03/30/16
ANSWER KEY
CHEM 101 Exam 1
Name: __________________________________
Open-End Questions and Problems
Please read the following. To receive full credit for a question or a problem, in addition to the
correct answer, you must show a neat, complete, and logical method of solution where each
number is labeled with the appropriate unit and the final answer is rounded to the correct
number of significant digits. The correct answer without any work shown will generally get zero
credit! When an explanation is required, it should be brief, but accurate and complete. There are
11 questions for a total of 99 points. (Part of the exam with multiple choice questions is worth
51 points. Both parts combined are 150 points total.
1. (9 points) Fill the blanks in the following table.
Diamond
Bromine
Sodium chloride
none
6.02×1023
none
6.02×1023
1.20×1024
1.20×1024
The number of elements in 1.00 mole
one
one
two
The number of substances in 1.00 mole
one
one
one
The number of compounds in 1.00 mole
none
none
one
The mass in grams of 1.00 mole
12.0
160
58.4
The volume in cm3 of 1.00 mole
3.41
51.5
26.9
The number of molecules in 1.00 mole
The number of atoms in 1.00 mole
2. (9 points) Perform the following conversions. Record the results either in decimal notation
or in scientific notation whichever of the two is more appropriate. (You don’t have to show
your work.)
2.20×10–4 km =
2.20
dm
786 nm =
7.0×1022 mm2 =
7.0×1040
pm2
0.003600 cm3 =
3.600×10−9
1.00 µL =
1.00
846 mL =
1 yd3 (exactly) =
8.46
dL
46,656
in3
1 L (exactly) =
7.86×10−5
0.001
cm
m3
mm3
m3
3. (9 points) Give two examples of each. Note: (i) do not to use the same substance as an
example more than once; (ii) paper, plastic, wood, a stick, gasoline, milk, food are not pure
chemical substances and should not be used as examples.
Physical Change (a change that involves a
specific pure substance)
Chemical Change (a change that involves a
specific pure substance)
Example 1:
ice melts
Example 1:
potassium reacts with water to
produce potassium hydroxide and
hydrogen gas
Example 2:
sulfur burns in air to produce sulfur
dioxide
Example 2:
sugar (sucrose) dissolves in water
Physical Property (specific property of a
specific pure substance)
Chemical Property (specific property of a
specific pure substance)
Example 1:
density of aluminum is 2.70 g/cm3
Example 1:
chlorine reacts with iron to produce
iron(III) chloride
Example 2:
Example 2:
ammonia is gas at room temperature neon does not react with other
and normal pressure
chemical elements to form any stable
compounds
4. (9 points) As elementary substances under normal conditions, both fluorine and chlorine are
gases made of diatomic molecules: F2(g) and Cl2(g).
What is the molecular weight of fluorine?
37.9968
amu
What is the molecular weight of chlorine?
70.9054
amu
What are the actual masses (in amu) of individual fluorine molecules? Briefly explain.
37.9968 amu for each F2 molecule. Fluorine has only one naturally occurring
isotope: 19F. All fluorine atoms (F) and all fluorine molecules (F2) of one type.
What are the actual masses (in amu) of individual chlorine molecules? Briefly explain.
Chlorine has two naturally occurring isotopes: 35Cl (75.9%) and 37Cl (24.2%).
Three types of molecules can be made up of atoms of those two isotopes: 35Cl2,
34.9689 × 2 = 69.9378 amu; 35Cl37Cl, 34.9689 + 36.9659 = 71.9348 amu; and
37
Cl2, 36.9659× 2 = 73.9318. 70.9054 amu is their weighted average.
5. (9 points) Briefly answer the following questions.
(a) Compare the passage of electricity through a wire and through a solution.
Through a wire electrons are moving; through a solution ions are moving.
(b) How can it be that aqueous solutions of all ionic compounds conduct electricity but
aqueous solutions of covalent molecular compounds may or may not conduct electricity?
