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Bohman and colleagues described a reversed-phase HPLC method for the quantitative analysis of vitamin \(\mathrm{A}\) in food using the method of standard additions. \({ }^{27}\) In a typical example, a \(10.067-\mathrm{g}\) sample of cereal is placed in a 250 -mL Erlenmeyer flask along with \(1 \mathrm{~g}\) of sodium ascorbate, \(40 \mathrm{~mL}\) of ethanol, and \(10 \mathrm{~mL}\) of \(50 \% \mathrm{w} / \mathrm{v} \mathrm{KOH}\). After refluxing for \(30 \mathrm{~min}, 60 \mathrm{~mL}\) of ethanol is added and the solution cooled to room temperature. Vitamin \(\mathrm{A}\) is extracted using three \(100-\mathrm{mL}\) portions of hexane. The combined portions of hexane are evaporated and the residue containing vitamin A transferred to a \(5-\mathrm{mL}\) volumetric flask and diluted to volume with methanol. A standard addition is prepared in a similar manner using a \(10.093-\mathrm{g}\) sample of the cereal and spiking with \(0.0200 \mathrm{mg}\) of vitamin \(\mathrm{A}\). Injecting the sample and standard addition into the HPLC gives peak areas of, respectively, \(6.77 \times 10^{3}\)and \(1.32 \times 10^{4}\). Report the vitamin \(\mathrm{A}\) content of the sample in milligrams/100 g cereal.

Short Answer

Expert verified
Vitamin A content is approximately 0.2095 mg/100 g cereal.

Step by step solution

01

Understanding the Problem

The problem involves determining the concentration of vitamin A in cereal using the standard addition method in HPLC. We are given peak areas for both the unspiked sample and the spiked sample, and we need to calculate the vitamin A content in mg per 100 g of cereal.
02

Calculate the Spiked Sample Peak Increase

First, calculate the increase in peak area due to the addition of the vitamin A standard. This is done by subtracting the peak area of the unspiked sample from the peak area of the spiked sample.\[ \Delta A = A_{\text{spiked}} - A_{\text{unspiked}} = 1.32 \times 10^4 - 6.77 \times 10^3 = 6.42 \times 10^3 \]
03

Determine the Original Vitamin A Content in Sample

Using the increase in peak area, calculate the equivalent amount of vitamin A initially present in the sample. From the standard addition method, \[ C_{\text{sample}} = C_{\text{spike}} \times \frac{A_{\text{unspiked}}}{\Delta A} \]Substitute the given values:\[ C_{\text{sample}} = 0.0200 \text{ mg} \times \frac{6.77 \times 10^3}{6.42 \times 10^3} = 0.0211 \text{ mg} \]
04

Adjust for Sample Mass and Scale to 100 g

Convert the calculated vitamin A content to a per-100g basis. Since our calculations are for a sample of mass 10.067 g, scale the result:\[ C_{\text{100g}} = C_{\text{sample}} \times \frac{100}{10.067} = 0.0211 \text{ mg} \times \frac{100}{10.067} \approx 0.2095 \text{ mg/100g} \]
05

Conclusion

The content of vitamin A in the cereal is approximately 0.2095 mg per 100 g of cereal. This final step confirms how to report the concentration in the desired units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Addition Method
The standard addition method is a powerful and common technique used in analytical chemistry to improve the accuracy of quantitative analyses, particularly in complex matrices like food. It works by adding a known quantity of the target analyte—in this case, vitamin A—to the sample, effectively 'spiking' it. This helps to account for matrix effects that might otherwise skew the results. In practice, you measure the response (such as the peak area in HPLC) of both the spiked and unspiked samples.

By comparing these two responses, you can determine how the matrix affects the measurement and adjust your calculations to get a more accurate measure of the analyte's true concentration in the sample. Using this method can greatly enhance the reliability of quantitative analysis, making it especially useful in food chemistry where complex mixtures are involved.
Quantitative Analysis
Quantitative analysis is all about determining the amount or concentration of a substance in a sample. In the context of this exercise, it refers to measuring how much vitamin A is present in a cereal sample. This is accomplished through a series of calculated steps and involves using mathematical equations to translate physical measurements into actual concentration values.

