Chemistry 1A Mock Exam #1 pp. 1
Moorpark College
Chemistry 1A Mock Exam #1 ANSWER KEY
Professor Torres
1.
A chemistry student has the idea of removing a decorative stone sphere with a
radius of 0.289 m from the front of a campus building. If the density of the stone
is 3.52 g/cm3 , what is the mass of the sphere?
4
Vsphere = πr 3
3
Volume = 4/3 π r3 = 4/3 x π x (28.9 cm) 3 = 1.01 x 105 cm3
Density = mass/volume ⇒ Mass = Density x Volume
Mass = (3.52 g/cm3 )(1.01 x 105 cm3 ) = 3.56 x 105 g
2.
Determine the mass (in grams) of 1.00 gal of cyclohexane. The density of
cyclohexane at 25 °C is 0.7781 g/mL. [NOTE: 1 L = 1.057 qt and 1 gal = 4 qt]
Mass (g) = 1.00 gal x 4 qt x 1L
x 1000 mL x 0.7781 g
1 gal 1.057 qt
1L
mL
= 2.94 x 103 g cyclohexane
3.
CFC’s, or chlorofluorocarbons, are damaging pollutants to the earth’s atmosphere
that are known as greenhouse gases. Above 30,000 feet, the atmosphere has a
density of 1.2 x 10-6 g/cm3 . Determine the mass of CFCl3 (a CFC) contained in
7.2 x 106 cm3 air if the CFCl3 concentration is 50. ppm (i.e. one gram of CFCl3 in
one million grams of air).
grams CFCl3 = 50 g CFCl3 x 1.2 x 10-6 g air x 7.2 x 106 cm3 air
106 g air
cm3
-4
= 4.32 x 10 g CFCl3
4.
The element potassium has two naturally occurring isotopes: 39 K and 41 K with a
mass of 40.962 amu. Potassium consists of 6.88% 41 K and has an average atomic
mass of 39.101 amu. Calculate the mass of 39 K.
x = mass of 39 K
39.101 = 40.962(0.0688) + 0.9312 x
x = 38.964 amu
5.
Nitrogen has two naturally occurring isotopes, 14 N and 15 N, whose atomic masses
are 14.0031 and 15.0001, respectively. The atomic mass of nitrogen is 14.0067.
Calculate the percentage abundance of each isotope of nitrogen.
x = % abundance of 14N; 1 – x = % abundance 15 N
14.0067 = 14.0031 x + 15.0001 (1 – x)
x = 0.9964 or 99.64% 14 N; therefore, 0.36% 15N
Chemistry 1A Mock Exam #1 pp. 2
6.
Name each of the following compounds:
A.
B.
C.
D.
E.
F.
G.
H.
7.
magnesium chlorate
chromium(II) hydroxide
dinitrogen penaoxide
aluminum phosphate
lithium permanganate
rubidium dihydrogen phosphate
dibromine pentaoxide
mercury(I) nitrate
Write formulas for the following compounds:
A.
B.
C.
D.
E.
F.
G.
H.
8.
Mg(ClO 3 )2
Cr(OH)2
N2O5
AlPO4
LiMnO 4
RbH2 PO4
Br2 O5
Hg2 (NO3 )2
ammonium dichromate
iron(III) carbonate
sodium oxalate
copper(II) acetate
potassium thiocyanate
sodium peroxide
iron(III) oxalate
potassium hypochlorite
(NH4 )2 Cr2 O7
Fe 2 (CO3 )3
Na2 C2 O4
Cu(C2 H3 O2 )2
KSCN
Na2 O2
Fe 2 (C2 O4 )3
KClO
Discuss the major points outlined in Dalton’s atomic theory. How does modern
atomic theory contrast with Dalton’s theory?
The following four points are outlined in Dalton’s atomic theory:
1.
2.
3.
4.
All elements are composed of atoms, which are indivisible and
indestructible particles.
All atoms of the same elements 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.
After Thomson’s discoveries, certain components of Dalton’s atomic theory
slowly became outdated as the atom was no longer the smallest indivisible
particle; it contained smaller particles called electrons. Modern atomic
theory contrasts with Dalton’s theory in that the former accounts for the
existence of isotopes as well as subatomic particles beyond the atom.
Chemistry 1A Mock Exam #1 pp. 3
9.
Discuss Rutherford’s gold- foil experiment (use a schematic for detail if necessary,
but also explain). Is this model depicting the atom consistent with modern
physics? Explain why or why not.