Ionic compounds are made up of ions. When an ionic compound dissolves in
water, ions go into the solution. In the solutions, the ions can move freely.
When most covalent compounds dissolve in water, the solution contains only
molecules that carry no net charge. However, all acids and ammonia that are
covalent compounds conduct electricity when dissolved in water. Molecules
of acids and ammonia react with water molecules to produce ions.
(c) Ammonia is a gas under normal conditions. Why is it considered to be an electrolyte?
Ammonia is highly soluble in water. When in an aqueous solution, ammonia
molecules react with water to produce ions:
NH3(aq) + HOH(l) ⇌ NH4+(aq) + OH−(aq)
6. (9 points) A 0.1005-g sample of an organic compound composed of C, H, and O is burned in
pure oxygen gas, producing 0.2829 g of CO2 and 0.1159 g of H2O.
(a) What is the mass percent of oxygen in the original compound?
0.2829 g CO2 ×
0.1159 g H2O ×
12.01 g C
44.01 g CO2
2×1.008 g H
18.01 g H2O
mass of O = 0.1005 g
% O =
0.0103 g
0.1005 g
= 0.07720 g C
= 0.01297 g H
– 0.07720 g – 0.01297 g H g = 0.0103 g
× 100% = 10.3%
(b) How many grams of oxygen gas have been consumed in the reaction?
0.1005 g CxHyOz + X g O2 = 0.2829 g CO2 + 0.1159 g H2O
X = 0.2829 + 0.1159 − 0.1005
0.2983 g O2
7. (9 points) An experiment is conducted in which varying amounts of solid iron are added to a
fixed volume of liquid bromine. The product of the reaction is a single compound, which can
be separated from the product mixture and weighed. The graph shows the relationship
between the mass of iron in each trial versus the mass of the product compound.
(a) Explain in four complete sentences why the graph has (i) a positive slope for low
masses of iron and (ii) a zero slope when the mass of iron added becomes larger.
For smaller than 2.0 g of Fe, Fe is the limiting reactant and the bromine is
in excess. The more iron is added to the fixed amount of Br2, the more
product is made. For the masses of Fe greater than 2.0 g, the bromine is
the limiting reactant. Adding 2.0 or more grams of Fe, results in the same
amount of product as the amount of bromine used through the entire
experiment is fixed.
(b) What is the formula of the product compound? Show calculations.
Fe
2.0 g
×
Br
9.0 g
×
1 mol
55.85
1 mol
79.90 g
= 0.0358 mol
= 0.1126 mol
Formula of the product: FeBr3
(c) Write a balanced chemical equation for the reaction.
Fe(s) + Br2(l) → FeBr3(s)
0.0358 mol
0.0358
0.1126 mol
0.0358
= 1.0 mol
= 3.1 mol
8. (9 points) In the presence of a catalyst, 44.01 grams of carbon dioxide enter a reaction with
10.08 grams of hydrogen gas that produces methyl alcohol, CH3OH, and water. Assuming
that the reaction yield was 25%, fill the blanks in the following table.
Balanced Chemical
Equation
CO2
L.R.
Mass before the reaction
44.01 g
Molar mass
44.01 g/mol
+
3 H2
→
CH3OH
10.08 g
+
0 g
2.016 g/mol 32.04 g/mol
0 g
18.02 g/mol
Moles before the reaction
1.0 mol
5.0 mol
Moles consumed
(–) or moles produced (+)
−0.25 mol
−0.75 mol
+0.25 mol
+0.25 mol
0.75 mol
4.25 mol
0.25 mol
0.25 mol
8.56 g
8.01 g
Moles after the reaction
Mass after the reaction
33.01 g
0 mol
H2O
0 mol
4.51
9. (9 points) Using the half-equation method, write the balanced net ionic equation and the
balanced full-formula equation for of the following reaction.
