Ni(NH3)xCl2 . zH2O

Chemistry 30AL Final Exam Review and Study Questions
Answer Key
1. Explain why the melting point of a crude product is lower than the melting point
of the product after recrystallization.
Answer:
Recrystallization purifies the product. Since the mole fraction of the impurity is
now smaller, there is less freezing point depression than in the crude product.
2. What observations would have differed in your ion exchange column experiment
if you had reversed the order of elution of the acid concentrations?
Answer:
You would obtain no separation. All the ions would be washed off the column at the
same time by the dilute HCl solution.
3. The stoichiometric equation for the aldol condensation of benzaldehyde with acetone
is given below.
2
MW: 102g/mol 58g/mol 234g/mol
(a) (If a reaction of 1.00 mL of benzaldehyde (density = 1.0415 g/mL) is reacted with
0.360 mL of acetone (density =0.7899 g/mL), what is the theoretical yield (in grams)
of product?
Answer:
Acetone is the limiting reagent with 0.00498 mol
0.00498 mol of acetone will theoretically produce 1.15g of product
(b) If the actual experimental yield for the reaction was 89%, how many mL of acetone
would be required to produce 4.00 grams of product?
Answer:
1.41 mL
O
H
+
O
NaOH/H 2O
C 2H 5OH
4. Ammonia, NH 3, forms complex ions with many metals. Like the iron oxalate
compound that you prepared this quarter, a variable number of the ammonia
ligands can replace water ligands on these metals. The general formula for a
nickel(II) chloride compound complexed with ammonia is
Ni(NH 3) xCl 2 .zH 2O
O

A student prepared this complex salt and decided to determine the weight of
compound that contained one mole of ammonia. A sample of the salt weighing
0.05932g was dissolved in water and the ammonia from it was titrated with 15.34
mL of 0.1000 M HCl.
The acid-base reaction is NH3 + H + +
à NH4 (a) How many moles of ammonia were in the sample?
Answer: (0.1534 mL)( 0.1000 M) = 1.534 x 10 -3 moles
(b)What weight of the complex salt contains one mole of ammonia?
Answer: (0.05932 g)/1.534 x 10 -2 mol = 38.67 g
Answer: 6
(c) What is the coefficient x?
5. Chloride, forms complex ions with many metals. Rather than making the oxalate
compound that you did this quarter, you could have prepared an iron chloride
complex replacing the water ligands with chloride. The general formula for an
iron (III) chloride complex ion is
[FeCl x .zH 2O] y+
(a) Iron predominantly forms complexes with 6 ligands. Considering this, what
stoichiometric situation would lead to a neutral chloride complex rather than an
ion?
Answer: x = 3, z = 3
(b) For your answer in (a), what weight of complex contains one mole of
chloride.
Answer: MW/3
(c) Explain why there is only one condition that satisfies part (a).
Answer: Cl- is the only negative ligand that can counter the charge on the Fe3+ ion.
6. The molecular structure of caffeine is given below. If an absorbance of 0.85
were recorded for a solution concentration of 1mg/40mL of water, what is the
extinction coefficient of caffeine at 270nm. (Assume a 1cm diameter UV cell).

Answer: A = ε bc or ε = 6,602
7. An unknown soil sample for a freshman chemistry class was determined to
contain both KCl and CoCl 2 as the only soluble inorganic compounds. In order to
determine how much of each compound was present, a 10.05-g sample of the soil
was extracted with three 30.0-mL portions of water. The extracts were
quantitatively transferred to a 100-mL volumetric flask and diluted to the mark.
A spectrum of the solution gave an absorbance of 0.238 at a λ max of 750 nm. For
comparison, a standard solution of 1.254 g of CoCl 2 . 6H20 dissolved in 100.00 mL
of water gave an absorbance of 0.639.
A 10.00-mL aliquot of the volumetric extract solution was evaporated in a tared
beaker. The resulting residue weighed 0.0542 g.
Calculate the percentages of KCl and CoCl 2 in the original soil sample.
Answer: There is 0.255 g CoCl 2 and 0.287 g KCl or 2.5% CoCl 2 and 2.9% KCl
8. Benzoic acid readily reacts with base to form a salt. A group of 30AL students
decided to quantitatively titrate the aqueous sodium benzoate phase extracted
from their “soils anaysis” lab with 0.0100M HCl. What volume of HCl would
have been required to titrate the benzoate from a 10.00-g sample of “soil” if the
“soil” contained 2.63% acid initially?
Answer: 21.6 mL
H 3C
O
N
C
O
C
N
CH 3
C
C
CH3 N
9. When a drop of black ink is placed in the center of a piece of filter paper and
drops of water are placed on the ink spot, capillary action wets the paper in an
ever larger circle and separates the dyes in the ink into concentric rings. A
schematic diagram is shown below.
N
CH
Aqua
Navy
Lilac
Rose
Yellow