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 zinc sulfide. 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. According to classical
physics of the time, since an electron loses energy by giving off light, it would
eventually fall toward the nucleus. This apparent anomaly goes contrary to
the fact that atoms appear very stable. Therefore, an adequate atomic model
must be postulated to explain why the electrons do not give off energy and
collapse into the nucleus.
10.
Write the balanced net ionic equation for each of the reactions given below. If no
reaction is expected to occur, state NR.
A.
Aqueous sodium sulfate is reacted with aqueous lead(II) nitrate.
Pb2+(aq) + SO4 2-(aq) → PbSO4 (s)
B.
Gaseous hexane (C 6 H14 ) is reacted with oxygen gas.
2C6 H14 (g) + 19O2 (g) → 12CO2 (g) + 14H2 O(g)
C.
Solid cobalt(II) chloride is electrolyzed.
elec
CoCl2 (s) → Co(l) + Cl2 (g)
D.
Solid calcium chloride is placed in a container of water.
CaCl2 (s) water
→ Ca2+(aq) + 2Cl-(aq)
E.
Solid aluminum reacts with oxygen gas.
4Al(s) + 3O2 (g) → 2Al2 O3 (s)
F.
Gaseous hydrofluoric acid reacts with solid silicon dioxide.
4HF(g) + SiO2 (s) → Si+4 (aq) + 4F-(aq) + 2H2 O(l)
G.
Methanol (CH3 OH) is reacted in air.
2CH3 OH(l) + 3O2 (g) → 2CO2 (g) + 4H2 O(g)
Chemistry 1A Mock Exam #1 pp. 4
11.
H.
Aqueous lead(II) nitrate reacts with aqueous sodium carbonate.
Pb+2 (aq) + CO3 -2 (aq) → PbCO3 (s)
I.
Solid magnesium oxide is reacted with hydrochloric acid.
MgO(s) + 2H+(aq) → H2 O(l) + Mg+2 (aq)
J.
Solid lead reacts with aqueous cadmium sulfate.
NR – No Reaction (lead metal CANNOT replace cadmium)
K.
An acidic solution is prepared containing chromate ions and iron(II) ions.
The products produced after reaction include the iron(III) ions and
chromium(III) ions. Balance this equation and label the
oxidizing/reducing agents.
8H+(aq) + CrO 4 2-(aq) + 3Fe +2 (aq) → Cr+3 (aq) + 3Fe +3 (aq) + 4H2 O(l)
***CrO 4 2-(aq) is the oxidizing agent; Fe +2 (aq) is the reducing agent.
L.
Liquid bromine reacts with aqueous sodium iodide.
Br2 (l) + 2I-(aq) → 2Br-(aq) + I2 (s)
M.
Tin(IV) oxide is synthesized from solid tin and nitric acid, producing
nitrogen dioxide gas as a by-product. Label the oxidizing and reducing
agents.
Sn(s) + 4H+ (aq) + 4NO3 -(aq) → SnO2 (s) + 4NO2 (g) + 2H2 O(l)
***Sn(s) is the reducing agent; HNO3 is the oxidizing agent.
N.
Phosphoric acid is reacted with excess sodium hydroxide.
H3 PO4 (aq) + 3OH-(aq) → 3H2 O(l) + PO4 3-(aq)
Consider the two compounds shown below. Determine the OXIDATION STATE
of the underlined atom.
A.
B.
12.
HClO3
S2 O3 -2
+5
+2
The density of an 18% solution of ammonium chloride solution is 1.05 g/mL.
What mass of ammonium chloride does 425 mL of this solution contain?
g NH4 Cl = 425 mL solution x 1.05 g solution x 18 g NH4 Cl
mL solution
100 g solution
= 80 g NH4 Cl
13.
Aluminum burns in bromine, producing aluminum bromide according to the
following unbalanced chemical equation:
Al(s) + Br2 (l) + heat → AlBr3 (s)
Chemistry 1A Mock Exam #1 pp. 5
When 6.02 g of aluminum reacted with 10.00 mL of bromine (density = 3.12
g/mL), 30.2 g of aluminum bromide was isolated.
14.
A.
Write the balanced chemical equation for this reaction.
2Al + 3Br2 → 2AlBr3
B.
Determine the mass (in g) of aluminum bromide that can be theoretically
produced.
34.71 g AlBr3 are obtained from 10.00 mL Br2 .
C.
Which reactant is limiting? Determine the mass (in g) of the reactant
which is left unreacted.
2.51 g Al remain at the conclusion of the reaction.
D.
What is the percentage yield of this reaction?
