Lab Report Fatty acid composition of oils_Aqilah Khairuddin
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CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID:z5432922 1 FATTY ACID COMPOSITION OF OILS: REPORT Results – Method A: GC-MS a. Report your sample information: Table 1.
Sample information Sample number 384 Sample mass / g 0.251g The following formula allows the calculation of the proportion of each methyl ester: Percent Fatty Acid = Peak Area (Fatty Acid methyl ester)
Total peak area × 100 b. Using your GC-MS data provided, and the formula above, calculate the percentage composition of fatty acids in the unknown oil. Total peak area = 72028+ 45927905 + 6268770 + 229964567 + 20831782 + 1703110 = 304768162 % C14 (mystiric)= Peak Area (Fatty Acid methyl ester)
Total peak area × 100
= 72028
304768162
× 100
= 0.02%
% C16 (palmitic) = Peak Area (Fatty Acid methyl ester)
Total peak area × 100
= 45927905
304768162
× 100 =15.07%
CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID:z5432922 2 % C18 (stearic) = Peak Area (Fatty Acid methyl ester)
Total peak area × 100
= 6268770
304768162
× 100 = 2.07% % C18:1 (oleic) = Peak Area (Fatty Acid methyl ester)
Total peak area × 100
= 229964567
304768162
× 100 =75.46% % C18:2 (linoleic) = Peak Area (Fatty Acid methyl ester)
Total peak area × 100
= 20831782
304768162
× 100 =6.84% % C18:3 (linolenic) = Peak Area (Fatty Acid methyl ester)
Total peak area × 100
= 1703110
304768162
× 100 =0.56%
CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID:z5432922 3 Figure 1: Gas chromatography and mass spectrum result
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CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID:z5432922 4 Results – Method B: Iodine value and uncertainty calculation 1. Report your raw data: Table 2.
Sample information Sample number 384 Mass / g Sample 0.168 0.174 Concentration of iodine monochloride in glacial acetic acid / mol L
-1 16.2g/L Table 3.
Titration results Volume of thiosulfate / mL Sample 29.5 30.3 Reagent blank 42.1 Concentration of sodium thiosulfate / mol L
-1 0.1 2. Calculate the iodine value (g(I) / 100g) ± uncertainty
from the formula below: iodine value = (b - a) × 1.269
mass of sample
where b = volume (mL) of thiosulfate in titration of reagent blank a = mean volume (mL) of thiosulfate in titration of sample 1. Calculate the mean of the two sample masses: s = 0.168 + 0.174
2
s = 0.171g
2. Calculate the percentage uncertainty of this sample mean: Use the largest mass as mass 1. χ
ୱ
= 0.174 - 0.168
2
÷ 0.171 ×100 χ
ୱ
= 1.754
CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID: z5432922 5 3. Calculate the mean of thiosulfate in the titration of sample: a = 29.5 + 30.3
2
a = 29.9ml 4. Calculate the percentage uncertainty of “a”: Use the largest volume as vol 1. χ
ୟ
= 30.3 - 29.5
2
÷29.9 ×100 χ
ୟ
= 1.338 5. Calculate the percentage uncertainty of “b”: χ
ୠ
= 0.05 42.1
×100 χ
ୠ
= 0.119 6. Calculate iodine value: iodine value = (42.1 -29.9) × 1.269
0.171
iodine value = 90.537 7. Calculate total absolute uncertainty: χ
୧୭ୢ୧୬ୣ ୴ୟ୪୳ୣ
= 1.754
+ 1.338
+
0.119
100
× 90.537 χ
୧୭ୢ୧୬ୣ ୴ୟ୪୳ୣ
= 2.907
8. Final value: 90.537 ± 2.907
CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID: z5432922 6 Scientific writing skills: Discussion Questions In this experiment, we analyze an unknown oil sample using two methods: iodine value and GC-MS. The first part of the experiment involves reacting a sample of unknown edible oil with sodium methoxide, leading to the transesterification of its fatty acids. Then, the resulting fatty acid methyl esters are analyzed by GC-MS (Gas Chromatography – Mass Spectrometry). Gas chromatography separates analyte components based on a principle similar to liquid chromatography or HPLC, except that the mobile phase is an inert carrier gas rather than a liquid solvent. GC presents peaks representing the generated current intensity from the detector plotted against retention time. In other words, it illustrates how much the sample exits the column, the