Question

Please help!

Ligand X forms a complex with both cobalt and copper, cach of which has a maximum absorbance at 510 nm and 645 nm, respectively. A 0.217 g sample containing cobalt and copper was dissolved and diluted to a volume of 100.0 mL. A solution containing ligand X was added to a 50.0 mL aliquot of the sample solution and diluted to a final volume of 100.0 mL. The measured absorbance of the unknown solution was 0.493 at 510 nm and 0.360 at 645 nm, when measured with a 1.00 cm cell. The molar absorptivities of the cobalt and copper complexes at each wavelength are shown in the table. Wavelength Molar Absorptivity (CM 'cm') Co 2., nm 510 645 36505 1247 Cu 5,550 x 10 17590 What is the concentration of cobalt and copper in the final diluted solution? What is the weight percent of cobalt (FM - 58.933 g/mol) and copper (FM = 63.546 g/mol) in the 0.217 8 sample? wt% Co?! wt% Cu?!

Answer 1

Question 1:

Let, C₁ be the concentration of Co and C₂ the concentration of Cu in the diluted solution.

Use Beer’s law.

A = ε₁C₁l + ε₂C₂l

where, ε₁ and ε₂ are the molar absorptivities of Co and Cu and l is the path length of the solution.

We have the following set of expressions.

0.493 = (36595C₁ + 5550C₂) x 1.00 …….(1)

0.360 = (1247C₁ + 17590C₂) x 1.00 …….(2)

Multiply (1) by 1247 and (2) by 36595 and subtract

0.493 x 1247 – 0.360 x 36595 = 36595 x 1247C₁ + 5550 x 1247C₂ – 1247x 36595C₁ – 17590 x 36595C₂

-12559.429 = -636785200C₂

C₂ = 1.9723 x 10^-5 M

Plug the value of C₂ in (1) and get,

0.493 = (36595C₁ + 5550 x 1.9723 x 10^-5) x 1.00

C₁ = 1.0480 x 10^-5 M

The molar concentrations of Co and Cu in the diluted sample are 1.0480 x 10^-5 M and 1.9723 x 10^-5 M respectively.

The volume of the dilute solution was 100 mL by taking 50 mL from the prepared sample solution and diluting with ligand X.

Find the number of moles of Co and Cu in the diluted sample.

Moles Co = (100 mL) x (1.0480 x 10^-5 M) = 1.0480 x 10^-3 mmole.

Moles Cu = (100 mL) x (1.9723 x 10^-5 M) = 1.9723 x 10^-3 mmole.

Since, 50 mL of the sample solution was diluted to a final volume of 100 mL, the dilution factor is (100 mL)/(50 mL) = 2; therefore, the millimoles of Co and Cu in the original sample are

mmoles Co = 1.0480 x 10^-3 mmole x 2 = 2.096 x 10^-3 mmole.

mmoles Cu = 1.9723 x 10^-3 mmole x 2 = 3.9446 x 10^-3 mmole.

Find out the masses of Co and Cu in the original sample.

Mass Co = 2.096 x 10^-3 mmole x (1 mole/1000 mmole) x (58.933 g/mol) = 1.2352 x 10^-4 g.

Mass Cu = 3.9446 x 10^-3 mmole x (1 mole/1000 mmole) x (63.546 g/mol) = 2.5066 x 10^-4 g.

Percentage Co in the original sample = (1.2352 x 10^-4 g)/(0.217 g) x 100 = 0.0569 ≈ 0.057 %

Percentage Cu in the original sample = (2.5066 x 10^-4 )/(0.217 g) x 100 = 0.1155 ≈ 0.116 %