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Chem 132 – Sample Lab Final
KEY
Organic Structures
1. Write the molecular formula for an alkane containing 5 carbons and the appropriate number of hydrogens.
Alkane = CnH(2n+2) = C5H12
2. Draw two different alcohols with the formula C3H7OH.
HO
H
H
H
C
C
C
H
H
H
H
H
H
OH H
C
C
C
H
H
H
H
3. Name the following compounds.
2-chloropropane
trans-2-butene or trans-but-2-ene
Intermolecular Forces and Solid-Liquid Transitions
4. Which of the following compounds will have a) the greater intermolecular forces (why?) and b) if we
evaporate them which will give the largest ∆t (why?). 8 pts
pentane ( CH3CH2CH2CH2CH3 ) or propane ( CH3CH2CH3)
a.
Pentane
Why?
Dispersion (London) forces are the only IMF governing the interactions between pentane or
propane molecules. Pentane is longer and has a larger molar mass than propane, so it should exhibit
greater dispersion forces.
b.
Propane
Why?
Greater ∆T values arose from molecules with weaker IMF – they evaporated faster!
5. Which would you expect to have a higher melting point, silicon or hexane? 4 pts
Silicon will have a higher melting point. Silicon is a covalent network solid (like diamond). That means
that the silicon atoms are held together in the solid through covalent bonds, which are very strong forces
(much stronger than intermolecular forces). The molecules in solid hexane are only held together
through weak dispersion forces, so less energy is required to disrupt these attractions and cause melting.
Sample Lab Final Ch. 132 W04 NF
KEY NF W09
pg. 1
6. Explain, in terms of changes in state and intermolecular forces, the labeled portions of the cooling curve for
sodium thiosulfate hydrate. 8 pts
a. This portion of the cooling curve represents
Temp. vs Heat for sodi um thi osul fate hydr ate
liquid (only) sodium thiosulfate cooling down.
a. As energy is released, the molecules are
slowing down their molecular movement.
55°
a
b
T (° C)
b. Portion b represents the crystallization of
sodium thiosulfate. As the molecules lock into
place on the crystal lattice, energy is released
to keep the system at a constant temperature.
b.
c
22
°
c.
c. The system as equilibrated with the
surroundings, so it has reached room
temperature.
Heat tak en away
Solid State Structures
7. Which structure has the same arrangement of atoms as face-centered cubic? 5pts
a. hexagonal close-packed
b. cubic close-packed
c. simple cubic
8. A cation occupies the corners of a cubic array. There is also 1 cation in the center of the cube. The anion
occupies sites at the edges of the cube. The cation to anion ratio in the ionic structure is: 5 pts
a. 4 to 4
b. 2 to 4
c. 4 to 2
d. 2 to 3
Cation: 1 center + 8 corners(1/8) = 2 cations
Anion: 6 edges(1/2) = 3 anions
9. Calculate Avogadro’s number for a face-centered cubic array of Cu atoms with a unit cell dimension of
0.3615 nm.
(A.W. for Cu = 63.55 g, density = 8.92 g/cm3)
8 pts
SHOW WORK OR NO CREDIT
Use the density to find the mass of the unit cell, and then the mass of one atom.
3
Vcell
1m
100 cm $
!
= # 0.3615 nm x
x
= 4.724 x 10 -23 cm 3
&
9
"
1 x 10 nm
1m %
8.92 g 4.724 x 10 -23 cm 3
4.214 x 10 -22 g
fcc cell
1.053 x 10 -22 g
m = DV =
x
=
x
=
cm 3
unit cell
unit cell
4 atoms
atom
The molar mass and the mass of one atom can be used to find Avogadro's number.
63.55 g
atom
6.032 x 10 23 atoms
x
=
mol
1.053 x 10 -22 g
mol
6.032 x 1023 atoms/mol
Sample Lab Final Ch. 132 W04 NF
KEY NF W09
pg. 2
10.
Why does graphite act as a lubricant?. 4 pts
The structure of graphite is in layers, with very weak intermolecular forces between layers. These
interactions are easily broken and the layers can slide over each other.
Acids and Bases
11. If the normality of a standard HCl solution is 0.0987 N, and you use 16.67 mL of it to titrate 20.00 mL of
an unknown NaOH solution, what is the normality of the NaOH? The molarity?
10 pt
N a Va = N bVb
( 0.0987 N ) (16.67 mL ) = N b ( 20.00 mL )
N b = 0.0823 N
Mb =
0.0823 equiv OH 1 mol
x
= 0.0823 M
L
1 equiv OH 0.0823 M
12. If the normality of a dilute acetic acid solution is 0.660 N and the density of the solution is 1.01 g/ml,
calculate the percent acetic acid in the solution. (Molar mass HC2H3O2 is 60.06 g/mol)
0.660 equiv H + 1 mol HC2 H 3O 2
60.06 g
L
mL
x
x
x
x
x 100% = 3.92%
+
L
1 equiv H
mol
1000 mL 1.01 g
3.92 %
Chemical Equilibria
13. List three factors that affect equilibrium concentrations. They were listed on the handout. 6 pts.
1) The concentration of reactants and/or products.
2) The temperature of the reaction.
3) The pressure at which the reaction is run.
14. The equilibrium reaction for ammonia in water is
NH3 + H2O  NH4+ + OHWhat happens to the equlibrium when NH4Cl is added to the solution? 6 pts.
shifts to the
left/reactants
the [OH-] (decreases, increases)
Sample Lab Final Ch. 132 W04 NF
decreases
KEY NF W09
pg. 3
15.