duration for a sample component to start exiting the column after the insertion of the entire 1ml sample volume, and the duration it takes to completely exit the column. The area under the peaks indicates the concentration of the sample, with larger peak areas corresponding to higher concentrations and vice versa (Haas, 2024). Further analysis of these peaks involves examining the mass spectrum generated by the mass spectrometer. The mass spectrometer serves as the detector, presenting data in a vertical bar graph format. The x-axis represents ions with specific mass-to-charge ratios (m/z), while the y-axis represents the relative intensity of each ion. The highest mass ion is presumed to be the entire molecular compound, while lower mass ions are assumed to be fragments of the molecular ion. Determining where this fragmentation occurred requires considering the strength of the bonds between different atoms within the molecules. Therefore, GC-MS data comprises both a chromatogram showing the retention times of all sample components and mass spectra recorded at each retention time. These mass spectra offer structural insights to aid in component identification (Haas, 2024). In the initial phase of the experiment, the GC graph revealed five significant peaks, which could potentially aid in identifying the oil. The sequence of compound elution hinges upon their affinity for the stationary phase and their interaction with the mobile phase, which can be elucidated by considering their chemical structures and polarity. For instance, among C18, C18:1, C18:2, and C18:3, the order of elution follows the trend: C18 < C18:1 < C18:2 < C18:3. C18, consisting of 18 carbon atoms and lacking double bonds (saturated), exhibits weak interaction with the polar stationary phase due to its nonpolar carbon-carbon and carbon-hydrogen bonds, causing it to elute early in the gas chromatograph (University of Costarica, 2019). Subsequently, C18:1, a monounsaturated fatty acid with one double bond, demonstrates a higher level of polarity owing to the presence of the double bond. While not
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CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID: z5432922 7 as polar as compounds with multiple double bonds, it still exhibits stronger interactions with the polar stationary phase compared to C18 (University of Costarica, 2019). As a result, oleic acid elutes later than stearic acid. Next, C18:2 represents a polyunsaturated fatty acid containing two double bonds, with the additional double bond further enhancing the molecule's polarity (Christie, 2023). Consequently, C18:2 fatty acid esters exhibit stronger interactions with the polar stationary phase compared to both C18 and C18:1. Lastly, C18:3 is another polyunsaturated fatty acid featuring three double bonds. Similar to C18:2, the presence of multiple double bonds intensifies the molecule's polarity even more. As a result, linolenic acid esters demonstrate the most robust interactions with the polar stationary phase among the compounds listed (Christie, 2023). Consequently, they elute last, following C18, C18:1, and C18:2. Thus, the elution order in GC-MS for these fatty acid esters aligns with the trend of increasing polarity and affinity for the stationary phase: C18 < C18:1 < C18:2 < C18:3. Then, the fatty acid concentration is calculated using this formula: Percent Fatty Acid = Peak Area (Fatty Acid methyl ester)
Total peak area × 100 The analysis of the unknown oil sample reveals that oleic acid is the predominant fatty acid, constituting 75.46% of the total. This is followed by palmitic acid at 15.07%, linoleic acid at 6.84%, stearic acid at 2.06%, linolenic acid at 0.56%, and myristic acid at 0.02%. Myristic acid, being the least abundant, may have minimal significance (Yanty Noorzianna Manaf, 2011). The mass spectrum indicates specific peaks, with the highest peaks occurring in the GC at a retention time of 11 minutes, corresponding to the concentration of oleic acid (C18:1). Additionally, it reveals that the compound's highest total molecular mass is 45, with a relative intensity of 100. In the second part of the experiment, iodine value is calculated using this formula: iodine value = (b - a) × 1.269
