I’m interested in quantitating lithium via NMR using a Magritek Spinsolve 60MHz bench top NMR spectrometer. Is there any evidence that suggests the volume of the liquid sample used in the NMR tube has any effect on the area of the spectra peaks? I’m interested in hearing opinions on the matter before running tests verifying if such correlations in-fact do exist.

I’ve seen in relevant literature that 0.6-1mL is generally a sufficient amount of solution for the 5mm tubes. However it’s not clear to me yet if this volume should be consistent between the calibrants and experimental solutions for analytical NMR. Thanks!

  • 2
    $\begingroup$ Is this spectrometer designed to detect lithium? As per their brochure, the standard configuration is for proton and fluorine nuclei. If this is not an pedantic academic exercise or an educational experiment, I would suggest to use a simple flame photometer to quantitate lithium for practical purposes. $\endgroup$
    – ACR
    Commented Jan 10, 2021 at 1:15
  • $\begingroup$ Yes it is designed to detect lithium @M.Farooq $\endgroup$
    – stuart
    Commented Jan 10, 2021 at 1:20
  • $\begingroup$ If you have too little volume there will be shimming problems. The actual volume is different for simple or Shigemi tubes, see e.g. sopnmr.ucsd.edu/sample-preparation.htm $\endgroup$
    – Karsten
    Commented Jan 10, 2021 at 4:35

1 Answer 1


No, sample volume will not effect quantitation if the following conditions are met: 1) you are comparing samples in tubes with the same inside diameter, 2) you have enough sample volume so that the sample liquid extends at least a few diameters past the ends of the RF coil, 3) your shim is good enough so that the peaks you want to measure do not significantly overlap other peaks. Field distortion due to susceptibility discontinuity at the ends of the sample liquid can degrade shim, and people often use samples that extend far beyond the RF coil for that reason. However, provided the above conditions are met, shim does not change peak integrals. This is because the integral of the line shape is exactly equal to the initial amplitude that the line contributes to the FID, and is independent of how it decays in time.


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