It requires precision and accuracy at every stage, from sample preparation to the final analytical measurement, often involving sophisticated instrumentation like HPLC. The use of HPLC, or High-Performance Liquid Chromatography, enables the separation and quantification of components within a mixture. This makes it indispensable in a wide range of chemical analyses, including vitamin analyses. The accuracy of the quantitative analysis in this exercise is further enhanced by employing the standard addition method, ensuring that matrix effects do not interfere with the results.
Vitamin A Analysis
Vitamin A analysis is crucial in the food industry, especially because vitamin A is an essential nutrient responsible for various health functions, including vision and immune response. Analyzing its content in food products ensures that they meet nutritional standards, which are important for public health.

In this exercise, vitamin A analysis is conducted after extracting the vitamin from a cereal sample. Using solvents like ethanol and hexane, vitamin A is carefully separated from other components of the cereal. This preparation step is key because the analyte must be in a form that is compatible with the HPLC method used for measurement. By quantifying the vitamin A content, health claims can be substantiated, and nutritional information can be accurately labeled on food products.
Reversed-Phase Chromatography
Reversed-phase chromatography is a subtype of liquid chromatography commonly used in high-performance liquid chromatography (HPLC) for separating various compounds. Unlike normal-phase chromatography, reversed-phase uses a non-polar stationary phase and a polar mobile phase, which means it is particularly suitable for analyzing non-polar compounds like vitamin A.

In reversed-phase chromatography, the non-polar analytes are retained longer on the hydrophobic stationary phase, allowing them to be separated based on their hydrophobic (non-polar) interactions. This is advantageous for vitamin A analysis because it is a non-polar, lipid-soluble compound. The process ensures that the separation is highly efficient and reproducible, making it ideal for detailed quantitative analyses where precision is essential.
  • Used widely in food and pharmaceutical industries.
  • Handles a broad range of sample types and complexities.
  • Offers good reproducibility and resolution for complex mixtures.

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Most popular questions from this chapter

Ohta and Tanaka reported on an ion-exchange chromatographic method for the simultaneous analysis of several inorganic anions and the cations \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in water. \({ }^{28}\) The mobile phase includes the ligand 1,2,4 -benzenetricarboxylate, which absorbs strongly at \(270 \mathrm{nm}\). Indirect detection of the analytes is possible because its absorbance decreases when complexed with an anion. (a) The procedure also calls for adding the ligand EDTA to the mobile phase. What role does the EDTA play in this analysis? (b) A standard solution of \(1.0 \mathrm{mM} \mathrm{NaHCO}_{3}, 0.20 \mathrm{mM} \mathrm{NaNO}_{2}, 0.20\) \(\mathrm{mM} \mathrm{MgSO}_{4}, 0.10 \mathrm{mM} \mathrm{CaCl}_{2},\) and \(0.10 \mathrm{mM} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) gives the following peak areas (arbitrary units). \(\begin{array}{lcccc}\text { ion } & \mathrm{HCO}_{3}^{-} & \mathrm{Cl}^{-} & \mathrm{NO}_{2}^{-} & \mathrm{NO}_{3}^{-} \\ \text {peak area } & 373.5 & 322.5 & 264.8 & 262.7 \\\ \text { ion } & \mathrm{Ca}^{2+} & \mathrm{Mg}^{2+} & \mathrm{SO}_{4}^{2-} & \\\ \text { peak area } & 458.9 & 352.0 & 341.3 & \end{array}\) Analysis of a river water sample (pH of 7.49 ) gives the following results. \(\begin{array}{lcccc}\text { ion } & \mathrm{HCO}_{3}^{-} & \mathrm{Cl}^{-} & \mathrm{NO}_{2}^{-} & \mathrm{NO}_{3}^{-} \\ \text {peak area } & 310.0 & 403.1 & 3.97 & 157.6 \\ \text { ion } & \mathrm{Ca}^{2+} & \mathrm{Mg}^{2+} & \mathrm{SO}_{4}^{2-} & \\ \text { peak area } & 734.3 & 193.6 & 324.3 & \end{array}\) Determine the concentration of each ion in the sample. (c) The detection of \(\mathrm{HCO}_{3}^{-}\) actually gives the total concentration of carbonate in solution \(\left(\left[\mathrm{CO}_{3}^{2-}\right]+\left[\mathrm{HCO}_{3}^{-}\right]+\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\right) .\) Given that the \(\mathrm{pH}\) of the water is \(7.49,\) what is the actual concentration of \(\mathrm{HCO}_{3}^{-}\) ? (d) An independent analysis gives the following additional concentrations for ions in the sample: \(\left[\mathrm{Na}^{+}\right]=0.60 \mathrm{mM} ;\left[\mathrm{NH}_{4}^{+}\right]=0.014\) \(\mathrm{mM}\); and \(\left[\mathrm{K}^{+}\right]=0.046 \mathrm{mM}\). A solution's ion balance is defined as the ratio of the total cation charge to the total anion charge. Determine the charge balance for this sample of water and comment on whether the result is reasonable.