Cr2(SO4)3 + K2S2O8 + H2O → K2Cr2O7 + K2SO4 + H2SO4
2 Cr3+
+ 7 H2O → Cr2O72− + 14 H+ + 6 e−
S2O82− + 2 e− → 2 SO42−
×1
×3
2 Cr3+(aq) + 3 S2O82−(aq) + 7 H2O(l) → Cr2O72−(aq) + 6 SO42−(aq) + 14 H+ (aq)
+ 3 SO42− + 6 K+(aq)
+ 3 SO42− + 6 K+(aq)
Cr2(SO4)3(aq) + 3K2S2O8(aq) + 7H2O(l) → K2Cr2O7(aq) + 2K2SO4(aq) + 7H2SO4(aq)
10. (9 points) 50.0 ml of 0.400 M solution of calcium nitrate and 50.0 mL of 0.200 M solution of
sodium phosphate are mixed in a 250-mL beaker.
(a) Write the full-formula, complete ionic, and net ionic equation (FFE, CIE, and NIE) for
the reaction occurred. In each equation, indicate physical state of each reactant and
product: (s), (l), (g), (aq).
(b) product: (s), (l), (g), (aq).
FFE:
3 Ca(NO3)2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaNO3(aq)
CIE:
3Ca2+(aq) + 6NO3−(aq) + 6Na+(aq) + 2PO43−(aq) → Ca3(PO4)2(s) + 6Na+(aq) +6NO3−(aq)
NIE:
3 Ca2+(aq) + 2 PO43−(aq) → Ca3(PO4)2(s)
(c) Complete the following table.
Calcium ion
mmol before the
reaction
20.0
Nitrate ion
Sodium ion
Phosphate ion
40.0
30.0
10.0
mmol consumed (–) or
moles produced (+)
- 15.0
0
0
- 10.0
mmol after the reaction
5.0
40.0
30.0
0
molar concentration in
the final solution (M)
0.050
0.400
0.300
0
(d) How many grams of a precipitate are formed? Show work.
If 10.0 mmol of PO43− are consumed, 5.00 mmol (or 0.00500 mol) of Ca3(PO4)2(s)
are produced.
0.00500 mol × (310.17 g/mol) = 1.55 g Ca3(PO4)2
11. (9 points) A titration experiment is being conducted. The buret is filled with a solution of
sodium hydroxide of unknown concentration. Oxalic acid dihydrate is used as a standard.
(a) A student transferred to an Erlenmeyer flask a carefully weighed amount of oxalic acid
dihydrate, but failed to completely dissolve it prior to titration. How will it affect the
calculated molarity of sodium hydroxide? Explain.
Smaller volume NaOH(aq) will be used. Moles of NaOH are calculated
from the weighed amount of H2C2O4∙2H2O. Therefore, the calculated
molarity of NaOH will be HIGHER: molarity = moles / liters of solution.
(b) A student transferred to an Erlenmeyer flask a carefully weighed amount of oxalic acid
dihydrate, but used 50% more water to dissolved it as compared to the recommended
amount. How will it affect the calculated molarity of sodium hydroxide? Explain.
Extra water used to dissolve H2C2O4∙2H2O WILL NOT AFFECT the
results of the experiment. The calculations are based on the moles of
H2C2O4∙2H2O, not on its concentration.
(c) A student transferred to an Erlenmeyer flask a carefully weighed amount of oxalic acid
dihydrate and completely dissolved it in the recommended amount of water, however the
student failed to notice that prior to the first run of titration the tip of the buret was not
completely filled with sodium hydroxide solution, but had an air bubble there. How will
it affect the calculated molarity of sodium hydroxide? Explain.
It would appear AS IF A LARGER VOLUME of NaOH(aq) was added to
the flask than the volume required to neutralize all of the
H2C2O4∙2H2O. Therefore, the calculated molarity of NaOH will be
LOWER: molarity = moles / liters of solution.