A 5.0 microliter sample containing aniline (C 6 H 5 NH 2 ) and anisole (C 6 H 5

Question

A 5.0 microliter sample containing aniline (C6H5NH2) and anisole (C6H5OCH3) together with other substances was injected into a gas chromatograph. The heights for the peaks of these two solutes in the resulting chromatogram were 4.22 (aniline) and 7.60 IAnisole) chart divisions. Another 5.0 microliter sample was injected together with 0.25 microliter of pure aniline (All in the same syringe), producing aniline and anisole peak heights of 8.73 and 7.60 chart divisions. Calculate the concentration, in volume percent, of the two components under the following conditions:

A) The detector responds equally to both compunds

B) The detector response (on a volume basis) is 1.35 times greater for anisole than the aniline.

Explanation / Answer

Solution:

1. For both sample 5.0 microliter,

Sample of aniline and anisole injected = 5.0 microliter

peak height of aniline = 4.22

peak height of anisole = 7.60

2. Additionally a sample of 0.25 microliter pure aniline is added,

Therefore, total aniline in the syringe is (5.0 + 0.25) 5.25 microliter.

Now the peak height of aniline = 8.73

the peak height of anisole = 7.60

Note:

1 Milliliter (ml) = 1000 Microliter (µl).

The concentration of the final solution with the concentration unit expressed in units of mass per volume of solution (e.g., mg/mL). m is the mass (i.e., weight) of solute that must be dissolved in volume V of solution to make the desired solution concentration.

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