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USP 41 General Information / á1160ñ 7451

PREPARATION

Type A medicated articles in dry form are typically produced by blending the drug substance(s) with carriers and other exci-
pients to promote uniform mixing when subsequently added to the animal feed. The drug substance(s) may first be mixed
with an excipient (e.g., starch or sodium aluminosilicate) that has a similar particle size and can help distribute the drug sub-
stance(s) uniformly throughout the final mixture. This pre-blend may then be mixed with bulk excipients (e.g., calcium carbo-
nate or soybean hulls). The product may be granulated and/or oil (e.g., mineral oil, soybean oil) may be added to aid uniform
distribution, to prevent particle segregation during shipping, and/or to minimize formation of airborne drug substance parti-
cles during production of another Type A medicated article or Types B or C medicated feeds.
Type A medicated articles in liquid form are produced by mixing the drug substance(s) with a suitable solvent (e.g., water or
propylene glycol). The drug substance(s) is usually dissolved to produce a solution, but suspension products could be pro-
duced also.
Types B or C medicated feeds are typically manufactured at feed mills or on-farm by livestock producers. To manufacture
Types B or C medicated feeds in dry form, Type A medicated articles are added to the feeds during the mixing process. Liquid
Type A medicated articles may be sprayed in at set rates, and dry Type A medicated articles are added using methods that
facilitate uniform distribution in the feeds. Types B and C medicated feeds in dry form may be further processed by heating,
steaming, and extruding into pellets. The pellets may be rolled or broken up to create crumbles. Types B and C medicated
feeds may also be prepared in liquid form. Liquid feeds are typically molasses based and contain an animal drug(s) dissolved or
suspended in the liquid matrix. The liquid feed may need to be recirculated or agitated on a routine basis to maintain a uni-
form distribution of the drug(s).
A Type B medicated feed may also be prepared by diluting another Type B medicated feed. A Type C medicated feed may
also be prepared by diluting a Type B medicated feed or another Type C medicated feed.

LABELING AND PACKAGING

Labeling for Type A medicated articles and Types B and C medicated feeds provides all information necessary for their safe
and effective use. The label for a Type A medicated article includes mixing directions for the manufacture of medicated feeds
from the Type A medicated article and feeding directions for the Type C medicated feeds. The label for a Type B medicated
feed provides mixing directions for the manufacture of medicated animal feeds. Labels for both Type A medicated articles and
Type B medicated feeds indicate that they are not to be fed directly to animals. The label for Type C medicated feeds includes
directions for feeding.
Type A medicated articles are packaged in bags (e.g., paper with polyethylene liners) for dry products or in appropriate con-
tainers (e.g., plastic) for liquids. Typical sizes are 50-lb bags or several-gallon containers. Dry Types B and C medicated feeds
may be packaged in bags for storage and delivery, or they may be shipped in bulk form for storage or immediate use. Free-
choice Type C medicated feeds may be packaged in bags (e.g., loose minerals), wrapped in film (e.g., compressed blocks), or
packaged in tubs (e.g., molded blocks). Liquid Types B and C medicated feeds are stored and shipped in bulk in tanks.

NOMENCLATURE

The drug product's established (non-proprietary) name consists of the drug substance (active moiety) and an appropriate
type of medicated article or feed. The following nomenclature options are available to indicate types of medicated articles or
feeds as part of the product established name:
[Drug] Type A medicated article

General Chapters
[Drug] Type A liquid medicated article
[Drug] Type B medicated feed
[Drug] Type B liquid medicated feed
[Drug] Type C medicated feed
[Drug] Type C liquid medicated feed
[Drug] Type C free-choice medicated feed
[Drug] Type C liquid free-choice medicated feed
[Drug] Type C top-dress medicated feed

á1160ñ PHARMACEUTICAL CALCULATIONS IN PHARMACY PRACTICE

Table of Contents
1. INTRODUCTION
2. CALCULATING AMOUNTS OF ACTIVE INGREDIENTS

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3. DOSAGE CALCULATIONS
4. USE OF POTENCY UNITS
5. VOLUME AND WEIGHT SUMS
6. DENSITY AND SPECIFIC GRAVITY
7. MILLIEQUIVALENTS AND MILLIMOLES
8. CONCENTRATION EXPRESSIONS (with Table 1)
9. ALCOHOL
10. ALLIGATION ALTERNATE AND ALGEBRA METHODS
11. ALIQUOT CALCULATIONS
12. POWDER VOLUME CALCULATIONS
13. INTRAVENOUS FLOW OR INFUSION RATES
14. ISOOSMOTIC SOLUTIONS
15. pH AND BUFFER CALCULATIONS
16. TEMPERATURE
17. ENDOTOXINS
18. STABILITY AND EXPIRATION DATE CALCULATIONS
19. APPENDIX 1: LOGARITHMS (with Table 9 and Table 10)

1. INTRODUCTION

The purpose of this general chapter is to provide general information to assist pharmacists and support personnel in per-
forming the necessary calculations for compounding and dispensing medications. This general chapter is not inclusive of all
the information necessary for performing pharmaceutical calculations. For additional information regarding pharmaceutical
calculations, consult a pharmaceutical calculations textbook. For additional information on pharmaceutical compounding and
drug stability, see: Pharmaceutical Compounding—Nonsterile Preparations á795ñ, Pharmaceutical Compounding—Sterile Prepara-
tions á797ñ, and Packaging and Storage Requirements á659ñ; Quality Assurance in Pharmaceutical Compounding á1163ñ and Stabil-
ity Considerations in Dispensing Practice á1191ñ.
Correct pharmaceutical calculations can be accomplished by using proper conversions from one measurement system to an-
other and properly placing decimal points (or commas, in countries where it is customary to use these in the place of decimal
points), by understanding the arithmetical concepts, and by paying close attention to the details of the calculations. Before
proceeding with any calculation, pharmacists should do the following: (a) read the entire formula or prescription carefully; (b)
determine the materials that are needed; and then (c) select the appropriate methods of preparation and the appropriate cal-
culations.
Logical methods that require as few steps as possible should be selected to ensure that calculations are done accurately and
correctly. A pharmacist should double-check each calculation or have someone else double-check, e.g., a technician, if another
pharmacist is not available, before proceeding with compounding the preparation. One expedient method of double-checking
is estimation, which consists of convenient rounding (e.g., 0.012 to 0.01, 0.44 to 0.5, 18.3 to 20, and 476 to 500) to approxi-
mate the magnitude of answers.

2. CALCULATING AMOUNTS OF ACTIVE INGREDIENTS

The pharmacist must be able to calculate the amount or concentration of drug substances in each unit or dosage portion of
a compounded preparation at the time it is prepared and again at the time it is dispensed. Pharmacists must perform calcula-
General Chapters

tions and measurements to obtain, theoretically, 100% of the amount of each ingredient in compounded formulations. Calcu-
lations must account for the active ingredient, or active moiety, and water content of drug substances, which includes that in
the chemical formulas of hydrates. Official drug substances and added substances must meet the requirements in general
chapter Loss on Drying á731ñ, which must be included in the calculations of amounts and concentrations of ingredients. The
pharmacist should consider the effect of ambient humidity on the gain or loss of water from drugs and added substances in
containers subjected to intermittent opening over prolonged storage. Each container should be opened for the shortest dura-
tion necessary and then closed tightly immediately after use.
The nature of the drug substance to be weighed and used in compounding a prescription must be known. If the substance
is a hydrate, its anhydrous equivalent weight may need to be calculated. On the other hand, if there is adsorbed moisture
present that is either specified on a Certificate of Analysis (CoA) or that is determined in the pharmacy immediately before the
drug substance is used in the preparation (see chapter á731ñ), this information must be used when calculating the amount of
drug substance that is to be weighed to determine the exact amount of anhydrous drug substance required.
There are cases in which the required amount of a dose is specified in terms of a cation (e.g., Li+), an anion (e.g., F–), or a
molecule (e.g., theophylline in aminophylline). In these instances, the drug substance weighed is a salt or complex, a portion
of which represents the pharmacologically active moiety. Thus, the exact amount of such substances weighed must be calcula-
ted on the basis of the required quantity of the pharmacological moiety.
The following formula may be used to calculate the theoretical weight of an ingredient in a compounded preparation:

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W = AB/CD
W = actual weighed amount
A = prescribed or pharmacist-determined weight of the active or functional moiety of drug or added substance
B = molecular weight (MW) of the ingredient, including waters of hydration for hydrous ingredients
C = MW of the active or functional moiety of a drug or added substance that is provided in the MW of the weighed ingre-
dient
D = the fraction of dry weight when the percent by weight of adsorbed moisture content is known from the loss on dry-
ing procedure (see chapter á731ñ) or from the CoA. The CoA should be lot specific.

2.1 Active Ingredients

2.1.1 CALCULATING DRUGS DOSED AS SALT FORM AND HYDRATE

Examples—Drugs dosed as salt form and hydrate

1. Drugs dosed as salt form and hydrate
Triturate morphine sulfate and lactose to obtain 10 g in which there are 30 mg of morphine sulfate for each 200 mg of the
morphine–lactose mixture. [NOTE—Morphine is dosed as the morphine sulfate, which is the pentahydrate.]
W = weight of morphine sulfate (g)
A = weight of morphine sulfate pentahydrate in the prescription, 1.5 g
B = MW of morphine sulfate pentahydrate, 759 g/mol
C = MW of morphine sulfate pentahydrate, 759 g/mol
D = 1.0
To solve the equation:
W = (1.5 g × 759 g/mol)/(759 g/mol × 1) = 1.5 g of morphine sulfate pentahydrate
2. Active drug moiety and correction for moisture
Accurately weigh an amount of aminophylline to obtain 250 mg of anhydrous theophylline. [NOTE—In this example, the
powdered aminophylline dihydrate weighed contains 0.4% w/w absorbed moisture as stated in the CoA received by the phar-
macy.]
W = AB/CD
W = weight of aminophylline dihydrate (mg)
A = weight of anhydrous theophylline, 250 mg
B = MW of aminophylline dihydrate, 456 g/mol
C = MW of anhydrous theophylline, 360 g/mol
D = 0.996
[NOTE—One mol of aminophylline contains 2 mol of theophylline. Theophylline has a MW of 180.]
To solve the equation:
W = (250 mg × 456 g/mol)/(360 g/mol × 0.996) = 318 mg of aminophylline dihydrate

2.2 Hydrates, Salts, and Esters

General Chapters
Frequently, for stability or other reasons such as taste or solubility, the base form of a drug is administered in another form
such as a salt or an ester. This altered form of the drug usually has a different MW, and at times it may be useful to determine
the amount of the base form of the drug in the altered form.

2.2.1 CALCULATING HYDRATES, SALTS, AND ESTERS

Examples—Hydrates, salts, and esters

1. Hydrates
If a prescription for 100 g of lidocaine hydrochloride 2% gel is to be made, 2 g of anhydrous lidocaine hydrochloride could
be used, or the equivalent amount of lidocaine hydrochloride monohydrate could be calculated as follows:
W = weight of lidocaine hydrochloride monohydrate (g)
A = weight of anhydrous lidocaine hydrochloride in the prescription, 2 g
B = MW of lidocaine hydrochloride monohydrate, 288.81 g/mol
C = MW of anhydrous lidocaine hydrochloride, 270.80 g/mol
D = 1.0
To solve the equation:
W = (2 g × 288.81 g/mol)/(270.80 g/mol × 1) = 2.133 g of lidocaine hydrochloride monohydrate

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2. Salts
A prescription calls for 10 mL of a fentanyl topical gel at a concentration 50 mg fentanyl/0.1 mL prepared from fentanyl cit-
rate. The amount of fentanyl citrate required for the preparation could be calculated as follows:
Amount of fentanyl needed for the preparation:
(50 mg fentanyl/0.1 mL) × 10 mL = 5000 mg of fentanyl
W = weight of fentanyl citrate in the prescription (mg)
A = weight of fentanyl in the prescription, 5000 mg
B = MW of fentanyl citrate, 528.59 g/mol
C = MW of fentanyl, 336.47 g/mol
D = 1.0
To solve the equation:
W = (5000 mg × 528.59 g/mol)/(336.47 g/mol × 1) = 7855 mg of fentanyl citrate
3. Esters
The amount of cefuroxime axetil contained in a single 250-mg cefuroxime tablet can be calculated as follows:
W = weight of cefuroxime axetil in tablet (mg)
A = weight of cefuroxime in the prescription, 250 mg
B = MW of cefuroxime axetil, 510.47 mg/mmol
C = MW of cefuroxime, 424.39 mg/mmol
D = 1.0
To solve the equation:
W = (250 mg × 510.47 g/mol)/(424.39 g/mol × 1) = 300 mg of cefuroxime axetil

3. DOSAGE CALCULATIONS

3.1 Dosing by Weight

Doses are frequently expressed as mg of drug per kg of body weight per a dosing interval.

3.1.1 CALCULATING DOSING BY WEIGHT

Example—Dosing by weight
A physician orders azithromycin for oral suspension at a dose of 15 mg/kg/day, divided every 12 h, for a child that weighs
36 lb. Calculate the volume of oral suspension, in mL, that should be administered for each dose of a 200-mg/5-mL suspen-
sion as follows:
(a) Calculate the child's weight in kg:
36 lb × kg/2.2 lb = 16.4 kg
(b) Multiply the weight, in kg, by the dosing rate:
16.4 kg × 15 mg/kg/day = 246 mg/day
(c) Divide the total daily dose by the number of doses/day:
General Chapters

246 mg/2 doses = 123 mg/dose

(d) Calculate the volume of each dose using ratio and proportion:
(123 mg/dose)/(200 mg/5 mL) = 3.1 mL/dose
Some calculations may also be completed using dimensional units analysis (DUA). The DUA should begin at the left end
with a factor containing the numerator answer units. All units other than those in the answer should cancel. If using DUA, the
preceding equation would be as follows:

3.2 Dosing by Body Surface Area (Humans)

Some medications, including chemotherapeutic agents, require dosing by body surface area (BSA). The dose is expressed as
amount of drug per meter squared (m2). BSA may be calculated using the following formulas:

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USP 41 General Information / á1160ñ 7455

3.2.1 CALCULATING BY BODY SURFACE AREA (HUMAN)

Example—Dosing by BSA (humans)

A physician orders rituximab at a dose of 375 mg/m2 every week for 6 weeks for a patient who is 6 ft 2 in tall and weighs
183 lb. Calculate the volume, in mL, of 10-mg/mL rituximab injection needed to make each IV infusion dose as follows:
(a) Calculate the patient’s BSA:

(b) Multiply the BSA by the dosing rate:

2.08 m2 × 375 mg/m2 = 780 mg/dose
(c) Calculate the volume of each dose using ratio and proportion:
(780 mg/dose)/(10 mg/mL) = 78 mL/dose
The preceding calculation may also be completed using DUA as follows:

3.3 Dosing By Body Surface Area (Animals)

BSA for cats and dogs may be calculated using the following formulas. For other animals, consult an appropriate veterinary
medicine reference.
Body surface area for cats:
BSA (m2) = {10 × [body weight (g)]0.667}/10,000
Body surface area for dogs:
BSA (m2) = {10.1 × [body weight (g)]0.667}/10,000

3.3.1 CALCULATING DOSING BY BODY SURFACE AREA (ANIMALS)

Example—Dosing by BSA (animals)

A veterinarian orders oral cyclophosphamide therapy at a dose of 50 mg/m2 for a cat who weighs 5.8 kg. Calculate the dose
of cyclophosphamide as follows:
(a) Calculate the cat’s BSA:
BSA (m2) = [10 × (5800 g)0.667]/10,000 = 0.324 m2
(b) Multiply the BSA by the dosing rate:
0.324 m2 × 50 mg/m2 = 16.2 mg

4. USE OF POTENCY UNITS

Because some substances cannot be completely characterized by chemical and physical means, it may be necessary to ex-

General Chapters
press quantities of activity in biological units of potency [see the USP General Notices and Requirements 5.50.10, Units of Potency
(Biological)].

4.1 Calculating by Use of Potency Units

Examples—Use of potency units

1. Potency units-to-milligrams conversion
A dose of penicillin G benzathine for streptococcal infection is 1.2 million units administered intramuscularly. If a specific
product contains 1180 units/mg, calculate the amount, in mg, of penicillin G benzathine in the dose as follows:
(1,200,000 units)/(1180 units/mg) = 1017 mg of penicillin G benazathine
2. Potency units-to-milligrams conversion
A prescription calls for 60 g of an ointment containing 150,000 units of nystatin per gram. Calculate the quantity of nystatin
with a potency of 4400 units/mg that should be weighed for the prescription as follows:
60 g × (150,000 units of nystatin/g) = 9,000,000 units
9,000,000 units/(4400 units/mg) = 2045 mg of nystatin

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5. VOLUME AND WEIGHT SUMS

Weights are additive in most mixtures of liquids, semisolids, and solids. Volumes in mixtures of miscible solutions and pure
liquids may or may not be additive, based primarily on the effects of volume proportions and intermolecular hydrogen bond-
ing. For example, mixtures containing equal or near-equal volumes of water and ethanol (and other miscible mono-hydroxy
alcohols) will be exothermic and result in a volume contraction of <5%, e.g., 50 mL water + 50 mL ethanol yield 97–98 mL at
20°–25°. Negligible volume contraction occurs between water and polyhydroxy or polyhydric alcohols, e.g., glycerin and pro-
pylene glycol. Volumes are additive with usually negligible error in aqueous mixtures that contain <10% of mono-hydroxy al-
cohols, i.e., there is <0.5% volume contraction.

6. DENSITY AND SPECIFIC GRAVITY

Density is defined as the mass of a substance in air at a specific temperature (typically 25°) per unit volume of that substance
at the same temperature. Density may be calculated with the following equation:
Density = (mass of substance/volume of substance) at a particular temperature and pressure
Specific gravity (SG) is the unitless ratio of the density of a substance to the density of water at 4°, or [(g of substance/mL)/
1.00 g/mL]. Alternatively, SG can be calculated at a particular temperature in some common units of density from density of
substance per density of water.
SG may be calculated with the following equation:
SG = (weight of the substance)/(weight of an equal volume of water)

6.1 Calculating Density and Specific Gravity

Examples—Density and specific gravity

1. Density calculation
2.3 g of activated charcoal powder occupies a bulk volume of 5.2 mL at 20° and 1 atm. The density of activated charcoal
powder can be calculated as follows:
Density = 2.3 g/5.2 mL = 0.44 g/mL
2. SG calculation
125 g of glycerin occupies a volume of 99 mL at 25°. [NOTE—The density of water at 25° is 0.997 g/mL.] The SG of glycerin
can be calculated as follows:
SG = (125 g/99 mL)/(0.997 g/mL) = 1.266
3. Concentrated acid calculation
Hydrochloric acid is approximately a 37% w/w solution of hydrochloric acid in water. Calculate the amount, in g, of hydro-
chloric acid contained in 75 mL of hydrochloric acid as follows. [NOTE—The SG of hydrochloric acid is 1.18.]
37% w/w × 1.18 = 43.7% w/v
(43.7 g/100 mL) × 75 mL = 32.8 g of hydrochloric acid
General Chapters

7. MILLIEQUIVALENTS AND MILLIMOLES

[NOTE—This section addresses milliequivalents (mEq) and millimoles (mmol) as they apply to electrolytes for dosage calcula-
tions. See also the 8. Concentrations Expressions section of this chapter.]
The quantities of electrolytes administered to patients are usually expressed in terms of mEq. Weight units such as mg or g
are not often used for electrolytes because the electrical properties of ions are best expressed as mEq. An equivalent (Eq) is the
weight of a substance that supplies 1 unit of charge. An equivalent weight is the weight, in g, of an atom or radical, divided by
the valence of the atom or radical. A mEq is 1/1000th of an Eq. The equivalent weight of a compound may be determined by
dividing its formula or MW in g by the valence of its largest valence ion.
A mole equals one gram-atomic weight or gram-molecular weight of a substance. A millimole equals 1/1000th of a mole.

