Science Buddies: "Ask an Expert"

Hi. I did the Which Orange Juice Has the Most Vitamin C? experiment using the kit I ordered. It was very helpful to have everything I needed; it is a great idea to offer those kits so thanks! Anyway, I had to change the experiment a little to check for the vitamin C in 5 different citrus fruits. Following the experiment instructions, I made 3 measurements of how much iodine solution it took to titrate the juice/starch mixture, writing down the volume of iodine solution at the start and end of the titration. I know that the instrument limit of error for a 50 mL buret is +/- 0.05 mL so the uncertainity of the starting and stopping reads would be +/-0.05 + +/- 0.05 = +/-0.10 mL. But that is where it stops, I am really not up on my chemistry yet (next year). So, I would really apppreciate some help in trying to figure out how to include this uncertainity when I have to take the average of the three measurements of the juices. Then as the final step to solving for the amount of vitamin C in the juices, I have to set up a proportion per experiment instructions of : (using the average amounts i just solved for) amount of iodine required to titrate vitamin C standard/ 20 mg of Vitamin C (the amt in the standard) = amount of iodine required to titrate the citrus juices/x - with x being the amt of vitamin C in the juices. I am just not grasping how to figure in that uncertainity in these last two steps. And I have searched the internet for further instructions, but I am not catching on. Thanks for any help you can offer.

Hi,

Welcome to Science Buddies! I think you are doing this really great project, and I’m glad the kit was helpful for you.

https://www.sciencebuddies.org/science- ... p044.shtml

It seems that you primary question is regarding the error in your measurements. The first consideration is regarding the accuracy or precision of the buret. According to this website, which includes a good discussion on this topic, the accuracy of a 50 ml buret or pipette is .05 or .1 ml.

http://dwb.unl.edu/Teacher/NSF/C14/C14L ... p01017.htm

The standard error is an estimate of the standard deviation for a group of sample results. Using the triplicate results you obtain from each sample, you can calculate the standard error of your measurements using the following online calculator. The Wikipedia article includes a good discussion of standard error.

http://www.miniwebtool.com/standard-error-calculator/

http://en.wikipedia.org/wiki/Standard_error

So you should go ahead and do the experiment and measure the results for each sample in triplicate as accurately as possible. After you have all of your results, you can calculate the standard error. I hope this answers your question. Please let us know if you have a question about actually calculating the concentration of vitamin C from the titrated results or if you need additional explanation.

Donna Hardy

Thanks for responding so quickly. The information you provided is a little confusing to me, but yes, that is the experiment that i am working on. I understand the idea of using the proportion to solve for the amount of vitamin C in the fruit juices when you have the average of the 3 volume measures of how much of the iodine titration solution it took to titrate the vitamin C standard and the average measure of how much it took to titrate a citrus fruit juice - and then knowing that the vitamin C tablet has 20 mg of vitamin C. I just don't know how to work in the uncertainty.

Example: in any trial, i am making 3 measures of how much iodine solution it takes to titrate a juice/starch mixture by finding the difference between the starting level of the solution in the buret and the ending level in the buret. I read that the instrument limit of error for a 50 mL buret is +/-0.05; so finding the difference of the start and finish levels would have double that.. an uncertainty of +/- 0.10 mL. So, if my measurements of lime juice were 2.45; 2.32; and 2.40 would i just find the average...2.39 and add the{ +/-0.10} ? But if this is correct so far, how would i then plug it {2.39 +/- 0.10 mL} into that proportion I mentioned above to find the amount of vitamin C in the lime juice ?

Thanks so much.

Hi,

I think I understand what you are asking now. I think that the accuracy of the buret is based on using it to measure a sample, so you don’t have to worry about the error of the beginning value and ending value. If you use the buret to titrate a sample and measure 2.39 ml of iodine solution to reach the endpoint, then the value could have been between 2.34 and 2.44.

Here’s one way to include standard error in your discussion section: use the 2.39 ml value in the proportion equation to calculate the concentration of vitamin C in the sample. Don’t worry about the limit of error on the buret at this point. Do the same for the second and third samples. Then use the 3 vitamin C values you have obtained for each sample to calculate the standard error. The standard error calculation automatically takes into consideration the error of the buret and any other errors in the method.

If you wanted to just consider the error caused by the accuracy of the buret, you could solve the proportion equation using 3 values for each sample, in this case, 2.34, 2.39, and 2.44. You could graph this by making the center point a solid point and then marking the upper and lower limits with a line.

Either method would be acceptable to a science fair judge as long as you explained what you were doing. Please do ask more questions if my explanation is not clear.


Donna Hardy

Hi Jane,

I have a question. I just read through the Vitamin C project idea again, and there are no specific directions for including error in the experimental protocol. Are you asking about this topic because of a specific assignment on propagating errors from the teacher?

Also, when is this project due?


Donna Hardy

Hi. Thanks for responding. Sorry in advance for the length of this, but I'm not understanding yet.

