Q

Universiti Malaysia PAHANG Engineering . Technology Creativity

FACULTY OF CHEMICAL & NATURAL RESOURCES ENGINEERING

FINAL EXAMINATION

COURSE : ANALYTICAL CHEMISTRY

COURSE CODE : BKF1243

LECTURER : EMAN N. AL!

FARHAN BINTI MOHD SAID

DATE : 17 JUNE 2O15

DURATION : 3 HOURS

SESSION/SEMESTER : SESSION 2014/2015 SEMESTER II

PROGRAMME CODE : BKB/BKC/BKG

INSTRUCTIONS TO CANDIDATE:

I. This question paper consists of FOUR (4) questions. Answer ALL questions. 2. All answers to a new question should start on new page. 3. All the calculations and assumptions must be clearly stated. 4. Candidates are not allowed to bring any material other than those allowed by

the invigilator into the examination room.

EXAMINATION REQUIREMENTS:

1. APPENDICES

DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO

This examination paper consists of NINE (9) printed pages including front page.

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BKB/BKC/BKG/1415111BKF1243

QUESTION 1

Analytical chemistry results need to be evaluated before writing the reports. Therefore,

data evaluation consider as a normal practice while doing research work or analysis of

any product.

a) The following data (Table 1) shows the content of the glucose content in a sample

using ultraviolet-visible methods (UV-VIS) and high performance liquid

chromatography (HPLC). Determine whether there is a significant difference in

variance between the two methods.

Table 1: Data of glucose content on two different methods

No Glucose content (mg/ml)

sample UV-VIS HPLC

1 13.8 13.0

2 13.6 15.2

3 14.6 14.0

4 13.5 13.3

5 15.0 13.5

(17 Marks)

b) In the titration of 10.00 ± 0.14 ml of 0.10 ± 0.05 M NaOH, 20.50 ± 0.05 ml of HC1 is

required for neutralization. Determine the moiarity of HCl and identify the

uncertainty of the value.

(8 Marks)

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QUESTION 2

Gravimetric methods are one of oldest techniques in analytical chemistry which is

considered the fundamental method.

a) Determine the mass of Cu (103) 2 can be formed from 0.500 g of CuSO4.51-120.

(4 Marks)

b) A 0.2 121 g sample of an organic compound was burned in a stream of oxygen, and

the CO2 produced was collected in a solution of barium hydroxide. Calculate the

percentage of carbon in the sample if 0.6006 g of BaCO 3 was formed.

(4 Marks)

c) Describe:

i) Inclusion

(1.5 Marks)

ii) Occlusion

(3 Marks)

d) Titration is a method of analysis using standard material as titrant to get the

concentration of the analyte. Describe the preparation of:

i) 500 mL of 0.0750 M AgNO 3 from the solid reagent.

(3 Marks)

ii) 2.00 L of 0.325 M HCL, starting with a 6.00 M solution of the reagent.

(2.5 Marks)

iii) A 50.00 mL of 0.100 M NaOH is titrated with 0.100 M HCL. Calculate

the pH of the solution after addition of 0.00, 10.00, 50.00, and 60 mL of

acid, and draw a titration curve from the data.

(7 Marks)

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QUESTION 3

Measurements based on light and other forms of electromagnetic radiation are widely

used throughout analytical chemistry. The interactions of radiation and matter are the

subject of science called spectroscopy.

a) Beryllium (II) forms a complex with acetylacetone (166.2glmol). Calculate the molar

absorptivity of the complex, given that a 1.34 ppm solution has atransmittance of

55.7% when measured in a 1.00 cm cell at 295 nm.

(6 Marks)

b) Express the following absorbance in terms of percent transmittance:

i) 0.494

(2 Marks)

ii) 0.229

(1 Mark)

c) Convert the following transmittance data to absorbance:

i) 0.013

(2 Marks)

ii) 55.5%

(1 Mark)

d) A typical simple infrared spectrometer covers a wavelength range from 3 to 15 .im.

express the range in wavenumbers.

