USP assay for denatonium benzoate

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Billie Smith
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USP assay for denatonium benzoate

I am performing the USP assay for denatonium benzoate in rubbing alcohol. This is my first solo project at my new job, so any help would be greatly appreciated. My analytical results are horrible and I think it may just be a math error. The procedure is given below.
The USP monograph indicates that the quantity of denatonium benzoate is calculated as shown below.
Calculate the quantity, in mg,of denatonium benzoate (C28H34N2O3H2O) in 100 mL of Rubbing Alcohol taken by the formula:
0.025 C (AU / AS)
If possible, could someone explain where the 0.025 multiplier is derived?
I think it comes from the evaporation of 200 mL of alcohol, reconstituted to 50 mL, 10 mL aliquot taken and extracted with a final extract volume of 100 mL.

50/200 * 10/100 = 0.025, but the standard is prepared at a concentration of 50 ug/mL and it is extracted concomitantly with the sample (10 mL aliquot, final extract volume of 100 mL), yet USP indicates that the standard concentration in the formula should be 50 ug/mL). To me, it looks like the standard concentration is 5 ug/mL.

NOTE: The assay preparation originates from a 200 mL volume of alcohol that is taken to dryness and then reconstituted in 50 mL of buffer. Ten (10) mL of the reconstituted solution is then transferred to separator funnel as described below. The limit is 1.4 mg / 100 mL. The concentration of the standard (C in the above equation is 54.9 ug/mL). My standard response is 0.2410 and my sample response is 0.0382 to produce a result of 0.22 mg/100 mL.

Procedure Treat the Standard preparation, Assay preparation, and blank similarly and concomitantly. Transfer 10.0 mL each of the Standard preparation, Assay preparation,and Buffer solution to individual 250-mL separators,and add to each 40 mL of Buffer solution, 10 mLof a 1in 1000solution of bromophenol blue in chloroform, and 60mLof chloroform. Shake the separators vigorously for 2 minutes,allow to stand for 15minutes, then withdraw the chloroform layers through chloroform-washed cotton into 100-mLvolumetric flasks. Repeat the extraction with 20 mLof chloroform,adding the filtered chloroform extracts to the respective volumetric flasks, dilute with chloroform to volume, and mix. Without delay, concomitantly determine the absorbances of the solutions in 1-cm cells at the wavelength of maximum absorbance at about 410 nm, with a suitable spectrophotometer, using the blank to set the instrument. Calculate the quantity,in mg,of denatonium benzoate (C28H34N2O3H2O)in 100 mLof Rubbing Alcohol taken by the formula: 0.025 C (AU / AS)

As mentioned the limit is 1.4 mg / 100 mL. The concentration of the standard (C in the above equation is 54.9 ug/mL). My standard response is 0.2410 and my sample response is 0.0382 to produce a result of 0.22 mg/100 mL.

The vendors CofA indicates >1.4 mg/100 mL

Thanks for any help. Billie

MS's picture


Your problem is a little bit of a mind teaser. Most folks define dilution factors as the final volume divided by the aliquot volume, not quite the same as your equation:

50/200 * 10/100 = 0.025

Your use of the term multiplier for the 0.025 value seems reasonable to me since it appears that the value contains the dilution factor as well as a conversion factor to express the calculated result as mg/100 mL.

The concentration (C) of the standard in the USP equation is expressed in g/mL
C mg/mL * 1000 equals C g/mL

Dilution Factor = 1/100 X 50/10 X 1/200

Without a conversion, the USP equation would produce a result reported in g/mL units.
To convert the sample result to mg/100 mL we need to multiply the result by
100 mL / 1000 (ug/mL to mg).

The conversion factor as used by USP

= 1/100 X 50/10 X 1/200 X 1000 X 100/1000 and after canceling

=5/200 = 0.025

Things to scope out:

1. Verify your C of A to make sure it meets your minimum requirements. You may even want to call your vendor to ensure that the C of A is correct. Often times the value is given as a wt/wt%, so dont forget to use the specific gravity to convert to mg/100 mL.

