Astandard solution containing 6.3×10-8Miodoacetone and 2.0×10-7Mp-dichlorobenzene (an internal standard) gave peak areas of 395 and 787, respectively, in a gas chromatogram. A 3.00-mlunknown solution of iodoacetone was treated with 0.100mLof 1.6×10-5Mp-dichlorobenzene and the mixture was diluted to. Gas chromatography gave peak areas of 633 and 520 for iodoacetone and p-dichlorobenzene, respectively. Find the concentration of iodoacetone in the 3.00mLof original unknown.

Short Answer

Expert verified

In3.00mloftheoriginalsolution,theconcentrationoflodoacetoneis0.41μM.

Step by step solution

01

Definition of iodoacetone

C3H5IOisthechemicalformulaforiodoacetone,anorganoiodinecompound. Undernormalcircumstances,thematerialisawhiteliquidthatissolubleinethanol.

02

Step 2: Determine the concentration of iodoacetone in the 3.00 mL of originalunknown

It is necessary to calculate the concentration of lodoacetone in 3.00ml of the original solution.

The formula for quantitative analysis with internal standard can be found here,

AxX=FAsS

Here Ax=signal region of analyte

As=area of internal standard

[X] = analyte's concentration

[S]= concentration of internal standard

F= response factor

To figure out how much lodoacetone is in 3.00ml of the original solution

Given,

Lodoacetone molarity =6.3×10-8M

P-dichlorobenzene molarity =2.0×10-7MPeakareas=395&787

Unknown volume of lodoacetone solutionrole="math" localid="1654853858008" =3.00ml

p-dichlorobenzene Volume and Molarity=0.100ml&1.6×10-5M

Lodoacetone has a peak area of 633 square metres.

p-dichlorobenzene has a peak area of =520.

The response factor is determined as follows,

39563nM=F787200nMF=1.59

When an internal standard is coupled with an unknown, the concentration is determined as,

0.100ml10.00ml1.6×10-5M=0.16μMLodoacetoneconcentrationisdeterminedas,633iodoacentone=1.595200.16μMiodoacentone=0.122μMIntheoriginalunknownsolution,theconcentrationoflodoacetoneisdeterminedas,iodoacentone=10.003.00-0.122μMiodoacentone=0.41μMIn3.00mloftheoriginalsolution,theconcentrationoflodoactone=0.41μMThe3.00mloftheoriginalsolution,theconcentrationoflodoacetonewasdeterminedandfoundtobe0.41μM

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Most popular questions from this chapter

This problem reviews concepts from Chapter 23. An unretained solute passes through a chromatography column in 3.7 min and analyte requires 8.4 min.

(a) Find the adjusted retention time and retention factor for the analyte.

(b) Find the phase ratio b for a 0.32-mm-diameter column with a 1.0-mm-thick film of stationary phase.

(c) Find the partition coefficient for the analyte.

(d) Determine the retention time on a similar length of 0.32-mm diameter column with a 0.5-mm-thick film of the same stationary phase at the same temperature.

(a) How can you improve the resolution between two closely spaced peaks in gas chromatography?

(b) What approach from (a) would be most cost effective (not involve a purchase)?


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?

Why is split less injection used with purge and trap sample preparation?

Nitric oxide (NO) is a cell-signaling agent in physiologic processes including vasodilation, inhibition of clotting, and inflammation. A sensitive chromatography-mass spectrometry method was developed to measure two of its metabolites, nitrite(NO-2)and nitrat(NO3),in biological fluids. Internal standards,NO-152and15NO-3, were added to the fluid at concentrations of 80.0 and, respectively.Natural14NO-2and14NO-3plus the internal standards were then converted to volatile derivatives:

Because biological fluids are so complex, the derivatives were first isolated by high-performance liquid chromatography. For quantitative analysis, liquid chromatography peaks corresponding to the two products were injected into a gas chromatograph, ionized by negative ion chemical ionization (giving major peaks forNO-2andNO3-and the products measured by selected ion monitoring. Results are shown in the figure on the next page. If the internal standards undergo the same reactions and same separations at the same rate as theN14analytes, then the concentrations of analytes are simply[14NOx-]=[15NOx-](R-Rblank)

where R is the measured peak area ratio(m/z46/47fornitriteandm/z62/63fornitrite)for nitrite andfor nitrate) andRblankis the measured ratio of peak areas in a blank prepared from the same buffers and reagents with no added nitrite or nitrate. The ratios of peak areas areand. The ratios for the blank werem/z46/47=0.062andm/z62/63=0.058. Find the concentrations of nitrite and nitrate in the urine.


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