Worksheet: UV-Vis Spectroscopy Worksheet: UV-Vis Spectroscopy Worksheet: UV-Vis

Question

Worksheet: UV-Vis Spectroscopy Worksheet: UV-Vis Spectroscopy Worksheet: UV-Vis Spectroscopy 4. Suppose you were not careful in preparing a sample for UV-Vis analysis, so that its actual concentration, c, in methanol was 0.00576 g/L rather than the 0 00600 g/L that you thought it was a. What is the error, in mg/L, in the mass of the sample present in solution b. Suppose that a maximum was observed at a wavelength of 278 nm for this solution and that the value of A at this maximum was 0.623 when a 1-cm cell was used i. Using the value of c that you thought you had for the solution, calcu- late e for the sample, assuming that its molar mass is 100 ii. Show how the information regarding thus maximum could be expressed. iii. What is the % error in associated with the inaccuracy in your know ing the actual concentration of the solution?

Explanation / Answer

a) The actual concentration of the analyte is Cactual = 0.00576 g/L; the theoretical concentration of the analyte required is Ctheo = 0.00600 g/L.

We need to express the absolute error in mg/L; therefore, absolute error = (Ctheo) – (Cactual) = (0.00600 g/L) – (0.00576 g/L) = 0.00024 g/L = (0.00024 g/L)*(1000 mg/1 g) = 0.24 mg/L (ans).

Percent error = (absolute error)/Ctheo*100 = (0.00024 g/L)/(0.00600 g/L)*100 = 4.00% (ans).

bi) The molar mass of the analyte is MW = 100 g/mol (molar masses are expressed in g/mol).

The assumed concentration of the analyte in the sample is Ctheo = 0.00600 g/L = (0.00600 g/L)*(1 mole/100 g) = 6.0*10-5 mol/L = 6.0*10-5 M (1 mol/L = 1 M).

Use Beer’s law as below.

A = *Ctheo*l where is the molar absorptivity of the analyte. Given A = 0.623 and l = 1.00 cm, plug in values and get

0.623 = *(6.0*10-5 M)*(1.00 cm)

====> = 0.623/(6.0*10-5 M).(1.00 cm) = 10383.33 M-1cm-1 10383.33 M-1cm-1 (ans).

ii) We can express Beer’s law at the wavelength of maximum absorption as

A = (10383.33 M-1cm-1)*C*l at 278 nm; A = absorbance of the sample, C is the concentration of the sample and l = path length of the solution through which the light passes (ans).

iii) The actual concentration of the analyte = Cactual = (0.00576 g/L)*(1 mol/100 g) = 5.76*10-5 M.

Put A = 0.623 and obtain

0.623 = *(5.76*10-5 M)*(1.00 cm)

====> = 10815.97 M-1cm-1.

We have actual = 10815.97 M-1cm-1 and theo = 10383.33 M-1cm-1; therefore, percent error in = theo – actual/(actual)*100 = (10383.33 – 10815.97) M-1cm-1/(10383.33 M-1cm-1)*100 = 4.1667% (the modulus value gives the positive value) (ans).

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