2. Does your extracted standard produce absorption responses consistent with historical values?

3. Make sure the source of your 200 mL sample is well mixed prior to sampling.

4. Although not required for the actual denatonium benzoate assay you should perform a weight verification of the evaporation dish before and after evaporation. If you do not have at least 2.8 mg of residue, you are probably below the assay limit; else your balance is off.

5. Evaporation of the 200 mL of alcohol, (as described in the limit of nonvolatile residue test) by placing no more than 25 mL of alcohol into the evaporation dish at a time to limit the dish surface area that the alcohol comes in contact with.

6. Effect residue transfer with 2 -25 mL portions of water rather than a single 50 mL portion.

7. Analyzed within 24 hours (beginning of evaporation to analysis by UV) to minimize possible residue binding to the dish.

Since the standard passes through the same partitioning stage as does the sample and since the concentration of the standard is within a reasonable range of the expected assay value, the resulting standard response is somewhat indicative of how well you have handled the sample during partitioning. But the standard does not undergo the evaporation process. You may want to attain an aliquot of ACS reagent grade Ethanol, absolute 200 proof, create a 70% ethanol solution and fortify with Denatonium Benzoate and see if how much you can recover. If you cannot get a good recovery on a spiked sample then most likely you have a procedural problem. If you get good recovery on the spike then you may have a problem with your original sample.

Hope this helps.


Billie Smith
Billie Smith's picture
Thanks for responding, I am

Thanks for responding, I am still having problems.

The C of A states that it meets USP specifications.

The extracted standard compares well with previous analyses. In the past (before Billie) there have been sample failures but a passing result was later determined.

I will reduce sample volume during evaporation and use 2 - 25 mL portions of water to effect the tranfer.

How would you perform the spiking?

We are ordering a new batch of alcohol since this batch is old and has caused so many problems during testing (this test and previous tests) but my boss wants a definitive answer to what the problem is, so I need to prove to him that I'm not the problem. I will be the only one here during the next week, so I would like to fiqure this out.

Thanks for your help. Billie

MS's picture


You can use your 50 ug/mL standard as the diluent. For a slight overage add 60 mL. By the way I assume that at some point during your investigation you have remade your analtyical standard!

Prepare the 70% solution as follows:
Transfer 140 mL of the 99.5% ethanol to a 200 mL volumetric flask. Add with swirling to ensure mixing 50 mL of a 50 ug/mL Denatonium Benzoate standard prepared in water to the flask. Allow the solution to sit for about 5 minutes to reach thermal equilibrium, then dilute to 200 mL with the final 10 mL of the standard solution, and mix well.

60 mL X 50 ug/mL of the standard solution when diluted to 200 mL and evaporated to dryness produces a theoretical mg residue value for Denatonium Benzoate of 3,000 ug or 3.0 mg. After evaporation, just prior to the partitioning phase of the procedure, the residue is reconstituted in 50 mL of water to produce a theoretical Denatonium Benzoate value of about 60 ug/mL.

Note: Check the weight of the evporation dish before and after evaporation. If the residue isn't at least 2.8 mg+ then you can't meet USP specification.

At this stage of the procedure, the sample, analytical standard (prepared at about 50 ug/mL) and the blank are all treated in the same fashion, i.e., 10 mL aliquots are partitioned against several volumes of chloroform prior to analysis of the combined chloroform extracts with final dilution to 100 mL in chloroform. For the sample and the standard, the ug amounts contained in the 10 mL aqueous aliquots that are subsequently partitioned is 50 ug/mL X 10 mL = 500 ug for the standard and 50 ug/mL X 10 mL = 600 ug for the sample. After partitioning, the extracts are diluted to a final volume of 100 mL. Assuming an extraction efficiency of 100%, and no substantial loss during evaporation the sample solution and the standard solution should each contain about 6 and 5 ug/mL of Denatonium Benzoate, repetitively. The UV absorption values of the analytical standard and the fortified sample should closely approximate one another. Similar absorption values will demonstrate your ability to accurately apply the USP test procedure to a 70% ethanol solution containing mg amounts of Denatonium Benzoate.

Billie, you should verify my calculations and make your own determination as to the soundness of any suggestions made by yours truly. Let me know how things turn out.

Is the level of Denatonium Benzoate critical to the performance of your product or is it critical because of the USP criteria?