Analytical chemistry, please help with the incorrect answers! a) A 0.90 L sample

Question

Analytical chemistry, please help with the incorrect answers!

a) A 0.90 L 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 0, 366 mg Mass of 1-nitropropane 0460 mg b) The peak areas produced on this injection were 1486 units for 2-pentanone and 1271 units for 1-nitropropane. Calculate the response factor for each compound as area per mg. 2-pentanone: 3950 units/mg 1-nitropropane: 2780 units/mg c) An unknown mixture of these two components produces peak areas of 1303 units (2-pentanone) and 1815 units (1-nitropropane). Use these areas and the response factors above to determine the weight of the components in the unknown sample. 2-pentanone: 33, 4 % 1-nitropropane: 66.6%

Explanation / Answer

0.9 microliter contains equal volumes of 2-pentanone and 1-nitropropane ie 0.45 microliters each.

from the density we can calculate mass of each in 45 microliter =

mass of 1-nitropropane = 1.0221*0.45 x 10 -3 = 0.460 mg

similarly mass of 2-pentonone = 0.8124*0.45 x 10 -3 = 0.366 mg

b)

respose of gas chromatography = area/ mass of the compound

for 2-pentonone = 1486/0.366 = 4060 units/mg

for nitropropane = 1271/0.46 =2763 units/mg

c) mass of uncknown = area/response function

for

2-pentanone = 1303/4060 = 0.3209 mg

1 -nitropropane = 1815/2763 =0.6568 mg

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