Practice Final Exam
1. Calculate the amount of heat, in kJ, needed to increase the temperature of
263.8 g of frozen phenol, C6H6OH, at -25.0°C (melting point is 40.5°C) to
liquid at 65°C. sl=1.43 J g-1 °C-1, Hfus=11.51 kJ mol-1,ss=0.306 cal g-1 °C-1.
Warm phenol to freezing point
q= ms∆T=
4.184 J 
 1 kJ 


 40.5 C − ( −25.0 C )  3 = 22.12224 kJ
 1 cal 
 10 J 
(263.8 g ) ( 0.306 cal g −1 C−1 ) 
Melt the phenol
(
 1 mol   11.51 kJ 
32.26328 kJ
q=
n∆H fus =
(263.8 g ) 
=

 94.1112 g   mol 
Warm the liquid to the final temperature
q = ms∆T = ( 263.8 g ) (1.43 J g −1 C −1 )( 65 C − 40.5 C ) = 9.242233 kJ
Total heat
qtotal = 22.12224 kJ + 32.26328 kJ + 9.242233 kJ = 63.6 kJ
)
2. Quantum Chemistry:
a. Give the electron configuration and orbital diagram of molybdenum
using the Noble gas shorthand.
Mo: [Kr]5s1 4d5
Mo: [Kr]  
5s
4d
b. Draw the electron dot structure of H2CO3 (hint: the hydrogens are
attached to the oxygens).
Give the electron group and molecular
geometries. Give the hybridization on the central atom. Is the ion polar?
H
O
C
O
H
O
EGG = MG = trigonal planar
Hybridization = sp2
Non-polar
3. Hydrogen is prepared from natural gas (mainly methane, CH4) by partial
oxidation.
2 CH4 + O2  2 CO + 4 H2
Calculate the enthalpy change H for this reaction, using standard
enthalpies of formation.
∆
=
H
∑
products
n∆H f −
∑
reactants
n∆H f
 2 mol CO  −110.525 kJ   4 mol H2   0.00 kJ    2 mol CH4
= 
  − 

+

 mol rxn  mol CO   mol rxn   mol H2    mol rxn
= −71.43 kJ
  −74.81 kJ   1 mol O2   0.00 kJ  

+


  mol CH4   mol rxn   mol O2  
4. A compound of carbon, hydrogen, and oxygen was burned in oxygen, and
1.000 g of the compound produced 1.434 g CO2 and 0.783 g H2O. In another
experiment, 0.1107 g of the compound was dissolved in 25.0 g of water. This
solution had a freezing point of –0.0894C. What is the molecular formula of
the compound? Kf = 1.86C m-1
Molar mass:
∆Tf =
−K f m
−∆Tf
( −0.0894 C ) =
=
−
m=
0.04806 m
Kf
1.86 C m−1

1 kg 0.04806 mol
×
= 0.0012015 mol
103 g
kg
0.1107 g
=
molar mass = 92.25 g mol−1
0.00120 mol
n = 25.0 g ×
Empirical formula:
1 mol CO2
1 mol C
×
=
0.03258387 mol C
44.0095 g CO2 1 mol CO2
12.0107 g C
×
=
0.3913 g C
1 mol C
1 mol H2O
2 mol H
H : 0.783 g H2O ×
×
=
0.08692611 mol H
18.0153 g H2O 1 mol H2O
1.00794 g H
×
=
0.0876 g H
1 mol H
1 mol O
= 0.03256895 mol O
O:1.000 g − 0.3913 g − 0.0876 g = 0.5211 g O ×
15.9994 g O
0.03258387 mol
= 1 C × 3= 3
0.03256895 mol
0.08692611 mol
= 2.66 H=
×3 8
0.03256895 mol
0.03256895 mol
= 1 O × 3= 3
0.03256895 mol
C : 1.434 g CO2 ×
Empirical mass = 92.0938 g mol−1
92.25
= 1
92.0938
molecular formula = C3H8O3
=
n
5. Polonium is the only metal that has a simple cubic unit cell structure. The
density of polonium is 9.31 g cm-3. Calculate the edge length of the unit cell
in pm. Calculate the atomic radius in pm.
3
3
cm3 209 g
mol
1 at  10 −2 m   pm 
7
3
×
×
×
×
 ×  −12 =
 3.727827 × 10 pm /cell
23
9.31 g mol 6.022 × 10 at cell  cm   10 m 
3.727827 × 107 pm3= 334.05 pm= 334 pm
e 334.05 pm
= 167.03 pm = 167 pm
e = 2r ⇒ r = =
2
2
edge length=
3
6. Ammonia will burn in the presence of a platinum catalyst to produce nitric
oxide, NO.
4 NH3 + 5 O2  4 NO + 6 H2O
What is the heat of reaction at constant pressure? Use the following
thermochemical equations:
2(N2 + O2  2 NO; H = 180.6 kJ)
-2(N2 + 3 H2  2 NH3; H = -91.8 kJ)
3(2 H2 + O2  2 H2O; H = -483.7 kJ)
∆H =( 2 )( +180.6 kJ) + ( −2 )( −91.8 kJ) + ( 3)( −438.7 kJ) =−771.3 kJ
7. A metal, M, was converted to the chloride, MCl2. Then a solution of the
chloride was treated with silver nitrate to give silver chloride crystals, which
were filtered from the solution.
MCl2 + 2 AgNO3  M(NO3)2 + 2 AgCl
If 2.434 g of the metal gave 7.964 g of silver chloride, what is the atomic
weight of the metal? What is the metal?
2.434 g M 143.321 g AgCl 2 mol AgCl 1 mol MCl2
×
×
×
=
87.61 g mol−1
7.964 g AgCl
1 mol AgCl
1 mol MCl2
1 mol M
The metal is strontium.
8. Lithium bromide and lithium chromate react in an oxidation-reduction
reaction in nitric acid. Two of the products of the reaction are the molecular
bromine and the chromium(II) ion. Write the balanced complete chemical
equation. If 24.55 mL of 0.05214 M lithium bromide solution reacts with
54.33 mL of 0.1244 M lithium chromate solution, what mass of bromine can
be obtained? If only 0.0996 g of bromine are produced, what is the percent
yield of the reaction?
LiBr + Li2CrO4  Br2 + Cr2+
2(2 LiBr  Br2 + 2 Li+ + 2 e-)
4 e- + 8 H+ + Li2CrO4  Cr2+ + 2 Li+ + 4 H2O
4 e- + 8 H+ + 4 LiBr + Li2CrO4  2 Br2 + Cr2+ + 6 Li+ + 4 H2O + 4 e8 H+ + 4 LiBr + Li2CrO4  2 Br2 + Cr2+ + 6 Li+ + 4 H2O
+8 NO3+ 8 NO38 HNO3 + 4 LiBr + Li2CrO4  2 Br2 + Cr(NO3)2 + 6 LiNO3 + 4 H2O
0.05214 mol LiBr 2 mol Br2 159.808 g Br2
×
×
=
0.1022 g Br2
1000 mL LiBr
4 mol LiBr
1 mol Br2
0.1244 mol Li2CrO4
2 mol Br2
159.808 g Br2
54.33 mL Na2Cr2O7 ×
×
×
=
2.160 g Br2
1000 mL Li2CrO4
1 mol Li2CrO4
1 mol Br2
24.55 mL LiBr ×
0.1022 g of bromine can be produced.
%yield
=
actual
0.0996
× 100
=
× 100
= 94.5%
theoretical
0.1022