mass of sample
where b = volume (mL) of thiosulfate in titration of reagent blank a = mean volume (mL) of thiosulfate in titration of sample
CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID: z5432922 8 The calculated iodine value is 90.537 with an uncertainty of ± 2.907
. Given the complexity of the sample used, manual interpretation of the graphs proves challenging. However, comparison with the provided standard GC graph and the table of oil (table 4), serving as a reference for iodine value and fatty acid composition, aids in analysis Oil
Iodine Value (g(I) / 100g)
C14
C16
C16:1
C18
C18:1
C18:2
C18:3
Avocado 65–95 <1 18 1 56 24 1 Canola 110-126 <1 5 2 60 22 11 Corn 107-128 13 4 29 54 Cottonseed 100-115 <1 29 2 4 24 40 Grapeseed 94-157 14 12 27 45 2 Macadamia 74–76 4 21 12 53 10 Olive 75-94 <1 14 2 1 72 11 Palm 49-55 1 48 4 38 9 Peanut 82-107 9 2 42 44 1 Safflower 135-150 <1 8 3 13 75 1 Sesame 100-120 10 5 40 45 Soybean 120-139 <1 11 4 25 51 9 Sunflower 110-145 11 6 30 53 Table 4: Iodine values and fatty acid composition of some edible oils. Those provided as sample unknown are in bold Based on the calculated data and comparison with the reference table, the unknown oil is predicted to be avocado oil. Several factors contribute to this conclusion, including the relative quantities of C18, C18:1, and C18:2, as well as the levels of C16 and C18:3, along with the presence of C14 and C16:1. Starting with the relative quantities of C18, C18:1, and C18:2, C18 obtained from the experiment is approximately 2%, which is expected to have a lower concentration of stearic acid according to the provided reference table, indicating stearic acid should only constitute 1%. Furthermore, C18:1 exhibits the highest peak among all peaks, indicating that oleic acid constitutes the highest concentration in the oil sample. The area under the peaks reflects the sample concentration, with larger peak areas corresponding to higher concentrations. According to the reference table, avocado exhibits the highest concentration of oleic acid compared to other fatty acids. Despite the calculated intensity being the highest, it remains too high in comparison to the reference table, which suggests it should be 56%, whereas the experimental result is approximately 75%. Compared to other fruits, the avocado, an oleaginous climacteric fruit known for its nutritious richness, has very low water content
CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID: z5432922 9 (varying from 65% to 75%). Its fat percentage ranges from 15% to 20%, making up around 60% to 70% of the dry pulp. Fruit loses water content as it ripens and gains fat content instead. Avocados and olive oil have comparable lipid compositions, with oleic acid and other monounsaturated fatty acids predominating over other fatty acids (Clemente Mendez Hernandez, 2023). Additionally, C18:2 should constitute the second-highest peak but ranks as the third most concentrated in the unknown oil sample. The calculated percentage of linoleic acid is only 6.84%, significantly lower than the standard reference, which indicates it should be around 24%. The sample's calculated value is approximately four times lower than expected. Myristic acid (C14) is included in the total peak area calculation, despite not showing any peak on the GC graph. Due to the compression of the graph into a small picture, it is not conclusive to deduce that C14 lacks peaks, as the quantitative results indicate a peak area at 6.816 retention time. This clarifies why C14 is incorporated into the total peak area. However, the percentage of C14 is merely 0.02%, rendering it practically absent and insignificantly small. Therefore, we infer that C14 is virtually absent, along with C16:1. This indicates that the oil sample is neither macadamia