Cooking in Vitamin C Content

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Sample: MicrowaveDCPIP 0.05% usedmass of the green peppers (samples): 13.78g
Titration 1Titration 2Titration 3average
3.3 (cm3)3.4(cm3)3.2(cm3)3.3(cm3)

Sample: steamedDCPIP 0.005% usedmass of the green peppers (samples): 12.41g
Titration 1Titration 2Titration 3average

Sample: boiledDCPIP 0.005% usedMass of the green peppers (samples): 14.30g
Titration 1Titration 2Titration 3average

Sample: normal (raw)
DCPIP 0.05% usedmass of the green peppers (samples): 11.63g
Titration 1Titration 2Titration 3average (ml)

Sample: bakedDCPIP 0.005% usedmass of the green peppers (samples): 12.87g
Titration 1Titration 2Titration 3average

How to work out mass of vitamin C per unit gram
1. Convert cm3 into dm3 by dividing the average titre by 1000 2. As I already standardised each concentration of DCPIP used I will work out number of moles using concentration x volume 3. As from the balanced equation from my background information, the DCPIP and the ascorbic acid react at a 1:1 molar ratio there no of moles of DCPIP = no of moles for ascorbic acid 4. As this number is no of moles in 25cm3 I will have to work out the no of moles in 250cm3 (as this is the sample I made up to for the extraction method) so I will multiple no of moles by 10 5. As we know no of moles and the Mr of ascorbic acid, we can now work out the mass of vitamin C in my pre weighed amount of green pepper (no of moles multiplied by the Mr of 176.13) 6. As we now know the mass of vitamin C in the pre weighed mass we just need to convert this into mass per unit gram by doing mass of vitamin C/pre weight mass

Raw DCPIP: 0.05 % Average titre 3.33cm3 Convert to dm3 – 3.33/1000 = 3.33x10-3dm3
No of mole (using C.V) = 1.699x10-3 X 3.3x10-3 =...
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