CHEMISTRY: ART, SCIENCE, FUN

PRACTICALEXAMINATIONSOLUTION andGRADING SCHEMEJULY 18, 2007
MOSCOW, RUSSIA
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Problem 1. Ion-exchange chromatography of amino acids.
1.1a 3 marks for each peak if at least two blank wells between peaks
9 marks maximum.
1.1b Not graded.
1.1.c 1 point for each proper choice. Combined mixtures should include all fractions identified as“2” and “3” and should be free of fractions without amino acids. Fractions identified as “1” maybe or may be not included.
3 marks maximum.
1.2-1.3

Al × n × V × M W,
e ×l
Al is the absorbance of the sample calculated from the spectra, l is the optical length (1.0 cm), n
is the dilution factor determined as a ratio of the aliquot of analyzing solution (0.1 mL) and thefinal volume of the sample in the cuvette, V is the volume of the combined fraction from thecorresponding peak, and MW is the molar mass of the amino acid.
3 marks for correct formula,
3 marks for correct determination of optical densities (1 mark for each assay)
1 mark penalty for miscalculation.
Content of an amino acid =Marks

Content of each amino acid is determined by using the following plot (values recalculated fromvolumes reported by students and absorbance values recorded by spectrophotometer)Pmax% mass determinedminmacc

min
mexp

max
mexp

max
macc

Official English version

Scoring:
m < min macc OR m > max macc

0 marks

min macc < m < min mexp

Pmax

(m - min macc )marks(min mexp - min macc )Pmax marks

min mexp < m < max macc
Amino acid

Minmexp/mexact
·100%
85%
85%
45%Minmacc/mexact
·100%
70%
60%
25%

Cys
HisArgPmaxMax
macc/mexact·100%
115%
115%
85%

Max
mexp/mexact·100%
110%
110%
65%

22222272 marks maximum
1.4 As it is given in the task text, mixed disulfide and 2-thio-5-nitrobenzoic acid are formed inthe reaction.
NO2+O-OO

NH3++O-O-

SHO

S

SO

OH-

H3N-H 2O

O-

SO

O-

S

S-O-+ONO2(I)

NO2(II)

NO2

Under slightly alkaline conditions, thiol group of (II) dissociates, and thiophenolate-anion isformed. Resonance structures can be realized for this compound:
O-OO-

COO-+COO

OO+N-

NSO--

SNO-+S-,
The electronic structure of asymmetrical disulfide (I) does not differ considerably from that of
the original Ellmann reagent. Therefore, it can be concluded that the compound responsible forcoloration is thiophenolate-anion (a form containing C=S bond).
O

2 marks for three correct ionized structures, one of which contains C=S bond
1 mark for three other structures, all without C=S bond
1 mark for less than three structures, one of which contains C=S bond.
2 marks maximum.

Official English version

Problem 2. Determination of carbonate and hydrogen phosphate in anabrasive sample.
2.1a, 2.2, 2.3 The values of the final volumes V1,f, V2,f, and V3,f, (as reported in the Answer
Sheet) are graded according to the following scheme:
DV = |Vreported – Vtrue|,
Vreported is either V1,f, V2,f, and V3,f, Vtrue is the corresponding master value (will be given with thecopies of students’ works).
Value of DV
DV£DVexpected
DV³DVacceptable
DVexpected£DV£DVacceptable

Marks250
æ DVacceptable - DV ö÷
25 ´ çç DV÷VD
exp ected øè acceptable

The values of DVexpected and DVacceptable (in mL) are listed in the table below.
DV, mL
expected
acceptable
V1,f
0.10
0.25
V2,f0.15
0.40
V3,f0.15
0.40
25 marks maximum for each titration
2.1b Calculation of NaOH concentration
Your workc( HCl ) ´ V ( HCl ) ´ V (aliquot ) 1.214 (mol / L) ´ 10.00(mL) ´ 10.00(mL)=c(NaOH) ==100.0(mL) ´ V1, f (mL)V ( flask ) ´ V ( NaOH )=с(NaOH) = __________ mol/L
2.4 Calculation of the mass of CO32Your workm(CO32-) (g) =c( NaOH ) ´ (V1, f - V3, f ) ´ V ( flask )M(CO32-)´1/2´=V (aliquot )c( NaOH ) (mol / L) ´ (V1, f - V3, f ) (mL) ´ 100.0 (mL)´ 0.001 ( L / mL) ==60.01 (g/mol)´1/2´10.00 (mL)
=m(CO32-) = __________ g

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2.5 Calculation of the mass of HPO42Your workm(HPO42-) (g) =c( NaOH ) ´ (V3, f - V2, f ) ´ V ( flask )= M(HPO42-)´=V (aliquot )c( NaOH ) (mol / L) ´ (V3, f - V2, f ) (mL) ´ 100.0 (mL)´ 0.001 ( L / mL) == 95.98 (g/mol)´
10.00 (mL)=m(HPO42-) = __________ g
2.1b, 2.4, 2.5
Correctness of calculations is graded by:1) comparing the numerical result (including the number of significant figures) reported by astudent with that obtained from the student's data using correct method of calculation;2) checking out the a student's way of calculation.
5 marks maximum for completely correct calculation of each value
Penalty for errors in calculations or data presentation.
Error type
1
2
3
4
56
78

Incorrect method of calculation
Mistakes in dilution factor
Confusion between units of measurement
Mistake in stoichiometric (equivalency) relationships
More or less than 4 significant figures in c(NaOH)
More than 4 or less than 3 significant figures in m(CO32-)
and m(HPO42-)
Rounding errors affecting the 1st or 2nd significant figuresin the final result
Miscalculations and slips

Penalty marks (forcalculation of each value)-5-1-1-1
-0.5*
-0.5*
-0.5**-0.2 per error*

Only the number of significant figures in the final answer is taken into account.
Not applied if the error originates from insufficient number of significant figures in previously
calculated c(NaOH). No double penalty!
**

2.6aCa2+ + H2PO4- ® CaHPO4 + H+or
3Ca2+ + 2HPO42- ® Ca3(PO4)2 + 2H+In course of these processes free protons evolve influencing the results of titration.
1 mark if the Answer Sheet contains at least one correct reaction equation. Incorrect equations
are not penalized.

Official English version

2.6bMistake
Below pH 4.5 (first endpoint, V2) H2CO3 is not titrated
Between pH 4.5 and 10 (second endpoint, V3) H2CO3 is titrated
Thus, increase of V3 ; V1 and V2 unchanged
No influence, C2O42- is too weak base
Increase of V1 and proportional decrease of c(NaOH) [asc(NaOH)V1 is equivalent to the initial amount of standard HCland so remains constant]; V2 and V3 unchanged
Losses of sample leads to proportional decrease of V2 and V3and therefore V3-V2; V1 unchanged
Increase of V2, V1 and V3 unchanged
Increase of V3, V1 and V2 unchanged

Error
CO3
HPO42content
content2--+00--+-0-+

Official English version