a) A 0.70 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 = 4.0mgmgMass of 1-nitropropane = 4.9b) The peak areas produced on this injection were 1402 units for 2-pentanone and 1181 units for 1-nitropropane.Calculate the response factor for each compound as area per mg.2-pentanone: 4.0units/mgunits/mg1-nitropropane: 4.0c) An unknown mixture of these two components produces peak areas of 1515 units (2-pentanone) and 1446 units (1-nitropropane).Use these areas and the response factors above to determine the weight % of the components in the unknown sample.2-pentanone: 4.01-nitropropane: 4.0%%

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) A 0.70 μL sample of an equal volume mixture of 2-pentanone and 1-nitropropane is injected into a gas chromatograph. he 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? lass of 2-pentanone = mg mg ass of 1-nitropropane = 49 ) The peak areas produced on this injection were 1402 units for 2-pentanone and 1181 units for 1-nitropropane. alculate the response factor for each compound as area per mg. -pentanone: 4.0 units/mg units/mg -nitropropane: 4.0 An unknown mixture of these two components produces peak areas of 1515 units (2-pentanone) and 1446 units (1- itropropane). se these areas and the response factors above to determine the weight % of the components in the unknown sample. -pentanone: 4.0 -nitropropane: 4.0 % %

) A 0.70 μL sample of an equal volume mixture of 2-pentanone and 1-nitropropane is injected into a gas chromatograph. he 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? lass of 2-pentanone = mg mg ass of 1-nitropropane = 49 ) The peak areas produced on this injection were 1402 units for 2-pentanone and 1181 units for 1-nitropropane. alculate the response factor for each compound as area per mg. -pentanone: 4.0 units/mg units/mg -nitropropane: 4.0 An unknown mixture of these two components produces peak areas of 1515 units (2-pentanone) and 1446 units (1- itropropane). se these areas and the response factors above to determine the weight % of the components in the unknown sample. -pentanone: 4.0 -nitropropane: 4.0 % %

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter5: Errors In Chemical Analyses

Section: Chapter Questions

Problem 5.12QAP

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