Okay, my interpretation of the experiment {which was changed by the teacher to test the juices of 5 citrus fruits is: I will make 3 volume measurements of the amount of iodine solution it takes to change color of the juice (reach the endpoint). I do this by solving for the difference from the beginning level to the ending level in the buret. I will do this 3 times according to the procedure. Then I will get an average of the 3 measurements. I just made up 3 numbers as an example of the measurements in the buret : 2.45 ml; 2.32 ml and 2.40 ml and the average of these 3 is 2.39. My understanding was that i was to use this amount (average amt of iodine solution to titrate lime juice) / a variable (to represent the amount of vitamin C in the lime juice sample) in a proportion equation with the average amt of iodine solution that it took to titrate the vitamin c standard/ 20 mg (because that is the known amount of vitamin C in the vitamin c tablet that was used). Then I would solve for the amount of vitamin C in the lime juice and compare it to the other citrus juices in that trial.

You wrote that I should use the 2.39 ml in the proportion to calculate the amt of vitamin C in the sample - and to do the same for the second and third samples..this is a point of confusion because 2.39 was not itself a measurement, but an average of the 3 measurements. Should I place each measurement I made..2.45; 2.32; and 2.40 in the proportion equation and find 3 vitamin C values... then place those 3 vitamin C values in that online error calculator that you provided in your previous response to me? If so, would I then use my average of the 3 measurements 2.39.. ALONG WITH.. the uncertainty provided by the online calculator and that would be my solution for the amt of vitamin C in the lime juice sample?

Thanks for your help.

Yes, the teacher directed us to consider errors and uncertainities even though the experiment does not directly address them.

The experiment is due later in the week and we have all our raw data, we just didn't know how to handle this uncertainity issue.

Thank you!

Hi Jane,

Yes, with my first suggestion, you would use the 3 volume numbers and put them in the proportion equation and find the 3 vitamin C values. Then put the 3 resulting vitamin C values in the standard error equation to calculation the standard error of the mean. You could then include a discussion of the sources of error. The following website includes a list of the source of errors in titration experiments, and you can tell which ones might apply to your experiment.

http://www.titrations.info/titration-errors

There’s more that you could do, but you can see that the calculations require college level math, so please do double check with your teacher to confirm exactly what is required. If you are taking a statistics class then you probably would be expected to include the propagation of error, but the math is even more complex, and I would not recommend including an analysis that you don't understand.

http://courses.washington.edu/phys431/p ... rs_UCh.pdf

Since the project is due this week, do focus and completing the project. Here is the information on the Science Buddies website for doing data analysis and graphs.

https://www.sciencebuddies.org/science- ... ysis.shtml


And here is information for the display board. Make sure you include all of the sections. The discussion on error should be included in your conclusion section.

Donna Hardy

Hi again. I'm home today with a bad cold and taking the chance to work on my project. I just thought of something else so I wanted to come and post. If I am on the right track about putting each of the 3 volume measurements of the iodine titration solution used in the lime juice in the proportion equation to find the 3 different amounts of Vitamin C for the lime juice ..Then... putting those 3 values in the error calculator to get a standard error, what do I do about the fact that in each trial I also took 3 volume measurements of the iodine titration solution that was used in the vitamin C standard? I couldn't just find the average of the 3 measurements of the iodine used in the vitamin C standard and then use that value in the 3 proportion equations in which i would be using each of the 3 separate measurements of the iodine used in the lime juice could I?

OOPS... just missed you!

Hi,

I'm sorry you are not feeling well, but I think it is smart to do as much as you can since it is due this week. I'm at my computer all day today, so please continue to ask questions.

Your question is a good one, of course; the same error that affected your sample results of course also affected the determination of the standard. The best value that you can use is the mean value of the standard for the proportion equation to determine the sample values. If you wanted to, you could also use the volume obtained for each replicate of the standard to recalculate the individual standard values and then use the 3 numbers in the standard error calculator. If you want to post your raw data for the standard and a sample or two, I could do the calculations also and we could compare result

Donna Hardy

Thank you! This is so kind; I really appreciate this.

Okay. In one trial the 3 measurements for my vitamin C standard were 6.41; 6.35; 6.48 - with an average measurement of 6.41

In the same trial, the measurements for my lime were 2.75;2.72;2.64 - with an average measurement of 2.70

Now I will use each one of the lime measurements in the proportion equation using the average measurement for the vitamin C standard, which makes for 3 separate equations

6.41/20= 2.75/x and 6.41/20 = 2.72/x and 6.41/20 = 2.64/x . When I solve, i get the values of 8.58; 8.48; and 8.23. I put these three values into the sample standard deviation calculator and get a result of 0.18

I find that my average lime measurement was 2.70, so with my uncertainty, the amount of vitamin C in my lime juice is 2.70 +/ 0.18 mg. And in my bar graphs, i could just draw an error bar above my lime bar graph and write +/ 0.18 beside the error bar?

Thanks again.

