A sample of coffee is analyzed to determine its caffeine concentration. Two port

Question

A sample of coffee is analyzed to determine its caffeine concentration. Two portions of this sample are prepared for analysis. The first portion contains 50 mL of brewed coffee, to which is added 10 mL of water. The second portion contains 50 mL of brewed coffee that has been spiked with 10 mL of an aqueous solution that contains 0.01 M caffeine. The first portion of the sample is found to give a measured absorbance of 243 units, and the second portion gives an absorbance of 387 untis. What is the concentration of caffeine in the brewed coffee?

Explanation / Answer

First portion total volume = 50 + 10 = 60 ml

Let caeefine moles = m1

Absorbance = e x C x L    ( where e , L are constanst)

243 = constant   x ( m1/0.06)    ,

constant x m1 = 14.58 ...................(1)

Now in second portion caeffiene moles added = M x V = 0.01 x 10 /1000 =0.0001

vol = 60 ml = 0.06 L

Now A = constant x C

387 = constant x ( m1+ 0.0001) / ( 0.06)

constant x ( m1+ 0.0001) = 23.22 .......................(2)

now eq ( 2) / eq( 1) = ( m1+ 0.0001) / m1 = 23.22 /14.58

m1 + 0.0001 = 1.5926 m1

m1 = 0.00016875 moles

Initial cofee has 2m1 moles = 2 x 0.00016875 = 0.0003375

initial volume = 100 ml = 0.1 L

Conc = moles/vol = 0.0003375/0.1 = 0.003375 M

Conc = m1/

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