a) A 0.70 μL sample of an equal volume mixture of 2-pentanone and 1-nitropropane is injected into a gas chromatograph. T

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

a) A 0.70 μL sample of an equal volume mixture of 2-pentanone and 1-nitropropane is injected into a gas chromatograph. T

Post by answerhappygod »

 1
1 (144.71 KiB) Viewed 11 times
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 4.0 284 Mass of 1-nitropropane 4.0 357 = mg mg b) The peak areas produced on this injection were 1105 units for 2-pentanone and 918 units for 1-nitropropane. = Calculate the response factor for each compound as area per mg. 2-pentanone: 4478.87 units/mg 1-nitropropane: 4.0 3495.79 X units/mg c) An unknown mixture of these two components produces peak areas of 1767 units (2-pentanone) and 1661 units (1-nitropropane). Use these areas and the response factors above to determine the weight % of the components in the unknown sample. 2-pentanone: 4.0 48.76 1-nitropropane: 4.0 51.23 Submit Answer X % X %
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply