a Sapling Learning macmillan learning Spectroscopic data for three organic dyes,

ID: 1049682 • Letter: A

Question

a Sapling Learning macmillan learning Spectroscopic data for three organic dyes, A, B, and C, are shown in the table. A solution containing a mixture of these three dyes in a 1.000 cm cuvet had absorbances of 0.4563 at 450 nm, 0.3520 at 480 nm and 0.6549 at 520 nm. Find the concentrations of the three dyes in the solutions. For each solution, the zero is set with a blank. Number Number nm 450 480 520 M EA EB EC cm-1 M-1cm-1 M TCm 3 200 10 900 19 300 9 400 10 100 5 900 27 500 12 200 3 500 Number [c]- M a

Explanation / Answer

(nm)

A (M-1cm-1)

B (M-1cm-1)

C (M-1cm-1)

450

3200

10900

19300

480

2900

9400

10100

520

27500

12200

3500

Let CA, CB and CC be the concentrations.

At 450 nm,

0.4563 = (3200.CA + 10900.CB + 19300.CC)*1.00 ……(1)

At 480 nm,

0.3520 = (2900.CA + 9400.CB + 10100.CC)*1.00 ……(2)

At 520 nm,

0.6549 = (27500.CA + 12200.CB + 3500.CC)*1.00 …..(3)

At 450 nm, A absorbs minimally and hence, we can ignore the contribution due to A.

Similarly at 480 nm, A absorbs minimally and we can ignore the contribution from A. Re-frame (1) and (2) as

0.4563 = 10900.CB + 19300.CC ……(4)

0.3520 = 9400.CB + 10100.CC …..(5)

Multiply (4) by 9400 and (5) by 10900 and subtract,

9400*(0.4563) – 10900.(0.3520) = 10900*9400*CB + 19300*9400*CC – 9400*10900*CB – 10100*10900*CC

====> 452.42 = 7.133*107*CC

====> CC = 6.343*10-6 6.34*10-6

Plug the value of Cc in (4) and get

0.4563 = 10900.CB + 19300*(6.34*10-6)

====> CB = 3.064*10-5 3.06*10-5

Plug the values of CB and CC in (3) to obtain,

0.6549 = 27500.CA + 12200*(3.06*10-5) + 3500*(6.34*10-6)

====> CA = 9.432*10-6 9.43*10-6

Ans:

[A] = 9.43*10-6 M

[B] = 3.06*10-5 M

[C] = 6.34*10-6 M