CIL NMR Reference Standards

CIL’s total quality assurance protocols and in-house manufacturing capabilities guarantee the highest standard of quality. Typically, these NMR Reference Standards have been evaluated and determined to exceed industry requirements. Furthermore, CIL welcome your bespoke requests for custom formulations of other reference standards as well as alternative fill heights of existing reference standards.

Can CIL offer reference standards with 40 mm fill height?

Yes, CIL is pleased to be able to offer 40 mm fill height on a custom basis. Most (but not all) other NMR reference standard fill heights are 2 inches. Please contact us with your request if you require this fill height.

Does CIL manufacture custom NMR reference standards?

Yes, CIL is pleased to be able to offer custom products such as larger or smaller diameter NMR reference standards as well as different reference solutions. Please contact us with your request if you do not see your product of interest on the website.

How do you calculate the molarity of DLM-45, Sodium deuteroxide (D, 99.5%)?

DLM-45, Sodium deuteroxide (D, 99.5 %),

Molarity Calculation from Weight Percentage

DLM-45 has a concentration of 40 % NaOD in D₂O, by weight (w/w).  This means 40 g NaOD per 100 g of solution.  The D₂O accounts for the remaining 60 g of the solution.

What is the molarity of this solution? 

Molarity (M) is defined as the number moles of solute, per liter (L) of solution.

𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 ,𝑀.= ,𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒-𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

Taking 40 g of solute (NaOD) and its molecular weight (MW = 41.01 g/mol),  we divide mass by MW to get 0.9754 mols for 40 g NaOD.

40 𝑔 𝑁𝑎𝑂𝐷 × ,1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐷-41.01 𝑔 𝑁𝑎𝑂𝐷.=0.9754 𝑚𝑜𝑙𝑠 𝑁𝑎𝑂𝐷

We need the density of deuterium oxide, D₂O, to find what volume 60 g will occupy.  The density is 1.103 g/mL, so 1L D₂O will weigh 1,103 g.  We get the volume of D₂O in L, by dividing the mass, 60 g D₂O, by the mass of one litre, to get 0.0544 L D₂O.

60 𝑔 ,𝐷-2.𝑂 ×,1 𝐿 ,𝐷-2.𝑂-1,103 𝑔 ,𝐷-2.𝑂 .=0.0544 𝐿 ,𝐷-2.𝑂

Assuming the NaOD occupies no additional volume in the solution,  the volume of D₂O is equal to the volume of the solution.  Dividing 0.9754 mols NaOD by 0.0544 L gives 17.93 moles NaOD per L, about 18 M.

,0.9754 𝑚𝑜𝑙𝑠 𝑁𝑎𝑂𝐷-0.0544 𝐿.=17.93 ,𝑚𝑜𝑙𝑠-𝑙𝑖𝑡𝑒𝑟. ≈18 𝑀

We have assumed that slight variations of MW due to isotopic enrichment, and density variations with temperature can be ignored.