# Chem 265 Assignment Four Answers

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```Chem 265
[1]
Assignment Four with Answers
1) If an analyte that is to be measured by UV-vis absorption spectroscopy has a molar
absorptivity of 5.6 x 105L/mol.cm at the wavelength that is chosen for its measurement. If
the smallest absorbance that can be measured is 0.002 and the sample cuvette has a
pathlength of 1.00 cm, what is the lowest concentration of the analyte that can be
measured?
π΄
[3]
0.002
= 5.6 π₯105 πΏ/πππ.ππ∗1 ππ = 4 π₯ 10−9 M
πΈπ΅
ππππ =
2) A geochemist wishes to estimate the volume of an irregularly shaped pool of water.
This task is accomplished by placing a known amount of highly coloured dye into this
solution, letting it mix throughout and then measuring the absorbance of the dye in a
sample of the pool water.
For example, a geochemist used 1.00 L of a dye solution that had an absorbance of 0.768
when it was measured in a 1.00 mm cuvette at 450 nm. They put all of the dye solution in
the pool and allowed it to be mixed around well. They then took a sample of the pool
water, which was then a very dilute solution of the dye, and measured the absorbance in a
10.00 cm cuvette. The absorbance was 0.142 at 450 nm.
What is the volume of the pool?
Initially:
A = Ebc = 0.768 = E. 0.100 cm . conc
0.768
E = 0.10 ππ.ππππ
A = Eb. Dil So 0.142 =
So
ππππ
π·ππ
=
0.768∗10.00 ππ
0.10 ππ∗0.142
0.768
0.10 ππ.ππππ
. 10.00 cm . dil
= 541
So the volume has changed by a factor of 541, so the volume of the pool must be 541 L.
[2]
3) What is the energy of a photon that has a wavelength of 500 nm, if the refractive index
of the medium it is travelling through is 1.00?
Given that the refractive index is 1, then c/v = 1 so speed = 3.0 x 108 m/s
E=hc/λ
h = 6.626 x 10-34J.s
E=
h = 6.626 x 10−34J.s∗3.0 x 108 m/s
500 x 109 m
= 3.98 x 10-19 J
What is the frequency of this light?
ο΅=
[3]
c
ο¬
ο½
3.00x10 8 m sec -1
ο½ 6.00 x1014 s ο­1
-9
500 x 10 m
4) The following results were obtained by a clinical chemist when measuring morphine
at 285 nm and using a square cuvette as the sample holder.
Morphine Concentration (M)
5.00 x 10-5
1.00 x 10-4
2.0 0 x 10-4
5.0 0 x 10-4
Sample measurement 1
Sample measurement 2
Sample measurement 3
Absorbance
0.229
0.308
0.467
0.942
0.615
0.617
0.613
Prepare a Beers law plot for these data using the standards. Use this plot to determine the
concentration of morphine in the sample. Give your result with a 95 % confidence
interval, as it would go to a client.
0.00005
0.0001
0.0002
0.0005
0.229
0.308
0.467
0.942
532
539
542 5.131601
0.615
0.617
0.613
AV
0.000294
0.000295
0.000292
0.000293667
1.52753E06 SD
CI
3.79489E06 CI
(2.94 &plusmn; 0.04)x 10-4 M
1
y = 1584.6x + 0.1498
R&sup2; = 1
0.9
0.8
0.7
0.6
0.5
Series1
0.4
Linear (Series1)
0.3
0.2
0.1
0
0
[0.5]
[0.5]
0.0001
0.0002
0.0003
0.0004
0.0005
0.0006
Is a solution of morphine coloured? No – it is a white powder and gives a
colourless solution
Would this be analyzed by visble or UV absorption?
UV – we used 285 nm here
```