2nd Order Terms in Purity Calculations of Reference Standards
1. Second Order Terms in Purity
Calculations of
Pharmaceutical Reference
Standards!
jbladon! 1!
2. CASE I!
Protocol requires using HPLC purity and!
the water content to determine analyte purity!
%Purity = 100% − %Water − % HPLC Impurities
?!
jbladon! 2!
3. CASE I!
Protocol requires using HPLC purity and!
the water content to determine analyte purity!
%Purity = F1 F2 100%
F1 is the fraction of entire sample that is not water.
F2 is the fraction of sample excluding water that is analyte.
jbladon! 3!
4. Let,
A = Analyte*
Im = Impurities*
W = Water
Then,
⎛ A + Im ⎞ ⎛ A ⎞
F1 = ⎜ ⎟ , F2 = ⎜
⎝ A + Im+ W ⎠ ⎝ A + Im ⎟
⎠
⎛ A + Im ⎞ ⎛ A ⎞ ⎛ A ⎞
F1F2 = ⎜ ⎟ =⎜
⎝ A + Im+ W ⎠ ⎝ A + Im ⎠ ⎝ A + Im+ W ⎟
⎟⎜ ⎠
____________________
• Where each response factor is known or it is assumed
they are identical jbladon! 4!
5. %HPLC Impurities ⎛ Im ⎞
=⎜ ⎟
100% ⎝ A + Im ⎠
Rearrange F2
⎛ A ⎞ ⎛ A + Im− Im ⎞ ⎛ Im ⎞
F2 = ⎜ ⎟ =⎜ ⎟ = ⎜1 − ⎟
⎝ A + Im ⎠ ⎝ A + Im ⎠ ⎝ A + Im ⎠
⎛ %HPLC Impurities ⎞
F2 = ⎜ 1 − ⎟
⎝ 100% ⎠
jbladon! 5!
6. %Water W
=
100% A + Im+ W
Rearrange F1
⎛ A + Im ⎞ ⎛ A + Im+ W − W ⎞
F1 = ⎜ ⎟ =⎜ ⎟
⎝ A + Im+ W ⎠ ⎝ A + Im+ W ⎠
⎛ A + Im+ W W ⎞
=⎜ − ⎟
⎝ A + Im+ W A + Im+ W ⎠
⎛ %Water ⎞
F1 = ⎜ 1 − ⎟
⎝ 100% ⎠
jbladon! 6!
9. Example
Let
%Water = 2.5%
%HPLC Impurities = 5.0%
Then
%Purity = {1- 0.025 - 0.050 + (0.025)(0.050)}100%
%Purity = 92.6%
If the second order term is omitted,
%Purity = 92.5%
jbladon! 9!
10. CASE II
Protocol requires using the HPLC purity,!
the water and solvent content, and!
the inorganic ion content !
to determine analyte purity!
jbladon! 10!
11. 1. Water Content by TGA!
2. Solvent Content by Gas
Chromatography!
3. Inorganic Ions by Ion Chromatography!
4. Fraction of what is left by HPLC Purity!
jbladon! 11!
13. %Purity = F1 F2 100%
F1 is the fraction of sample that is
not water/solvent/inorganic
F2 is the fraction of sample
excluding water/solvent/inorganic
that is analyte.
jbladon! 13!
14. Let,
A = Analyte*
IMP = Impurities*
W = Water
S = Solvent
IN = Inorganic
Then,
⎛ A + IMP ⎞ ⎛ A ⎞
F1 = ⎜ ⎟ , F2 = ⎜
⎝ A + IMP + IN + W + S ⎠ ⎝ A + IMP ⎟
⎠
____________________
• Where each response factor is known or
it is assumed they are identical
jbladon! 14!
15. ⎛ A + IMP ⎞⎛ A ⎞
F1F2 = ⎜
⎝ A + IMP + W + S + IN ⎟ ⎜ A + IMP ⎟
⎠⎝ ⎠
⎛ A ⎞
F1F2 = ⎜
⎝ A + IMP + W + S + IN ⎟
⎠
jbladon! 15!
16. %HPLC Impurities IMP
=
100% A + IMP
Rearrange F2
⎛ A ⎞ ⎛ A + IMP − IMP ⎞
F2 = ⎜ ⎟ =⎜ ⎟
⎝ A + IMP ⎠ ⎝ A + IMP ⎠
⎛ A + IMP IMP ⎞
F2 = ⎜ −
⎝ A + IMP A + IMP ⎟⎠
%HPLC Impurities
F2 = 1 −
100%
jbladon! 16!
17. %Water / Solvent W +S
=
100% A + IMP + W + S + IN
%IN IN
=
100% A + IMP + W + S + IN
Rearrange F1
⎛ A + IMP ⎞
F1 = ⎜ ⎟=
⎝ A + IMP + W + S + IN ⎠
⎛ A + IMP + W + S + IN − W − S − IN ⎞
F1 = ⎜ ⎟
⎝ A + IMP + W + S + IN ⎠
⎛ A + IMP + W + S + IN W +S ⎞
−
⎜ A + IMP + W + S + IN A + IMP + W + S + IN ⎟
F1 = ⎜ ⎟
⎜− IN ⎟
⎝ A + IMP + W + S + IN ⎠
%Water / Solvent %IN
F1 = 1 − −
100% 100%
jbladon! 17!
20. Example
Let
%Water/Solvent = 2.5%
%Inorganic = 5.0%
%HPLC Impurities = 5.0%
Then
%Purity = {1- 0.025 - 0.050 - 0.050 + (0.025)(0.050)
+ (0.050)(0.050)}100%
%Purity = 87.9%
If both second order terms are omitted,
%Purity = 87.5%
jbladon! 20!
21. Meet the Boys of the FDA
Dr. Wiley and the Division of Chemistry Staff in 1883.
Wiley, third from right, was 37.
Courtesy Wallace F. Janssen
FDA Historian jbladon! 21!
22. If you have a question about an
equation to calculate the purity of a
reference standard in a protocol or
method, see your supervisor and
bring my handouts for this
presentation
jbladon! 22!