7.1 Calculating Milliequivalents and Millimoles

Examples—Milliequivalents and millimoles

1. Calculate the mEq weight of calcium. [NOTE—Calcium has a MW of 40.08, and the valence of calcium is 2+.]
Eq weight = 40.08 g/2 = 20.04 g

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mEq weight = 20.04 g/1000 = 0.02004 g = 20.04 mg

2. Calculate the quantity, in mEq, of potassium in a 250-mg Penicillin V Potassium Tablet. [NOTE—Penicillin V potassium has
a MW of 388.48 g, there is one potassium atom in the molecule, and the valence of potassium is 1+.]
Eq weight = 388.48 g/1 = 388.48 g
mEq weight = 388.48 g/1000 = 0.38848 g = 388.48 mg
(250 mg/tablet)/(388.48 mg/mEq) = 0.644 mEq of potassium/tablet
3. Calculate the mEq of magnesium and sulfate in a 2-mL dose of 50% Magnesium Sulfate Injection. [NOTE—Magnesium
sulfate (MgSO4 · 7H2O) has a MW of 246.47, and the highest valence ion is magnesium 2+ and sulfate 2−.]
(50 g/100 mL) × (2 mL/dose) = 1 g/dose
Eq weight = 246.47 g/2 = 123.24 g/Eq
(1g/dose)/(123.24 g/Eq) = 0.008114 Eq = 8.114 mEq of both magnesium and sulfate per dose
This problem may also be worked using DUA as follows:

4. A vial of sodium chloride injection contains 3 mEq/mL of sodium chloride. Calculate the strength, in % w/v, of the injec-
tion. [NOTE—Sodium chloride has a MW of 58.44.]

(0.1753 g/mL) × 100 mL = 17.53 g in 100 mL = 17.53% w/v

5. Calculate the weight of potassium in mmol. [NOTE—Potassium has a MW of 39.1.]
The weight of 1 mol is 39.1 g and the weight in mmol is:
39.1 g/1000 = 0.0391 g or 39.1 mg
6. Calculate the mmol of penicillin V potassium in a 250-mg Penicillin V Potassium Tablet. [NOTE—Penicillin V potassium has
a MW of 388.48.]
The weight of 1 mol is 388.48 g, and the weight of 1 mmol is:
388.48 g/1000 = 0.38848 g or 388.48 mg

8. CONCENTRATION EXPRESSIONS

The concentration expressions in this section refer to homogeneous mixtures of the following states of matter at a tempera-
ture of 20°–30° and pressure of 1 atm (29.92 in Hg, 760 mm Hg, 101.3 kPa, 1013.3 mb): gas in gas, gas in liquid, liquid in

General Chapters
liquid, liquid in semisolid, solid in liquid, solid in semisolid, and solid in solid. Concentration expressions used in pharmacy
practice and pharmaceutical research include, but are not limited to, those listed in Table 1. Common metric drug strength
and clinical concentrations include, for example, mg/mL, mg/dL, g or mg per L, and ng/mL (see General Notices and Require-
ments 8.240, Weights and Measures).
Table 1
Title Abbreviation Definition
Mass of a dispersed or dissolved ingredient per volume
Mass in volume ratios None is standard amount of mixtures containing that ingredient
mEq of an electrolyte or salt per unit of volume of solu-
mEqa per volume mEq/volume unit tions containing that electrolyte or salt
a 1 mEq = Eq/1000.
b The abbreviation for mole is mol.
c 1 mol of solute per 1 kg of solvent is a 1 molal (1 m) solution.
d 1 mol of solute per 1 L of solution of that solute is a 1 molar (1 M) solution.
e Normality = (Molarity × largest valence ion of a compound), e.g., (18 M H SO × 2) = 36 N H SO , where 2 derives from the 2− valence of SO .
2 4 2 4 4
f Eq of a compound = (1 mol × largest valence ion of a compound), e.g., 1 mol of lithium citrate = 3 Eq of lithium citrate; 1 mol of Ca(gluconate) = 2 Eq of
2
Ca(gluconate)2; and 1 mol of KCl = 1 Eq of KCl.
g 1 Eq of solute per 1 L of solution of that solute is a 1 normal (1 N) solution.
h R, X, and Y are whole numbers.

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Table 1 (Continued)
Title Abbreviation Definition
Molality m molb of a solute/kg of a solvent containing that solutec
Molarity M mol of a solute/L of a solvent containing that soluted
Equivalents (Eqf) of a solute/L of a solvent containing that
Normalitye N soluteg
Parts of a gas, liquid, or solid per 1 million part of another
gas, liquid, or solid containing the first gas, liquid, or sol-
Parts per million ppm id
% Volume in volume % v/v mL of liquid per 100 mL of a solvent containing that liquid
g of a solute per 100 mL of a solvent containing that sol-
% Weight in volume % w/v ute
% Weight in weight % w/w g of a solute per 100 g of a mixture containing that solute
1 part of an ingredient per Rh parts of a mixture contain-
1:R ing that ingredient
1 part of an ingredient in Rh parts of a mixture containing
1 in R that ingredient
Xh parts of one ingredient per Yh parts of another ingredi-
Ratio strength X:Y ent in a mixture
a 1 mEq = Eq/1000.
b The abbreviation for mole is mol.
c 1 mol of solute per 1 kg of solvent is a 1 molal (1 m) solution.
d 1 mol of solute per 1 L of solution of that solute is a 1 molar (1 M) solution.
e Normality = (Molarity × largest valence ion of a compound), e.g., (18 M H SO × 2) = 36 N H SO , where 2 derives from the 2− valence of SO .
2 4 2 4 4
f Eq of a compound = (1 mol × largest valence ion of a compound), e.g., 1 mol of lithium citrate = 3 Eq of lithium citrate; 1 mol of Ca(gluconate) = 2 Eq of
2
Ca(gluconate)2; and 1 mol of KCl = 1 Eq of KCl.
g 1 Eq of solute per 1 L of solution of that solute is a 1 normal (1 N) solution.
h R, X, and Y are whole numbers.

8.1 Calculating Normality

Example—Normality
Calculate the amount of sodium bicarbonate powder needed to prepare 50 mL of a 0.07 N solution of sodium bicarbonate
(NaHCO3). [NOTE—Sodium bicarbonate has a MW of 84.01.]
In an acid or base reaction, because NaHCO3 may act as an acid by giving up one proton, or as a base by accepting one pro-
ton, one Eq of NaHCO3 is contained in each mole of NaHCO3.

8.2 Calculating Percentage Concentrations

Percentage concentrations of solutions and other homogeneous mixtures are usually expressed in one of three common
forms in which numerator and denominator quantities are in g and mL measurement units.
1. Volume percent (% v/v) = (volume of liquid solute/volume of solution or suspension) × 100
General Chapters

or % v/v = mL of liquid solute in 100 mL of solution or suspension

2. Weight percent (% w/w) = (weight of solute/weight of mixture) × 100
or % w/w = g of ingredient in 100 g of mixture
3. Weight in volume percent (% w/v) = (weight of solute/volume of solution or suspension) × 100
or % w/v = g of solute in 100 mL of solution or suspension
The preceding three equations may be used to calculate any one of the three values (i.e., weights, volumes, or percentages)
in a given equation if the other two values are known (see also General Notices and Requirements 8.140, Percentage Concentra-
tions).
Examples—Percentage concentrations
1. Weight percent
A prescription order reads as follows (see Table 2):
Table 2
Zinc oxide 7.5 g
Calamine 7.5 g
Starch 15 g
White petrolatum 30 g

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Calculate the percentage concentration for each of the four components using the preceding weight percent equation as
follows:
(a) The total weight of ointment = 7.5 g + 7.5 g + 15 g + 30 g = 60.0 g
(b) The weight percent of zinc oxide = (7.5 g of zinc oxide/60 g of ointment) × 100% = 12.5%
(c) The weight percent of calamine = (7.5 g of calamine/60 g of ointment) × 100% = 12.5%
(d) The weight percent of starch = (15 g of starch/60 g of ointment) × 100% = 25%
(e) The weight percent of white petrolatum = (30 g of white petrolatum/60 g of ointment) × 100% = 50%
2. Volume percent
A prescription order reads as follows:
Rx: Eucalyptus Oil 3% v/v in Mineral Oil.
Dispense 30 mL.
Calculate the quantities of ingredients in this prescription using the volume percent equation as follows:
(a) The amount of eucalyptus oil.
3% v/v = (volume of oil in mL/30.0 mL) × 100%
volume in oil = 0.9 mL of eucalyptus oil
(b) The amount of mineral oil.
30 mL − 0.9 mL = 29.1 mL of mineral oil

8.3 Conversions of Concentration Expressions

8.3.1 SOLID-IN-LIQUID SOLUTION CONVERSIONS

The calculations used to convert from percent weight in volume, % w/v, to other concentrations and vice versa, using the
same densities and formula or molecular weights, are illustrated as follows for calcium chloride, magnesium sulfate, and potas-
sium chloride solutions in water.
8.3.1.1 Calculating solid-in-liquid conversions
Examples—Solid-in-liquid conversions
1. Convert 10% w/v calcium chloride (CaCl2 · 2H2O) to molality (m). [NOTE—Calcium chloride has a MW of 147.01 g; 10%
w/v solution has a density of 1.087 g/mL.]
10% w/v = 10 g of calcium/100 mL of solution
Using the density of the solution:
100 mL of solution × 1.087 g/mL = 108.7 g of solution
108.7 g of solution − 10 g of calcium chloride = 98.7 g of water = 0.0987 kg of water
10 g of calcium chloride/(147.01 g of calcium chloride/mol of calcium chloride) = 0.068 mol of calcium chloride
0.068 mol of calcium chloride/0.0987 kg of water = 0.689 m
2. Convert 50% w/v magnesium sulfate (MgSO4 · 7H2O) to molarity (M). [NOTE—Magnesium sulfate has a MW of 246.47
g.]

3. Convert 10% w/v calcium chloride (CaCl2 · 2H2O) to normality (N). General Chapters

*2 Eq/mol derived from the 2+ valence of calcium

4. Convert 10% w/v calcium chloride (CaCl2 · 2H2O) to mEq/mL.

5. Convert 0.1% w/v calcium chloride (CaCl2 · 2H2O) to ppm.

(0.1 g/100 mL) × (1 × 106 ppm) = 1000 ppm
6. Convert 33% w/v potassium chloride (KCl) to 1:R ratio strength.

