a) A 0.90 mu L sample of an equal volume mixture of 2-pentanone and 1-nitropropa

Question

a) A 0.90 mu 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 = mg Mass of 1-nitropropane = 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:units/mg 1-nitropropane: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:% 1-nitropropane:%

Explanation / Answer

a) Given-

0.9µl sample of 2-pentanone and 1-nitro propane

Convert microliter to ml

1 µl = 0.001ml

0.9µl sample = 0.0009 ml

Now divide 0.0009 ml into two equal parts

Volume of 2-pentanone = 0.0009/2 = 0.00045 ml

Volume of 1-nitro propane = 0.00045 ml

Density of 2-pentanone = 0.8124g/ml

Density of 1-nitro propane = 1.0221 g/ml

Mass of 2-pentanone = volume of 2-pentanone x density of 2-pentanone

Mass of 2-pentanone = 0.00045 ml x 0.8124g/ml = 0.0003655g

Now convert g to mg by multiplying g to 1000

0.0003655g x 1000 mg/1g = 0.366 mg

Mass of 2-pentanone = 0.366mg

Mass of 1-nitro propane = volume of 1-nitro propane x Density of 1-nitro propane

Mass of 1-nitro propane = 0.00045 ml x 1.0221 g/ml = 0.0004599 g

Now convert g to mg by multiplying g to 1000

0.00046 g x 1000 mg/1g = 0.460 mg

Mass of 1-nitro propane = 0.460 mg

b)

The peak areas produced on this injection were 1446 units for 2-pentanone and 1271 units for 1-nitropropane.

Calculate the response factor for 2-pentanone as area per mg.

Response Factor = 1446 units/0.366 mg = 3950.819 units/mg

Response Factor for2-pentanone = 3951units/mg

Calculate the response factor for 1-nitropropane as area per mg

Response Factor = 1271 units/0.460 mg = 2763.04 units/mg

Response Factor for 1-nitropropane = 2763 units/mg

c)

An unknown mixture of these two components produces peak areas of 1303 units (2-pentanone) and 1815units (1-nitropropane).

Now calculate the weight of 2-pentanone

Weight of 2-pentanone = units/ response factors

Weight of 2-pentanone= 1303 units/3951units/mg = 0.3297mg

(0.3297mg/0.9865mg) x 100 = 33.42%

Now calculate the weight of 1-nitropropane

Weight of 1-nitropropane = 1815units/ 2763 units = 0.6568mg

Add the weight of two components

0.6568mg + 0.3297mg = 0.9865 mg

Weight % of 2-pentanone (0.3297mg/0.9865mg) x 100 = 33.42%

Weight % of1-nitropropane = 0.6568mg /0.9865 mg) x 100 = 66.58 %

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