 12.0107 g mol-1 = 2.3443 g C
The number of moles of water produced was
Moles H2O
= mass H2O / molar mass H2O = 3.520 g / 18.0152 g mol-1
= 0.19539 mol H2O
Since each H2O contains two hydrogen atoms, the original butyric acid sample contained
0.19539 mol x 2 = 0.39078 mol of H. The mass of hydrogen atoms in the butyric acid
sample was
Mass H = moles H x molar mass H = 0.39078 mol x 1.0079 g mol-1 = 0.39387 g H
Since we’re told that butyric acid only contains C, H and O atoms, the mass of the O
atoms in the sample will be
Mass O
= Mass butyric acid – mass C – mass O = 4.300 g - 2.3443 g - 0.39387 g
= 1.5618 g O
The number of moles of O atoms is
Moles O
= mass O / molar mass O = 1.5618 g / 15.9994 g mol-1 = 0.097617 mol O
For an empirical formula divide the number of moles of each element by the number of
moles of the element present in the smallest amount (here this is the 0.097617 mol of O) :
Relative C : Relative H : Relative O
(0.1952 mol C/0.09762 mol):(0.3908 mol H/0.09762 mol):(0.09762 mol O/0.09762 mol)
2.000 : 4.003 : 1.000
The empirical formula must contain whole integer numbers of the relative amounts.
Therefore the empirical formula is C2H4O.
b)
If the molar mass of butyric acid is found to be 88.11 g mol-1 then what is the
molecular formula of butyric acid?
The molar mass of the empirical formula is
Molar mass C2H4O = (2 x molar mass C) + (4 x molar mass H) + (molar mass of O)
= (2 x 12.0107 g mol-1) + (4 x 1.0079 g mol-1) + (15.9994 g mol-1)
= 24.0214 g mol-1 + 4.0316 g mol-1 + 15.9994 g mol-1 = 44.05 g mol-1
Since the molecular formula must be an integer multiple of the empirical formula, and we
see the molar mass of butyric acid is twice the molar mass of the empirical formula, then
the molecular formula should be twice the empirical formula: C4H8O2
The End
Page 6 of 7
Chemistry 1010 Midterm Exam 1
Page 7 of 7