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(1/R) = (33 g/100 mL)

R = 3.03
1:R = 1:3

8.3.2 LIQUID-IN-LIQUID SOLUTION CONVERSIONS

The calculations used to convert from percent weight in weight, % w/w, and volume in volume, % v/v, to other concentra-
tions and vice versa using the same densities and formula or MWs, are illustrated for glycerin and isopropyl alcohol in water.
Besides liquid-in-semisolid, solid-in-semisolid, and solid-in-solid mixtures, % w/w is used for viscous liquids, such as coal tar,
glycerin, and concentrated acids.
8.3.2.1 Converting liquid-in-liquid solutions
Examples—Liquid-in-liquid conversions
1. Convert 50% w/w glycerin to % w/v. [NOTE—50% w/w glycerin has a density of 1.13 g/mL.]
(50 g/100 g) × (1.13 g/mL) = 0.565 g/mL
56.5 g/100 mL = 56.5% w/v
2. Convert 70% v/v isopropyl alcohol to % w/w. [NOTE—Isopropyl alcohol has a density of 0.79 g/mL, and 70% v/v iso-
propyl alcohol has a density of 0.85 g/mL.]
70 mL of isopropyl alcohol × (0.79 g/mL) = 55.3 g of isopropyl alcohol
100 mL of solution × (0.85 g/mL) = 85 g of solution
(55.3 g of isopropyl alcohol/85 g of solution) × 100 = 65.06% w/w
3. Convert 70% v/v isopropyl alcohol to % w/v. The following values are from example 2.
55.3 g of isopropyl alcohol/100 mL of solution = 55.3% w/v
4. Convert 50% w/w glycerin to molality (m). [NOTE—Glycerin has a MW of 92.1.]
50 g of glycerin/(92.1 g/mol) = 0.543 mol of glycerin
100 g of solution − 50 g of glycerin = 50 g of water = 0.05 kg of water
(0.543 mol of glycerin/0.05 kg of water) = 10.86 m
5. Convert 70% v/v isopropyl alcohol to molality (m). [NOTE—Isopropyl alcohol has a density of 0.79 g/mL and a MW of
60.1; 70% v/v isopropyl alcohol has a density of 0.85 g/mL.]
70 mL of isopropyl alcohol × (0.79 g/mL) = 55.3 g of isopropyl alcohol
100 mL of solution × (0.85 g/mL) = 85 g of solution
(85 g of solution − 55.3 g of isopropyl alcohol) = 29.7 g of water = 0.0297 kg of water
55.3 g of isopropyl alcohol/(60.1 g/mol) = 0.92 mol of isopropyl alcohol
General Chapters

(0.92 mol of isopropyl alcohol/0.0297 kg of water) = 30.98 m

6. Convert 50% w/w glycerin to molarity (M). [NOTE—Glycerin has a MW of 92.1 g.]
From example 1, 50% w/w glycerin = 56.5% w/v glycerin
(56.5 g/100 mL) × (mol/92.1 g) × (1000 mL/L) = 6.13 M
7. Convert 50% w/w glycerin to % v/v. [NOTE—50% w/w of glycerin has a density of 1.13 g/mL; 100% glycerin has a densi-
ty of 1.26 g/mL.]
50 g of glycerin/(1.26 g/mL) = 39.7 mL of glycerin
100 g of solution/(1.13 g/mL) = 88.5 mL of solution
(39.7 mL of glycerin/88.5 mL of solution) × 100% = 44.8% v/v
9. Convert 50% w/w glycerin to 1 in R ratio strength.
1/R = (50 g of glycerin/100 g of solution)
R=2

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1 in R = 1 in 2

8.3.3 SOLID AND SEMISOLID IN SOLID AND SEMISOLID MIXTURE CONVERSIONS

The calculations used to convert from percent weight in weight (% w/w) to ppm and ratio strengths are illustrated as fol-
lows for fluocinonide and tolnaftate in topical semisolids and powders.
8.3.3.1 Calculating solid and semisolid in solid and semisolid mixture conversions
Examples—Solid and semisolid in solid and semisolid mixture conversions
1. Convert 0.05% w/w fluocinonide ointment to ppm.
(0.05 g/100 g) × (1 × 106 ppm) = 500 ppm
2. Convert 1.5% w/w tolnaftate powder to 1:R ratio strength.
1/R = (1.5 g of tolnaftate/100 g of powder)
R = 67
1:R = 1:67
3. Convert 1% w/w tolnaftate in talcum powder to X:Y ratio strength.
100 g of powder − 1 g of tolnaftate = 99 g of talcum
X:Y = 1 g of tolnaftate:99 g of talcum

8.4 Dilution and Concentration

A more concentrated solution can be diluted to a lower concentration to obtain appropriate strength and precision when
compounding preparations. Powders and semisolid mixtures can be triturated or mixed to achieve lower concentrations. The
amount of an ingredient in the diluted mixture is the same as that in the portion of the more concentrated source used to
make the dilution; thus, the following equation can be applied to dilution problems (Q1)(C1) = (Q2)(C2), where Q1 and Q2 are
the quantity of solutions 1 and 2, respectively, and C1 and C2 are concentrations of solutions 1 and 2, respectively. Any quanti-
ties and concentration terms may be used but the units of those terms must be the same on both sides of the equation.

8.4.1 CALCULATING DILUTION AND CONCENTRATION

Examples—Dilutions and fortifications

1. Semisolid dilution
Calculate the quantity (Q2), in g, of diluent that must be added to 60 g of a 10% w/w ointment to make a 5% w/w oint-
ment.
(Q1) = 60 g, (C1) = 10% w/w, and (C2) = 5% w/w

60 g × 10% w/w = (Q2) × 5% w/w

(Q2) = 120 g

General Chapters
120 g − 60 g = 60 g of diluent to be added
2. Solid dilution
Calculate the amount of diluent that should be added to 10 g of a trituration (1 in 100) to make a mixture that contains 1
mg of drug in each 10 g of the final mixture.
Convert mg to g: 1 mg of drug = 0.001 g of drug
10 g of mixture should contain 0.001 g of drug
(Q1) = 10 g, (C1) = (1 in 100), and (C2) = (0.001 in 10)

10 g × (1/100) = (Q2) × (0.001/10)

(Q2) = 1000 g

Because the final mixture of 1000 g contains 10 g of the trituration, 990 g (or 1000 g − 10 g) of diluent is required to pre-
pare the mixture at a concentration of 0.001 g of drug in each 10 g.
3. Liquid dilution
Calculate the percentage strength (C2) of a solution obtained by diluting 400 mL of a 5.0% w/v solution to 800 mL.

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(Q1) = 400 mL, (C1) = 5.0% w/v, and (Q2) = 800 mL

400 mL × 5% w/v = 800 mL × (C2)

(C2) = 2.5% w/v

4. Liquid fortification
Calculate the additional amount, in g, of codeine phosphate that need to be added to 180 mL of a 12 mg/5 mL elixir of
acetaminophen with codeine to have a final concentration of 30 mg/5 mL of codeine phosphate.
Amount to add = Total amount required − Amount present
Total amount required: (30 mg/5 mL) × 180 mL = 1080 mg of codeine phosphate
Amount present = (12 mg of codeine/5 mL) × 180 mL = 432 mg of codeine phosphate
Amount to add: 1080 mg − 432 mg = 648 mg of codeine phosphate

9. ALCOHOL

To achieve compliance with the statements in the General Notices and Requirements about Alcohol and the USP monograph
for Alcohol, some conventions and special calculations are needed. See General Notices and Requirements 5.20.20.1 In Com-
pounded Preparations, 8.30 Alcohol Content, and Labeling á7ñ, Alcohol for information. The USP monograph for Alcohol states
that it contains 92.3%–93.8% by weight corresponding to 94.9%–96.0% by volume of alcohol (C2H5OH) at 15.56°. The per-
cent concentration for alcohol is generally taken to be 95% v/v of alcohol (C2H5OH) in water.
In summary:
• When the word alcohol is written on a prescription order or in a formula, as for example “alcohol 10 mL” or “dissolve in 5
mL of alcohol”, the compounder should use the Alcohol, USP [that is 95% alcohol (C2H5OH)].
• When the word alcohol is written with a percent, for example “alcohol 20%”, this means 20% v/v of alcohol (C2H5OH). If
this percent is on a label of a commercial product, it means the product contains 20% v/v alcohol (C2H5OH). If this is part
of a compounding formula, it means the compounder must add the equivalent of 20% v/v alcohol (C2H5OH), which may
require special calculations.
• Labels of products and compounded preparations are to include the content of alcohol (C2H5OH) in % v/v. For compoun-
ded preparations, this value must often be calculated based on the volume(s) of alcohol-containing ingredients added.
For calculations when preparing compounded drug preparations using Alcohol, USP, the first step is to determine the quan-
tity, in mL, of alcohol needed, and the second step is to determine the % v/v of alcohol (C2H5OH) in the final preparation so
that it can be properly labeled.

9.1 Calculating Alcohol

Examples—Alcohol
1. Determine the quantity of alcohol needed for the prescription (see Table 3):
Table 3
Clindamycin 1%
Alcohol 15%
General Chapters

Propylene glycol 5%
Purified water, a sufficient quantity to make 60 mL

(a) In this prescription order, the alcohol 15% means the preparation contains 15% v/v of alcohol (C2H5OH).
(b) For 60 mL of preparation, calculate the quantity of alcohol (C2H5OH) needed:
15% v/v × 60 mL = 9 mL of alcohol
(c) Because the source of alcohol (C2H5OH) is Alcohol, USP, calculate the volume, in mL, of Alcohol, USP needed to give 9 mL
of alcohol (C2H5OH):
9 mL alcohol/95% v/v Alcohol, USP = 9.5 mL of Alcohol, USP
Therefore, add 9.5 mL of Alcohol, USP to this preparation.
(d) Determine the % v/v alcohol content for labeling.
Because labeling of alcohol is in % v/v of alcohol (C2H5OH), the alcohol content of this preparation would be labeled:
Alcohol 15%.
2. Determine the alcohol content, in % v/v, for the prescription (see Table 4):

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Table 4
Castor oil 40 mL
Acacia As needed
Alcohol 15 mL
Cherry syrup 20 mL
Purified water, a sufficient quantity to make 100 mL

Because the USP monograph for Alcohol is to be used when alcohol is called for in formulas, measure 15 mL of Alcohol, USP.
95% Alcohol, USP × 15 mL = 14.25 mL of alcohol
14.25 mL of alcohol/100 mL of preparation = 14.25% alcohol

10. ALLIGATION ALTERNATE AND ALGEBRA METHODS

10.1 Alligation Alternate

Alligation is a method of determining the proportions in which substances of different strengths are mixed to yield a desired
strength or concentration. Once the proportion is found, the calculation may be performed to find the exact amounts of sub-
stances required.
Set up the problem as follows.
1. Place the desired percentage or concentration in the center.
2. Place the percentage of the substance with the lower strength on the lower left-hand side.
3. Place the percentage of the substance with the higher strength on the upper left-hand side.
4. Subtract the lower percentage from the desired percentage, and place the obtained difference on the upper right-hand
side.
5. Subtract the desired percentage from the higher percentage, and place the obtained difference on the lower right-hand
side.
The results on the right side determine how many parts of the two different percentage strengths should be mixed to pro-
duce the desired percentage strength of a drug mixture. The total parts will equal the final weight or volume of the prepara-
tion.

