Los Angeles City College Chemistry 101
Mock Examination #1 ANSWER KEY
1.
(10 pts. total) Consider the combustion of acetylene gas, C2H2(g).
A.
(2 pts.) Write the balanced chemical equation for this process.
2C2H2 + 5O2 → 4CO2 + 2H2O
B.
(4 pts.) If 175 g of acetylene gas is reacted with 175 g of oxygen,
determine the theoretical yield (in g) of carbon dioxide. Clearly label
BOTH your limiting reagent and excess reagent.
g CO2 = 175 g C2H2 x 1 mol C2H2 x 4 mol CO2 x 44 g CO2 = 592 g CO2
26 g C2H2 2 mol C2H2 1 mol CO2
g CO2 = 175 g O2 x 1 mol O2 x 4 mol CO2 x 44 g CO2 = 193 g CO2
32 g O2 5 mol O2 1 mol CO2
Therefore, the limiting reagent is O2 and the excess reagent is C2H2!
C.
(2 pts.) Determine the mass (in g) of unreacted reactant at the conclusion
of this reaction.
g reacted C2H2 = 193 g CO2 x 1 mol CO2 x 2 mol C2H2 x 26 g C2H2
44 g CO2 4 mol CO2 1 mol C2H2
= 57.0 g reacted C2H2
g unreacted C2H2 = 175 g – 57 g = 118 g unreacted C2H2
D.
(2 pts.) If 52.5 L of carbon dioxide (density = 1.80 g/L) are produced,
what is the percentage yield of this reaction?
% Yield = 100% x (Actual/Theoretical)
= 100% x (52.5 L x 1.80 g/L)/193 g
= 49.0%
2.
(5 pts. total) A certain hydrate of potassium aluminum sulfate has the formula
KAl(SO4)2 • xH2O. When a hydrate sample weighing 5.459 g is heated to remove
all the water, 2.583 g of the anhydrous salt remains.
A.
(2 pts.) What is the mass percent of water in the hydrate?
%H2O = 100% x (5.459 g – 2.583 g)/5.459 g = 52.68%
B.
(3 pts.) Determine the value of x, and clearly write the complete formula
for the hydrate.
X = mol H2O/mol anhydrous compound
(5.459 g – 2.583 g) x 1 mol H2O
=
18 g H2O = 16
2.583 g alum x 1 mol alum
258 g alum
Therefore, the complete formula of the hydrate is KAl(SO4)2 • 16H2O
3.
(6 pts.) The insecticide dieldrin contains carbon, hydrogen, chlorine, and oxygen.
Upon complete combustion, a 1.510 g sample yields 2.094 g CO2 and 0.286 g
H2O. The compound has a molecular mass of 381 amu and has half as many
chlorine atoms as carbon atoms. What is the molecular formula of dieldrin?
mol C = 2.094 g CO2 x 1 mol CO2 x 1 mol C = 0.0476 mol C (0.5712 g)
1 mol CO2
44 g CO2
mol H = 0.286 g H2O x 1 mol H2O x 2 mol H = 0.0318 mol H
18 g H2O
1 mol H2O
mol Cl = 0.0476 mol C/2 = 0.0238 mol Cl (0.8437 g)
mol O = (1.510 – 0.5712 – 0.0318 – 0.8437) g x 1 mol O = 0.00396 mol O
16 g O
Divide by the smallest number of moles (i.e. 0.00396) to obtain the empirical
formula C12H8Cl6O.
The molar mass of C12H8Cl6O is 381 amu; therefore, the empirical formula
and the molecular formula are the same!
4.
(27 pts. total) REACTIONS! Consider the sets of reactants below. Predict the
product(s), balance the overall reactions, and include net ionic equations where
appropriate. Write NR if no reaction occurs.
A.
(3 pts.) Solid mercury(I) bromide is strongly heated.
Hg2Br2(s) → 2Hg(l) + Br2(l)
B.
(3 pts.) Hydrogen sulfide gas is bubbled into a solution of mercury(II)
chloride.
H2S(g) + Hg+2(aq) → 2H+(aq) + HgS(s)
C.
(3 pts.) Solid zinc reacts with an aqueous solution of iron(II) chloride.
Zn(s) + Fe+2(aq) → Zn+2(aq) + Fe(s)
D.
(3 pts.) Bromine gas is reacted with aqueous sodium chloride.
NR; bromine CANNOT replace chlorine E.
(3 pts.) A solution of copper(II) sulfate is added to a solution of barium
hydroxide.
Cu+2(aq) + SO4-2(aq) + Ba+2(aq) + 2OH-(aq) → Cu(OH)2(s) + BaSO4(s)
F.
(3 pts.) A solution of phosphoric acid is reacted with excess potassium
hydroxide solution.
H3PO4(aq) + 3OH-(aq) → 3H2O(l) + PO43-(aq)
G.
(3 pts.) An aqueous solution of rubidium phosphate is mixed with an
aqueous solution of titanium(IV) carbonate.
3Ti+4(aq) + 4PO4-3(aq) → Ti3(PO4)4(s)
H.
(3 pts.) Tetraphosphorus decoxide powder is sprinkled over distilled
water.
P4O10(s) + 6H2O(l) → 4H3PO4(aq)
I.
(3 pts.) Calcium metal is heated strongly in nitrogen gas.
3Ca(s) + N2(g) → Ca3N2(s)
5.
