A compound with molecular mass = 292.16 g/mol was dissolved in a 5.000 mL volumetric flask....

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A compound with molecular mass = 292.16 g/mol was dissolved in a5.000 mL volumetric flask. A 1.000 mL aliquot was withdrawn, placedin a 10.000 mL flask and diluted to volume. The apparent absorbanceof 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 dilutedby exactly a factor of 2; the apparent absorbance was now 0.421. Areagent blank showed negligible absorbance. The (true) molarabsorptivity 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 thecuvette.
c) What mass of the compound was used to make the solution inthe 5.000 mL flask?

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3.6 Ratings (656 Votes)

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.827/2.130 x 10^4

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3.88 x 10^-5 M

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 5.67 x 10^-4 g

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.57 mg

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