Hi Jane,

Thanks for posting the data; this is helpful. I obtained the same values that you did for the samples, except I rounded up the 8.48(673) value to 8.49, but it did not make a difference in the calculations. Using the calculator, the standard deviation is 0.18 and the standard error of the mean is .10. The standard deviation calculation number assumes that your data has a normal Gaussian distribution, so a standard deviation of 0.18 means that 68% of your sample values will be 8.42 +/-.18 and 95% of your results will be between 8.42 + .36 and 99 percent will 8.42 +/- .54. Normally, the +/- 2 standard deviation value is used for data like this.

http://www.miniwebtool.com/standard-error-calculator/

Look at the graph under "standard deviation and confidence intervals” in the Wikipedia article below to see what plus or minus 2 standard deviations looks like:

http://en.wikipedia.org/wiki/Normal_distribution

The standard error of the mean (SEM) is the standard deviation of the sample mean, so you use this value on you graphs to show that the mean value could actually have been 8.42 +/- .20 (2 SEM), and I would mark this value on your graph.

For your standard sample, usually the individual titration volumes, the values are 20, 19.8, and 20.21 and the standard deviation is 0.2 and the SEM is .118.

Your data is actually quite impressive and reproducible. You were obviously very meticulous is doing your titrations. With the low standard deviations, you should be able to compare results between the samples and determine if there is a difference in vitamin C between samples. What samples are you comparing? Do you know how to calculate if there is significantly significant difference in your results between the various samples?

Donna Hardy

Hi. Thanks for getting back to me. You asked what samples I am comparing - In each trial, i am working with vitamin C standard, orange juice, grapefruit juice, lemon juice, lime juice and tangelo juice with the intent of finding which citrus fruit has the most vitamin C. Okay, I'm going to type a bit just to clear my thoughts and make them flow easier (so i hope!). Procedure (slightly changed by teacher) says to take 3 volume measurements of the amt of iodine tritation solution required to oxidize vitamin C for each of the 6 (standard, orange, grapefruit, etc) in each trial. Then the old procedure said to take an average of every set of 3 measurements. But have you been telling me that I only have to find one average initially - the vitamin C standard - because that will be used in a proportion equation/ 20 mg (the amt of vitamin C known to be in the vitamin C tablet used in the standard)? Then, in turn, i will take the 3 measurements from each fruit and use each of the 3 in a proportion equation/x (the amount of vitamin C in the lime juice). We did that in an above post and you agreed with my values (with exception of 8.49 instead of 8.48)

Then I think i get confused. You tell me in your post that "SEM is the standard deviation of the sample mean, so you use this value on your graph to show that the mean value could actually have been 8.42 +/ .20 (2 SEM), and I would mark this value on your graph." Where did the 8.42 come from? I get 8.43 when i get the average of the vitamin- C -in - lime - measurements (that we did in the above post). I'm not trying to be smart alecky; I am just not sure if maybe you wrote a 2 instead of a 3. Actually, i hope so; otherwise I am more confused than i thought! Then why did you use +/- .20 (2 SEM)? Is it because you stated in the first paragraph that the standard error of the mean was .10 (when using these 3 lime values in the online standard error calculator).. and then when you consider .10 error 1 way and .10 error another way would be a total of .20?

If this is somehow correct, how would I graph? Make a bar graph extending to 8.43(?) for the lime juice and draw error bars over it and write in ... +/- .20 or 2 SEM?

Thank you so much. I appreciate you helping me all day.

Oh, one more thing. Did you tell me it would be acceptable to just use the average measurement of the amt of iodine required to oxided vitamin C in the vitamin C standard in all of my proportion equations? I saw that you did some computations with those (and think i actually see how you did it , but is it really necessary? Or since I am using this ratio on all the fruit juices, i should use it on the vitamin C standard as well??

Hi Jane,

I'm so sorry. Yes, you are correct. The average of 8.58, 8.49, and 8.23 is 8.43, so this is the correct value to use for your standard. I do apologize for my error; I was definitely not trying to add to the confusion on this complicated topic.

Here is additional explanation for the SEM. The SEM is a standard deviation, so +/- one standard deviation (68% of the values) would be 8.43 plus or minus 0.1; two standard deviations (95% of the values) would be 8.43 plus or minus 0.2. You could mark either value on your graph as long as your explained what you were doing. Your suggestion of making the bar graph with a value of 8.43 and drawing in the error bars either +/- 0.1 or +/- 0.2 sounds like a good way to present the data.

For the last question, you can only use one standard value to calculate the unknown values, so you have to use the average value. It is not necessary to calculate the standard deviation of the standard values; I was just trying to show you that the standard values have error also, but to minimize confusion, you should omit this.

Donna Hardy

Hi
As I read your second paragrah and think back to the standard deviation graph that you referred me to last night, it's clicking into place, and I'm ready to get this project finished. Thanks for being patient with me yesterday. You're awesome!

Hi Jane,

Congratulations on completing your project! This was a really great project, and I am completely impressed that you were able to understand the statistics on your first attempt. You will be able to use these tools in any project in the future. We will look forward to hearing from you next year. You will be amazed at how much you will understand after you have had chemistry.

Donna Hardy