The selenium content in soil can be determined fluorometrically upon conversion

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 0.141 g soil sample was converted to the fluorescent product and extracted in 50.0 mL of Cyclopean. The method of standard additions was used to determine the selenium content in the soil sample. 3.00 mL of the Cyclopean solution were placed in a cuvette for fluorescence determination. Several small additions of a 2.25 yg 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

In the method of standard addition,for each addition

Cx/Cx+Cs=signal x/signal x+s

Cx= concentration of analyte without std addition

Cx+s= concentration of analyte with std addition

A plot of signal of Cx+s vs concentration of added sample yields a straight line .The conc of unknown is determined from the point where the extrapolated line meets the concentration axis at zero signal.

Total added Se (micro L)

Signal(Y)

Conc of added se(X)

1

0

54.3

0

2

20

64.5

2.98*10^-5 microgm/micro L

3

40

76.7

2.96*10^-5 microgm/micro L

4

60

91.1

4.4*10^-5 microgm/micro L

5

80

108.3

5.8*10^-5 microgm/micro L

Initial volume=3.00ml=3000 microL

Concentration of standard Se solution=2.25 microgram/ml=2.25microgm/1000 micro L=2.25*10^-3 microgm/micro L

Conc of added calculated by =(Volume added *2.25*10^-3 microgm/micro L)/3000+ added V

2)20*2.25*10^-3 microgm/micro L)/3000+ 20=2.98*10^-5microgm/micro L

3) 40*2.25*10^-3 microgm/micro L)/3000+ 40=2.96*10^-5 microgm/micro L

4) 60*2.25*10^-3 microgm/micro L)/3000+ 60=4.4*10^-5 microgm/micro L

5) 80*2.25*10^-3 microgm/micro L)/3000+ 80=5.8*10^-5 microgm/micro L

Conc of unknown=1.6*10^-5 micro g/microL

Mass in 3.00 ml=1.6*10^-5 micro g/microL=1.6*10^-5 g/L

Mass in 50.0 ml(soil sample)=0.05 L*1.6*10^-5 g/L=8.0*10^-7 g

Wt %=8.0*10^-7 g/0.141g*100=56.73*10^-5 %

Actual WT % calculations

Absorbance(Y)=X*e*l (beer’s law)

E=molar absorption coefficient

L=path length of light

Y=k*X ,k=constant=eL

K=Y/X

K can be calculated and average taken to calculate X for solution with zero volume of standard solution.

K1=64.5/2.98*10^-5 =21.5*10^5

K2=76.7/2.96*10^-5=25.5*10^5

K3=91.1/4.4*10^-5=20.7*10^5

K4=108.3/5.8*10^-5=18.7*10^5

Kaverage=k1+k2+k3+k4/4=21.6*10^5

X(actual)-=Y/K=54.3/21.6*10^5=2.5*10^-5 g/L

%Uncertainty= Actual WT %-CALCULATED WT %/ Actual WT % *100

           =2.5*10^-5 -1.6*10^-5/2.5*10^-5 *100

           =36%

Total added Se (micro L)

Signal(Y)

Conc of added se(X)

1

0

54.3

0

2

20

64.5

2.98*10^-5 microgm/micro L

3

40

76.7

2.96*10^-5 microgm/micro L

4

60

91.1

4.4*10^-5 microgm/micro L

5

80

108.3

5.8*10^-5 microgm/micro L

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