Percentage Yield = 87.0% AlBr3
The aluminum sulfate hydrate, [Al2 (SO4 )3 • xH2 O], contains 8.10 percent Al by
mass. Calculate x, that is, the number of moles of water associated with one mole
of Al2 (SO4 )3 .
g Al2 (SO4 )3 = 8.10 g Al x 1 mol Al x 1 mol Al2 (SO4 )3 x 342 g Al2 (SO4 )3
27 g Al
2 mol Al
1 mol Al2 (SO4 )3
= 51.3% Al2 (SO4 )3
Therefore, %H2 O = 48.7%
mol H2 O = 48.7 g H2 O x 1 mol H2 O = 2.71 mol H2 O
18 g
mol Al2 (SO4 )3 = 51.3 g Al2 (SO4 )3 x 1 mol Al2 (SO4 )3 = 0.15 mol Al2 (SO4 )3
342 g
x = mol H2 O/mol Al2 (SO4 )3 = 2.71 mol H2 O/0.15 mol Al2 (SO4 )3 = 18
Therefore, the formula of the hydrate is Al2 (SO4 )3• 18H2 O.
15.
Hexamethylenediamine, a compound made up of carbon, hydrogen, and nitrogen
atoms, is used in the production of nylon. When 6.315 g of
hexamethylenediamine is burned in oxygen, 14.36 g of carbon dioxide and 7.832
g of water are obtained.
A.
B.
Determine the empirical formula of hexamethylenediamine.
What is the molecular formula of this compound if the molar mass is
116.2 g/mol?
A.
mol C = 14.36 g CO2 x 1 mol CO2 x 1 mol C = 0.326 mol C
44 g CO2
1 mol CO2
Chemistry 1A Mock Exam #1 pp. 6
mol H = 7.832 g H2 O x 1 mol H2 O x 2 mol H = 0.870 mol H
18 g H2 O
1 mol H2 O
mol N = (6.315 – 3.912 – 0.870) g x 1 mol N = 0.109 mol N
14 g N
Divide by the smallest number of moles (i.e. 0.109) to obtain the
empirical formula C3 H8 N.
B.
16.
The empirical mass for C3 H8 N is 58 g/mol, which is half the molecular
mass. Therefore, you multiply the empirical formula by two and
obtain C6 H16 N2 .
An acidic solution of potassium permanganate reacts with oxalate ions to form
carbon dioxide and manganese(II) ions according to the following unbalanced
REDOX equation:
MnO4 -(aq) + C2 O4 -2 (aq) → CO2 (g) + Mn+2 (aq)
A.
Write the balanced chemical equation for this reaction. Identify the
species that is oxidized and reduced.
2[5e - + 8H+ + MnO4 - → Mn+2 + 4H2 O]
5[C2 O4 2- → 2CO2 + 2e -]
2MnO4 - + 5C2 O4 -2 + 16H+ → 10CO2 (g) + 2Mn+2 + 8H2 O
B.
If 38.4 mL of 0.150 M potassium permanganate is required to titrate 25.2
mL of sodium oxalate solution, what was the concentration of oxalate ion?
mol oxalate = 0.150 mol MnO4 - x 0.0384 L x 5 mol C2 O4 2- x 1
L oxalate
L
2 mol MnO4 - 0.0252 L
= 0.571 M
C.
Explain how one would prepare 50 mL of a 0.150 M potassium
permanaganate solution from solid potassium permanganate crystals in the
laboratory setting?
Dissolve 1.19 grams of KMnO4 (this can be calculated from the
molarity/volume given) with approximately 50 mL of water in a
volumetric flask.
D.
Using the stock solution from part C, determine how much water MUST
BE ADDED to dilute the solution to 0.025 M?
M1 V1 = M2 V2
(0.150 M)(50 mL) = (0.025 M)(x)
x = 300 mL
Add (300 – 50) mL or 250 mL of water to 50 mL of the stock solution
from part C in order to generate 300 mL of 0.025 M solution.
Chemistry 1A Mock Exam #1 pp. 7
17.
An element X forms both a dichloride, XCl2 , and a tetrachloride, XCl4 .
Treatment of 10.00 g of XCl2 with excess chlorine forms 12.55 g of XCl4 .
Calculate the atomic mass of X and identify X.
g X = 10.00 g – 2.55 g = 7.45 g X
mol X = 2.55 g Cl2 x 1 mol Cl2 x 1 mol XCl2 x 1 mol X = 3.591 x 10-2 mol X
71 g Cl2
1 mol Cl2 1 mol XCl2
X (g/mol) = 7.45 g/(3.591 x 10-2 mol) = 207.4 g/mol
X = Pb