A 0.80 mu L sample of an equal volume mixture of 2-pentanone and 1-nitropropane
ID: 509128 • Letter: A
Question
A 0.80 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 The peak areas produced on this injection were 1447 units for 2-pentanone and 1238 units for 1-nitropropane. Calculate the response factor for each compound as area per mg. 2-pentanone: _______ units/mg 1-nitropropane: _______ units/mg An unknown mixture of these two components produces peak areas of 1761 units (2-pentanone) and 1278 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
Solution:-
Density =Mass/Volume
Volume= 0.8 uL= 0.8 * 10-6 L
Densityof 2 pantanone= 0.8124 g/ml=0.8124 * 103 g/L=8124g/L
Density of 1-nitopropane=1022.1 g/L
Mass of 2-Pantanone =Density * Volume
=812.4 * 0.8 * 10-6
=0.6499* 10-3 gm
Mass of 1-Nitro Propane= 1022.1 * 0.8*10-6
=0.81768* 10-3 gm
b) Response Factor= Peak Area/ Concentration
Peak Area of 2-Pentanone=1447 unit
Peak Area of 1- Nitropropane=1238 unit
Response Factor of 2-Pentanone= 1447 unit *0.6499* 10-3 gm
= 0.9404 units/gm
=0.9404 *10-3 units / mg ..........(gm=103mg)
Response Factor of 1- Nitropropane=1238 unit *0.81768* 10-3 gm
= 1.01228 units/gm
= 1.01228 * 10-3 units/ mg
c)
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