Question

Advanced Organic Chemistry

Q 1: Explain how you would separate K₄[Fe(CN)₆] from K3[Fe(CN)₆] ?

Q 2: Determine the mole fraction of CrCl₂(OH₂)₄⁺ and CrCl(OH₂)₅⁺ if 5mL of it was collected had an absorbance of 0.534 and 0.257. Assume that the extinction coefficients of the two coplexes are the same

Answer 1

Q 1: Explain how you would separate K₄[Fe(CN)₆] from K3[Fe(CN)₆] ?

The color of K₄[Fe(CN)₆] is bright blue and the color of K3[Fe(CN)₆] is a bright red. So to separate them first we will crystalize them from solution and pick the crystals according to their colors.

Q 2: Determine the mole fraction of CrCl₂(OH₂)₄⁺ and CrCl(OH₂)₅⁺ if 5mL of it was collected had an absorbance of 0.534 and 0.257. Assume that the extinction coefficients of the two coplexes are the same

Using Beer's Law (A = ebc, where A = absorbance, e=molar absorption coefficient, b is path length, and c is concentration); rearrange to solve for c:

c = A/(eb)

Since we're given its absorbance, the path length, and molar absorption coefficient in the problem, we can calculate concentration (in units of mol/L).

Assume that the extinction coefficients of the two coplexes are the same = 1.0 and length also 1.0 cm

Than first calculate the number of moles of both as follows:

c = A/(eb)

mol / L = A/(eb)

moles = A/(eb) *L

here volume = 5mL 0.005 L

CrCl₂(OH₂)₄⁺

Moles of CrCl₂(OH₂)₄⁺ = 0.534 /(1*1   *0.005

Moles of CrCl₂(OH₂)₄⁺ = 2.67*10^-3

CrCl(OH₂)₅⁺

Moles of CrCl(OH₂)₅⁺ = 0.257 /(1*1   *0.005

Moles of CrCl₂(OH₂)₄⁺ = 1.285*10^-3

Total mole = 3.955*10^-3

Mole fraction of CrCl(OH₂)₅⁺ = 1.285*10^-3/3.955*10^-3

=0.325

Mole fraction of CrCl₂(OH₂)₄⁺ = 2.67*10^-3/3.955*10^-3

=0.675