Vitamin C 1 (1)

BCH 367 Results Six different dilutions of Vitamin C in a 1% phosphate-citrate buffer were created by using the equation C1V1=C2V2 with the total volume of 1 mL. Example of equation for the dilution of 3 ug/mL is as demonstrated below: C1= 50 ug/mL V1=? C2= the dilution ( 3) V2= 1 ml total 50 ug/mL x V1 = 3x 1 mL = 0.06 mL ( This is what V1 equals to) . The same steps were completed for all 6 dilutions in the chart below (figure 1). Figure 1: Standard curve Vitamin C (ug/ mL ) DCPIP (mL) Phosphate citrate buffer (mL) Vitamin C stock ( mL) Total Volume (mL) Absorbance 0 0.67 0.33 0 1.00 0.625 A 3 0.67 0.27 0.06 1.00 0.624 A 6 0.67 0.21 0.12 1.00 0.575 A 9 0.67 0.15 0.18 1.00 0.574 A 12 0.67 0.09 0.24 1.00 0.445 A 15 0.67 0.03 0.3 1.00 0.384 A The dilutions created with the values obtained, the spectrophotometric assay was taken of each sample at an absorbance of 518 nm ( 0.683 A). DCPIP absorbance vs the concentration of Vitamin C is shown in the graph below: Figure 2: Absorbance vs. Concentration of Vitamin C

The first point ( 0.625 ) and the fourth point ( 0.574) were excluded from the linear equation used to calculate the Vitamin C concentration. This was done in order to get a linear trendline as you can see in figure 2. Figure 3: Spectrophotometric analysis of samples Unknown sample Background absorbance (no DCPIP ) Concentration Absorbance at 518 nm Actual Absorbance Vegetable ( green bell pepper ) 0.024 A 0.368 A 0.344 A Fruit ( orange ) 0.004 A 0.423 A 0.419 A The concentration of Vitamin c was calculated using the following equations: Calculation of concentration of cuvette: ● Vegetable ( green bell pepper)

Starting solution = 58.10 ug/mL = 0.0581 mg/ mL x 25 ml = 1.45 mg of vitamin C / 1g of vegetable Discussion The purpose of the experiment is to determine the amount of ascorbic acid in foods and beverages by use of a spectrophotometric assay. In the lab experiment for part 1 is to use DCPIP investigations and generation of a standard curve to determine the wavelength of the maximum absorbance by performing a wavelength scan using a spectrometer. DCPIP was prepared by 0.67 ml of solution and 0.33ml of phosphate- citrate buffer. In Figure 1 you can see the wavelength and the values of the absorbance reading for the unknown samples and the volumes for Vitamin C. In figure 2 it shows the absorbance of the Vitamin C Dilutions graph goes on a decreasing range and results in y =-0.0166x + 0.6623. According to the USDA website there is 0.8 g of vitamin c in a green bell pepper. Based on our calculations there is 1.45 mg of Vitamin C in 1 gram of a green bell pepper. There is a 33 % error in our calculation. When comparing the amount of the vitamin C in the green pepper (vegetable ) was calculated in the experiment to an actual amount in the certain article provided, they were inaccurate when comparing both of them. On the other hand when comparing the fruit ( orange) calculations in the experiment to an actual amount given in the certain article it is very similar. According to the USDA website there is 0.5 mg of Vitamin C in 1 mL of orange juice. Our calculations were 0.4442 mg. Possible errors in our experiment could be inaccurate values on the samples. Also, there were errors when conducting Part 2. There was confusion on how to complete the procedure to obtain a color purple. However, the TA explained multiple times the procedure and it was completed successfully and approved by the TA. This experiment can be improved by taking more samples between different pieces of sampled produce, or taking samples across different varieties. This would expand the range of measured Vitamin C concentrations, and theoretically lower the error percentage. Questions 1. Compare the amount of vitamin C per gram of vegetable with the amount per ml of freshly squeezed fruit juice. Estimate how much of the vegetable (used in your experiment) you would have to eat to equal the vitamin C content of 5 ml of fruit juice. Vegetable used: 0.9893 g Fruit juice 0.5 mL

0.9893 x 0.8 mg/ g ( green bell peppers ) = 0.79144 mg . Amount of vitamin c present in 5ml of fruit juice ( orange juice) = 0.5 ml. Vs vitamin c in 5 ml juice 5 mL x 0.5 mg/mL = 2.5 mg so 2.5 / 0.8 = 3.125 grams. 3.125 g of vegetables will need to be eaten to equal the vitamin c content of 5 mL of fruit juice. 2. Use a literature source to find typical vitamin C values for the foods that you analyzed. Compare these values with what you obtained. Based on USDA there is 14.2 mg of Vitamin C in 1 oz of orange juice. In our calculations above there is 0.4442 mg of orange juice in 1 ml. Which means that 13.13 mg of Vitamin C are in 1 oz based on our calculations. This is a 67 % error. 1 mL divide it by 0.033814= 29.57. So 29.57 x 0.4442 mg of orange juice = 13.13 mg of Vitamin C. 14.66 x 330 ml = 4837.8/1000= 4.8378 ug/ mL 14.66 - 4.8378 /14.66 = 67 % error There is 0.8 mg of Vitamin C in a gram of green bell peppers. My calculated Vitamin C for the vegetable was 14. 525 mg of Vitamin C per 1 g of vegetable. This is a 33 % error. 19.17 ug/ ml x 670 mL = 12843.9/ 1000 = 12.84 ug/ mL 19.17 -12.84 / 19.17= % error = 33% 3. Why is it important that DCPIP is present in access in your measurement? Is important that DCPIP is present in access in our measurement because if the Vitamin C concentration increases then DCPIP absorbance would decrease. 4. Why should the instrument be zeroed using the phosphate-citrate buffer as opposed to using a solution containing DCPIP? If the instrument is not zeroed and if the instrument is previously used for DCPIP then it might be contaminated with the DCPIP. This can lead to incorrect reading of the PH buffer therefore it has to be zeroed to record the exact PH and find the correct concentration of the solutions. 5. If a fruit or vegetable sample contained a molecule that absorbed light close to the λmax for DCPIP, why is it important to perform an analysis of the sample without DCPIP present?