Chemistry 1A Spring 2005
Examination #1 ANSWER KEY
1.
(6 pts.) A stock solution of potassium dichromate is made by dissolving 189.3 g
of the compound in 2.00 L of solution. How many milliliters of this solution are
required to prepare 750 mL of 0.250 M K2 Cr2O7 ? How much water (in mL)
needs to be added to dilute the more concentrated solution to form the less
concentrated solution?
M1 V1 = M2 V2 where M1 is calculated from 189.3 g (convert to moles using
MW = 294.2 amu) divided by the 2.0 L volume!
(0.322 M)(V1 ) = (0.250 M)(750 mL)
V1 = 582 mL
Therefore, one would need 582 mL of the 0.321 M potassium dichromate
solution to prepare the less concentrated 0.250 M solution. Moreover, 750
mL – 582 mL or 168 mL of H2 O would be needed for the dilution.
2.
(6 pts.) Two elements, R and Q, combine to form two binary compounds. In the
first compound, 14.0 g of R combines with 3.00 g of Q. In the second compound,
7.00 g of R combines with 4.50 g of Q. Show that these data are in accord with
the law of multiple proportions. If the formula of the second compound is RQ,
what is the formula of the first compound?
From compound 1 data:
14.0 g R
3.00 g Q
From compound 2 data:
7.00 g R = 1R
4.50 g Q = 1Q
One can manipulate the compound 1 data to match the 7.00 g = 1R from
compound two data by multiplying both compound 1 data masses by ½:
Modified compound 1 data: 7.00 g R
1.50 g Q
Since Q for compound 1 is one -third the mass of compound 2, the formula
for the first compound becomes R3 Q.
3.
(6 pts.) A 1.013 g sample of ZnSO4 · xH2O is dissolved in water, and the sulfate
ion is precipitated as BaSO4 . The mass of pure, dry BaSO4 obtained is 0.8223 g.
What is the formula of ZnSO4 · xH2 O?
x = mol H2 O/mol ZnSO4
To determine the mol ZnSO4 , one must use the dry BaSO4 data, recognizing
that the sulfate ion reacts one -to-one with the barium ion to form the BaSO4
precipitate:
mol ZnSO4 = 0.8223 g BaSO4 x 1 mol BaSO4 x 1 mol SO4 -2 x 1 mol ZnSO4
233.43 g BaSO4 1 mol BaSO4 1 mol SO4 -2
= 0.003523 mol ZnSO4 (carry out one more conversion to
determine the mass of ZnSO4 to be 0.5688 g ZnSO4 )
Therefore, x =
(1.013 – 0.569)g H2 O x 1 mol H2 O
18 g H2 O
0.003523 mol ZnSO4
=7
The empirical formula of the hydrate is ZnSO4 · 7H2 O
4.
(8 pts.) What are the percent natural abundances of the two naturally occurring
isotopes of boron, 10 B and 11 B? These isotopes have masses of 10.012937 amu
and 11.009305 amu, respectively.
x = % abundance of 10 B; 1 – x = % abundance 11 B
10.811 = 10.012937 x + 11.009305 (1 – x)
x = 0.199 or 19.9% 10 B; therefore, 80.1% 11B
5.
(10 pts.) Adipic acid is used in the manufacture of nylon. It consists of carbon,
hydrogen, and oxygen. Combustion of 100.06 mg of adipic acid gives 180.01 mg
CO2 and 62.10 mg H2 O. In a separate experiment, the molecular mass of the
compound was found to be 146 amu. Determine the empirical and molecular
formula of adipic acid.
mol C = 180.01 mg CO2 x 1 mol CO2 x 1 mol C = 4.091 mmol C
44 g CO2
1 mol CO2
mol H = 62.10 mg H2 O x 1 mol H2 O x 2 mol H = 6.90 mmol H
18 g H2 O
1 mol H2 O
mol O = (100.06 – 49.092 – 6.90) mg x 1 mol O = 2.754 mmol O
16 g O
Empirical formula = C4.091 H6.90 O2.754 = C1.5 H2.5 O = C3 H5 O2 (73 amu)
There fore, the molecular formula = 2(C3 H5 O2 ) = C6 H10 O4
6.
REACTIONS! (32 pts. total; 4 pts. each) Write balanced chemical equations (net
ionic where appropriate) for the following laboratory situations described below.
Assume that solutions are aqueous unless otherwise indicated. Write NR if no
reaction occurs.
A.
A strip of magnesium is added to a solutio n of silver nitrate.
Mg(s) + 2Ag +(aq) ?
B.
Lithium metal is dropped into a container of nitrogen gas.
6Li(s) + N2 (g) ?
C.
2Li3 N(s)
Solid calcium carbonate is strongly heated.
CaCO3 (s) ?
D.