Considering this paper chromatographic separation of the dyes, identify the
stationary phase in the system, the mobile phase in the system, the color of the
most polar dye, and the color of the least polar dye. Explain your reasoning for
each of the choices.
Answer: paper, (The dye is initially adsorbed onto the paper, and moved with the
water. The yellow dye moves the least, therefore is most tightly bound to the paper
and is the most polar; the aqua dye moves the furthest.
10. Sketch the solid-liquid phase diagram for an ideal solution of the two component
system A and B. Assume the melting point of A is 120 o C, the melting point of B is
135 o C, and the eutectic has a composition that is 0.67 mole fraction in B at a
temperature of 95 o C. Label the components that exist in each phase of the diagram.
Describe the observations that would occur if you had a 1:1 solution of A and B and
cooled it from 140 o C to 90 o C
Answer:
150
140
130
melting point
of pure A
110
100
90
solution of
A+ B and solid A
solution of A + B
eutectic composition
solid A + B
0 0.20 0.4 0.6 0.8 1.00
mole fraction of B
solution of
A+ B and
solid B
melting point
of pure B
Eutectic
melting point
Answer: In cooling a mixture of A and B from 140 o C to 90 o C, several changes will
be observed. Between 140 o C and 105 o C the solution will cool at a steady rate and
remain liquid. At 105 o C the first solid will begin to form. The rate of cooling will
decrease as solid forms. At 95 o C solid will continue to form but there will be no
change in the temperature until all the liquid has solidified. Between 95 o C and 90
o C cooling of the solid will continue

11. An organic chemistry lab book gives the following solubility data for oxalic acid
(IUPAC name - ethanedioic acid)
9.5 g/100mL of water 23.7 g/100 mL of ethanol 16.9 g/100 mL of ether
(a) A scientist wanted to extract 40 g of oxalic acid that was in 1000 mL of water
into an organic solvent for further experiments. Which solvent above is the best
choice for the extraction? Explain your choice.
Answer: Since ethanol and water are miscible, you must use ether.
(b) Calculate the partition coefficient for your choice. (Be sure to define the terms of
your equation.)
Answer: Defining the partition constant Kp, as solubility in ether/solubility in water
gives Kp = 16.9/9.5 = 1.78
(c) Calculate the weight of acid remaining in the water phase if you extracted the 40
g/1000 mL of water with 1000 mL of the solvent you chose in part (a).
Answer: Kp = 1.78 = x/(40-x) and x = 25.6 g and 14.4g left in the water.
(Since the volumes are the same, the ratio of the weights are the same as the
ratio of the (wt/vol)’s
12. When a compound melts the intermolecular forces between the molecules are
broken. The following pairs of compounds have approximately the same molecular
weight. Predict which of each pair has the higher melting point. Explain your
reasoning.
(a) ethanol (C 2H 6O) and ethyl amine(C 2H 8N)
Answer: Ethanol. Alcohols form stronger hydrogen bonds than amines do.
(b) acetic acid ethyl ester and propanoic acid (both C 4H 8O 2)
Answer: Propioic acid. Hydrogen bonds are possible here also, but not in the ester.
(c) benzene (C 6H 6) and cyclohexane (C 6H 12)
Answer: Benzene. The polarizable pi bonds of the aromatic ring provide stronger
van der Waals forces between the molecules than are possible in cyclohexane
where sigma bonds exist.
(d) ammonium chloride (NH 4Cl) and methyl chloride (CH 3Cl)

