a) A 0.90 muL sample of an equal volume mixture of 2-pentanone and 1-nitropropan

Question

a) A 0.90 muL 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-penta none = mg Mass of 1-nitropropane = mg b) The peak areas produced on this injection were 1395 units for 2-pentanone and 1264 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 1397 units (2-pentanone) and 1774 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

I'm not quite sure, but to the response factor, The calculation would be like this:

2-pentanone: 1395 / 0.366 = 3811.48 units/mg

1-nitropropane: 1264 / 0.46c= 2747.83 units/mg

Then for part c:

f = A / m ---> m = A / f

2-pentanone: m = 1397 / 3811.48 = 0.367 mg

1-nitropropane: m = 1774 / 2747.83 = 0.646 mg

mt = 0.646 + 0.367 = 1.013 mg

%m1 = 0.367 / 1.013 x 100 = 36.23%

%m2 = 0.646 / 1.013 x 100 = 63.77%

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