Question

A compound with molecular mass = 292.16 g/mol was dissolved in a 5.000 mL volumetric flask. A 1.000 mL aliquot was withdrawn, placed in a 10.000 mL flask and diluted to volume. The apparent absorbance of the diluted solution at 340 nm was 0.827 in a 1.000 cm cuvette. To check for a possible stray light effect, this sample was diluted by exactly a factor of 2; the apparent absorbance was now 0.421. A reagent blank showed negligible absorbance. The (true) molar absorptivity at 340 nm = 2.130 x 104 M-1 cm-1.

a) Estimate the stray light in the spectrophotometer.

b) Calculate the concentration of the compound in the cuvette.

c) What mass of the compound was used to make the solution in the 5.000 mL flask?

Answer 1

a) Stray light in the spectrophotometer = 0.421 - 0.827/2 = 0.0075

b) For the given data

x g of compound in 5 ml gave a solution

1 ml of above solution diluted to 10 ml gave an absorbance = 0.827

concentration of compound = absorbance/molar absorptivity

                                            = 0.827/2.130 x 10^4

                                            = 3.88 x 10^-5 M

concentration in original 5 ml sample solution = 3.88 x 10^-5 x 10

                                                                         = 3.88 x 10^-4 M

c) Mass of compound used to make 5 ml solution = 3.88 x 10^-4 M x 0.005 L x 292.16 g/mol

                                                                                = 5.67 x 10^-4 g

                                                                                = 0.57 mg