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### Topic: Need to multiply Dilution Factor to find Concentration?  (Read 33306 times)

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#### a confused chiral girl

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• Mole Snacks: +2/-1 ##### Need to multiply Dilution Factor to find Concentration?
« on: July 23, 2010, 04:02:40 PM »
Hii,

Let's say I am preparing a set to make a calibration curve. I'm making this in a 50mL Erlenmeyer flask, and I have added in 1ml of 1mg/mL of my compound and 5mL of 5ug/mL of the deuterated compound (internal standard). Then I diluted this to the mark with MeOH.

So when I calculate my compound's concentration, I have to multiply it by the dilution factor (which is 1mL/50mL). However, I don't understand why I have to multiply it by the dilution factor because my internal standard also got diluted as well right?

Thank you!! #### a confused chiral girl

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• Mole Snacks: +2/-1 ##### Re: Need to multiply Dilution Factor to find Concentration?
« Reply #1 on: July 28, 2010, 12:45:05 AM »
does anyone have any explanations as to when you have to multipy by the Dilution Factor to find the Concentration of a compound relative to the Internal Standard?

#### el13

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• Mole Snacks: +0/-1 ##### Re: Need to multiply Dilution Factor to find Concentration?
« Reply #2 on: July 28, 2010, 07:52:45 AM »
An internal standard in analytical chemistry is a chemical substance that is added in a constant amount to samples, the blank and calibration standards in a chemical analysis. This substance can then be used for calibration by plotting the ratio of the analyte  signal to the internal standard signal as a function of the analyte concentration of the standards. This is done to correct for the loss of analyte during sample preparation or sample inlet. The internal standard is a compound that matches as closely, but not completely, the chemical species of interest in the samples, as the effects of sample preparation should, relative to the amount of each species, be the same for the signal from the internal standard as for the signal(s) from the species of interest in the ideal case. Adding known quantities of analyte(s) of interest is a distinct technique called standard addition, which is performed to correct for matrix effects.

This ratio for the samples is then used to obtain their analyte concentrations from a calibration curve. The internal standard used needs to provide a signal that is similar to the analyte signal in most ways but sufficiently different so that the two signals are readily distinguishable by the instrument. For example deuterated chlorobenzene (C6D5Cl) is an internal standard used in the analysis of volatiles on GC-MS because it is similar to Chlorobenzene but does not occur naturally.
From Wikipedia

#### Train

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• Mole Snacks: +2/-0 ##### Re: Need to multiply Dilution Factor to find Concentration?
« Reply #3 on: July 28, 2010, 09:32:12 PM »
The calibration curve (with or without internal standard) will give you the ability to estimate the concentration of your test solution, but that is not the same as your sample concentration because you diluted the sample in order to get the test solution.  The dilution factor tells you how much more concentrated your sample was than your test solution.  In your case, the amount of analyte that used to fill 1 mL now fills 50 mL, which means that your sample is 50 times more concentrated than your test solution.

#### a confused chiral girl

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• Mole Snacks: +2/-1 ##### Re: Need to multiply Dilution Factor to find Concentration?
« Reply #4 on: July 29, 2010, 01:54:50 AM »
Thanks for explaining.

Train- but concentration is found by the ratio of the analyte to internal standard. In this case where 1mL of the compound is added and diluted up to 50mL, the internal standard also gets diluted 50 times. So both the sample and the internal standard were 50 times more concentrated than the test solution. Then why do I need to multiply the Dilution Factor, since the INt std. already compensated for that?

thanks!!

#### Train

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• Mole Snacks: +2/-0 ##### Re: Need to multiply Dilution Factor to find Concentration?
« Reply #5 on: July 29, 2010, 09:53:36 PM »
Train- but concentration is found by the ratio of the analyte to internal standard.

Can you elaborate?  What does your calculation look like?