PrintCalculatorPeriodic Table Question 16 of 16 Map Sapling Learning Spectroscop

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PrintCalculatorPeriodic Table Question 16 of 16 Map Sapling 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 0.5415 at 450 nm, 0.4003 at 480 nm and 0.5969 at 520 nm. Find the concentrations of the three dyes in the solutions. For each solution, the zero is set with a blank EA ?? Ec Mcm-1M 450 480 520 19 300 9 40010 100 10 900 5 900 27 500 12 200 3 Number Number Number [A-D MI

Explanation / Answer

Let CA, CB and CC are concentrations of A, B and C respectively.

Since A= ebC, e = molar extinction coefficient, b is path length and C is concentration

For the mixture at 450 nm, 0.5415= 3200*CA*1+ 10900*CB+19300*CC

Dividing by 3200, the equation becomes 0.5415/3200 = CA+10900CB/3200+19300CC/3200

0.000169= CA+3.41CB+6.03CC (1)

At 480 nm, 0.4003= 5900CA+9400CB+ 10100CC

Dividing with 5900, 0.4003/5900 = CA+9400/5900CB+10100CC/5900

6.78*10-5= CA+1.59CB+1.711CC (2)

Eq.1-Eq.2 gives 0.000101 = 1.82CB+4.32CC (3)

Dividing by 1.82, 0.000101/1.82= CB+4.32CB/1.82

5.6*10-5= CB +2.4CC (3A)

At 520 nm, 0.5969= 27500CA+12200CB+3500 CC

Dividing by 27500, 0.5969/27500 = CA+12200CB/27500 +3500CC/27500

2.17*10-5= CA+0.443CB+0.13CC (4)

Eq.1-Eq.4 gives 0.000148= 2.967CB+5.94CC (5)

Dividing by 2.967, the equation becomes 4.97*10-5= CB+2CC (5A)

Eq.3A- Eq.5A, 0.63*10-5= CC*(2.4-2), CC=1.575*10-5M

From Eq.5A, 4.97*10-5= CB+2*1.575*10-5

CB= 0.0000182M

From Eq.1, .000169= CA+3.41*0.0000182+6.03*1.575*10-5 (1)

CA= 1.196*10-5M

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