oil nor olive oil. However, since avocado oil contains a minimal percentage of C14, less than 1%, we still consider the 0.02% presence in the oil sample. Palmitic acid (C16) has somewhat exchanged positions with linoleic acid to become the second most concentrated fatty acid. Despite this, the experiment's calculated percentage of C16 is approximately 15%, which is nearly comparable to the standard reference of 18%. Additionally, linolenic acid (C18:3) from the experiment is also comparable to the standard reference, with calculated data of 0.56% compared to the reference of 1%. When rounded off, the linolenic acid content is 1%. In conclusion, these factors collectively affirm the prediction that the unknown oil is avocado oil. Both the iodine value and gas chromatography-mass spectrometry (GC-MS) are commonly employed methods for analyzing fatty acids, each serving distinct purposes and possessing unique strengths. The iodine value measures the degree of unsaturation in fats and oils by quantifying the iodine (in grams) absorbed per 100 grams of fat or oil. This method rapidly estimates unsaturation levels in a sample through halogen uptake. Given that melting point and oxidative stability correlate with unsaturation, the iodine value offers an estimate of these quality factors. Higher iodine values indicate greater unsaturation and increased
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CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID: z5432922 10 susceptibility to oxidation (Sanders, 2003). However, it does not offer detailed insights into individual fatty acid composition and is somewhat imprecise due to the need for accounting for human and apparatus errors. GC-MS, a highly sensitive and accurate technique, analyzes the composition of fatty acids in a sample. It separates individual fatty acids based on their retention times in a gas chromatograph and then identifies each fatty acid using mass spectrometry. GC-MS enables the identification and quantification of individual fatty acids present in a sample, offering detailed information about the fatty acid profile. Each method has its own specialty: while the iodine value is useful for quickly estimating the degree of unsaturation in a sample, for detailed information regarding the composition of fatty acids, including their identification and quantification, GC-MS is the preferred method. However, it's important to note that GC-MS analysis may be more time-consuming and require specialized equipment and expertise compared to iodine value determination (Labio Scientific, 2022).
CHEM2921: Food Chemistry Lab Manual Student name: Aqilah Khairuddin zID: z5432922 11 References Christie, W. W. (2023, November 8). GC-MS of Fatty Acid Derivatives . Retrieved from The Lipid Webs: https://www.lipidmaps.org/resources/lipidweb/lipidweb_html/ms/basics/GCcolumn/ind
ex.htm
Clemente Mendez Hernandez, D. R.-G.-R. (2023). Study of environmental factors on the fat profile of Hass avocados . Journal of Food Composition and Analysis
. Haas, K. (2024, January 11). 5.3: Quantitative and Qualitative GC and GC-MS
. Retrieved from LibreTexts Chemistry: https://chem.libretexts.org/Courses/Duke_University/CHEM_401L%3A_Analytical_C
hemistry_Lab_Manual/05%3A_Determination_of_cocaine_on_bills_using_Gas_Chro
matography_Mass_Spectrometry_(GC-
MS)/5.03%3A_Quantitative_and_Qualitative_GC_and_GC-MS Labio Scientific. (2022, December 30). Limitations and Disadvantages of GC-MS
. Retrieved from Labio: https://labioscientific.com/limitations-and-disadvantages-of-gc-ms/ Sanders, T. (2003). Ground Nut Oil. Encyclopedia of Food Sciences and Nutrition (Second Edition)
, 2967-2974. University of Costarica. (2019). Variability of the total oil content and fatty acid profile of creole avocados from Nuevo Leon, Mexico. Agronomía Mesoamericana
, 705-719. Yanty Noorzianna Manaf, K. L. (2011). Effect of Varietal Differences on Composition and Thermal Characteristics of Avocado Oil. Journal of the American Oil Chemists' Society
, 1998 - 2003.
