Expressed values are considered significant to the last digit shown (see Significant Figures and Tolerances in the General Notices). Significant figures are digits with practical meaning. The accuracy of the determination is implied by the number of figures used in its expression. In some calculations zeros may not be significant. For example, for a measured weight of 0.0298 g, the zeros are not significant; they are used merely to locate the decimal point. In the example, 2980 g, the zero may also be used to indicate the decimal point, in which case the zero is not significant. Alternately, however, the zero may indicate that the weight is closer to 2981 g or 2979 g, in which case the zero is significant. In such a case, knowledge of the method of measurement would be required in order to indicate whether the zero is or is not significant. In the case of a volume measurement of 298 mL, all of the digits are significant. In a given result, the last significant figure written is approximate but all preceding figures are accurate. For example, a volume of 29.8 mL implies that 8 is approximate. The true volume falls between 29.75 and 29.85. Thus, 29.8 mL is accurate to the nearest 0.1 mL, which means that the measurement has been made within ±0.05 mL. Likewise, a value of 298 mL is accurate to the nearest 1 mL and implies a measurement falling between 297.5 and 298.5, which means that the measurement has been made within ±0.5 mL and is subject to a maximum error calculated as follows:

A zero in a quantity such as 298.0 mL is a significant figure and implies that the measurement has been made within the limits of 297.95 and 298.05 with a possible error calculated as follows:

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The degree of accuracy in the last example is greatest. Thus, the number of significant figures provides an estimate both of true value and of accuracy.

Measurement	Number of Significant Figures
2.98	3
2.980	4
0.0298	3
0.0029	2

Sums and Differences— When adding or subtracting, the number of decimal places in the result shall be the same as the number of decimal places in the component with the fewest decimal places.

Products and Quotients— When multiplying or dividing, the result shall have no more significant figures than the measurement with the smallest number of significant figures entering into the calculation.

The logarithm of a number is the exponent or the power to which a given base must be raised in order to equal that number.

pH = –log [H+], and

pKa = –log Ka

pH = pKa + log [salt]/[acid]

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 dispensed. Pharmacists must perform calculations and measurements to obtain, theoretically, 100% of the amount of each ingredient in compounded formulations. Calculations 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 under Loss on Drying 731, 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 duration necessary and then closed tightly immediately after use.

The nature of the drug substance that is to be weighed and used in compounding a prescription must be known exactly. 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 or that is determined in the pharmacy immediately before the drug substance is used by the procedure under Loss on Drying 731, this information must be used when calculating the amount of drug substance that is to be weighed in order 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+, netilmicin (n+)], 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 calculated on the basis of the required quantity of the pharmacological moiety.

The following formula may be used to calculate the exact theoretical weight of an ingredient in a compounded preparation:

in which W is the actual weighed amount; a is the prescribed or pharmacist-determined weight of the active or functional moiety of drug or added substance; b is the chemical formula weight of the ingredient, including waters of hydration for hydrous ingredients; d is the fraction of dry weight when the percent by weight of adsorbed moisture content is known from the loss on drying procedure (see Loss on Drying 731); and e is the formula weight of the active or functional moiety of a drug or added substance that is provided in the formula weight of the weighed ingredient.

Equation Factor	Numerical Value
W	weight, in g, of Morphine Sulfate USP
a	1.5 g of morphine sulfate pentahydrate in the prescription
b	759 g/mole
d	1.0
e	759 g/mole

Equation Factor	Numerical Value
W	weight, in mg, of Aminophylline USP (dihydrate)
a	250 mg of theophylline
b	456 g/mole
d	0.996
e	360 g/mole

Equation Factor	Numerical Value
W	weight, in g, of Lithium Citrate USP (tetrahydrate)
a	200 mEq of Li+ or 1.39 g of Li+
b	282 g/mole
d	0.975
e	3 × 6.94 g/mole or 20.8 g/mole

Equation Factor	Numerical Value
W	weight, in g, of Netilmicin Sulfate USP
a	2.5 g
b	1442 g/mole
d	0.88
e	951 g/mole

(0.01/0.25)/0.50 = 0.08(Eq/L)/(pH change)

(mg of drug per kg of body weight ) × (kg of body weight) = dose for an individual for a 24-hour period

(amount of drug per m2 of body surface area) × (body surface area in m2) = dose for an individual for a 24-hour period

BSA (m2) = square root of {[Height (in) × Weight (lb)] / 3131}

BSA (m2) = square root of {[Height (cm) × Weight (kg)]/ 3600}

Percentage concentrations of solutions are usually expressed in one of three common forms:

See also Percentage Measurements under Concentrations in the General Notices. The above 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.

