Chemistry 1A Fall 2007
Examination #1 ANSWER KEY
1.
(5 pts.) Combustion analysis of a 13.42 g sample of equilin (which contains only
carbon, hydrogen, and oxygen) produced 39.61 g CO2 and 9.01 g H2O. The
molar mass of equilin is 268.34 g/mol. Determine the molecular formula of
equilin.
mol C = 39.61 g CO2 x 1 mol CO2 x 1 mol C = 0.900 mol C
44 g CO2
1 mol CO2
mol H = 9.01 g H2O x 1 mol H2O x 2 mol H = 1.001 mol H
1 mol H2O
18 g H2O
mol O = (13.42 – 10.81 – 1.00) g x 1 mol O = 0.101 mol O
16 g O
Empirical formula = C0.900 H1.001 O0.101 = C9H10O (134 amu)
Therefore, the molecular formula = 2(C9H10O) = C18H20O2
2.
(3 pts.) A hydrate of copper(II) chloride has the following formula: CuCl2 · xH2O.
The water in a 3.41 g sample of the hydrate was driven off by heating. The
anhydrous salt had a mass of 2.69 g. Find the number of waters of hydration
in the hydrate.
x = mol H2O/mol CuCl2
Therefore, x =
(3.41 – 2.69)g H2O x 1 mol H2O
18 g H2O = 2
0.0200 mol CuCl2
The empirical formula of the hydrate is CuCl2 · 2H2O
3.
(4 pts.) How would one prepare 400.0 mL of a 1.15 M NaNO3 solution from solid
sodium nitrate crystals in the laboratory setting. Once this stock solution is made,
how much water must be added to dilute the ENTIRE solution to 0.350 M?
g NaNO3 = 1.15 mol NaNO3 x 0.4000 L x 85 g NaNO3 = 39.1 g NaNO3
L
1 mol NaNO3
Therefore, dissolve 39.1 g NaNO3 in water and dilute up to 400.00 mL
M1V1 = M2V2; solve for V2 to obtain V2 = 1.31 L
Therefore, volume (H2O) added = 1.31 L – 400.0 mL = 914.3 mL H2O
4.
(14 pts. total) Consider the combustion of liquid ethanol (CH3CH2OH). After
4.62 mL of ethanol (density = 0.789 g/mL) was allowed to burn in the presence of
15.55 g of oxygen gas, 3.72 mL of water vapor was collected.
A.
(3 pts.) Write the balanced chemical equation for this combustion.
C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
B.
(6 pts.) Determine the maximum mass (in g) of water vapor that can be
obtained. Label both the limiting and excess reagents.
g H2O = 4.62 mL C2H6O x 0.789 g x 1 mol C2H6O x 3 mol H2O x 18 g
mL
46 g C2H6O 1 mol C2H6O 1 mol
= 4.28 g H2O from C2H6O
g H2O = 15.55 g O2 x 1 mol O2 x 3 mol H2O x 18 g
3 mol O2
1 mol
32 g O2
= 8.75 g H2O from O2
Therefore, 4.28 g of H2O result from the limiting reactant—ethanol.
C.
(3 pts.) Determine the mass (in g) of unreacted reactant left over upon
conclusion of the reaction.
g unreacted O2 = (8.75 – 4.28) g H2O x 1 mol H2O x 3 mol O2 x 32 g O2
18 g H2O 3 mol H2O 1 mol
Therefore, there are 7.95 g of unreacted O2
D.
(2 pts.) What is the percent yield of the reaction?
% yield = 100 (actual/theoretical) = 100 (3.72 g/4.28 g) = 86.9%
5.
(12 pts. total) APPLICATION! Nanotechnology, the field of building
microscale structures one atom at a time, has progressed in recent years. One
potential application of nanotechnology is the construction of artificial cells. The
simplest cells could mimic red blood cells, the body’s oxygen transporters. For
example, nanocontainers, perhaps constructed of carbon, could be pumped full of
oxygen and injected into a person’s bloodstream. If the person needed additional
oxygen—perhaps due to a heart attack or for the purpose of space travel—these
containers could slowly release oxygen into the blood, allowing tissues that would
otherwise die to remain alive. Suppose that nanocontainers were cubic and had an
edge length of 25 nanometers.
