Probably the most common question I get asked is "How much sample do I need for a carbon?" The answer is - it depends on the probe being used, the molecular weight of the compound, and how long you are prepared to run the experiment for. For most people, however, "it depends" is not a very satisfactory answer, so I've tried to find a way to get some numbers to answer the question.
When people ask "How much sample do I need?" they want to know how many milligrams of their compound to put in an NMR tube to obtain a spectrum with acceptable signal-to-noise. They're normally wanting to be given a particular mass to use but, since each NMR signal is directly proportional to the number of atoms, larger molecules need more mass in the tube to obtain the same size signal for their atoms. For this reason expressing how much sample is needed in terms of number of moles, or a concentration, allows a more generally useful answer. Beyond amount of material the other major contributor to the size of an NMR signal is the number of scans. Its well known that NMR is a non destructive technique and that increasing the number of scans will increase the size of the signal. Also well known, however, is that the signal-to-noise only increases as the square root of the number of scans. These facts can be expressed in the following equation,
S = k.C/V.√NS
where S is the signal-to-noise, k is a scaling constant, C is the sample concentration, V is the volume in the NMR tube, and NS is the number of scans. The scaling constant, k, will depend on the probe and magnet used and will also account for the parameters used to acquire the spectrum. If k can be determined for a particular hardware and parameter setup then we can relate the concentration to the number of scans needed to obtain a given signal-to-noise.
To determine a value for k I recorded a spectrum of 100 mg/ml (233.26
mM) cholesteryl acetate in chloroform-d at 599 MHz with a room temperature 5
mm BBI probe. The parameters used were the standard SSPPS NMR Facility ones; a
30o pulse, a 0.5s relaxation delay, an acquisition time of 1.748s,
128 scans, and 8 dummy scans. This experiment takes 5 minutes and 14 seconds.
Calculating the signal-to-noise on the peak at 11.9 ppm gave a value of 76.60.
Using this result and a volume of 500µL in the equation above gives a value
for k of 0.014512877. With this value in hand we can rearrange the
equation to express NS in terms of C and create a graph showing
the number of scans needed to obtain a certain signal-to-noise at a given
concentration, as shown below. Note that the vertical NS axis is
logarithmic. The red line shows the correlation between number of scans and concentration for signal-to-noise of 100, while the blue line is for signal-to-noise of 10.
The graph indicates that for the 233 mM cholesteryl acetate sample a spectrum with signal-to-noise of 10 will require 2 scans, and signal-to-noise of 100 will require 218 scans. Note how a ten-fold increase in sensitivity requires a 100-fold increase in NS. Spectra recorded with these number of scans are shown below. Happily, the signal to noise calculated on the 11.9 ppm peak was close to the predicted (9.15 and 102.65).
The value of k determined here provides a way to determine how much sample is needed to get a 1D 13C spectrum, but this value is specific for the probe and parameters used here. If other experimental setups are used then a new k value needs to be determined. It also requires that the probe be tuned and matched for the sample. Poor tuning and matching reduces the sensitivity.
Well timed reminder re S/N! Also, for those of us working with heavily halogenated natural products (Cl, Br, I) it’s worth keeping in mind halogen adds ‘mass’. Putting this another way, the 13C content is ‘diluted’. The S/N of your 5 mg sample of, say halomon (C10H15Br2Cl3, 66.7% of its molar mass halogen!) will be much lower than 5 mg of, say, myrcene (C10H16). And for the same mole amounts of myrcene and cholesteryl acetate, the S/N of any single comparable C signal will be lower in the latter than the former, because of simple arithmetic: dividend diminishes by 10 C vs 29 C, respectively. Other factors are operative (e.g relaxation times). Bottom line: be cognizant of your sample properties. At a time when NMR 13C probe sensibilities were much lower than today, but hourly rates comparable (corrected for inflation), I would say to my students, “Don’t run overnight 13C NMR experiments on a ~1-5 mg sample in your NMR tube when another -50-100 mg is siting back on your bench!”
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