Write the molecular, ionic and net ionic reaction for 2 moles of NaOH reacting with Zn(NO3)2 6 pts
molecular
2 NaOH(aq) + Zn(NO3)2(aq)  Zn(OH)2(s) + 2 NaNO3(aq)
total ionic
2 Na+(aq) + 2 OH-(aq) + Zn+2(aq) + 2 NO3-(aq)  Zn(OH)2(s) + 2 Na+(aq + 2 NO3-(aq)
net ionic
2 OH-(aq) + Zn+2(aq)  Zn(OH)2(s)
Reaction Rate: Iodination of Acetone
16.
2NO (g) + O2 (g)  2 NO2 (g) (8 pts)
Exp. 1
Exp. 2
Exp. 3
[NO] M
[O2] M
Initial Rate M/s
0.0126
0.0252
0.0252
0.0125
0.0250
0.0125
1.41 x 10-2
1.13 x 10-1
5.64 x 10-2
Obtain the rate law from the data above. What is the value of k?
k [ 0.0252 M ] [ 0.0125 M ]
Exp 3 5.64 x 10 -2 Ms-1
=
m
n
-2
-1
Exp 1 1.41 x 10 Ms
k [ 0.0126 M ] [ 0.0125 M ]
m
n
4 = 2m
m=2
k [ 0.0252 M ] [ 0.0250 M ]
Exp 2 1.13 x 10 -1 Ms-1
=
2
n
-2
-1
Exp 3 5.64 x 10 Ms
k [ 0.0252 M ] [ 0.0125 M ]
2
n
2 = 2n
n=1
Rate = k [ NO ] [ O 2 ]
2
1
5.64 x 10 -2 Ms-1 = k [ 0.0252 M ] [ 0.0125 M ]
2
1
k = 7.11 x 10 3 M -1s-1
Rate = 7.11 x 103 M-1s-1 [NO]2[O2]1
17. What is the effect on the rate of reaction if you decrease the temperature of the reaction by 10°C? (3 pts)
The rate will typically double when the temperature increases by 10°C.
Sample Lab Final Ch. 132 W04 NF
KEY NF W09
pg. 4
18. If you mix 40 mL of 4 M acetone + 10 mL of 1 M HCl + 30 mL of 0.005 M I2 + 20 mL of water in a
beaker, what are the following concentrations:
(8 pts)
C1V1 = C2V2
Acetone = 2M
! 40 mL $
Acetone: 4 M #
= 1.6 M = 2 M
" 100 mL &%
H+ ion = 0.1 M
! 10 mL $
H + ion: 1 M #
= 0.1 M
" 100 mL &%
Iodine = 2 x 10-3 M
! 30 mL $
Iodine: 0.005 M #
= 1.5 x 10 -3 M = 2 x 10 -3 M
&
" 100 mL %
19. The Arrhenius equation is a relationshp between the rate constant and temperature:
ln k = -Ea + ln A
RT
If the slope of a straight plot of ln k vs 1/T gives a value of - 12.6 K, what is the value of Ea?
R=8.31J/mol*K? 5pts
E
slope = - a
R
8.314 J
kJ
E a = -slope x R = -12.6 K x
= -105
moliK
mol
-105 kJ/mol
Potentiometric Titrations
20. How many equivalence points would you expect in a titration of H2SO3 with NaOH. 6pt
2 equivalence points (it is a diprotic acid)
21. Write the reaction for HPO42- with NaOH.
6pt
+
HPO-24(aq) + NaOH(aq) ! PO-34(aq) + Na (aq)
+ H 2O(l)
Sample Lab Final Ch. 132 W04 NF
KEY NF W09
pg. 5
22. On the following graph, a) what do the dots on the graph represent? b) If the volume from 0 to the first dot
is 18.6 mL, what is the total volume from zero volume to the second dot.?
a.
Equivalence Points
b.
2 x 18.6 mL = 37.2 mL
6 pts
Solutions
23. A solution is prepared from 1.576 g of benzoic acid dissolved in 16.32 g of lauric acid. Both are molecular
solids. The mixture has a freezing point of 42.4°C, and the normal freezing point of lauric acid is 45.6°C.
Kf for lauric acid is 3.9°C/m. Determine the molecular weight of benzoic acid.
!Tf = K f im
!Tf
( 45.6°C - 42.4°C) = 0.82 m
=
Kfi
! 3.9°C $
#"
& (1)
m %
0.82 mol benzoic acid
x 0.01632 kg lauric acid = 0.013 mol benzoic acid
kg lauric acid
1.576 g benzoic acid
MW =
= 1.20 x 10 2 g mol
0.013 mol benzoic acid
m=
1.20 x 102 g/mol
Sample Lab Final Ch. 132 W04 NF
KEY NF W09
pg. 6
24. The solubility of NaCl in water is 36g/100g at 25°C.
a. if you have 5 g in 15 g at 25°C, is the solution:
i. Unsaturated
5g
36 g
<
15 g 100 g
ii. Saturated
b. if you have 82 g in 200 g at 25°C, is the solution:
i. Unsaturated
ii. Supersaturated
82 g
36 g
>
200 g 100 g
ii. Saturated
ii. Supersaturated
c. Do you expect 45 g of NaCl to be soluble in 100 g of H2O at 50°C? (Y/N)
Yes. Solubility increases as
temperature increases
Sample Lab Final Ch. 132 W04 NF
KEY NF W09
pg. 7