(4 Marks)

e) Explain wave property that is measured by Fourier Transform JR and the units for

this property?

(2 Marks)

f) Draw a simple spectrum using FTIR showing the following with naming x-y axis.

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i) Transmittance and,

(2 Marks)

Absorbance

(2 Marks)

g) List the names of three equipments in applying spectroscopy technology.

(3 Marks)

QUESTION 4

a) An HPLC separation of two components pharmaceutical product yielded the

following results (Table 2).

Table 2: Retention time and peak width of caffeine and aspirin on HPLC

Compound Retention time

(mm)

Peak width at base

(mm)

Peak width at V2

height (mm)

Solvent 30

Aspirin 75 6.5 3.0 Caffeine 86 8.1 3.25

Determine the following items:

i) The capacity factor, k, for each compound.

(4 Marks)

ii) The plat number of each compound using base widths.

(5 Marks)

iii) The resolution of the two compounds, R.

(3 Marks)

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iv) Given the column used in this analysis is 25 cm long, determine the height of the

theoretical plate?

(4 Marks)

b) The separation of two compounds on a packed and capillary column gave the

following data in Table 3:

Table 3: Data of two compounds on a packed and capillary column

Component Packed column Capillary column

Dead time (s) 15 30

Retention time of Component A (s) 160 75

Retention time of Component B (s) 170 86

Number of theoretical plates 6400 1967

Identify which column is giving the better resolution and, justify your answer.

(9 Marks)

END OF QUESTION PAPER

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a (tR)B tM

(tR)A tM

N = 16(t/w)2

u — Xx -J exp

S

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APPENDICES

±xi

N

hc - E = h V = - = hcv

A = —logT = —log- = log-- = abC = sbC po PT

R = 2Z 2[(tR)B—(tR)A]

WA + WB WA+WB

k' tRtM

(x )2

=i=1

N—i

F°2 exp

pX = - Jog [X]

SR _1JSA +SB

SR =

r(L;

2

R)

H

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Table 1: F-table for one-tailed test at 95% confidence level

i:TabIe for One-Tailed Test at cc= 0.05 (95% Confidence Level)

-1 2 3 4 5 6 7 8 9 10 15 20 co

i 161 4 1993 215.7 2244 2302 234.0

.---- .--

236.

--

238.9 2405 241,9 2459 248.0 2543 2 18.51 19.00 19.16 1925 1930 1933 1935 19.37 -19.38 19.40 19.43 19.46 19.50 10.13 91552 9.277 9.117 9013 8 8,845 8.812 8.786 8.703 S..660 8.526 4 7.709 6.944 5.591 6388 6.256 6.163 6.094 6.041 5.999 5964 5.858 5.803 5628 5 6.608 5,786 5.409 5 192 5050 4.950 4.876 4818 4.772 4,735 4.619 4.558 4365 6 5.987 5.143 4.757 4.534 4387 4284 4.207 4.147 4.099 4.060 3.938 3.874 3.669

5.591 4.737 4.347 4.120 3.972 3.666 3.787 1726 3677 3.637 3.511 3,445 3230 8 5,318 4.459 4.066 3.838 3.587 3.581 3.500 3438 3.358 3.347 1218 3.150 2.928 9 5.117 4256 3.863 3.63-3 3482 3374 3,293 3.230 3.179 1137 3.006 2.935 2307

