I have been provided the following table of calibration standards from an ICP-MS experiment. (Some text is in German, I am operating in a location where DIN standardization is ubiquitous) enter image description here

I understand that I must subtract the provided method blank from each of the provided data points before plotting them and drawing a regression line, but the sample solution I am working with (500 mg manganese slag in 10mL of a mixture of hydrogen peroxide and nitric acid) has also been diluted in a 1:10 ratio with water. How do I account for this dilution when calculating the mass concentration of manganese of the sample?


1 Answer 1


There is a very simple dilution formula:

C$_{initial}$ x V$_{initial}$=C$_{final}$ x V$_{final}$

where C and V refer to concentration and volume in consistent units.

When you analyze the sample (=Probe in your table) using the regression equation, the concentration unit of the sample will be the same as your standards ng/mL.

Lets say your diluted sample conc. from the regression equation 12 ng/mL (= C$_{final}$ see the above equation), all the variables are known to you except C$_{initial}$. You know the sample volume (=V$_{final}$), you must also know how much initial volume was used for dilution.

Once you find C$_{initial}$, one can determine the mass from the original sample volume, because analyte mass, Mn = concentration x volume.

Note: In certain methods multiple dilutions are made, then one must apply the dilution formula repeatedly for each dilution.

  • $\begingroup$ Thanks for pointing out that multiple dilutions could happen. In this case it was just the one, however. $\endgroup$ Jun 17, 2021 at 15:17

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