Question

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Spectrophotometry Extra Credit Question:

a)A 0.00004 M sample of Compound X in a solvent has an absorbance of 0.309 at 528 nm in a 1.000-cm cuvet. The solvent alone has an absorbance of 0.065. What is the molar absorptivity of Compound X?

b)A different sample of Compound X in the same solvent has an absorbance of 0.4 at 528 nm when measured with the same cuvet. What is the concentration of this second sample?

c)What absorbance value corresponds to 45% Transmission?

d)If a 0.015 M solution of a certain analyte exhibits 42% T at a wavelength of 280 nm, what will the percent transmittance be for a 0.030 M solution of the same analyte? Assume the pathlengths for each measurement are equal.

Answer 1

a) The concentration of sample (c) = 0.00004 M = 4*10^-5 M

The absorbance of solution (A) = 0.309

The length of cuvet (l) = 1 cm

According to the Beer's law, one can write as shown below.

A = (/2.303)cl, where is the molar absorptivity of the compound

i.e. = 2.303*A / (cl)

= 2.303*0.309 / (4*10^-5 M * 1 cm)

~ 17791 M^-1 cm^-1

Therefore, ~ 17791 M^-1 cm^-1

b) Here, c = 2.303*A / (l)

i.e. c = 2.303*0.4 / (17791 M^-1 cm^-1 * 1 cm)

i.e. c = 5.18*10^-5 M

c) If %T = 45, i.e. T = 45/100 = 0.45

Formula: A = -LogT

i.e. A = -Log(0.45)

i.e. A = 0.347

d) Formula: A1/c1 = A2/c2

i.e. A2 = A1*(c2/c1)

i.e. A2 = -Log(0.42) * (0.03/0.015)

i.e. A2 = 0.754

i.e. -Log(T2) = 0.754

i.e. T2 = 10^-0.754

i.e. T2 = 0.1764

i.e. %T2 = 17.64