a) A 1.00 ??sample of an equal volume mixture of 2-pentanone and 1-nitropropane

Question

a) A 1.00 ??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 = Mass of 1-nitropropane b) The peak areas produced on this injection were 1883 units for 2-pentanone and 1421 units for 1-nitropropane mg 492 mg Calculate the response factor for each compound as area per mg. 2-pentanone: 49 1-nitropropane: c) An unknown mixture of these two components produces peak areas of 1975 units (2-pentanone) and 1366 units (1-nitropropane) units/mg units/mg Use these areas and the response factors above to determine the weight % of the components in the unknown sample. 2-pentanone: 1-nitropropane:

Explanation / Answer

ans)

a)

Volume of 2-pentanone = volume of 1-nitropropane

= 1/2 = 0.5 ?L = 0.5 x 10-3 mL

Mass of 2-pentanone = volume x density

= 0.5 x 10-3 x 0.8124

= 0.406m g

Mass of 1-nitropropane = volume x density

= 0.5x 10-3 x 1.0221

= 0.511x 10-3 ? 0.5 11m g

b)

f = A/W

f1 = 1883 / 0.406 = 4637.93 units/mg

f2 = 1421 / 0.511= 2780.82 units/mg

c)

W = A/f

W1 = 1975 / 4637.93 = 0.426 mg

W2 = 1366/ 2780.82 = 0.491 mg

Wt = W1 + W2 = 0.426 + 0.491 = 0.92 mg

%W1 = (0.426 / 0.92) x 100 = 46.3 %

%W2 = 100 - 46.3 = 53.7%

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