(10 pts. total) Consider the species 98Mo+3. Determine each of the following with
respect to the given species:
(1 pt.) Total number of protons: 42
(1 pt.) Total number of neutrons: 56
(1 pt.) Total number of electrons: 39
(1 pt.) Mass number: 98 amu
(1 pt.) Atomic number: 42
(1 pt.) Atomic mass: 95.94 amu
(2 pts.) Is this an ion? Briefly explain why or why not. Yes; the +3 charge
reduces the electron count by three and is called a cation.
(2 pts.) Is this an isotope? Briefly explain why or why not. Yes; species that
contain the same number of protons but differ in the number of neutrons are
called isotopes.
6.
(14 pts. total; 2 pts. each) Briefly define/explain each of the following terms listed
below:
A.
distillation: separation of a homogeneous mixture by boiling points,
where the more volatile substance vaporizes at a lower temperature,
travels down a condenser, and is collected in a receiving flask
B.
oxidation: involves the loss of electrons in a REDOX process
C.
element: a substance that cannot be separated into simpler substances
by ordinary chemical means; each element is composed of only one
kind of atom
D.
lanthanides: inner transition metals that are difficult to extract from
their natural sources and separate from one another
E.
beta (β) radiation: comprised of negatively charged particles, possess
the same properties as electrons, and are attracted to the positive pole
of a magnet when injected into a magnetic field
F.
deposition: phase change which involves conversion of a gas directly to
a solid
G.
Dalton’s Law of Atomic Theory:
1.
2.
3.
4.
All elements are composed of atoms, which are indivisible and
indestructible particles.
All atoms of the same element are exactly alike; in particular,
they all have the same mass.
Atoms of different elements are different; in particular, they
have different masses.
Compounds are formed by the joining of atoms of two or more
elements.
7.
(10 pts.) LABORATORY APPLICATION
One of the most widely used means of performing analytical separations in the
laboratory setting involves chromatography, a method that finds application in
many branches of science. Chromatography encompasses a diverse and important
group of methods that permit the scientist to separate closely related components
of complex mixtures. Discuss the basic principles of elution or column
chromatography, making sure to mention the following key terms in your
description: mobile phase, stationary phase, adsorption, and retention time.
In chromatography, mixtures can be separated by the differences in the
degree to which they are adsorbed onto a surface. The mixture is dissolved
in a mobile phase and injected into the chromatography column, where it
passes through a stationary phase. As the substances pass over the adsorbing
surface or stationary phase, the components that stick to the surface with
greater attraction will move SLOWER (through the column) than the
substances that are less attracted to the surface. This difference in speeds is
recorded by a detector, and as the components of the mixture elute out of the
column, their respective retention times are recorded.
8.
(5 pts.) Gold is a very soft metal that can be hammered into extremely thin sheets.
If a 1.10 g piece of gold (density = 19.32 g/cm3) is hammered into a sheet whose
area is 40.0 ft2, what is the average thickness of the sheet (in cm)?
Note: 1 ft = 0.3048 m
Density = mass/volume ⇒ Volume = mass/Density = 0.0569 cm3
Area = 40.0 ft2 x 0.3048 m x 0.3048 m x 100 cm x 100 cm = 37161 cm2
1 ft
1 ft
1m
1m
thickness = volume/area = 0.0569 cm3 / 37161 cm2 = 1.53 x 10-6 cm
9.
(4 pts.) Naturally occurring rubidium consists of two isotopes: 85Rb (84.9117
amu) and 87Rb (86.9092 amu). Determine the percent abundances of the two
isotopes in rubidium.
85.4678 amu = 84.9117 amu (x) + 86.9092 amu (1 – x)
Solve to obtain x = 72.1%
Therefore, Rb-85 is 72.1% abundant, and Rb-87 is 27.9% abundant.
10.
(10 pts. total) FIGURE IT OUT! While in the laboratory, your instructor asks
you to determine the weight percent of copper in a copper-containing alloy. After
dissolving a 0.251 g sample of the alloy in acid, an excess of KI is added, and the
Cu2+ and I- ions undergo the following reaction:
Cu+2(aq) + I-(aq) → CuI(s) + I3-(aq)
The liberated I3- is titrated with sodium thiosulfate (Na2S2O3) according to the
equation:
I3-(aq) + S2O3-2(aq) → S4O6-2(aq) + I-(aq)
A.
(4 pts.) Balance each REDOX reaction above.
2Cu+2(aq) + 5I-(aq) → 2CuI(s) + I3-(aq)
I3-(aq) + 2S2O3-2(aq) → S4O6-2(aq) + 3I-(aq)
B.
(2 pts.) Designate the oxidizing and reducing agents in the two reactions
above.
Oxidizing agents: Cu+2 and I3-
C.
Reducing Agents: I- and S2O3-2
(4 pts.) If 26.32 mL of 0.101 M Na2S2O3 is required for titration to the
equivalence point, what is the weight percent of Cu in the 0.251 g sample
of the alloy?
g Cu = 0.101 mol S2O3-2 x 0.02632 L x 1 mol I3- x 2 mol Cu+2 x 63.55 g
L
2 mol S2O3-2 1 mol I3mol
g Cu = 0.169 g Cu
Weight % Cu in alloy = (0.169 g/0.251 g) x 100 = 67.3% Cu in alloy