Chemistry 3410
1.
January 18th, 2008
Test #1 Answers
Convert the wavelength 432.5 nm into frequency (Hz) and wavenumbers (cm–1).
[4]
υ=
2.
c
λ
=
2.998 × 108 m / s
−9
432.5 × 10 m
= 6.932 × 1014 Hz
υ=
1
λ
=
1
= 2.312 × 104 cm −1
−7
432.5 × 10 cm
One mole of photons is called an Einstein of radiation. Calculate the energy in S.I. units of one Einstein of
monochromatic radiation with a wavelength of 354.7 nm.
[4]
E=
3.
hc
λ
NA =
6.626 ×10-34 J s × 2.998 ×108 m / s × 6.0221420 ×1023 mol -1
= 337.3 kJ / mol
354.7 ×10-9 m
Titanium is reacted with hydrogen peroxide in 1 M sulphuric acid to form a coloured complex. If a 2.00 × 10–5 M
solution absorbs 31.5% of the incident radiation at 415 nm in a standard path-length cell (1.00 cm), what would be (a)
the molar absorptivity, ε, and (b) what would be the absorbance measured for a 6.00 × 10–5 M solution?
[5]
(a)
(b)
4.
ε=
P
1
1
1
log o =
log
= 8.22 × 103 M −1cm −1
−5
1 − 0.0.315
cb
P 2.00 ×10 M ×1.00 cm
log
Po
P1
A2
=
ε c1b
⇒ A2 =
ε c2b
log
Po
P1
c1
log
× c2 =
1
(1 − 0.315)
× 6 = 0.493
2
Consider a molecule that can fluoresce from the S1 state and phosphoresce from the T1 state. Which is emitted at longer
wavelength, fluorescence or phosphorescence? Make a sketch showing absorption, fluorescence, and phosphorescence
(from the same molecule) on a single spectrum axis (Intensity vs. Wavelength.)
[5]
Phosphorescence is always emitted at longer wavelength (for any given luminescent molecule).
5.
(a) Iron(II) is determined spectrophotometrically by reacting with 1,10-phenanthroline to produce a complex that
absorbs strongly at 510 nm. A stock standard iron(II) solution is prepared by dissolving 0.0702 g of ferrous ammonium
sulphate, Fe(NH4)2SO4·6H2O, in water in a 1.000 L volumetric flask, adding 2.5 mL H2SO4, and diluting to volume. A
series of working standards is prepared by transferring 1.00-, 2.00-, 5.00- and 10.00 mL aliquots of the stock solution
to separate 100.0 mL volumetric flasks and adding hydroxylammonium chloride solution to reduce any iron(III) to
iron(II), followed by 1,10-phenanthroline solution and then dilution to volume with water. A sample is added to a 100.0
mL volumetric flask and treated the same way. A blank is prepared by adding the same amount of reagents to a 100.0
mL volumetric flask and diluting to volume. The absorbance readings listed below are obtained after subtracting the
intensity of the blank. Determine the amount of iron in the unknown sample (as mg of elemental Fe).
[22]
Solution
Standard 1
Standard 2
Standard 3
Standard 4
Unknown sample
Fe
1.0079
Solution
Standard 1
Standard 2
Standard 3
Standard 4
Unknown sample
32.066
Stock Concentration:
55.847
N
14.0067
H
Absorbance
0.078
0.169
0.435
0.856
0.463
Absorbance
0.078
0.169
0.435
0.856
0.463
Aliquot,mL
Aliquot, mg Fe in 100 mL
1.00
0.013241288
2.00
0.026482576 Parameter
5.00
0.066206441 Std Dev
10.00
0.132412881 R^2
S
O
15.9994
Absorbance =
0.0702
g FeAmSulf
296.0784MM FeAmSuf
13.24128812mg Fe in 1000 mL
Slope
Intercept
0.1533509 0.00062238
0.00164243
0.0008032
0.99977063 0.00099261
Note: the regression formula was NOT
taken from the graph but from the
LINEST function, which includes errors!
0.153350898*mgFe +
0.000622376 mg(Fe)=
Amount
Error
0.07162384 0.00076045
mg(Fe) in unknown =
Best answer = 0.072±0.002 mg Fe
(b) Does Beers Law hold in this experiment?
Provide a brief answer!
m g(Fe) vs. Absorbance
y = 0.1534x + 0.0006
R2 = 0.9998
0.14
0.12
We note that the line in the linear regression is very
straight, implying linear relationship between
concentration of standard and the absorbance; this
implies that Beer’s law holds. Note also that the
concentrations are in the desired low range.
(c) Estimate the error in your unknown answer and
explain briefly how you obtained this.
0.0716±.0008
0.1
0.08
0.06
0.04
0.02
0
0
0.2
0.4
0.6
0.8
1
A b so r b ance o f so lut io n
The error analysis performed here is described on
page 68-69 of Harris, 7th Edition. It uses the Excel
function LINEST to perform the linear regression analysis with errors.
This calculation does not take into account the inherent errors of each measurement, so it is most likely a significant
underestimate of the true errors of the experiment. I chose to round up the error to the nearest 10, making it ±0.001 mg.
This is thus a conservative estimate of the error of measurement using the statistics of the four standard measurements
as a basis of the error estimate.
Note: by graphing mg(Fe) on the y axis and A on
the x, the parameter I want is the direct result of
the L.R. equation as written!
Be sure to review the text book treatment on error analysis (Chapter 4) which will now make
much more sense to you than when you were just starting Chem2410!