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### Topic: Internal Standards & Dilution  (Read 329 times)

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#### confusedcollegestudent

• Regular Member
•   • Posts: 18
• Mole Snacks: +0/-0 ##### Internal Standards & Dilution
« on: April 27, 2019, 05:57:59 PM »
We are doing a lab with chocolate samples to figure out the concentration of caffeine and theobromine in single-origin chocolates with internal standards.

We added the internal standard to all of our samples, including the calibration curve and our chocolate samples.
I'm confused about how to back calculate to find the original concentration in our original sample.

Our professor said to not use response factor and to just plot and use linear regression.

I can calculate to find the concentration of analytes in the samples, but I'm not sure how to calculate back to the signal of just the sample and not with the internal standard.

First we added 0.500 g of chocolate to a centrifuge tube and added 20.00 mL of Milli-Q water and melted.

Then we added a 5.00 mL of internal standard and ran our columns.

To run our columns, we only used 0.5 mL of our sample from the 25.00 mL.

How many steps is this? Where do I even begin?

The linear regression plot is with x as [concentration of analyte] and y as (Ianalyte*[IS])/IIS
Analyte being either theobromine or caffeine.

So I assume that the slope of the linear regression equation is the response factor, but I'm not understanding how to calculate to the original chocolate sample since we had several volume changes.

#### mjc123

• Chemist
• Sr. Member
• • Posts: 1685
• Mole Snacks: +237/-11 ##### Re: Internal Standards & Dilution
« Reply #1 on: April 30, 2019, 08:18:40 AM »
Just think it through slowly, carefully and logically. Suppose you have z mg of analyte per g of chocolate. After each step, work out the concentration in terms of z. (A step is whenever you do anything to your sample.) Eventually, you should get an expression for y in terms of z. From this, and your measured value of y, you can calculate z.
To start off: you take 0.500 g of chocolate. This contains 0.5 g x z mg/g = z/2 mg analyte.
This is dissolved in 20.00 mL water. What is the concentration of z?
What is the concentration after adding the internal standard? (You must take account of both the amount of internal standard and the change in volume.)
Etc.