Answer: Ammonium chloride. Methyl chloride is a covalent molecule. The only
intermolecular forces are van der Waals attractions. Ammonium chloride is an
ionic salt composed of amonium ions and chloride ions.
13. Explain why heating a sample rapidly during a melting point determination may lead
to the erroneous conclusion that
(a) the compound is impure
Answer: If the rate of temperature change is faster than the rate of transfer of heat
across the capillary tube, then the thermometer, which is measuring the
temperature of the environment, not the sample, will register more change than
actually occurs in the sample. The result will be a larger temperature range
than should be obtained, and the conclusion could be drawn that the compound
is more impure that it really is.
(b) the sample is a different compound
Answer: Again the same phenomenon will occur as above. It is possible that the
disappearance of the last solid will occur when the temperature of the
environment is higher than the temperature of the sample in the tube. Since the
impurities depress the melting point, then a final melting point above the
expected one, would imply a different compound.
14. Potassium dichromate can frequently be used as an oxidizing agent in place of
2-
potassium permanganate. In the redox reaction, the dichromate ion, Cr2O7 is
reduced to Cr 3+ .
(a) Write a balanced half reaction for this reduction.
Answer: First balance the masses of the chromium (the coefficient is 2)
Then balance the mass of oxygen (7 molecules)
Then balance the mass of hydrogen (14 protons)
Then balance for electrical neutrality (6 electrons on the left need to be added)
Check everything is balanced.
2- +
Cr2O7 + 14H + 6e- = 7H2O + 2Cr 3+
(b) Write a balanced redox equation for the reaction of dichromate with the oxalate
ion.
Answer: The half reaction for oxalate is
2-
C2O4 = 2CO2 + 2e-
Since the reaction has 2 electrons and the dichromate in part a has 6, the oxalate
reaction must be multiplied by 3 so that the sum of the two has both electrical
and mass balance.

2- + 2-
Cr2O7 + 14H + 3C2O4 = 7H2O + 2Cr 3+ + 6CO2 (c) What weight of potassium dichromate would be required to react with the
oxalate in a 1.00 g sample of potassium tris(oxalate)ferrate(III)?
Answer: One mole of dichromate reacts with 3 moles of oxalate. There are
three moles of oxalate/mole of iron compound. Thus, one mole of potassium
dichromate is required for each mole of potassium tris(oxalate)ferrate(III).
Calculate the moles of iron salt in 1.00 g of this compound, then multiply this
by the molecular weight of potassium dichromate.
15. Calculate the equivalent weight of each of the following salts.
(a) Ni(NH 3) 6Cl 2
(c) Ni(NH 3) 5(H 2O) 3Cl 2
(e) Ni(NH 3) 3(H 2O) 3Cl 2
(b) Ni(NH 3) 5(H 2O)Cl 2
(d) Ni(NH 3) 4(H 2O) 2Cl 2
(f) Ni(H 2O) 6Cl 2
Answer: The equivalent weight of a compound is that weight of the compound that
reacts with one mole of acid or base. Since one mole of acid reacts with each
mole of ammonia, the equivalent weight is the weight of compound that contains
one mole of ammonia. In other words the equivalent weight is the molecular
weight divided by the coefficient indicating the moles of ammonia in the
compound. Note compound (f) does not contain ammonia, there is no
equivalent weight for the compound.
16. What volume of 0.100 M HCl would be required to directly titrate one gram of each
of the salts in question 6.
Answer: If you know the equivalent weight (g/equivalent) from question 6, you
can calculate the equivalents/ in a 1-gram sample for each of the salts. Since
equivalents of acid = equivalents of base, and HCl has one proton/mole or one
equivalent/mole, then the volume of acid needed = equivalents of base/molarity
of acid.
17. In the aspirin experiment you recrystallized your product from an ethanol solution to
which you added water. In the iron oxalate experiment you recrystallized the product
from aqueous solution to which you added ethanol. Explain why these two
recrystallizations were carried out in opposite directions.
Answer: In the case of aspirin, you had a saturated ethanol solution for an organic
compound. To decrease the solubility you added a more polar miscible solvent -
water. In the case of the iron oxalate complex, you had an aqueous solution of
the salt. When you added the ethanol, you decreased the polarity of the solvent
and thereby the solubility of the salt.

