A 1.00 mu L sample of an equal volume mixture of 2-pentanone and 1-nitropropane
ID: 1017786 • Letter: A
Question
A 1.00 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-nitopropane. What mass of each compound was injected? The peak areas produced on this injection were 1466 unit for 2-pentanone and 1420 units for 1-nitropropane. Calculate the response factor for each compound as area per mg. An unknown mixture of these two components produces peak areas of 1835 units (2-pentanone) and 1745 units (1-nitropropane). Use these areas and the response factors above to determine the weight % of the component in the unknown sample.Explanation / Answer
a) 1.00 microL sample of equal volume ratio gives 0.50 microL of 2-pentanone and 1-nitropropane
Hence mass of 2-pentanone = 0.50 * 10^-3 mL * 0.8124 g/mL = 4.062 * 10^-4 g
Mass of 1-nitropropane = 0.50 * 10^-3 mL * 1.0221 g/mL = 5.11 * 10^-4 g
b) response factor for 2-pentanone = 1466 units/4.062 * 10^-4 g = 3609 units/mg
response factor for 1-nitropropane = 1420 units/5.11 * 10^-4 g = 2779 units/mg
c) Not all the data provided.
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