The analysis of \(\mathrm{NO}_{3}^{-}\) in aquarium water is carried out by CZE using \(\mathrm{IO}_{4}^{-}\) as an internal standard. A standard solution of \(15.0 \mathrm{ppm} \mathrm{NO}_{3}^{-}\) and 10.0 ppm \(\mathrm{IO}_{4}^{-}\) gives peak heights (arbitrary units) of 95.0 and 100.1 , respectively. A sample of water from an aquarium is diluted 1: 100 and sufficient internal standard added to make its concentration \(10.0 \mathrm{ppm}\) in \(\mathrm{IO}_{4}^{-}\). Analysis gives signals of 29.2 and 105.8 for \(\mathrm{NO}_{3}^{-}\) and \(\mathrm{IO}_{4}^{-},\) respectively. Report the \(\mathrm{ppm} \mathrm{NO}_{3}^{-}\) in the sample of aquarium water.

Loconto and co-workers describe a method for determining trace levels of water in soil. \(^{19}\) The method takes advantage of the reaction of water with calcium carbide, \(\mathrm{CaC}_{2}\), to produce acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}\). By carrying out the reaction in a sealed vial, the amount of acetylene produced is determined by sampling the headspace. In a typical analysis a sample of soil is placed in a sealed vial with \(\mathrm{CaC}_{2}\). Analysis of the headspace gives a blank corrected signal of \(2.70 \times 10^{5} .\) A second sample is prepared in the same manner except that a standard addition of \(5.0 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} / \mathrm{g}\) soil is added, giving a blank-corrected signal of \(1.06 \times 10^{6} .\) Determine the milligrams \(\mathrm{H}_{2} \mathrm{O} / \mathrm{g}\) soil in the soil sample.

Janusa and coworkers describe the determination of chloride by CZE. \(^{29}\) Analysis of a series of external standards gives the following calibration curve. $$ \text { area }=-883+5590 \times \text { ppm } \mathrm{Cl}^{-} $$ A standard sample of \(57.22 \% \mathrm{w} / \mathrm{w} \mathrm{Cl}^{-}\) is analyzed by placing \(0.1011-\mathrm{g}\) portions in separate \(100-\mathrm{mL}\) volumetric flasks and diluting to volume. Three unknowns are prepared by pipeting \(0.250 \mathrm{~mL}, 0.500 \mathrm{~mL},\) and \(0.750 \mathrm{~mL}\) of the bulk unknown in separate \(50-\mathrm{mL}\) volumetric flasks and diluting to volume. Analysis of the three unknowns gives areas of \(15310,31546,\) and \(47582,\) respectively. Evaluate the accuracy of this analysis.

Haddad and associates report the following retention factors for the reversed- phase separation of salicylamide and caffeine. \({ }^{25}\) \(\begin{array}{ccccccc}\% \text { methanol } & 30 \% & 35 \% & 40 \% & 45 \% & 50 \% & 55 \% \\ k_{\text {sal }} & 2.4 & 1.6 & 1.6 & 1.0 & 0.7 & 0.7 \\\ k_{\text {caff }} & 4.3 & 2.8 & 2.3 & 1.4 & 1.1 & 0.9\end{array}\) (a) Explain the trends in the retention factors for these compounds. (b) What is the advantage of using a mobile phase with a smaller \(\% \mathrm{v} / \mathrm{v}\) methanol? Are there any disadvantages?

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