The following calibration data for a chromatograhpic method for the determinatio

Question

The following calibration data for a chromatograhpic method for the determination of isooctane in a hydrocarbon mixture was obtained:

Mol percent isooctane: 0.352, 0.803, 1.08, 1.38, 1.75

peak area: 1.09, 1.78, 2.60, 3.03, 4.01

The calibration curve determined was used for the chromatographic determination of isooctane in a hydrocarbon mixture. A peak area of 2.65 was obtained. Calculate the standard deviation (to 2 sig figs) for the mole percent of isooctane if the peak area was the result of the mean of 4 measurements.

Explanation / Answer

From the given data

mean mole percent isooctane (X) = 1.073

variance = sum (x-X)^2/7

= (0.5198 + 0.0729 + 4.9 x 10^-5 + 0.0942 + 0.45833)/5

= 0.2290

standard deviation of mol percent isooctanne = sq.rt.(variance)

= 0.48

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