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5. Food Industrial Products (Testing). Exercise 5.1 (a)What is the difference between unprocessed and processed foods? unprocessed – no cooking, cleaning/cutting.

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Presentation on theme: "5. Food Industrial Products (Testing). Exercise 5.1 (a)What is the difference between unprocessed and processed foods? unprocessed – no cooking, cleaning/cutting."— Presentation transcript:

1 5. Food Industrial Products (Testing)

2 Exercise 5.1 (a)What is the difference between unprocessed and processed foods? unprocessed – no cooking, cleaning/cutting only raw meats fresh fruit & vegetables (b)What should unprocessed foods be tested for? freshness contaminants (especially by organisms – moulds, insects) (c)Is there anything else that processed foods should be tested for? additive levels required physical properties (d)Are all food product types tested before consumption? yes, but unprocessed foods only get visual inspection

3 Shelf life Exercise 5.2 (a) less tasty baked products, e.g. bread, biscuits (b) unsafe raw meat dairy products baked-for date – a date not later than 12 hours after the time the bread was baked baked-on date – the date on which the bread was baked best-before date – will retain any specific qualities use-by date – not safe afterwards

4 Water C LASS E XERCISE 5.3 Why is it necessary to test for water? measure of freshness/processing C LASS E XERCISE 5.4 What are the limitations of the oven drying method? not good for samples with: high viscosity (eg honey) high levels of other volatiles (eg essences) low water content C LASS E XERCISE 5.5 (a) bread (b) honey (c) olive oil (d) vanilla essence oven KF, D&S KF GC

5 Triglycerides and other lipids major component of fats in food Exercise 5.6 What is the structure of a typical triglyceride?

6 Analysis of fats total – solvent extraction (Soxhlet extractor) individual triglycerides can’t be analysed too many different types too similar to each other for chrom. separation to obtain a "fat profile" by the Fatty Acid Methyl Ester Method (FAME) 1.hydrolysis of the triglycerides in methanolic KOH to produce the fatty acid salts 2.conversion of the fatty acids to their corresponding methyl esters using methanol and the catalyst BF 3 3.analysis of the esters by GC

7 C LASS E XERCISE 5.7 (a) Write the equations for steps 1 & 2. (b) Explain why this gets around the problem described for the triglycerides. fatty acid methyl esters differ by FW only 10-20 of them

8 Simpler QC tests saponification value => average chain length hydrolysis of the triglycerides under reflux with a known amount of KOH excess KOH is back-titrated with standardised HCl SV is the mass of KOH in mg that reacts with 1 g of food shorter chains mean more triglyceride molecules per gram, therefore more reaction with KOH iodine value => degree of unsaturation reaction of the triglycerides with iodine monochloride (Wij’s reagent) in the dark for 1 hour C=C undergoes an addition reaction with the ICl excess ICl is then back-titrated with standardised sodium thiosulfate IN is the mass of iodine that reacts with 100 g of sample more C=C => more reaction

9 Proteins C LASS E XERCISE 5.8 What is the structure of a typical protein?

10 Kjeldahl N analysis oxidation of the sample in hot concentrated sulfuric acid converts the bound nitrogen into ammonium ions addition of strong base to the solution converts the ammonium ions to ammonia gas collected by distillation in a known volume of standard acid, then back titrated or in a solution of excess unstandardised boric acid borate ion produced is a stronger base than ammonia and is titrated with standardised acid 1 mole of N gives 1 mole NH 3 amount N related to amount of protein by factor which is the proportion of N in typical protein of given material, eg flour, milk

11 Exercise 5.9/10 Calculate the %N & %protein in a sample of meat if 1.0589 g of sample, after the Kjeldahl procedure, gave a titre of 15.8 mL of 0.103 M HCl. FW of N is 14. Protein factor is 6.25. moles HCl = 15.8 x 10 -3 x 0.103 = 1.63 x 10 -3 = moles NH3 = moles N mass N = 1.63 x 10 -3 x 14 = 0.0228g %N = 0.0228 x 100 ÷ 1.0589 = 2.15% %protein = 2.15 x 6.25 = 13.4% Read remainder of chapter yourself


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