:(a) Describe the steps in QuEChERS and explain their purpose.

(b) Why is an internal standard used in QuEChERS?

(c) What is displayed in the total ion chromatogram in Figure 28-22?

(d) What is displayed in the extracted ion chromatogram in Figure 28-22? What is the difference between an extracted ion chromatogram and a selected ion chromatogram? Which would have greater signal-to-noise ratio?

(e) What mass spectrometric method could be used to obtain even greater signal-to-noise ratio from the same QuEChERS extract?

Short Answer

Expert verified
  1. The name is abbreviated as "Quick, Easy, Cheap, Effective, Rugged, and Safe."
  2. Because the internal standard suffers from the same losses as the analyte, it is employed in QuFChFRS.
  3. The reconstructed total ion chromatogram shows all of the ions that appear in the column
  4. .Ion chromatogram has a superior signal-to-noise ratio since more time is given to detect signal at the indicated mass.
  5. To obtain a greater signal-to-noise ratio, the selective ion chromatogram is used.

Step by step solution

01

Concept used

QuEChERS is a solid-phase extraction method for assessing or identifying pesticide residues in food.

02

Describe the steps in QuEChERS

(a)

QuEChERS is a solid-phase extraction method for assessing or identifying pesticide residues in food. The name is abbreviated as "Quick, Easy, Cheap, Effective, Rugged, and Safe."

- In this process, the sample can be taken as a homology solution, centrifuged with the reagent, and agitated for 1 minute.

- Pesticide residue in food samples is identified using this method.

03

Step 3:Why is internal standard used in QuEChERS

(b)

- An internal standard solution is one that has documented results and will not alter the QuEChERS technique. We will use it in conjunction with the analyte solution to record the readings. Whose value is the analyte solution's reference value, i.e. we wanted to get results.

- The internal standard is used in QuFChFRS because the internal standard experiences the same losses as the analyte.

04

Chromatogram of total ions

(c)

The current from all ions over a given mass is shown as a function of time in this chromatogram.

Rebuilt total ion chromatogram is defined as a total ion chromatogram that has been reconstructed by adding or summing the intensities of all observed values of \(m/z\)

The reconstructed total ion chromatogram shows all of the ions that appear in the column. By looking at the chromatogram in figure 28-22, we can see that the complete signal from all ions was seen and recorded as a chromatogram at any given time.

05

Extracted ion chromatogram

(d)

We may state that an extracted ion chromatogram displays a value of \({\rm{m}}/{\rm{z}}312\) that is obtained from the complete spectrum, while a chosen ion chromatogram shows a value of \({\rm{m}}/{\rm{z}}312\)that is recorded continuously and has a greater signal-to-noise ratio.

The following is an explanation of the difference between an extracted ion chromatogram and a chosen ion chromatogram.

Ion chromatogram after extraction:

This chromatogram depicts detector current as a function of time for one or more values of \({\rm{m}}/{\rm{z}}\)

The depicted intensities are taken from the entire mass spectrum obtained at each interval period.

- This is for analyte interest only.

- Which means there isn't a good signal-to-noise ratio.

Ion chromatogram using a few selected ions:

This chromatogram, similar to an extracted ion chromatogram, can display detector current for a single or small number of \({\rm{m}}/{\rm{z}}\)values.

The signal for all \({\rm{m}}/{\rm{z}}\)values at each time interval is not measured by the selected ion chromatogram.

It may be used to gather data for specified or desired \({\rm{m}}/{\rm{z}}\) values over a period of time.

This is for analyte interest only.

- Because the most time is provided to detect signal at the designated mass, this chromatogram has a better signal-to-noise ratio.

06

Mass spectrometric method

(e)

- Because the longest time is given to detect signal at the targeted mass, the selective ion chromatogram is employed to acquire a higher signal-to-noise ratio.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Referring to Table 28-7, explain how an anion-exchange resin can be used for absorption and analysis of \({\bf{S}}{{\bf{O}}_2}\)released by combustion.

EXAMPLE- Particles designated \(50/00\)mesh pass through a 50 mesh sieve bou are retained by a lo0 mesh sieve. Their size is in the range 0.150-0.300 mm.

does not pass is retained for your sample. This procedure gives particles whose diameters are in the range \(0.85 - 1.18\;{\rm{mm}}.\) We refer to the size range as \(16/20{\rm{mesh}}.\)

Suppose that much finer particles of \(80/120\)mesh size (average diameter \( = 152\mu {\rm{m}},\) average volume\( = 1.84\;{\rm{nL}}\)) were used instead. Now the mass containing \({10^4}\) particles is reduced from \(11.0to0.0388\;{\rm{g}}.\) We could analyze a larger sample to reduce the sampling uncertainty for chloride.

If 105 particles are taken, what is the relative standard deviation of each measurement?

In analyzing a lot with random sample variation, you find a sampling standard deviation of \({\bf{65}}\% .\)Assuming negligible error in the analytical procedure, how many samples must be analyzed to give \(9{\bf{5}}\% \)confidence that the error in the mean is within\(64\% \)of the true value? Answer the same question for a confidence level of \(90\% \).

The following wet-ashing procedure was used to measure arsenic in organic soil samples by atomic absorption spectroscopy: A 0.1- to \({\bf{0}}.{\bf{5}} - \)g sample was heated in a \({\bf{150}} - {\bf{mL}}\) Teflon bomb in a microwave oven for \(2.5\;{\rm{min}}\) with \(3.5\;{\rm{mL}}\)of\(70\% \,\,\,{\rm{HN}}{{\rm{O}}_3}\). After the sample cooled, a mixture containing \(3.5\;{\rm{mL}}\)of \(70\% \,\,\,{\rm{HN}}{{\rm{O}}_3},1.5\;{\rm{mL}}\) of\(70\% \,\,{\rm{HCl}}{{\rm{O}}_4}\), and \(1.0\;{\rm{mL}}\) of \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\)was added and the sample was reheated for three \({\bf{2}}.{\bf{5}} - {\bf{min}}\) intervals with 2 -min unheated periods in between. The final solution was diluted with \(0.2{\rm{M}}\,\,\,{\rm{HCl}}\)for analysis. Why was \({\rm{HCl}}{{\rm{O}}_4}\) not introduced until the second heating?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free