Background:
Lanreotide is considered as an example. Lanreotide is peptide molecule and it is active ingredient of Lanreotide injection. It is available in acetate salt form.
(C54H69N11O10S2).X(C2H4O2)
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Data required:
Water is likely to be present as one of component in Lanreotide acetate salt.
The analysis data (in % W/W) of Lanreotide, acetic acid and water are required and same is used to derive X value of acetic acid.
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Assume Lanreotide is diacetate and dihydrate molecule, then molecular formula will be (C54H69N11O10S2).2(C2H4O2).2(H2O).
and molecular mass will be 1252.34 gm/mole and details as below
Molecular Weight of Lanreotide is 1096.34
Molecular weight of acetic acid is 60
Molecular weight of water is 18
Molecular weight of Lanreotide diacetate and dihydrate is
1096.34+(60*2)+(18*2) = 1,252.34
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Theoretical mass (in %W/W) of three components are as below
Acetic acid : (60*2)*100/1252.34=9.6%
Water : (18*2)*100/1252.34=2.9%
Lanreotide : 100-9.6-2.9=87.5%
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Approach-1 to characterise X value of acetate:
Moles of acetic acid and water per one molecule of lanreotide from %W/W of components
Moles of Acetic acid/ Moles of Lanreotide: (9.6/87.5)*(1096.34/60)=2
Moles of Water/ Moles of Lanreotide : (2.9/87.5)*(1096.34/18)=2
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Approach-2 to characterise X value of acetate:
Calculation of Concentration of components in Molal
(One of mole of component (analyte) per 1000gm of mass) from % W/W
Lanreotide:
(87.5/100)*(1000/1096.34) = 0.798
Acetic acid :
(9.6/100)*(1000/60) = 1.6
Water :
(2.9/100)*(1000/18) = 1.611
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Select component of lowest Molal concentration (i.e.#0.798) out of three components and use the same as denominator for molal concentration of each component(analyte) to derive Mole ratio.
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Moles of Lanreotide :(0.798/0.798)=1
Moles of Acetic acid :(1.6/0.798)=2
Moles of Water :(1.611/0.798)=2
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Inference of Approach-1 and 2:
X value for acetic acid is 2 for Lanreotide 87.5% W/W, acetic acid 9.6 % W/W and water 2.9% W/W.
Read also: Solid Fraction Calculation Formula
Resource Person: Dilip Patil
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