Question

1. [spectroscopy and chemical dyes] A compound with molecular mass 292.16 g/mol was dissolvedin a 5 mL volumetric flask. A 1.00 mL aliquot was withdrawn, placed in a 150.0 mL volumetricflask, and diluted to the mark. The absorbance at 340 nm was 0.427 in a 1.00 cm cuvette. The molar absorptivity at 340 nm is equal to 6130 M-1 cm-1.

a. Calculate the concentration of the compound in the cuvette.

b. What was the concentration of the compound in the 5 mL flask?

c. How many milligrams of compound were used to make the 5 mL solution?

Answer 1

a) We can calculate the concentration of the compound in the cuvette by using the Beer-Lambert Law

A=ebc

Where A is absorbance = 0.427

e is the molar absorptivity = 6130 M-1 cm-1.

b is the path length of the sample =1.00 cm

c is the concentration of the compound in solution

A=ebc

c=A/eb

c = 0.427/ (6130 M-1 cm-1. x 1.00cm)

c = 6.97e-5M

The concentration of the compound in the cuvette = 6.97e-5M

b. The concentration of the compound in the 5 mL flask-

Only 1.00 ml of the stock solution was diluted up to 10 ml

C₁ V₁=C₂V₂

C₁ (1.00mL)=(6.966e-5)(10.00ml)

C₁= 6.97e-4 M

The concentration of the compound in the 5 mL flask is 6.97e-4 M

c.

Calculate number of moles- n = cv

C= 6.966e-4M

V = 5 ml = 0.005 litres

We could calculate the moles

0.005 litres x 6.966e-4 mol/litre = 3.483e-6 moles

Now use the molar mass to calculate the grams used

Molar mass = 292.16 g/mol

3.483e-6 moles x 292.16 g/mol = 1.0176e-3 grams

= 0.00102 grams

= 1.02 milligrams

1.02 milligrams of compound were used to make the 5 ml solution