18. The Merck Index lists the solubility of cholesterol in alcohol as 1.29% (w/w) at 20 o
and 28g/100g at 80 o )
(a) What volume of alcohol would you use to recrystallize 5 g of cholesterol?
Answer: You want to dissolve the cholesterol in the minimum volume of hot
solution. If 28 g dissolve in 100 g of ethanol, then by direct proportions, you
will require 17.9 g of ethanol to dissolve 5 g of cholesterol. The density of
ethanol is 0.79 g/mL. You will need 22.6 mL of ethanol.
(b)What is the maximum weight of product you could recover in a recrystallization?
Answer: At 20 o C, the solubility is 1.3 g/100g. The difference in solubility between
80 o and 20 o is 26.7 g. The percent recovery is therefore, 26.7/28 = 95.4%. In the
situation in part (a) you could recover 95.4% of 5 g = 4.7 g.
19. The four compounds below form a mixture of products from an oxidation reaction
of an alcohol. (Note: The C 6H 5 group is an aromatic phenyl ring.)
O
C 6 H 5 -C-C(CH 3 ) 3
CH 3
CH 3 -C-OH
CH 3
C 6 H 5 C
O
H
C 6 H 5 C
I II III IV
(a) What IR absorptions would easily distinguish these four molecules?
Answer: (I) will have a distinct C=O stretch and both aliphatic and aromatic C-H
stretches. (II) will have a broad O-H stretch. (III) will have a distinct C=O stretch
and only aromatic C-H stretches. (IV) will have both a broad O-H stretch and a
distinct C=O stretch.
(b) Sketch the 13 C NMR spectra of each of these molecules. Sketch the 13 C DEPT
spectra for these molecules.
Answer: (I) When there is only one group substituted on the 6-membered aromatic
ring you have symmetry on the aromatic ring and only 4 lines will be seen. The
three CH 3’s of the trimethyl group are also identical. You will get 7 lines. The
ketone will occur around 200ppm, there will be 4 lines in the 130 - 180 ppm region
due to the phenyl ring, and two lines in the aliphatic region around 35 and 25.
In the DEPT 90 spectrum you will see three of the lines in the 130 - 180 ppm range
due to the three types of CH group. In the DEPT 135 spectrum, you will see the line
attributed to the CH 3’s (25ppm) as well as the three lines due to the CH’s that were
present in the DEPT 90.
O
OH

(II) This compound will show two signals in the 20 – 60 ppm range in the full 13 C
spectrum. The DEPT 90 spectrum will show no signals; the DEPT 135 spectrum
will show only the signal due to the three identical methyl carbon atoms.
(III) and (IV) The 13C spectrum of these two compounds will differ mainly in the
position of the carbonyl carbon. Conjugated aldehydes absorb in the 175- 195 ppm
range; conjugated acids absorb in the 160- 175 range. Both compounds will show
the characteristic four signals of a mono-substituted aromatic ring in the 130- 180
ppm region. The DEPT spectra will clearly distinguish between these compounds
based only on the number of signals. Since the aldehyde carbon has one hydrogen
atom attached, it will give a signal in both the DEPT 90 and DEPT 135 spectrum.
The carboxylic acid carbon has no attached hydrogen atoms, and will not appear in
either of these two spectra.
(b) Rank the compounds in order of least to highest R f, if the mixture were analyzed by
TLC.
Answer:
The most polar compound will have the smallest Rf and the least polar will move the
furthest giving the largest Rf. The order is IV < II < III < I
20. (a) Sketch the 13 C NMR spectrum of caffeine, oxalic acid, and salicylic acid.
Answer: You should have the structures of all of these compounds, which you
worked with in lab, in your notebook.
Caffeine: There is no symmetry in the molecule, so you will expect 8 lines in the 13 C
spectrum. There are three different methyl groups. These absorptions occur at
about 25, 29 and 34 ppm. The remaining 5 sp 2 carbon atoms occur at 108, 142,148,
152 and 158 ppm.
Oxalic acid: This compound has symmetry so there will only be one signal. The
carbon is a carbonyl of an acid group. The line will appear around 170 ppm.
Salicylic acid: There is no symmetry in this molecule. All the carbon atoms are sp 2
hybridized. The absorptions will all occur between 100 and 200 ppm.
(b) Explain how you could separate these three compounds.
melting point solubility in
water
solubility in
ether
Caffeine 238 (sublimes) 1 g/46 mL 1 g/530 mL
Oxalic acid 101 (sublimes) 1 g/7 mL insoluble
Salicylic acid 157 - 158 1g/460 mL 1 g/3 mL