A.
(4 pts.) What is the volume (in L) of one nanocontainer?
V = (25 nm)3 x (1 m)3 x (100 cm)3 x 1 mL x 1 L = 1.6 x 10-20 L
(109 nm)3 (1 m)3 1 cm3 1000 mL
B.
(2 pts.) Suppose that each nanocontainer could contain pure oxygen
pressurized to a density of 85 g/L. How many grams of oxygen could be
contained by each nanocontainer?
g O2 = 85 g x 1.6 x 10-20 L = 1.4 x 10-18 g O2
L
C.
(3 pts.) Normal air contains about 0.28 g of oxygen per liter. An average
human inhales about 0.50 L of air per breath and takes about 20 breaths
per minute. How many grams of oxygen does a human inhale per hour?
g O2 = 0.28 g O2 x 0.50 L x 20 breaths x 60 min = 168 g O2/hr
L
breath
min
1 hr
D.
(3 pts.) What is the minimum number of nanocontainers that a person
would need in their bloodstream to provide 1 hour’s worth of oxygen?
# = 168 g O2 x 1 L x 1 nanocontainer x 1 hr = 1.2 x 1020 nanocont.
hr
85 g O2 1.6 x 10-20 L
6.
(21 pts. total; 3 pts. each) Write BALANCED equations (net ionic where
appropriate) for the following seven laboratory situations described below. Write
NR if no reaction occurs.
A.
Aqueous mercury(I) nitrate is reacted with aqueous sodium chloride.
Hg2+2(aq) + 2Cl-(aq) → Hg2Cl2(s)
B.
A solution of lead(II) acetate is mixed with a solution of potassium sulfate.
Pb+2(aq) + SO4-2(aq) → PbSO4(s)
C.
Solid calcium carbonate is added to a solution of hydrofluoric acid.
CaCO3(s) + 2HF(aq) → Ca+2(aq) + 2F-(aq) + H2O(g) + CO2(g)
D.
Sulfur trioxide gas is bubbled in water.
SO3(g) + H2O(l) → H+(aq) + HSO4-(aq)
E.
Strontium metal is reacted with nitrogen gas.
3Sr(s) + N2(g) → Sr3N2(s)
F.
Hydrogen gas is reacted with solid iron(III) oxide.
3H2(g) + Fe2O3(s) → 3H2O(l) + 2Fe(s)
G.
A solution of nitrous acid is mixed with aqueous magnesium hydroxide.
2HNO2(aq) + Mg(OH)2(aq) → Mg+2(aq) + 2NO2-(aq) + 2H2O(l)
7.
(12 pts. total) Consider the unbalanced REDOX reaction shown below:
P4S3(aq) + NO3-(aq) → H3PO4(aq) + SO4-2(aq) + NO(g)
A.
(6 pts.) Write the balanced net-ionic equation for this REDOX process
under acidic conditions. Demonstrate how this equation changes under
basic conditions.
3 x [28H2O + P4S3 → 4H3PO4 + 3SO4-2 + 44H+ + 38e-]
38 x [4H+ + NO3- + 3e- → NO + 2H2O]
84H2O + 3P4S3 → 12H3PO4 + 9SO4-2 + 132H+ + 114e152H+ + 38NO3- + 114e- → 38NO + 76H2O
8H2O + 3P4S3 + 20H+ + 38NO3- → 12H3PO4 + 9SO4-2 + 38NO
(acidic)
For this equation to be expressed under basic conditions, one must
add the equivalent number of hydroxides (i.e. 20 OH-) to both sides of
the equation and simplify:
28H2O + 3P4S3 + 38NO3- → 12H3PO4 + 9SO4-2 + 38NO + 20 OHB.
(basic)
(2 pts.) Label both the oxidizing and reducing agents in the balanced
reaction.
Oxidizing agent: NO3-; Reducing agent: P4S3
C.
(4 pts.) Suppose 68.6 mL of P4S3(aq) is titrated to the stoichiometric point
with 25.67 mL of 0.327 M HNO3(aq). What is the molarity of P4S3?