10.1.1 CALCULATING BY USING THE ALLIGATION ALTERNATE

Examples—Alligation alternate
1. Determine the amount of ointment containing 12% drug concentration and the amount of ointment containing 16%
drug concentration must be used to make 1 kg of a preparation containing a 12.5% drug concentration.

General Chapters
In a total of 4 parts of 12.5% preparation, 3.5 parts of 12% ointment and 0.5 parts of 16% ointment are needed.
4 parts correspond to 1 kg or 1000 g.
1 part corresponds to 250 g.
3.5 parts correspond to 3.5 × 250 g or 875 g of 12% ointment.
0.5 parts correspond to 0.5 × 250 g or 125 g of 16% ointment.
2. Determine the volume, in mL, of 20% dextrose in water and 50% dextrose in water needed to make 750 mL of 35%
dextrose in water.

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In a total of 30 parts of 35% dextrose in water, 15 parts of 50% dextrose in water and 15 parts of 20% dextrose in water are
required.
30 parts correspond to 750 mL.
15 parts correspond to 375 mL.
Thus, use 375 mL of the 20% solution and 375 mL of the 50% solution to prepare the preparation.

10.2 Algebra Method

The following algebraic equation may be used instead of alligation to solve problems of mixing two different strengths of
the same ingredient:
(Cs × Qs) + (Cw × Qw) = (Cf × Qf), where C is concentration or strength, Q is the quantity; and the subscript s identifies the
strongest strength, w identifies the weakest strength, f represents the final mixture with a strength less than s and greater than
w, (Qs + Qw) = Qf, Qs = (Qf − Qw), and Qw = (Qf − Qs).

10.2.1 CALCULATING BY USING THE ALGEBRA METHOD

Examples—Algebra method
1. Determine the amount, in g, of 16% w/w drug ointment and 12% w/w drug ointment required to prepare 1 kg of 12.5%
w/w drug ointment.
(16% × Qs) + [12% × (1000 g − Qs)] = 12.5% × 1000 g

16% Qs + 120 g − 12% Qs = 125 g

4% Qs = 5 g

Qs = 5 g/4% = 125 g of 16% ointment

Qw = 1000 g − 125 g = 875 g of 12% ointment

2. Determine the volume, in mL, of 10% dextrose injection and 50% dextrose injection needed to make 750 mL of 35%
dextrose injection.
(50% × Cs) + [10% × (750 mL − Cs)] = 35% × 750 mL

50% Cs + 75 mL − 10% Cs = 262.5 mL

40% Cs = 187.5 mL

Cs = 187.5 mL/40% = 468.75 mL (470 mL practically)

Cw = 750 mL − 468.75 mL = 281.25 mL (280 mL practically)

11. ALIQUOT CALCULATIONS

General Chapters

When the quantity of drug desired requires a degree of precision in measurement that is beyond the capability of the availa-
ble measuring devices, the pharmacist may use the aliquot method of measurement. It applies when potent drug substances
are compounded, or when the total amount of the active drug in a single dose or individualized doses is less than the mini-
mum accurately weighable quantity (MAWQ). Even if the amount of drug needed is greater than the MAWQ per unit, an ali-
quot will provide more material per unit, which will aid in handling and administration. Aliquot means “containing an exact
number of times in something else”; the aliquot must be a proportional part of the total. Therefore, 5 is an aliquot part of 15,
because 5 is contained exactly 3 times in 15. Both the total volume of solution or weight of powder triturate and the aliquot
volume/weight should be easily and accurately measurable. If the solution or powder triturate is highly concentrated and a
small error is made in measuring the aliquot, a large error can occur in the quantity of drug brought to the final formulation.
Aliquots can be: solid–solid, when the active drug and the diluents are solids; solid–liquids, when the active drug is solid and
is to be incorporated into a liquid preparation, such as a solution, an emulsion, or a suspension; and liquid–liquid, when the
active drug is liquid and the diluents are liquids. It can be a pure liquid or a concentrated solution of a drug. Aliquots of pure
liquids are relatively uncommon because few drugs are liquid in their pure state. Aliquots involving concentrated solutions are
more common.
There are two general methods to prepare aliquots:
1. Aliquot method 1 is applicable to drugs or substances that have to be within the degree of accuracy provided by the
measuring device. It is the simplest method and can be applied to solid and liquid aliquots.

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2. Aliquot method 2, also known as the dilution factor method, is useful when there is more flexibility in the amount of drug
that may be measured.
Aliquot Method 1:
(a) The MAWQ amount of drug is measured.
(b) The drug is diluted with an arbitrary amount of diluent.
(c) The amount of dilution that will give the desired amount of drug is calculated, and the amount is measured.
Aliquot Method 2:
(a) The quantity of drug to be measured is determined by multiplying the amount of drug needed by an appropriately deter-
mined factor, called the dilution factor. The dilution factor must be a whole number more than or equal to the MAWQ
divided by the amount of drug needed.
(b) An arbitrary amount of diluent is measured and added. The amount of diluent used can be determined by different meth-
ods, provided the amount of diluent chosen will give an aliquot greater than or equal to the MAWQ.
(c) The amount of aliquot needed is determined by multiplying the weight or volume of the dilution by the inverse of the
dilution factor. Dilution factors are usually chosen to be whole numbers.
The general calculations can be shown as:
A/B = C/D
A = amount of drug desired
B = amount of drug measured
C = amount of drug in aliquot
D = aliquot total amount

11.1 Calculating Aliquots

Examples—Aliquots
1. Solid-in-liquid dilution (Aliquot Method 1)
Prepare 100 mL of a solution containing 0.2 mg/mL of clonidine using water as the diluent. To prepare this solution, 20 mg
of clonidine is needed.
(a) Select the weight of drug desired (A) to be equal to or greater than the MAWQ. In this situation, the MAWQ of the bal-
ance is 120 mg.
(b) Select the aliquot volume (D) in which the desired amount of drug (C) will be contained. This establishes the concentra-
tion of the solution to be prepared. Clonidine solubility is 1 g/13 mL, so if 5 mL is selected as the aliquot volume, the
concentration in that solution will be 20 mg/5 mL. Therefore, solubility will not be a problem in this aqueous solution.
(c) Using the preceding formula, calculate the volume of solution (B) to be prepared.
120 mg of clonidine/B = 20 mg of clonidine/5 mL of aliquot
B = 30 mL
(d) Prepare the solution containing 120 mg of clonidine in 30 mL of Purified Water. Transfer a 5-mL aliquot from this solution
to a final container, and add sufficient Purified Water to bring the formulation to a final volume of 100 mL.
2. Solid-in-solid dilution (Aliquot Method 2)
Prepare an individual dose of codeine phosphate 20 mg.
(a) Select a dilution factor that will yield a quantity that is greater than or equal to the MAWQ, and weigh this amount. In
this case, the dilution factor may be greater than or equal to 6 because 6 × 20 mg = 120 mg. The smallest dilution factor

General Chapters
that may be chosen is 6 if the MAWQ of the balance is 120 mg.
(b) Weigh an amount of diluent that will give an aliquot greater than or equal to the MAWQ. In this example, 600 mg of
diluent is weighed.
(c) Mix the two powders thoroughly by geometric trituration in a mortar.
(d) Calculate the total weight of the dilution: 120 mg codeine phosphate + 600 mg diluent = 720 mg.
(e) Calculate the aliquot part of the dilution that contains 20 mg of codeine phosphate by multiplying the total weight of the
dilution by the inverse of the dilution factor: 720 mg × (1/6) = 120 mg.
(f) Weigh this calculated amount of the dilution (120 mg) to get the desired 20 mg of codeine phosphate per dose.

12. POWDER VOLUME CALCULATIONS

12.1 Displacement in Suspension

12.1.1 CALCULATING POWDER VOLUME

Examples—Powder volume
1. Powder displacement in suspension

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The directions to reconstitute a 150 mL bottle of an amoxicillin for oral suspension of 250 mg/5 mL require 111 mL of
Purified Water. The physician has requested that the product be reconstituted at a concentration of 500 mg/5 mL. Calculate
the amount of Purified Water required for the higher concentration.
(a) Calculate the volume of the suspension occupied by the amoxicillin powder:
150 mL − 111 mL = 39 mL
(b) Calculate the quantity of amoxicillin present in the entire bottle:
150 mL × (250 mg/5 mL) = 7500 mg
(c) Calculate the total volume of the suspension at the requested concentration (500 mg/5 mL):
7500 mg/(500 mg/5 mL) = 75 mL
(d) Calculate the volume of Purified Water needed to reconstitute the powder by subtracting the powder volume calculated in
step a:
75 mL − 39 mL = 36 mL of Purified Water
[NOTE—Such formulations may be too viscous to flow freely.]
2. Powder volume in drugs for injection
If the powder volume of 250 mg of ceftriaxone for injection is 0.1 mL, calculate the amount of diluent that should be added
to 500 mg of ceftriaxone for injection to make a suspension with a concentration of 250 mg/mL.
(a) Calculate the total volume of injection:
500 mg/(250 mg/mL) = 2 mL
(b) Calculate the volume occupied by 500 mg of ceftriaxone for injection:
500 mg/(250 mg/0.1 mL) = 0.2 mL
(c) Calculate the volume of the diluent required:
(2 mL of suspension) − (0.2 mL of ceftriaxone for injection) = 1.8 mL of diluent

13. INTRAVENOUS FLOW OR INFUSION RATES

Intravenous (IV) solutions and emulsions may be administered by gravity flow and infusion or syringe pumps. Gravity-flow IV
sets are regulated by an adjustable clamp on the tubing, and the approximate flow rate is determined by counting the num-
ber of drops per 10–15 seconds, then adjusting that to a per minute rate. Manufactured IV sets are typically calibrated to deliv-
er from 15 to 60 drops/mL, depending on the particular set.

13.1 Solving by Multiple or Separate Steps

As in previous sections, the following examples may be solved by multiple separate steps, or a single-DUA procedure.

13.1.1 CALCULATING INTRAVENOUS FLOW OR INFUSION RATES

Examples—IV or infusion rates

1. An IV infusion of dextrose 5% in water with 20 mEq of potassium chloride is to be administered to a 6-year-old child at
the rate of 12 mL/hour. An IV administration set that delivers 60 drops/mL is available. Calculate the flow rate in drops/minute:
General Chapters

2. A 63.6-kg patient is admitted to the Emergency Department and requires a dopamine hydrochloride infusion to maintain
an adequate blood pressure. The drug is ordered at an initial rate of 2 mg/kg/minute. A 400-mg/250-mL dopamine hydro-
chloride injection is available. Calculate the flow rate in mL/hour to be administered by infusion pump:

14. ISOOSMOTIC SOLUTIONS

The following discussion and calculations have therapeutic implications in preparations of dosage forms intended for oph-
thalmic, subcutaneous, intravenous, and intrathecal administration as well as for neonatal use.