2Ag(s) + Mg+2 (aq)
CO2 (g) + CaO(s)
Liquid butane is burned in air.
2C4 H10 (l) + 13O2 (g) ?
E.
Sulfuric acid is added to excess potassium hydroxide.
H+(aq) + OH-(aq) ?
F.
8CO2 (g) + 10H2 O(g)
H2 O(l)
A solution of hydrochloric acid is added to a solution of mercury(I)
nitrate.
Hg2 2+(aq) + 2Cl-(aq) ?
G.
Hg2 Cl2 (s)
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)
H.
Concentrated hydrobromic acid is added to solid manganese(II) sulfide.
2H+(aq) + MnS(s) ?
7.
H2 S(g) + Mn+2 (aq)
(8 pts.) The aluminum in a package containing 75 ft2 of kitchen foil weighs
approximately 12 oz. Aluminum has a density of 2.70 g/cm3 . What is the
thickness of the aluminum foil in millimeters?
(Note: 1 oz = 28.4 g; 1 ft = 0.3048 m)
mass = 341 g; area = 6.97 x 104 cm2 ; volume = 126 cm3
This identical problem appeared on one of the old quizzes (answers posted
previously); please refer to these notes to obtain that the thickness of the foil
is equal to 1.81 x 10-2 mm.
8.
SHORT ANSWER! (9 pts.) Compare and contrast alpha (a), beta (ß), and
gamma (?) radiation. How do these forms of radiation behave in an electric field?
Ernest Rutherford identified two types of radiation from radioactive
materials: alpha and beta. Alpha particles carry two fundamental units of
positive charge and have essentially the same mass as helium atoms (identical
to He +2 ions). Beta particles are negatively charged particles produces by
changes occurring within the nuclei of radioactive atoms and have the same
properties as electrons. Gamma rays are NOT made up of particles
(although they do contain slight mass); they are electromagnetic radiation of
extremely high penetrating power with no charge (neutral).
In an electric field, alpha particles are attracted to the negative plate, beta
particles are attracted to the positive plate, and gamma rays are unaffected
by any charged plates.
9.
(15 pts. total) Azobenzene, an intermediate in the manufacture of dyes, can be
prepared from nitrobenzene by reaction with triethylene glycol in the presence of
Zn and KOH. In one reaction, 0.10 L of nitrobenzene (density = 1.20 g/mL) and
0.30 L of triethylene glycol (density = 1.12 g/mL) yield 55 g of azobenzene.
2 C6 H5NO2
+
Nitrobenzene
A.
, Zn
4 C6 H14O4 KOH


→ (C 6 H5N)2 +
triethylene
azobenzene
glycol
4 C6 H12 O4
+
4 H2 O
(6 pts.) Determine the theoretical yield (in grams) of azobenzene.
g azo = 0.10 L nitro x 1000 mL x 1.20 g x 1 mol nitro x 1 mol azo x 182 g azo = 89 g
1L
mL
123 g nitro 2 mol nitro mol azo
g azo = 0.30 L tg x 1000 mL x 1.12 g x 1 mol tg x 1 mol azo x 182 g azo = 102 g
1L
mL
150 g tg 4 mol ntg
mol azo
Therefore, the limiting reagent is nitrobenzene, resulting in 89 g
azobenzene.
B.
(6 pts.) Determine the mass (in grams) of unreacted reactant left over at
the conclusion of the reaction.
g unreacted tg = (102 – 89) g azo x 1 mol azo x 4 mol tg x 150 g tg = 43 g leftover tg
182 g azo 1 mol azo 1 mol tg
Therefore, there are 43 g of unreacted triethylene glycol
C.
(3 pts.) Determine the percent yield of this reaction.
% yield = 100 (actual/theoretical) = 100 (55 g/89 g) = 62%
BONUS! (7 pts.)
A sample of a mixture containing only sodium chloride and potassium chloride has a
mass of 4.000 g. When this sample is dissolved in water and excess silver nitrate is
added, a white precipitate (silver chloride) forms. After filtration and drying, this
precipitate has the mass 8.590 g. Calculate the mass percentage of sodium chloride in the
mixture.
There are many different ways to solve this particular problem! One way (shown
below) involves using algebra and a system of two equations with two unknowns:
x = sodium chloride; y = potassium chloride
x + y = 4.000 g (Law of Conservation of Mass)
Ag+(aq) + Cl-(aq) ?
AgCl
x
y
8.590 g
+
=
107.868amu 35.453amu 143.321amu
Simplify to obtain: x + 3.0426 y = 6.4656
Use the following algebraic system of two equations to solve for x and y respectively:
x + y = 4.000
x + 3.0426 y = 6.4656
x = 2.793 g NaCl; y = 1.207 g KCl
Therefore, % NaCl in mixture = 100 (2.793 g/4.000 g) = 69.8%