Question

1. A 10.0 mL aliquot of a flouride sample solution was diluted to 50 mL in a volumetric flask. The F- concentration of this DILUTED sample was determined with a F- ion selective electrode and found to be 8.7 mg/L. What is the concentration of F- in the original, undiluted sample?

2. a) A 0.80 µL sample of an equal volume mixture of 2-pentanone and 1-nitropropane is injected into a gas chromatograph. The densities of these compounds are 0.8124 g/mL for 2-pentanone and 1.0221 g/mL for 1-nitropropane.

What mass of each compound was injected?
Mass of 2-pentanone = mg
Mass of 1-nitropropane =  mg
b) The peak areas produced on this injection were 1469 units for 2-pentanone and 1423 units for 1-nitropropane.

Calculate the response factor for each compound as area per mg.
2-pentanone: units/mg
1-nitropropane: units/mg
c) An unknown mixture of these two components produces peak areas of 1097 units (2-pentanone) and 1744 units (1-nitropropane).

Use these areas and the response factors above to determine the weight % of the components in the unknown sample.

2-pentanone:   %
1-nitropropane:  %

Answer 1

1) For flouride sample, Concentration and volume are related to

2) a)Both 2-pentanone and 1-nitropropane are in 1:1 ratio, hence of each is the volume used

Concentration =density= mass/Volume

mass of 2-pentanone= denstity * Volume

Similarly mass of 1-nitropropane=

b) Response factor = Peak area/concentration

Response factor for 2-pentanone=1462units/0.8124(g/ml)=1799.6 units. g/ml

Response factor for 1-nitropropane=1423units/1.0221(g/ml)=1392.2 units. g/ml

c) For unknown mixture of 2-pentanone and 1-nitropropane

Area of 2-pentanone=1097units

Area of 1-nitropropane=1744units

Total area =1097+1744=2841units

%A= =38.61%

%B==61.38%