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I.85cm
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Dashboard
O NAMING AND DRAWING ORGANIC MOLECULES
Identifying organic functional groups
compound
CH₂
|
0=C=0— CH,
CH,—NH– CH3
Explanation
H
CH3-C- CH3
1
OH
CH₂=CH-C=0
CH3
Check
family
ester
0
0
0
ALEKS
A ALEKS-Kimberly Cedillos - Learn
Ⓒ2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Priv
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- In Example 12, the crude petroleum had an overall octane rating of 52. What is the composition of a solution of MTBE and n-heptane that has this octane rating?arrow_forwardPlease answer as soon as possibly and I will leave a like.arrow_forwardbad drove Use the following information to answer numerical-response question 9. Eight Common Compounds 1 2 3 4 5 6 7 8 C6H₂CH₂OOCCH3(1) Numerical Response 9. CO₂(g) C10H8(s) CaCO3(s) NaCIO(1) C3H6C1₂(1) NaHCO3(s) CH3COOH(aq) toodu The organic compounds above are numbered (Record all four digits of your answer in any order in the numerical-response section on the answer sheet.) andarrow_forward
- Using family suffixes to name organic compounds Name these organic compounds: structure name H. H–C– OH H. CH, — СН,— СH, –C – H || Н— С — ОН Check Explanation O 2021 McGraw Hill LLC. AlI Rights Reserved. Tern ||arrow_forwardFor both images do the following: a) Give the full name for the molecule shown above: b) In 1 sentence, What is the physiological importance of this molecule?arrow_forwardAnswer part b and C please.arrow_forward
- Table 1. Physical Properties of Selected Compounds Density (g/mL) Melting point (°C) Boiling point (°C) Substance Acetanilide 1.22 114 304 Acetone 0.79 -95 56 Benzophenone Bromoform 1.15 48 306 150 2.89 0.98 0.79 2.46 8 -2.4 2,3-Butanedione t-Butyl alcohol Cadmium nitrate 4H;O Chloroform Cyclohexane p-Dibromobenzene p-Dichlorobenzene Diphenyl Diphenylamine Diphenylnethane Ether, ethyl propyl 88 83 25 59 132 1.49 0.78 1.83 -63.5 6.5 86.9 61 81.4 219 1.46 53 70 53 27 174 0.99 255 1.16 302 265 1.00 1.37 -79 64 Hexane 0.66 69 Isopropyl alcohol Lauric acid 0.79 0.88 1.63 -98 83 225 330 43 Magnesium nitrate 6H2O Methyl alcohol Methylene chloride Naphthalene a-Naphthol Phenyl benzoate Propionaldehyde Sodium acetate 3H2O Stearic acid Thymol Toluene p-Toluidine Zinc chloride 89 79 1.34 -98 -97 65 40.1 1.15 1.10 80 218 94 288 1.23 71 314 0.81 1.45 0.85 0.97 -81 58 48.8 123 70 291 232 52 0.87 0.97 2.91 -95 111 45 200 283 732 MAR ...arrow_forwardNow, on top of the structure of methyl amine (CH3NH₂) below, select the region where the partial positive charge (8+) would be located within the structure. Keyboard controls Add selection & I H H C-N. H H 8arrow_forwardI need these two!! Plzarrow_forward
- Unknown Mixture #1 Unknown Mixture #2 Sample caffeine ibuprofen aspirin acetaminoph en Distance to solvent front 14,70 cm 14,70cm 14.70cm1u.70 cm 1나,구 cm |14.구0cm 5. 70 cm 11.90 cm I.85cm Distance Spot(s) traveled 175 cm 13.75cm 11.9 5 cm 6.4 0 cm 13,75 cm 13.75 cm 38 78 1908 .8095 Calculated R:(s) 119 ,9354 812 9 43 S .9354 9354 #1:Aspirin an d lbuprofen Components of Unknowns #2: Ace taminophen, aspirin an d ibu profen D. Additional Exercises 1. Which of the substances tested is most polar? What parts of the structure of this substance is polar? The Sul 2. Which of the substance tested is most non-polar? What parts of the structure of this substance in non-polar?arrow_forwardUse the following image All compounds belong to the alcohol or ester family because they contain a hydroxyl or carboxyl functional group?arrow_forwardConcentration of Reagents (NH4)2S2O8 KI Na2S2O3 0.25 M 0.22 M 0.19 M Volumes of Solutions S₂082 (mL) Starch soln (mL) KNO3 (mL) EDTA soln (mL) KI (mL) S₂O32- (mL) x 5 9.7 0.9 26.4 0.5 9.9 1.1 Using the above data determine the concentration of I at the moment the reaction begins. HINT: This is a mixing problem. Determine the total number of moles of I and divide this amount by the total volume (in L). Be careful... If you go back and try this exercise again the values will change. Read the question carefully! Answer:arrow_forward
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