Answer: Extract the mixture with water and ether. The caffeine and oxalic acid will
go into the aqueous layer, the salicylic acid will move into the ether layer. Add salt
to the aqueous layer and extract with propanol as you did in class. The caffeine will
transfer into the organic phase, the oxalic acid will stay in the aqueous phase.
(c) What is the equivalent weight of each of the compounds?
Answer: Caffeine is a weak base, and can accept one proton/per molecule.
Therefore, equivalent weight = molecular weight.
Oxalic acid has two acidic groups; EW = 1/2 MW
Salicylic acid has one acidic proton; EW = MW
21. The pain reliever phanacetin (Tylenol) has a solubility of 1.0 g/1310 mL in water and
1.0 g/mL in diethyl ether.
(a) Calculate the partition coefficient Kether/water.
Answer: (1310 g/1310 mL) ether / (1.0 g/1310 mL) water = 1310
(b) If 50 mg of phenacetin were dissolved in 100 mL of water, how much ether
would be required to extract 90% of the phenacetin in a single extraction?
Answer: 90% of the phenacetin = 45 mg.
1310 = (0.045 g/x mL) ether / (0.005 g/100mL) water = 0.68 mL.
22. An analytical lab was given a sample of a white powder of an unknown organic solid
for analysis as a suspected controlled substance. The person the sample was taken
from claimed it was over-the-counter pain medication. The lab carried out a
traditional elemental analysis to determine the composition of the unknown and then
obtained the mass spectrum, IR spectrum, and 13 C and DEPT NMR spectra.
(a) The elemental analysis report indicated that the compound consisted of 63.5% carbon,
5.9% hydrogen, 21.2% oxygen, and 9.3% nitrogen, What is the empirical formula of
the compound?
Answer: Assume 100g of compound
Atom Weight moles ratio of whole numbers
C 63.5 5.29 8
H 5.9 5.9 9
O 21.2 1.33 2
N 9.3 0 66 1
Empirical formula = C8H9O2N
(b) What is the empirical formula weight of the compound?
Answer: 151.2g

The mass spectrum of the compound is shown below.
(c) What is the molecular formula for the compound?
Answer: Molecular ion peak at 151. Therefore, empirical formula = molecular
formula - C8H8O2N
(d) Calculate the number of unsaturations and/or rings in the compound.
Answer: 5
(e) Based on the molecular formula and the number of unsaturations and/or rings, what
functional groups could be present in the compound.?
Aromatic ring, ketone, amine, alcohol
Aromatic ring, ketone, amine, ether
Aromatic ring, ester, amine
Aromatic ring, acid, amine
Aromatic ring, amide, alcohol
Aromatic ring, aldehyde, alcohol, amine
Aromatic ring, nitro group
Two triple bonds, nitro group
etc
(f) The IR spectrum of the compound is shown below.

What functional groups or structural information can be confirmed or deduced about the
compound from this spectrum?
Answer:
Amide, alcohol, aromatic and aliphatic C-H stretches
(g) The 13 C and DEPT NMR spectra of this same compound are shown below.

(g) Is there symmetry in the compound? How many quaternary, methyne, methylene and
/or methyl compounds are there in the compound?
Symmetry yes (8 C atoms in formula, 6 signals in 13 C NMR)
Quaternary 3 (no signals in DEPT spectra)
( C, no H)
Methyne (-CH) 2
Methylene (-CH 2) 0
Methyl (-CH 3) 1
(h) Draw two structures for the compound that are consistent with all the spectra and the
chemical formula.
HO NH
O
23. A scientist was given an unknown organic compound for analysis. She obtained its
mass spectrum, IR spectrum, 13 C and DEPT NMR spectra, and an elemental analysis
report.
(a) The elemental analysis report indicated that the compound consisted of 78.48%
carbon, 6.59% hydrogen, and 14.94% oxygen. What is the empirical formula of
the compound?
(b)
Answer: Assume 100g of compound
Atom Weight moles ratio of whole numbers
C 78.48 6.53 7
H 6.59 6.54 7
O 14.94 0.933 1
Empirical formula = C7H7O
Empirical formula weight: 107
(c) The mass spectrum of the compound is shown below. What is the molecular
formula for the compound?
HO
Answer:
Molecular weight = weight of molecular ion = 214
O
NH

(c) (5 points) Calculate the number of unsaturations and/or rings in the compound.
Answer = 8
(d) (10 points) A 10 mg sample of the compound was dissolved in 100 mL of ethanol
and the UV spectrum obtained using a 1-cm diameter quartz cell. The maximum
absorbance at 270 nm of 0.765. Calculate the molar absorptivity, ε, of the
compound.
Answer:
Concentration of solution = 4.67 x 10 -4 M
ε = 1638 ∼ 1640 M -1 cm -1

(e) The IR spectrum of this compound is shown below.
What can be deduced about functional groups in the compound from this spectrum and
the elemental composition?
Answer:
OH stretch, aromatic and aliphatic CH, no carbonyl
(f) The 13 C and DEPT NMR spectra of this same compound are shown below.
13 Carbon NMR Spectrum

13 C DEPT Spectra
Complete the following table based on these NMR spectra.
Compound
Is there symmetry in the compound? Yes ___X__ _
# of kinds of quaternary carbons 2
# of kinds of methyne carbons 3
# of kinds of methylene carbons 0
# of kinds of methyl carbons 1
(g) Deduce a structure for the compound. Write it in the answer box. No work outside
of the box will be graded.
HO
CH 3
H
OH