Note that weights are always additive, i.e., 50 g plus 25 g = 75 g. Volumes of two different solvents or volumes of solvent plus a solid solute are not strictly additive. Thus 50 mL of water + 50 mL of pure alcohol do not produce a volume of 100 mL. Nevertheless, it is assumed that in some pharmaceutical calculations, volumes are additive, as discussed below under Reconstitution of Drugs Using Volumes Other than Those on the Label.

examples—

The definition of specific gravity is usually based on the ratio of weight of a substance in air at 25 to that of the weight of an equal volume of water at the same temperature. The weight of 1 mL of water at 25 is approximately 1 g. The following equation may be used for calculations.

examples—

A concentrated solution can be diluted. Powders and other solid mixtures can be triturated or diluted to yield less concentrated forms. Because the amount of solute in the diluted solution or mixture is the same as the amount in the concentrated solution or mixture, the following relationship applies to dilution problems.

Almost any quantity and concentration terms may be used. However, the units of the terms must be the same on both sides of the equation.

examples—

See Units of Potency in the General Notices.

Because some substances may not be able to be defined by chemical and physical means, it may be necessary to express quantities of activity in biological units of potency.

examples—

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

examples—

Occasionally it may be necessary to reconstitute a powder in order to provide a suitable drug concentration in the final product. This may be accomplished by estimating the volume of the powder and liquid medium required.

examples—

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 desired percentage from the lower percentage, and place the obtained difference on the upper right-hand side.
5. Subtract the higher percentage from the desired percentage, and place the obtained difference on the lower right-hand side.

1. How much ointment having a 12% drug concentration and how much ointment having a 16% drug concentration must be used to make 1 kg of a preparation containing a 12.5% drug concentration?
   
   In a total of 4.0 parts of 12.5% product, 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.
   0.5 parts correspond to 0.5 × 250 g or 125 g.
2. How many mL of 20% dextrose in water and 50% dextrose in water are needed to make 750 mL of 35% dextrose in water?
   
   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 product.

1. How much ointment having a 12% drug concentration and how much ointment having a 16% drug concentration must be used to make 1 kg of a preparation containing a 12.5% drug concentration?
   Let 1 kg be the total quantity of ointment to be prepared, let x be the quantity, in kg, of the 12% ointment, and let (1 – x) be the quantity in kg of the 16% ointment. The equation is as follows:
   (12/100) x + (16/100)(1 – x) = (12.5/100)(1)
   Solving the equation, x equals 0.875 kg of the 12% ointment and (1 – x) equals (1 – 0.875) or 0.125 kg of the 16% ointment.
2. How many mL of 20% dextrose in water and 50% dextrose in water are needed to make 750 mL of 35% dextrose in water?
   Let x be the volume, in mL, of the 20% solution, and let (750 – x) be the volume in mL of the 50% solution. The equation is as follows:
   (20/100)x + (50/100)(750 – x) = (35/100)(750)
   Solving the equation, x equals 375 mL of the 20% solution and (750 – x) equals (750 – 375) or 375 mL of the 50% solution.

See Concentrations in the General Notices.

1. How much sodium bicarbonate powder is needed to prepare 50.0 mL of a 0.07 N solution of sodium bicarbonate (NaHCO3)? (MW of NaHCO3 is 84.0 g per mol.)
   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 proton, one Eq of NaHCO3 is contained in each mole of NaHCO3. Thus the equivalent weight of NaHCO3 is 84 g. [note—The volume, in L, × normality of a solution equals the number of equivalents in the solution.]
   The number of equivalents of NaHCO3 required = (0.07 Eq/L)(50.0 mL/1000 mL /L) = 0.0035 equivalents.
   1 equivalent weight is 84.0 g.
   0.0035 equivalents equals 84.0 g/Eq × 0.0035 Eq = 0.294 g.
2. A prescription calls for 250 mL of a 0.1 N hydrochloric acid (HCl) solution. How many mL of concentrated hydrochloric acid are needed to make this solution? [note—The specific gravity of concentrated hydrochloric acid is 1.18, the molecular weight is 36.46 and the concentration is 37.5% (w/w). Because hydrochloric acid functions as an acid and reacts by giving up one proton in a chemical reaction, 1 Eq is contained in each mole of the compound. Thus the equivalent weight is 36.46 g.]
   The number of equivalents of HCl required is 0.250 L × 0.1 N = 0.025 equivalents.
   1 equivalent is 36.46 g.
   0.025 equivalents correspond to 0.025 Eq × 36.46 g/Eq = 0.9115 g.
   37.5 g of pure HCl are contained in 100 g of concentrated HCl.
   Thus 1 g of pure HCl is contained in (100/37.5) g = 2.666 g of concentrated acid, and 0.9115 g is contained in (0.9115 × 2.666) g or 2.43 g of concentrated acid.
   In order to determine the volume of the supplied acid required, use the definition for specific gravity as shown below.
   Specific gravity = (weight of the substance)/(weight of an equal volume of water).
   1.18 = 2.43 g/(weight of an equal volume of water).
   The weight of an equal volume of water is 2.056 g or 2.06 g, which measures 2.06 mL. Thus, 2.06 mL of concentrated acid is required.

