If the internal standard was incorporated into the sample before the extraction, then the dissolution volume (and injection volume) are irrelevant since the ratio of internal standard to analyte was established at that initial point. The standard curve should be concentration ratio (analyte to standard) versus peak area ratio (analyte to standard). The samples and calibration points on the standard curve should both be made the same way (i.e. 50ul of analyte standard solution plus the same amount of internal standard solution which is added to the samples). Only the initial sample volume has any meaning. Once you know the concentration ratio from the sample, then you can calculate the amount of analyte in the original 50ul of sample based on the concentration of the internal standard used (i.e. multiply the concentration ratio determined in the analysis by the internal standard concentration and divide by 50ul, the initial volume of sample).

Sir, 

Thank you for your response. I am still confused.

Lets say,

I have standard dissolved in 1000ul and injected 5ul in MS. However thinking that signal will be very low, I dissolved the sample analyte in 500ul. Now, to quantitate my analyte (that was dissolved in 150uL solvent) based on standard (that was dissolved in 1000uL solvent), should I need to divide my calculated concentration (analyte) by 2 (as my standard is two-fold higher in volume than my analyte)?

Thank you again. It has been a great help for the beginners like me to learn.

Sincerely,

Sangita

Note: I am referring to the external standard, not the internal standard.

I completely agree with Bruce Levison. However, I would like to put forth some more information about your question directly as to where would you use that 200uL. Following is an example.

You dissolved Entire extract in 200uL and then use directly this 200uL (out of this 200uL whether you use 10 or 20 or entire 200uL is irrelevant as bruce has already pointed out), then this 200uL is of no use to you. However, now if you use this 200uL as a stock solution and then used 50uL of this and then diluted to 200uL and then used this, then basically your answer from calibration curve is 1/4th of an actual answer since you used 1/4th of the stock solution. If you keep diluting this solution to reach to a proper dilution, you need to keep adding this (rather multiplying this) into answer and factor the volume. Now, only thing you need to make sure that the ratio of Internal standard (if you are using IS) and the analyte should fall in the region where your calibration curve is drawn. If your min on calibration curve is 0.2 and maximum is 0.5 you cannot use the ratio that has a value of 0.5. In this case you need such dilutions series what I just talked about. Hope this helps understand more. Bruce's answer was just perfect anyways.

 In the case you are using External standard as well, dilution effect would be exactly same. First you need to just get the value in terms of mmol/ml (or concentration) and then factor in all the dilutions as and when you diluted it.

Thank you Puskar. It is good explanation. Can you please refer to my recent question and help me to get over it.

I have external standard and analyte. I dissolved external standard in 1000uL and analyte in 500uL. Now, when I calculate the concentration of my analyte, do I need to change anything? Lets say concentration of std is 1000ug/mL. If I use this concentration to calculate response factor and use that response factor to calculate concentration of analyte (dissolved in 500uL), do I need to do anything extra? What I am missing?

Thank you for your time.

Whether you use 1000uL or 500uL the concentration term that you get at the end of the day is mmol/Vol. I am assuming that you have carried out the calibration curve using your external standard and the compound of interest. In order for you to draw the line, you need two axis. One is the ration of response (e.g. intensity in your case) and other axis is the ration of concentration of External standard and analyte (known). From such calibration factor, you can get a straight line that passes through an origin. If you have such line it should give you an equation Y=mX. where Y is intensity ratio (dependent variable) and x is concentration (independent variable). by knowing Y for your unknown and m you can calculate X and from X you can calculate the amount of unknown. Up until now there is no factorization needed since this is only pure value. Now in order to get this particulat sample (the one that you injected) if you have done any dilutions, you need to factor in that here (after getting the value). It is irrelevant whether you use 1000uL standard and 500uL analyte, the ratio of them is what you are interested in and not the actual volume, This is the beauty of calibration curve. Hope that helped

By the way it's Pushkar and not Puskar....!!!!