M (P4S3) = 0.327 mol HNO3 x 0.02567 L x 3 mol P4S3 x (0.0686 L-1)
L
38 mol HNO3
= 9.63 x 10-3 M P4S3(aq)
8.
(5 pts.) Consider the following ion: 98Mo+5. Determine each of the following.
# protons: 42
# neutrons: 56
# electrons: 37
Atomic Mass: 95.94 amu
Isotope (circle yes or no)? Briefly explain your circled response.
Yes, this ion is an example of an isotope. The atomic mass rounds up to a
mass number of 96 amu, which is different than the provided mass number
in this particular problem.
9.
(20 pts. total) SHORT ANSWERS! Respond to each of the questions below.
A.
(6 pts.) Briefly contrast the properties of metals and nonmetals. Are they
solids, liquids, gases? How are the various group of elements generally
found in nature? Give examples of each.
Metals are generally malleable (capable of being flattened into thin
sheets), ductile (capable of being drawn into fine wires), good
conductors of heat and electricity, possess moderate to high melting
points, and have a lustrous or shiny appearance. Most metals are
solids at room temperature (exception is mercury, a liquid). Metals
are generally found as monatomic species (e.g. Na metal).
In contrast, nonmetals can be bent and are poor conductors of heat
and electricity. Several of the nonmetals such as nitrogen, oxygen,
and chlorine are gases at room temperature. Bromine is a liquid, and
some exist as solids such as silicon and sulfur. This particular group
of elements can be monatomic, diatomic, or even greater (e.g. Si, Cl2,
S8, respectively).
B.
(4 pts.) Discuss Rutherford’s gold-foil experiment. What important
conclusions were drawn from this experiment?
Rutherford used a thin sheet of gold as a target, and alpha particles
emitted from a radioactive substance were directed towards the foil.
This foil is surrounded by a fluorescent screen coated with zincsulfide. Each time an alpha particle hits this zinc-sulfide coating, a
flash of light was produced at the point of contact. By observing these
flashes of light, it was possible to see whether the alpha particles that
passed through the foil had been deflected from their straight-line
path.
Rutherford’s model depicted the atom as having a positively charged
nucleus of relatively great mass. Moreover, traveling around the
nucleus were one or more negatively charged electrons of very small
mass.
C.
(6 pts.) List the postulates of Dalton’s atomic theory. Which points are
not valid today and have been corrected?
1.
2.
3.
4.
D.
All elements are composed of atoms, which are indivisible and
indestructible particles. – FALSE! Atoms can be subdivided
into further subatomic particles.
All atoms of the same element are exactly alike; in particular,
they all have the same mass. – FALSE! This statement does
not acknowledge the existence of isotopes.
Atoms of different elements are different; in particular, they
have different masses. – TRUE!
Compounds are formed by the joining of atoms of two or more
elements. – TRUE!
(4 pts.) What is chromatography? How is this separation technique
utilized in the laboratory setting? Describe the process in sufficient detail.
As discussed in lecture, column chromatography involves packing a
column with an adsorbing material, referred to as the stationary
phase, which will stick to the various components of the injected
mixture differently. The mixture, known as the mobile phase, passes
through the column and is thus separated into its various components,
which elute out of the column with a certain retention time that is
recorded by the detector.
10.
(4 pts.) For the last problem, consider the generation of KClO4(aq) via the
following sequential reactions, where the percentage yield of each reaction is
shown above the reaction arrows:
Cl2(g) + 2KOH(aq)
3KClO(aq)
4KClO3(aq)
92.1%
86.7%
75.3%
KCl(aq) + KClO(aq) + H2O(l)
2KCl(aq) + KClO3(aq)
3KClO4(aq) + KCl(aq)
What mass of Cl2(g) is required to produce 284 kg of KClO4(aq)?
kg Cl2 = 284 kg KClO4 x 1 mol KClO4 x (0.753)-1 x 4 mol KClO3 x (0.867)-1
138.6 g KClO4
3 mol KClO4
x 3 mol KClO x (0.921)-1 x 1 mol Cl2 x 70.9 g Cl2
1 mol KClO3
1 mol KClO 1 mol Cl2
= 966 kg Cl2 or 9.66 x 105 g Cl2