Question

an experiment was preformed under identical conditions as yours. The absorbance of the penny solution was recorded as 0.231 absorbance units. A calibration plot of absorbance vs concentration of cu(II) (mM) yielded the following trendily equations y= 11591x +.50

a. What is the concentration of the original penny solution?

b. How many grams of Cu are in this solution?

c. if the percent Cu was determined to be 2.70 percent what was the mass of the penny?

Answer 1

This problem is based on Beer-Lambert law whose mathematical equation is ,

A = e x b x C ...................(1)

Where A = absorbance

e (epsilone) = molar absorptivity in unit L/ cm, mol

b = path length of cell(cuvette) = 1.00 cm,

C = concentration of the sample mol/L.

The plot of aborbance (A) against the Concentration(C) gives a straight line with slope = e= molar absrorptivity.

The equation for plot is, y = 11591x +50 (compare with y=mx +c(y-interceppt)

Hence e= molar aborptivity = 11591 L/cm.mol

a) Concentration of [Cu(II)]in penny solution c =?

we have e = 11591 L/cm.mol, b=1.00 cm, A = 0.231

by putting all the known values in eq. (1)

0.231 = 11591 x 1.00 x C

0.231 = 11591 x C

C = 0.231 / 11591

C = 1.993 x 10^-5 mol/L

Hence, concentration of penny solution = 1.993 x 10^-5 mol/L.

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b) Atomic mass of Cu is 63..5 g/mol

i.e. 1 mol Cu = 63.5 g og Cu

so, 1.993 x 10^-5 mol Cu = say 'Z' g Cu

so, 'Z' = 63.5 x 1.993 x 10^-5 .

'Z' = 1.266 x 10^-3 g

'Z' = 1.266 mg.

Hence the mass of C(II) in given sample is 1.266 x 10^-3 g or 1.266 mg.

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c) Present Cu % in given penny is 2.70 %

that means 2.70 mg Cu(II) = 100 mg of penny.

so, 1.266 mg Cu(II) = say 'M' mg of penny.

'M' = (100 x 1.266) / 2.70

M = 46.89 mg

M = 46.89 x 10^-3 g = 4.689 x 10^-2 g.

Hence mass of penny = 46.89 mg = 4.689 x 10^-2 g.

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