The quantities of electrolytes administered to patients are usually expressed in terms of mEq. This term must not be confused with a similar term used in quantitative chemical analysis as discussed above. 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 is the weight of a substance (equivalent weight) that supplies one 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 milliequivalent is one-thousandth of an equivalent (Eq). Because the ionization of phosphate depends on several factors, the concentration is usually expressed in millimoles, moles, or milliosmoles, which are described below. [note—Equivalent weight (Eq.wt) = wt. of an atom or radical (ion) in g/valence (or charge) of the atom or radical. Milliequivalent weight (mEq.wt) = Eq.wt. (g)/1000.]

examples—

1. Potassium (K) has a gram-atomic weight of 39.10. Calculate its weight in millimoles (mmol).
   The weight of one mole is 39.10 g and the weight in millimoles is 39.10 g/1000 = 0.0391 g or 39.1 mg.
2. How many millimoles of Penicillin V are in a tablet that contains 250 mg of Penicillin V Potassium? (Molecular weight of penicillin V potassium is 388.48 g per mol.)
   The weight of one mole is 388.48 and the weight in millimoles is 388.48/1000 = 0.3848 g or 388.48 mg. Thus there are 250 mg/388.48 mg/mmol = 0.644 mmol of Penicillin V ion per tablet.

The following discussion and calculations have therapeutic implications in preparations of dosage forms intended for ophthalmic, subcutaneous, intravenous, intrathecal, and neonatal use.

Cells of the body, such as erythrocytes, will neither swell nor shrink when placed in a solution that is isotonic with the body fluids. However, the measurement of tonicity, a physiological property, is somewhat difficult. It is found that a 0.9% (w/v) solution of sodium chloride, which has a freezing point of – 0.52, is isotonic with body fluids and is said to be isoosmotic with body fluids. In contrast to isotonicity, the freezing point depression is a physical property. Thus many solutions that are isoosmotic with body fluids are not necessarily isotonic with body fluids, e.g., a solution of urea. Nevertheless many pharmaceutical products are prepared using freezing point data or related sodium chloride data to prepare solutions that are isoosmotic with the body fluids. A closely related topic is osmolarity (see Osmolality and Osmolarity 785).

Freezing point data or sodium chloride equivalents of pharmaceuticals and excipients (see Table 1 below) may be used to prepare isoosmotic solutions, as shown in the examples below.

Table 1. Sodium Chloride Equivalents (E) and Freezing Point (FP) Depressions for a 1% Solution of the Drug or Excipient

example—

Determine the amount of sodium chloride required to prepare 60 mL of an isoosmotic solution of atropine sulfate 0.5% using the sodium chloride equivalent values and also the freezing point depression values.

Some calculations concerning flow rates in intravenous sets are provided below. [note—Examples below are not to be used for treatment purposes.]

examples—

The relationship between Celsius degrees (C) and Fahrenheit degrees (F) is expressed by the following equation:

in which C and F are the numbers of Celsius degrees and Fahrenheit degrees, respectively.

examples—

The relationship between the Kelvin and the Celsius scales is expressed by the equation:

in which K and C are the numbers of Kelvin degrees and Celsius degrees, respectively.

See Pharmaceutical Stability 1150 for the definition of mean kinetic temperature (MKT). MKT is usually higher than the arithmetic mean temperature and is derived from the Arrhenius equation. MKT addresses temperature fluctuations during the storage period of the product. The mean kinetic temperature, TK, is calculated by the following equation:

in which DH is the heat of activation, which equals 83.144 kJ per mol (unless more accurate information is available from experimental studies); R is the universal gas constant, which equals 8.3144 × 10–3 kJ per degree per mol; T1 is the average temperature, in degrees Kelvin, during the first time period, e.g., the first week; T2 is the average temperature, in degrees Kelvin, during the second time period, e.g., second week; and Tn is the average temperature, in degrees Kelvin during the nth time period, e.g., nth week, n being the total number of temperatures recorded. The mean kinetic temperature is calculated from average storage temperatures recorded over a one-year period, with a minimum of twelve equally spaced average storage temperature observations being recorded (see Pharmaceutical Stability 1150). This calculation can be performed manually with a pocket calculator or electronically with computer software.

examples—