Consider the molecule anthanthrene dissolved in benzene. The first electronic tr
ID: 956830 • Letter: C
Question
Consider the molecule anthanthrene dissolved in benzene. The first electronic transition in thismolecule occurs near an energy of 23,090 cm?1. A series of measurements of its absorbance A (at23,090 cm?1) as a function of concentration c gives the following data:
The path length of the optical cell was 1.0 cm.
A) Determine the molar decadic absorption coefficient ?, in units of M?1cm?1, of the molecule at thiswavelength.
B) Determine the absorption cross-section ? of the molecule at this wavelength, where ?, in units ofcm2, is defined by:
C) Use bond increments (C-C: 154 pm, C=C: 134 pm, C-H: 109 pm) and van-der-Waals radii of thehydrogen atoms of 120 pm to estimate the physical size of the molecule. Compare your absorptioncross-section to the cross-sectional area of the molecule.
8. Consider the molecule anthanthrene dissolved in benzene. The first electronic transition in this molecule occurs near an energy of 23,090 cm1. A series of measurements of its absorbance A (at 23,090 cm') as a function of concentration c gives the following data 0.0830.414 0.820 4.16 c [106 M1.05.010. 50. The path length of the optical cell was 1.0 cm A) Determine the molar decadic absorption coefficient , in units of M-1cm-, of the molecule at this wavelength B) Determine the absorption cross-section of the molecule at this wavelength, where , in units of cm, ; is defined by 1000-In (10) . = (3.8 x 10-21 ) . C) Use bond increments (C-C: 154 pm, C-C: 134 pm, C-H: 109 pm) and van-der-Waals radii of the hydrogen atoms of 120 pm to estimate the physical size of the molecule. Compare your absorption cross-section to the cross-sectional area of the moleculeExplanation / Answer
a) we know that
Abosrption = X concentration X length
so = Absroption / concentration X length = 0.083 / 10^-6 Molar X 1 = 0.083 X 10^6 M-1 cm-1
b) Absroption cross section = = 3.8 X 10^-21 X (as given above)
=3.8 X 10^-21 X 0.083 X 10^6 = 0.3154 X 10^-15 = 3.154 X 10^-16