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14.1 Tonicity

Cells of the body, such as erythrocytes, will neither swell nor shrink when placed in a solution that is isotonic with body
fluids. The measurement of tonicity, however, which is a physiological property, is somewhat difficult. A 0.9% w/v sodium
chloride injection, which has a freezing point (FP) of −0.52°, is both isotonic and isoosmotic with body fluids. In contrast to
isotonicity, FP depression is a physical property. Some solutions that are isoosmotic with body fluids are not isotonic, because
they contain solutes to which cells are freely permeable rather than semipermeable. Freely permeable solutes (e.g., boric acid
and urea) can cause erythrocyte lysis, i.e., behave as if they were hypotonic in concentrations that are hyperosmotic relative to
body fluids. Nevertheless, many pharmaceutical products are prepared using FP data or related sodium chloride data to pre-
pare solutions that are isoosmotic with body fluids. A closely related topic is osmolarity (see chapter Osmolality and Osmolarity
á785ñ).
FP data or sodium chloride equivalents of pharmaceuticals and excipients (see Table 5) may be used to prepare isoosmotic
solutions, as shown in the following examples.

14.1.1 CALCULATING TONICITY

Example—Tonicity
Determine the amount of sodium chloride (NaCl) required to prepare 60 mL of an isoosmotic solution of atropine sulfate
injection 0.5% using the E values and the FP depression values in Table 5.
Table 5. Sodium Chloride Equivalents (E) and FP Depressions for a 1% Solution of the Drug or Excipient
Drug or Excipient E FP Depression
Atropine sulfate 0.13 0.075
Sodium chloride 1.00 0.576

Using the E values:

(a) The total amount of substances equivalent to a 0.9% sodium chloride injection = (0.9 g/100 mL) × 60 mL = 0.54 g.
(b) The amount of atropine sulfate required = (0.5 g/100 mL) × 60 mL = 0.3 g.
(c) 1 g of atropine sulfate is equivalent to 0.13 g of sodium chloride.
(d) 0.3 g of atropine sulfate is equivalent to 0.3 × 0.13 g = 0.039 g of sodium chloride.
(e) Thus, the required amount of sodium chloride is 0.54 g − 0.039 g = 0.501 g or 0.5 g.
Using FP depression values:
(a) The FP depression required is 0.52°.
(b) A 1% solution of atropine sulfate causes an FP depression of 0.075°.
(c) A 0.5% solution of atropine sulfate causes an FP depression of 0.5 × 0.075° = 0.0375°.
(d) The additional FP depression required is 0.52° − 0.0375° = 0.483°.
(e) A 1% solution of sodium chloride causes an FP depression of 0.576°.
(f) Therefore, an FP depression of 1° is caused by a 1%/0.576 = 1.736% solution of sodium chloride.
(g) 1.736% × 0.483 = 0.838% solution of sodium chloride causes an FP depression of 0.482°.
(h) The required amount of sodium chloride is (0.838%) × 60 mL = 0.502 g or 0.5 g.

15. PH AND BUFFER CALCULATIONS

15.1 pH Calculations

General Chapters
See Appendix 1 for logarithmic definitions and applications.
pH = −log [H3O+], and pKa = −log ([H3O+][A−])/[HA], where [H3O+] is the hydrodium ion concentration in an aqueous solu-
tion, [A−] is the ionic form of the relevant acid, and Ka is the ionization constant of either a monoprotic acid or a particular
proton from a polyprotic acid in aqueous solution. The [H+] = the antilogarithm of (−pH) or 10−pH; and Ka = the antilogarithm
of (−pKa) or 10−pKa.
The pH of an aqueous solution containing a weak acid may be calculated using the Henderson–Hasselbalch equation:
pH = pKa + log [(base form)/(acid form)]
The buffer equation symbol (↔) represents the equilibrium between conjugate base and acid forms or pairs of the same
molecule. It is called the buffer equation, because small changes in the ratio of concentrations of the conjugate forms result in
a logarithmically smaller change in pH. The salt form can be an acid or base, depending on structure; thus, its conjugate form
is a base or acid, respectively.
Example 1:
B and BH+ represent a nonionized or “free” base and cationic acid pair, BH+ ↔ B+H+
Example 2:
HA and A− represent a nonionized or “free” acid and anionic base pair, HA ↔ A− + H+

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Example 3:
HnA− and Hn−1A2−, such as H2PO4− and HPO42−, represent an anionic acid and anionic base relative to each other; the pKa =
7.2 for H2PO4− ↔ HPO42− + H+.

15.1.1 CALCULATING pH

Example—pH
A solution contains 0.020 mol/L of sodium acetate and 0.010 mol/L of acetic acid, which has a pKa value of 4.76. Calculate
the pH and the [H+] of the solution as follows:
pH = 4.76 + log (0.020/0.010) = 5.06
[H+] = antilogarithm of (−5.06) = 8.69 × 10−6

15.2 Buffer Solutions

15.2.1 DEFINITION

A buffer solution is an aqueous solution that resists a change in pH when small quantities of acid or base are added, when
diluted with the solvent, or when the temperature changes. Most buffer solutions are mixtures of a weak acid and one of its
salts, or mixtures of a weak base and one of its salts. Water and solutions of a neutral salt, such as sodium chloride, have very
little ability to resist the change of pH and are not capable of effective buffer action.

15.2.2 PREPARATION, USE, AND STORAGE OF BUFFER SOLUTIONS

Buffer solutions for Pharmacopeial tests should be prepared using freshly boiled and cooled water (see Reagents, Indicators
and Solutions). They should be stored in containers such as Type I glass bottles and used within 3 months of preparation.
Buffers used in physiological systems are carefully chosen so as to not interfere with the pharmacological activity of the med-
icament or the normal function of the organism. Commonly used buffers in parenteral products, for example, include: the
nonionized acid and base salt pairs of acetic acid and sodium acetate, citric acid and sodium citrate, glutamic acid and sodium
glutamate, and monopotassium or monosodium phosphate and dipotassium or disodium phosphate; and the acid salt and
nonionized base pair tris(hydroxymethyl)aminomethane hydrochloride and tris(hydroxymethyl)aminomethane. Buffer solu-
tions should be freshly prepared.
The Henderson–Hasselbalch equation, noted in 15.1 pH Calculations, allows calculation of the pH and concentrations of con-
jugate pairs of weak acids and their salts and weak bases and their salts in buffer solutions when the pKa of the acid form of
the buffer pair is known. Appropriately modified, this equation may be applied to buffer solutions composed of a weak base
and its salt.

15.2.3 BUFFER CAPACITY

The buffer capacity of a solution is the measurement of the ability of that solution to resist a change in pH upon addition of
small quantities of a strong acid or base. An aqueous solution has a buffer capacity of 1 when 1 L of the buffer solution re-
quires 1 g equivalent of strong acid or base to change the pH by 1 unit. Therefore, the smaller the pH change upon the addi-
General Chapters

tion of a specified amount of acid or base, the greater the buffer capacity of the buffer solution. Usually, in analysis, much
smaller volumes of buffer are used to determine the buffer capacity. An approximate formula for calculating the buffer capaci-
ty is g equivalents of strong acid or base added per L of buffer solution per unit of pH change, i.e., (g equivalents/L)/(pH
change).

15.2.4 CALCULATING BUFFER CAPACITY

Example—Buffer capacity
The addition of 0.01 g equivalents of sodium hydroxide to 0.25 L of a buffer solution produced a pH change of 0.50. The
buffer capacity of the buffer solution is calculated as follows:
(0.01 Eq/0.25 L)/0.50 pH change = 0.08(Eq/L)/(pH change)

16. TEMPERATURE

The relationship between Celsius or Centigrade (°C) and Fahrenheit (°F) temperature scale is expressed by the following
equations:
°C = (°F − 32) × (5/9)

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°F = (°C × 1.8) + 32

16.1 USP Temperatures

According to the General Notices and Requirements 8.180 Temperatures, temperatures are expressed in centigrade (Celsius)
degrees, and all other measurements are made at 25° unless otherwise indicated. For instructional purposes, °F is shown in the
examples.

16.1.1 CALCULATING TEMPERATURES

Examples—Temperatures
1. Convert 77°F to Celsius degrees.
°C = (77°F − 32) × (5/9) = 25°C
2. Convert 30°C to Fahrenheit degrees.
°F = (30°C × 1.8) + 32 = 86°F
The relationship between the Kelvin or absolute (K) and the Celsius (°C) scales is expressed by the equation:
K = °C + 273.1

17. ENDOTOXINS

An endotoxin is a lipopolysaccharide that comes from a particular source, where species and strain number are usually indi-
cated.

17.1 Endotoxin Concentrations

For more information concerning endotoxins, see Bacterial Endotoxins Test á85ñ.

17.1.1 CALCULATING ENDOTOXINS

Example—Endotoxins
A 71.8-kg patient is to receive an intrathecal infusion of morphine sulfate at a rate of 0.3 mg/hour. The solution will be pre-
pared by diluting preservative-free morphine sulfate injection, which contains 10 mg/mL of morphine sulfate, with 0.9% so-
dium chloride injection to produce an infusion rate of 2 mL/h.
1. Determine the volume, in mL, of morphine sulfate injection (10 mg/mL) and 0.9% sodium chloride injection needed to
prepare a 24-hour infusion.
0.3 mg of morphine sulfate/hour × 24 hours = 7.2 mg of morphine sulfate
7.2 mg of morphine sulfate/(10 mg/mL) = 0.72 mL of morphine sulfate injection
2 mL of infusion/h × 24 h = 48 mL of total volume

General Chapters
48 mL total volume − 0.72 mL morphine sulfate injection = 47.28 mL of 0.9% sodium chloride injection
2. Calculate the maximum potential endotoxin load/hour for this preparation. [NOTE—USP monographs specify upper limits
of 14.29 USP Endotoxin Units (EU)/mg of morphine sulfate in injections for intrathecal use, and 0.5 EU/mL for injections
containing 0.5%–0.9% sodium chloride.]
7.2 mg of morphine sulfate injection × 14.29 EU/mg of morphine sulfate = 102.89 EU from morphine sulfate
47.28 mL of sodium chloride injection × 0.5 EU/mL = 23.64 EU from 0.9% sodium chloride injection
Endotoxin load = 102.89 EU + 23.64 EU = 126.53 EU
126.53 EU/24 hour = 5.27 EU/hour
3. Determine if the endotoxin load in step 2 exceeds the allowable USP limit for this patient. [NOTE—The maximum endo-
toxin load by intrathecal administration is 0.2 EU/kg/hour (see chapter á85ñ).]
Maximum endotoxin load = (0.2 EU/kg/hour) × 71.8 kg-patient = 14.36 EU/hour
The endotoxin load of 5.27 EU/hour does not exceed the allowable limit of 14.36 EU/hour.

