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The selenium content in soil can be determined fluorometrically upon conversion

ID: 940668 • Letter: T

Question

The selenium content in soil can be determined fluorometrically upon conversion of the selenium to the fluorescent product shown below, followed by extraction in cyclohexane. Emission from the fluorescent product is observed at 518 nm. The selenium content in a 0.200 g soil sample was converted to the fluorescent product and extracted in 50.0 ml. of cyclohexane. The method of standard additions was used to determine the selenium content in the soil sample. 3.00 mL of the cyclohexane solution were placed in a cuvette for fluorescence determination. Several small additions of a 1.51 mu g Se/mL standard solution containing the fluorescent product were added to the 3.00 mL solution in the cuvette as shown in the table below. The fluorescence was measured at 518 nm. Determine the weight percent (wt%) of selenium in the soil sample and find the absolute uncertainty in the wt% Se. Report the weight percent and absolute uncertainty with three significant figures.

Explanation / Answer

Total added Se std

Total added Se std

Sample volume

Fl. intensity

Corrected fl. Int.

mL

g

mL

au

au

0

0

3

54.3

54.3

0.02

0.0302

3.02

64.5

64.9

0.04

0.0604

3.04

76.7

77.7

0.06

0.0906

3.06

91.1

92.9

0.08

0.1208

3.08

108.3

111.2

Corrected fluorescence intensity = Fluorescence intensity x sample volume(mL)/3 mL

E.g.,     64.9 = 64.5x3.02/3

Use Excel to calculate

469.4

slope

51.9

intercept

for Corrected intensity vs. added Se mass

Intensity = 51.9 + 469.1 · (Added Se mass + x)

If x is the unknown mass of Se in the 3 mL sample

The calibration curve without addition is

Intensity = 469.1· x

If the sample intensity (without additions) is 54.3

54.3 = 469.1· x

x = 0.115 g = 1.15x10-4 g

% Se in sample = 100x 1.15x10-4 g/ 2x10-1 g = 0.0575%

Use the function STEYX in Excel to calculate the standard deviation for the pairs x,y in your calibration curve.

Std dev = 0.00580 = 0.0058 g

In % = 100x0.0058 x10-6 g/ 2.9 x10-1 g = 0.00290%

Thus the reported result will be

(0.0575 ± 0.0029) % or

0.0575 % ± 0.0029 %

Total added Se std

Total added Se std

Sample volume

Fl. intensity

Corrected fl. Int.

mL

g

mL

au

au

0

0

3

54.3

54.3

0.02

0.0302

3.02

64.5

64.9

0.04

0.0604

3.04

76.7

77.7

0.06

0.0906

3.06

91.1

92.9

0.08

0.1208

3.08

108.3

111.2