Where would an unknown with a retention index of 936 be eluted in Figure 23-7?

Short Answer

Expert verified

The eluted unknown with a retention index of 936 would be found after nonane.

Step by step solution

01

Step 1: Concept used

Kovats retention index:

For linear alkanes, the retention index, I , equals 100times the number of carbon atoms.

For a molecule of Octane, I=800 calculated using the formula

I=100[n+N-nlogL'unknown-logL'nlog4N-log4'n]

where

n=number of carbon atoms in smaller alkane

N=number of carbon atoms in larger alkane

tr'(n)=adjusted retention time of smaller alkane

localid="1663581965257" tr'(N)=adjusted retention time of larger alkane.

02

Position of the unknown

Nonane has a retention value of 900, whereas the unknown has a retention index of 936. Therefore, the eluted unknown with a retention index of 936 would be found after nonane.

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Most popular questions from this chapter

(a) Use Trouton's rule, Hvap°(88Jmol-1K-1)Tbp, to estimate the enthalpy of vaporization of octane (b.p. 126°).

(b) Use the form of the Clausius-Clapeyron equation below to estimate the vapor pressure of octane at the column temperature in Figure 24-9(70°C)

In(P1P2)=-(HvapR)(1T1-1T2)

(c) Calculate the vapor pressure for hexane (b.p. 69°C) at70°C

(d) What is the relationship between solute vapor pressure and retention?

(e) Why is the technique called "gas chromatography” if retained analytes are only partially vaporized?


In the analysis of odorants in tequila in Figure 24-21, tequila was diluted with water and extracted four times with dichloromethane(CH2Cl2) , b.p. 40CThe400mLof(CH2Cl2)was evaporated down to1mLand1μLof the extract was injected on-column onto a poly(ethylene glycol) open tubular column (30m×0.53mm,filmthickness=1μm), film thickness ) initially at and then ramped to 230C

(a) Why was the diluted tequila extracted four times with dichloromethane instead of once with a larger volume?

(b) Why was on-column injection used?

(c) Why was a poly(ethylene glycol) column chosen for this application?

(d) What was the phase ratio of the column?

(e) Why was a wide-bore 0.53-mm-diameter column chosen for this application?

Efficiency of solid-phase microextraction. Equation24-9gives the mass of analyte extracted into a solid-phase microextraction fiber as a function of the partition coefficient between the fiber coating and the solution.

(a) A commercial fiber with a100-μm-thickcoating has a film volume of6.9×10-4mL. Suppose that the initial concentration of analyte in solution is

c0=0.10μg/mL(100ppb).Use a spreadsheet to prepare a graph showing the mass of analyte extracted into the fiber as a function of solution volume for partition coefficients of 10000,5000,1000and100and. Let the solution volume vary from 0to100mL.

(b) Evaluate the limit of Equation24-9asVr gets big relative to KVf. Does the extracted mass in your graph approach this limit?

(c) What percentage of the analyte fromof solution is extracted into the fiber when and whenK=100andwhenk=10000?

This problem reviews concepts from Chapter 23 using

Figure 24-15.

(a) Calculate the retention factor for peak 11 given tm 5 6.7 min.

(b) Calculate the number of theoretical plates (N in Equation 23-31)

and the plate height (H) for peak 11.

(c) Find the resolution (Equation 23-23) between peaks 16 and 17.

Heptane, decane, and an unknown had adjusted retention times of 12.6min(heptane),22.9min(decane), and20.0min(unknown). The retention indexes for heptane and decane are 700 and 1000 , respectively. Find the retention index for the unknown.

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