10 4.965 4.103 3.708 3,478 3.326 3.217 3.135 3.072 3.020 2.978 2.845 2374 2338 ii 4.844 3.982 3.587 3.357 3204 3.095 3.012 1948 2896 2.854 2.719 2 646 2,404 12 4.747 3885 1490 3.259 3.196 2.996 2.913 2.849 1796 2.753 - 2.617 2.544 1296 13 4.667 3.806 3.411 3.179 3.025 2,915 2.832 2767 2.714 2671 2.533 2.459 2.206 14 4.600 3.739 3.344 3.112 1958 2.848 2.764 2.699 2.546 2,602 2.463 2.388 2.131 15 4.534 3 682 3.287 3056 2.901 2.790 2707 2.641 2588 2.544 2403 2.328 2.066 16 4.494 3.634 3.239 3.007 2.852 2.741 2.657 2.591 2.538 2.494 2.352 2.276 2.010 17 4.451 3.592 3.197 2.965 2.810 2.699 3614 2.548 2.494 2.450 2.308 2.230 1.960 18 4.414 3.555 3.160 2.928 2.773 2661 2.577 2.510 2.456 2,412 2.269 2.191 1.917 19 4.381 -3.552 3.127 2.895 2.740 2.628 2.544 2477 2.423 2.378 2134 2.155 1.878 20 4.351 3.493 3.098 2.866 2.711 2.599 2.514 2.447 2.393 2.348 2.203 2.124 1.843 00 3.842 2.996 2.605 2.372 2.214 2.099 2.010 1.938 1.880 L831 1.666 1570 1000

'e degrees of freedom in numerator v 2 = degrees of freedom in denominator

Table 2: F-table for two-tailed test at 95% confidence level

F-Table for Two-Tailed Test at a = 0.05 (95% Confidence Level)

V31V11 1 2 3 4 5 6 7 8 9 10 15 20

1 647.8 799.5 864.2 899.6 921.8 937.1 948.2 956.7 963.3 968.6 984.9 993.1 1018

2 38.51 39,00 39.17 39.25 39.30 39.33 39.36 39.37' 39.39 39.40 39,43 39-45 39498

3 17.44 16.04 15.44 15.10 14.88 14.73 14.62 14.54 14.47 14,42 14.25 14.17 13.902

4 12.22 10.65 9.979 9,605 9.364 9.197 9.074 8.980 8.905 8.844 8.657 8.560 8.257

5 10.01 8,434 7.764 7.388 7.146 6.978 6.853 6.757 6.681 6.619 6.428 6.329 6.015

6 .8.813 7.260 6.599 6.227 5.988 5.820 5.695 5.600 5.523 5.461 5,269 5.168 4.849

7 8.073 6.542 5.890 5.523 5.285 5.119 4.995 4.899 4.823 4.761 4.568 4.467 4,142

8 7.571 6.059 5.416 5.053 4.817 4.652 4.529 4.433 4.357 4,295 4.101 3.999 1670

9 7.209 5.715 5.078 4.718 4.484 4.320 4.197 4.102 4.026 3.964 3.769 1667 3.333

10 6.937 5.456 4,826 4.468 4.236 4.072 3,950 3.855 3,779 3.717 3.522 3.419 3.080

11 6.724 5.256 4.630 4275 4.044 3.881 3,759 3.664 3.588 3.526 3330 3.226 2283

12 6.544 5.096 4.474 4.121 3.891 3.728 3.607 3.512 3.436 3.374 3.177 3.073 2.725

13 6,414 4.965 4.347 3.996 3.767 3.604 3.483 3.388 3.31-2 3.250 3.053 2.948 2.596

14 6.298 4.857 4.242 3.892 3.663 3.501 3.380 3.285 1209 3.147 2.949 2.844 2.487

15 6.200 4.765 4.153 3.804 3.576 3,415 3.293 3.199 3.123 3.060 2.862 2.756 2.395

Ov, = degrees of freedom In numerator v3 degrees of freedom in denominator

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Table 3: Table of critical value of Q test

Ntmber of Obsrvation 90% Confidence

Q (Rtji.ci ii > Q.

5% ConiIdence 99% Confidence

3 0.941 0.970 04994 4 0.765 0.829 0.926 5 (o.oi 0.710 0.821 6 0.560 0.625 0.740 7 0.507 0L56 0.68()

O46 0.526 0.634 9 0.437 0.493 0,598

JO 0.412 0.466 0.568

Table 4: Periodic Table

The Periodic Table of the Elements

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