To determine the iron concentration in a California Burgundy

To determine the iron concentration in a California Burgundy wine, a 10-ml aliquot of the wine was added to each of four 50ml beakers. To the first beaker was added 10ml of water. To the second beaker was added 10ml of 4ppm Fe2; 10ml of 8ppm Fe? was added to the third; and 10ml of 12ppm Fe? was added to the fourth. If the measured absorbances were 0.028, 0.045, 0.063, and 0.077, what was the concentration of iron in the original wine? Briefly describe how a monochromator separates light into its component wavelengths. (Hint: it may be helpful to show an equation.) . 5. I want to isolate the 600 nanometer (nm) line from polychromatic light (400nm-800nm) using a blazed diffraction grating. The angle of incidence of the polychromatic relative to the grating normal is 40 degrees, and the angle of diffraction for the 600nm line 30 degrees. What is the distance between adjacent groves in the blazed grating necessary to achieve the desired result? (10pts) What is higher order diffraction? How does one go about eliminating higher order diffraction in a spectrometer? 6. 7. Consider a diffraction grating. which serves as the monochromator for a Spec 20.

Solution

3) Use the dilution equation to find the concentrations of the added Fe²⁺ standard in the 50 mL beakers. The dilution equation is

C₁*V₁ = C₂*V₂

where C₁ = concentration of stock Fe²⁺ added in ppm; V₁ = volume of stock Fe²⁺ added = 10 mL; V₂ = final volume of the solutions = 50 mL and C₂ = concentration of the added Fe²⁺ in the final solution. Therefore,

C₂ = C₁*V₁/V₂

Prepare the following table.

Solution

Concentration of added Fe²⁺ standard in the final solution

C₂ = C₁*V₁/C₂ (ppm)

Absorbance of the solution

1

0*(10)/(50) = 0

0.028

2

(4)*(10)/(50) = 0.8

0.045

3

(8)*(10)/(50) = 1.6

0.063

4

(12)*(10)/(50) = 2.4

0.077

Plot absorbance vs concentration of the added Fe²⁺ standard.

Plot of absorbance vs concentration for standard Fe²⁺ added

Use the regression equation to determine the concentration of Fe²⁺ in the original wine sample. The original sample produced negligible absorbance; hence put y = 0 in the regression equation and obtain

0 = 0.0041x + 0.0285

====> -0.0041x = 0.0285

====> x = -0.0285/0.0041 = -6.9512 -6.9

The concentration of Fe²⁺ in the original wine sample cannot be negative; hence, the concentration in the 50 mL diluted sample is 6.9 ppm.

We took 10 mL sample and diluted to 50 mL; hence, the concentration of Fe²⁺ in the original wine sample = (6.9 ppm)*(50 mL/10 mL) = 34.5 ppm (ans).

Solution	Concentration of added Fe2+ standard in the final solution C2 = C1*V1/C2 (ppm)	Absorbance of the solution
1	0*(10)/(50) = 0	0.028
2	(4)*(10)/(50) = 0.8	0.045
3	(8)*(10)/(50) = 1.6	0.063
4	(12)*(10)/(50) = 2.4	0.077