Hi all,
Attached below is the proton nmr spectra of biodiesel. The integration value of peak at 5.2 ppm is set to 1. and corresponding all other integration values are obtained. Is that right .
From there. I'm calculating the yield of biodiesel by using integration values at 3.6 and 2.3 ppm.
Dear Alisha.
The integral depends on both, the concentration of the substance and how many protons are forming the peak. For instance Ethanol has a CH3 group and a CH2 group. The integral of the CH3 peak is 1.5 times larger than the one of the CH2 peak. If you know how many protons your reference peak consisits of, you know the concentration of the corresponding reference substance and you know at least for one peak of Biodiesel how many protons lead to this peak. The ratio of concentrations are equal to the ratio of integral normalized by the respective number of protons forming the peak.
Many Greetings,
Kristof
The integration values of different peaks at different chemical shifts just give the ratio of different types of protons which originate those peaks. This information may be used in several ways depending on the material analyzed, provided for a reference peak, number of protons is known.
Dear Alisha,
to answer the question I (we) need to know the reaction-scheme.
Was it a trans-esterification of glycerol-esters? If this is the case I'd guess that the signal at 3.6 is the O-CH3 of the product (which you have divide by 3, as described above) and the signal at 5.2 might be one hydrogen of the glycerol in the educt. And I(O-CH3)/3/I(educt) would be the molar ratio of product/educt (which is not the necessarily the yield!). But this might be wrong if my "hand-wavy" assignment is wrong.
Alfred
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