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18. STABILITY AND EXPIRATION DATE CALCULATIONS

18.1 Stability Based on Rate Calculations

Calculation of a predetermined minimum percentage of initial drug strength or other quality parameter, e.g., in vitro disso-
lution of active pharmaceutical ingredients (APIs) or active drugs in solid oral dosage forms, is based on component-specific
assays and other validated scientific testing. The expiration date or time elapsed until such minimum acceptable limits are
reached for a specific manufactured product is exclusive to the specific formulation, packaging, and environmental conditions,
e.g., temperature, humidity, and illumination, to which the item is subjected. See also chapters á659ñ, á795ñ, á797ñ, á1163ñ,
and á1191ñ.
The degradation or concentration loss rates or kinetics of most APIs can be accurately represented or modeled by either zero
order (constant) or first order (mono-exponential) rate equations. Zero order calculations are generally applicable to solids,
semisolids, suspensions in which a majority of the drug strength is present as solid particles, and auto-oxidation in solutions.
First order calculations are generally applicable for drug hydrolysis in solutions.

18.2 Zero Order Rate Calculations

The isothermal zero order or constant rate equation for a particular formulation is C = C0 − kt, where C is the concentration
of API at any time, C0 is the concentration at origination or time zero, k is the reaction rate constant, and t is any time after
origination or zero. The values and units of the rate, dC/dt, and rate constant, k, are the same for zero order processes, i.e., the
units are concentration/time, such as mg/mL/day.

18.2.1 ZERO ORDER RATE EQUATION DERIVED FROM ORIGINAL DATA

The following examples illustrate calculations of the zero order rate equation from original concentration assay and time da-
ta, and an expiration date using that equation.
18.2.1.1 Calculating zero order rate
Examples—Zero order rate
1. Calculate the zero order rate equation based on the assay results for a drug suspension at 25° (see Table 6):
Table 6
C (mg/mL) t (days)
49 3
47.5 8
44.8 17
42.3 26

Linear regression of the C (ordinate) versus t (abscissa) values yields the equation, C = 49.84 − 0.292t with a correlation coef-
ficient of 0.9996.
2. Calculate the time when C = 0.9 × C0, i.e., the expiration date where the concentration will be 90% of the original con-
centration (t90):
C = 49.84 − 0.292t
General Chapters

0.9 × 49.84 = 49.84 − 0.292(t90)

t90 = (44.86 − 49.84)/−0.292 = 17.05 days

3. Using the previous linear regression equation, calculate the C of the drug suspension at 25° when t = 12 days:
C = 49.84 − (0.292 × 12) = 46.34 mg/mL
4. Calculate t when C = 45 mg/mL:
45 = 49.84 − 0.292t
t = (45 − 49.84)/−0.292 = 16.6 days

18.2.2 ZERO ORDER VALUES CALCULATED FROM A RATE EQUATION

The following are examples of expiration dates calculated from a rate equation derived from original concentration assay
and time data.

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18.2.2.1 Calculating zero order values from a rate equation

Examples—Zero order from a rate equation
1. Calculate the t80 expiration date of a drug cream at 25° using the equation, C = 0.05 − 0.0003t, where the C unit is %
w/w and the t unit is months. At t80, C = 0.8C0.
0.8 × 0.05 = 0.05 − 0.0003(t80)

t80 = 33.3 months

2. Calculate the t80 expiration date of the drug cream formulation in example 1, but for which C0 is 0.1:
0.8 × 0.1 = 0.1 − 0.0003(t80)

t80 = 66.7 months

18.3 First Order Rate Calculations

The isothermal first order rate equation for a particular formulation in exponential form is C = C0e−kt, and in linear form is
ln(C) = ln(C0)−kt, where C is the concentration of an API at any time, C0 is the concentration at origination or time zero, k is the
reaction rate constant, and t is any time after origination or zero. The constantly changing rate, dC/dt, and rate constant, k, are
not the same for first order processes. The rate units are concentration/time, e.g., mg/mL/hour, but the rate constant unit is
reciprocal time, time−1, e.g., hour−1.

18.3.1 FIRST ORDER LINEAR RATE EQUATION DERIVED FROM ORIGINAL DATA

The following examples illustrate calculation of the linear first order rate equation from original concentration assay and time
data and calculation of an expiration date using that equation.
18.3.1.1 Calculating first order linear rate equations
Example—First order linear rate
1. Calculate the linear first order rate equation based on the assay results for a drug solution at 27° (see Table 7):
Table 7
C (mg/mL) t (hour)
12.3 2
11.9 6
11.5 14
10.6 24

Linear regression of the ln(C) (ordinate) versus t (abscissa) values yields the equation, ln(C) = 2.522 − 0.0065t with a correla-
tion coefficient of 0.992.
2. From the linear regression equation, calculate the time when 95% of the original concentration is reached, t95, when C =
0.95C0, which is the predetermined expiration date:
ln(C0) = 2.522; thus, C0 = e2.522 = 12.45 mg/mL

ln(0.95 × 12.45) = 2.522 − 0.0065(t95)

General Chapters
t95 = (2.470 − 2.522)/−0.0065 = 8 hours

18.3.2 FIRST ORDER VALUES CALCULATED FROM A LINEAR EQUATION

The following are examples of an expiration date, concentration, and time calculated for the same drug solution at 22° from
the rate equation, ln(C) = 4.382 − 0.076t, where the C units are mg/mL and the t unit is days, derived from the original con-
centration assay and time data.
18.3.2.1 Calculating first order values from a linear rate equation
Examples—First order from a linear rate
1. Calculate the t90 expiration date:
ln(C0) = 4.382; thus, C0 = e4.382 = 80

ln(0.9 × 80) = 4.382 − 0.076(t90)

t90 = (4.277 − 4.382)/−0.076 = 1.4 days

2. Calculate the time at which C = 75 mg/mL:

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ln(75) = 4.382 − 0.076t

t = (4.317 − 4.382)/−0.076 = 0.86 day
3. Calculate whether C = 70 mg/mL occurs before or after t90:
ln(70) = 4.382 − 0.076t
t = (4.248 − 4.382)/−0.076 = 1.8 days
C = 70 mg/mL occurs at 1.8 days, after a t90 of 1.4 days

18.3.3 FIRST ORDER EXPIRATION DATE CALCULATED FROM TWO VALUES OF CONCENTRATION AND TIME

When degradation or other cause of concentration loss is known from experience or reference information to obey first or-
der kinetics, the rate constant can be accurately estimated from accurate assays of only two concentrations at their respective
times. In this case, the linear first order rate equation, ln(C) = ln(C0) − kt, may be transformed or integrated as ln(C2) = ln(C1) −
k(t2 − t1), which when rearranged is k = ln(C1/C2)/(t2 − t1). The following examples apply these equations to calculate expiration
dates, concentrations, and times.
18.3.3.1 Calculating first order expiration date from two values
Examples—First order expiration date from two values
1. At 25°, the concentration of an antibiotic in solution was 89 mg/mL after 3 hours and 74 mg/mL after 8 hours. Calculate
the initial concentration at time zero:
k = ln(89/74)/(8 − 3) = 0.037 hour−1
ln(89) = ln(C0) − (0.037 hour−1 × 3 hour)

ln(C0) = 4.489 + 0.111 = 4.6

C0 = e4.6 = 99.5 mg/mL

2. Calculate the t90 expiration date using the data in example 1. At t90, C = 0.9C0.
ln(0.9 × 99.5) = ln(99.5) − 0.037(t90)

t90 = (4.495 − 4.600)/−0.037 hour−1 = 2.8 hour

3. Calculate the concentration at 6 hour using the data in example 1:

ln(C) = ln(99.5) − (0.037 × 6)
ln(C) = 4.378
C = e4.378 = 79.7 mg/mL

18.3.4 FIRST ORDER TIMES, tn, FOR 0.n FRACTION OR n% OF REMAINING ORIGINAL CONCENTRATION

The two most common first order pharmaceutical tn values are the t50, which is a primary parameter factor in clinical phar-
macokinetics, and the t90, which is the most common stability shelf life or expiration date. Values of any tn, where 0 < n < 100,
General Chapters

are derived from the linear first order equation, ln(C) = ln(C0) − kt. The equations for t50 and t90 in particular are derived in the
following examples. The value of k by definition is constant for a specific drug chemical in a specific formulation at a specific
temperature; thus, tn values derived from such values of k are also constant.
18.3.4.1 Calculating first order times for remaining original concentrations
Examples—First order times for remaining original concentrations
1. At tn, C = 0.n × C0.
ln(0.n × C0) = ln(C0) − ktn

tn = [ln(0.n × C0) − ln(C0)]/−k = ln[(0.n × C0)/C0]/−k = ln(0.n)/−k

tn = ln(0.n)/−k

2. At t50, C = 0.5(C0).
ln(0.5 × C0) = ln(C0) − kt50

t50 = [ln(0.5 × C0) − ln(C0)]/−k = ln[(0.5 × C0)/C0]/−k = ln(0.5)/−k = −0.693/−k = 0.693/k

t50 = 0.693/k

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3. At t90, C = 0.9(C0).
ln(0.9 × C0) = ln(C0) − kt90

t90 = [ln(0.9 × C0) − ln(C0)]/−k = ln[(0.9 × C0)/C0]/−k = ln(0.9)/−k = −0.105/−k = 0.105/k

t90 = 0.105/k

18.4 Stability Prediction Based on Arrhenius Theory

The basis of the Arrhenius theory is that reaction rates and rate constants change exponentially in the direction of arithmetic
temperature change. The pharmaceutical application of the Arrhenius theory is based on scientifically accurate and statistically
valid assay data obtained at three or more temperatures that are ³10° warmer than the intended drug storage temperature
and each other. The Arrhenius equation may be expressed in an exponential form, k = Ae−(Ea/RT), a linear form, ln(k) = ln(A) −
(Ea/RT), and an integrated form, ln(k2/k1) = Ea(T2 − T1)/[R(T2 × T1)], where k, k1, and k2 are isothermal rate constants, A is a
thermodynamic factor, Ea is energy of activation for the degradation reaction, R is the gas constant (1.987 × 10−3 kcal mol−1K −1
or 8.314 × 10−3 J K−1 mol−1), and T, T1, and T2 are absolute or Kelvin temperatures.

