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Measurement of the vitamin C content of fruit juices

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Introduction

James Hobbs 10IP Measurement of the vitamin C content of fruit juices Plan: Aim: To investigate the effect of heating on the concentration of vitamin C in fruit juice. Introduction: DCPIP is a purple dye that is decolourised by adding vitamin C or a fruit juice that contains vitamin C. If a standard solution of DCPIP is used then the vitamin C content of different fruit juices can be compared, the more juice it takes to decolourise a standard of DCPIP than the smaller the concentration of the vitamin C in the juice. If the DCPIP solution is first tested with a known concentration of Vitamin C (ascorbic acid) then it is possible to calculate the vitamin C content of other juices. Vitamin C is found in foods such as oranges, lemons, grapefruits, tomatoes, fresh green vegetables and potatoes. A lack of vitamin C causes a disease known as scurvy. The symptoms of this are; Fibres in connective tissue of skin and blood vessels do not form properly, leading to bleeding under the skin, particularly at the joints, swollen, bleeding gums and poor healing of wounds. Vitamin C cannot be stored in the body, so there needs to be a daily intake of vitamin C. Vitamins are group of organic substances quite unrelated to each other in their chemical structure. ...read more.

Middle

I.e. you have added 0.6 cm�. Safe Test: Do not carry the syringe around the room. Hold the syringe with the needle pointing down. Wear safety goggles at all times. If the syringe appears blocked, do not force it. Prediction: I thick that the longer the vitamin C is heated the more solution it will take for the DCPIP to decolourise. Observing Time the fruit juice was boiled for (Hours) Vitamin C added -Test 1 (cm�) Vitamin C added -Test 2 (cm�) Vitamin C added -Test 3 (cm�) Vitamin C added -Average (cm�) Standard solution - 0.9 0.95 0.9 0.92 A Fresh 1.0 1.0 - 1.0 B 1/2 hour 1.08 1.1 - 1.09 C 1 hour 1.15 1.12 - 1.14 D 2 hours 1.5 1.4 - 1.45 E 5 hours 2.25 2.2 - 2.23 Analysis I took the averages of my results and used them in the equation mentioned earlier. Percentage of Vitamin C = Average volume of vitamin C used x 0.1 in fruit juice Average volume of fruit juice used Percentage of Vitamin C = 0.92 x 0.1 in fruit juice A 1.0 = 0.092 = 0.09% Percentage of Vitamin C = 0.92 x 0.1 in fruit juice B 1.09 = 0.084403669 = 0.08% Percentage of Vitamin C = 0.92 x 0.1 in fruit juice C 1.14 = 0.080701754 = 0.08% Percentage of Vitamin C = 0.92 x 0.1 ...read more.

Conclusion

I had one anomalous result, the test of fruit juice B. The amount of vitamin C added should have been about 0.12 cm� instead of the 0.09 cm� that I got. This means I did not put enough fruit juice into the DCPIP, which shows a human error as I said before. This was the only anomaly, which shows the results were quite accurate, but also points out the importance of repeating the experiment three times, so that you can see any anomalies. As I said before, you could use a more precise syringe to get more accurate results. Also the more times you repeat the experiment the more accurate the experiment, which means the more accurate the results. Ensure that the Standard solution is tested well as that result effects the end overall results. Another experiment, which could be done, is; to compare the vitamin C contents of Oranges, Lemons and grapefruits, to see which one had the highest amount of vitamin C. The method would be: * Draw up 2 cm� fresh orange juice into a syringe. Add this drop by drop to 1 cm� of DCPIP solution. * Record how much was added and repeat the experiment to obtain averages. * Repeat the experiment with orange and grape fruit juices * Compare the results of all three samples. The one, which uses the least amount of juice, contains the most vitamin C in the juice. ...read more.

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A fairly good account but more detail need in places and errors made in display of results.

Marked by teacher Adam Roberts 16/07/2013

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