18.4.1 ARRHENIUS LINEAR EQUATION DERIVED FROM ORIGINAL DATA

The following examples illustrate derivation of a linear Arrhenius equation from original assay data and its application to pre-
dicting a drug stability expiration date at a cooler or lower storage temperature.
18.4.1.1 Calculating Arrhenius equations
Examples—Arrhenius equations
1. Calculate the linear Arrhenius equation based on the rate constants and temperatures for a beta-lactam antibiotic that
decomposes in solution at a first order rate (see Table 8):
Table 8
T (°C) T (K) 1/T (K−1) k (hour−1) ln(k)
40 313 3.195 × 10−3 0.0014 −6.571
50 323 3.096 × 10−3 0.005 −5.298
60 333 3.003 × 10−3 0.016 −4.135

Linear regression of the ln(k) (ordinate) versus 1/T (abscissa) values yields the equation, ln(k) = 33.977 − (12,689/T) with a
correlation coefficient of 0.99997.
2. Calculate the t90 shelf life expiration date, in days, at 25°C (298 K) using the equation in example 1:
ln(k25) = 33.977 − (12,689/298) = 33.977 − 42.581 = −8.604

k25 = e−8.604 = 1.834 × 10−4 hour−1 = 4.402 × 10−3 day−1

t90 = 0.105/k = 0.105/4.402 × 10−3 day−1 = 23.85 days

t90 at 25°C = 23.85 days

18.4.2 STABILITY PREDICTIONS USING THE INTEGRATED ARRHENIUS EQUATION

General Chapters
The following examples illustrate stability predictions based on one accurately determined isothermal rate constant and ad-
herence to the same degradation rate order, e.g., first order, at temperatures at which stability is to be calculated from the
equation, ln(k2/k1) = Ea(T2 − T1)/[R(T2 × T1)].
18.4.2.1 Calculating stability prediction using integrated Arrhenius equation
Example—Stability using integrated Arrhenius equations
1. Calculate the t85 stability expiration date at 4°C (277 K) for an ester hydrolysis with an Ea = 15 kcal/mol and k = 0.0045
hour−1 at 23°C (296 K):
ln(k277/0.0045) = 15(277 − 296)/[1.987 × 10−3 (277 × 296)]

ln(k277) − ln(0.0045) = −285/162.92

ln(k277) = −1.749 + ln(0.0045) = −7.153

k277 = e−7.153 = 7.825 × 10−4 hour−1

t85 = ln(0.850)/k277 = −0.163/−7.825 × 10−4 hour−1 = 208.3 hours

t85 at 4°C = 208.3 hours (t85 at 23°C is 36.2 hours)

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18.4.3 ARRHENIUS-BASED Q10 STABILITY ESTIMATION

The temperature coefficient (Q10) represents the multiplicative factor by which a chemical reaction rate constant changes in
the same direction as the temperature for each 10°C change. For drug molecules, Q10 ranges from 2 to 5, corresponding to an
Ea range of 10–25 kcal/mol or 42–105 kJ/mol. A Q10 of 3 yields reasonable estimates of drug stability in the equation, (tn at T2)
= (tn at T1)/{Q10 }, where n is a percentage of remaining C0, T1 is the temperature at which tn is known, and T2 is the tem-
perature at which tn is to be estimated. Calculations using Q10 values of both 2 and 4 may be used to obtain the shortest or
most conservative stability estimate, but Q10 = 3 is applied in the following two examples.
18.4.3.1 Calculating Arrhenius-based Q10 stability estimation
Example—Arrhenius-based Q10 stability
1. Estimate the t90 expiration date in hours of an antibiotic suspension stored in a closed automobile at 57° for which the 8°
refrigeration t90 is 14 days.
t90 at 57° = [14 days × (24 hours/day)]/{3[(57 − 8)/10]} = 336 hours/34.9 = 336 hours/217.7 = 1.54 hours

t90 at 57° = 1.54 hours

APPENDIX 1: LOGARITHMS

The logarithm of a number is the exponent or power to which a given base number must be raised to equal that number.
Thus, the logarithm of Y to the base, b, equals X, or logb(Y) = X. The logarithm of 0 and all negative numbers is undefined or
nonexistent. The logarithm of 1 is 0 and of numbers <1 is negative in all systems (see Table 9).
Table 9. Common (or Briggsian) and Natural (or Napierian) Logarithms
Antilogarithm
or Inverse
Logarithmic System Abbreviation or Symbol Base Number Format Logarithm
Common Log 10 log Y = X 10X = Y
Natural Ln e or 2.7183a ln Y = X eX = Y
a e is an irrational number derived from an infinite series of reciprocal whole number factorials, e = 1 + 1/1! + 1/2! + 1/3! + 1/4! ... + 1/n!, where n = infinity. e
rounds to 2.7183 when n ³ 8.

2. The relationships between common and natural logarithms are the following:
(a) log Y = ln Y/ln 10 = ln Y/2.303
(b) ln Y = ln 10 × log Y = 2.303 × log Y
3. Rules for some common calculations with logarithms are shown in Table 10.
Table 10. Rules for Calculating with Logarithms
Formula Example
ln(0.62) + ln(1.73) = ln(0.62 × 1.73) =
ln(1.0726) = 0.070
ln(A) + ln(B) = ln(A × B) log(5.7) + log(0.43) = log (5.7 × 0.43) =
Additions and multiplications log(A) + log(B) = log(A × B) log(2.451) = 0.389
ln(0.5) − ln(4) = ln(0.5/4) = ln(0.125) = −2.079
ln(A) − ln(B) = ln(A/B) log(1.57) − log(2.48) = log(1.57/2.48) =
Subtraction and quotients log(A) − log(B) = log(A/B) log(0.6330645) = −0.199
General Chapters

13.6−Z = 1.25
ln(1.25) = −Z × ln(13.6)
Z = −ln(1.25)/ln(13.6) = −0.223/2.610 =
−0.085
0.57Z = 2.3
log(2.3) = Z × log(0.57)
ln(Yz) = Z × ln(Y) Z = log(2.3)/log(0.57) = 0.362/−0.244 =
Simple non-base exponentials log(Yz) = Z × log(Y) −1.484
67 × 10b = 15.1
log(67) + b = log(15.1)
In (a × e±b) = ln(x) ↔ ln(a) ± b = ln(x) 1.826 + b = 1.179
Base exponentials log (a × 10±b) = log(x) ↔ log(a) ± b = log(x) b = 1.179 − 1.826 = −0.647

Official from August 1, 2018

Copyright (c) 2018 The United States Pharmacopeial Convention. All rights reserved.
Accessed from 64.233.173.20 by regis22 on Tue Aug 21 12:51:11 EDT 2018
USP 41 General Information / á1163ñ 7475

á1163ñ QUALITY ASSURANCE IN PHARMACEUTICAL COMPOUNDING

INTRODUCTION

The need for a quality assurance system is well documented in United States Pharmacopeia (USP) chapters for compounded
preparations (see Quality Control under Pharmaceutical Compounding—Nonsterile Preparations á795ñ and Quality Assurance (QA)
Program under Pharmaceutical Compounding—Sterile Preparations á797ñ). A quality assurance program is guided by written pro-
cedures that define responsibilities and practices that ensure compounded preparations are produced with quality attributes
appropriate to meet the needs of patients and health care professionals. The authority and responsibility for the Quality Assur-
ance program should be clearly defined and implemented and should include at least the following nine separate but integra-
ted components: (1) training; (2) standard operating procedures (SOPs); (3) documentation; (4) verification; (5) testing; (6)
cleaning, disinfecting, and safety; (7) containers, packaging, repackaging, labeling, and storage; (8) outsourcing, if used; and
(9) responsible personnel.
The definition of compounding for the purpose of this chapter is defined in general test chapter á795ñ.
The safety, quality, and efficacy and/or benefit of compounded preparations depend on correct ingredients and calculations;
accurate and precise measurements; appropriate formulation, facilities, equipment, and procedures; and prudent pharmaceuti-
cal judgment. As a final check, the compounder shall review each procedure in the compounding process. To ensure accuracy
and completeness, the compounder shall observe the finished preparation to ensure that it appears as expected and shall in-
vestigate any discrepancies and take appropriate corrective action before the prescription is dispensed to the patient.
The water used in all aspects of compounding should meet the requirements of Waters for Pharmaceutical Purposes á1231ñ.
Radiopharmaceuticals and radiolabeled materials have unique characteristics requiring additional quality assurances descri-
bed in Positron Emission Tomography Drugs for Compounding, Investigational, and Research Uses á823ñ and the Radiopharmaceuti-
cals as CSPs section under á797ñ.
The responsibilities of the compounder and compounding personnel can be found in chapters á795ñ and á797ñ.

TRAINING

Personnel involved in nonsterile or sterile compounding require additional, specific training and periodic retraining beyond
the training needed for routine dispensing duties. A thorough quality assurance program for compounded preparations re-
quires documentation of both training and skill competency. In addition, the authority and responsibility for the QA program
should be clearly defined as implemented. Training for nonsterile compounders should meet or exceed the standards set forth
in á795ñ, and personnel training for sterile preparation compounders should meet or exceed the standards set forth in á797ñ.

STANDARD OPERATING PROCEDURES

SOPs for pharmaceutical compounding are documents that describe how to perform routine and expected tasks in the com-
pounding environment, including but not limited to procedures involving:
• Beyond-Use dating
• Chemical and physical stability
• Cleaning and disinfecting
• Component quality evaluation

General Chapters
• Compounding methods
• Dispensing
• Documentation
• Environmental quality and maintenance
• Equipment maintenance, calibration, and operation
• Formulation development
• Labeling
• Materials and final compounded preparation handling and storage
• Measuring and weighing
• Packaging and repackaging
• Patient monitoring, complaints, and adverse event reporting
• Patient or caregiver education and training
• Personnel cleanliness and garb
• Purchasing
• Quality Assurance and Continuous Quality Monitoring
• Safety
• Shipping
• Testing

Official from August 1, 2018

Copyright (c) 2018 The United States Pharmacopeial Convention. All rights reserved.