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:

1. 29.8 mL = 29.8 ± 0.05 mL (accurate to the nearest 0.1 mL)
2. 29.80 mL = 29.80 ± 0.005 mL (accurate to the nearest 0.01 mL)
3. 29.800 mL = 29.800 ± 0.0005 mL (accurate to the nearest 0.001 mL)

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

EXAMPLE—

(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}.

EXAMPLE—

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.

1. Calculate the percentage concentrations (w/w) of the constituents of the solution prepared by dissolving 2.50 g of phenol in 10.00 g of glycerin. Using the weight percent equation above, the calculation is as follows.
   Total weight of the solution = 10.00 g + 2.50 g = 12.50 g.
   Weight percent of phenol = (2.50 g × 100%)/12.50 g = 20.0% of phenol.
   Weight percent of glycerin = (10 g × 100%)/12.50 g = 80.0% of glycerin.
2. A prescription order reads as follows:
   Eucalyptus Oil 3% (v/v) in Mineral Oil.
   Dispense 30.0 mL.
   What quantities should be used for this prescription? Using the volume percent equation above, the calculation is as follows.
   Amount of Eucalyptus Oil:
   3% = (Volume of oil in mL/30.0 mL) × 100%
   Solving the equation, the volume of oil = 0.90 mL
   Amount of Mineral Oil:
   To 0.90 mL of Eucalyptus Oil add sufficient Mineral Oil to prepare 30.0 mL.
3. A prescription order reads as follows:
   Zinc oxide	7.5 g
   Calamine	7.5 g
   Starch	15 g
   White petrolatum	30 g

   Calculate the percentage concentration for each of the four components. Using the weight percent equation above, the calculation is as follows.
   Total weight = 7.5 g + 7.5 g + 15 g + 30 g = 60.0 g.
   Weight percent of zinc oxide = (7.5 g zinc oxide/60 g ointment) × 100% = 12.5%.
   Weight percent of calamine = (7.5 g calamine/60 g ointment) × 100% = 12.5%.
   Weight percent of starch = (15 g starch/60 g ointment) × 100% = 25%.
   Weight percent of white petrolatum = (30 g white petrolatum/60 g ointment) × 100% = 50%.

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.

1. A liquid weighs 125 g and has a volume of 110 mL. What is the specific gravity?
   The weight of an equal volume of water is 110 g.
   Using the above equation, specific gravity = 125 g/110 g = 1.14.
2. Hydrochloric Acid NF is approximately 37% (w/w) solution of hydrochloric acid (HCl) in water. How many grams of HCl are contained in 75.0 mL of HCl NF? (Specific gravity of Hydrochloric Acid NF is 1.18.)
   Calculate the weight of HCl NF using the above equation.
   The weight of an equal volume of water is 75 g.
   Specific Gravity 1.18 = weight of the HCl NF g /75.0 g.
   Solving the equation, the weight of HCl NF is 88.5 g.
   Now calculate the weight of HCl using the weight percent equation.
   37.0 % (w/w) = (weight of solute g/88.5 g ) × 100.
   Solving the equation, the weight of the HCl is 32.7 g.

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.

1. Calculate the amount (Q2), in g, of diluent that must be added to 60 g of a 10% (w/w) ointment to make a 5% (w/w) ointment.
   Let (Q1) = 60 g, (C1) = 10%, and (C2) = 5%.
   Using the above equation,
   60 g × 10% = (Q2) × 5% (w/w)
   Solving the above equation, the amount of product needed, Q2, is 120 g. The initial amount of product added was 60 g, and therefore an additional 60 g of diluent must be added to the initial amount to give a total of 120 g.
2. How much diluent 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 final mixture?
   Determine the final concentration by first converting mg to g. One mg of drug in 10 g of mixture is the same as 0.001g in 10 g.
   Let (Q1) = 10 g, (C1) = (1 in 100), and (C2) = (0.001 in 10).
   Using the equation for dilution, 10 g × (1/100) = (Q2) g × (0.001/10).
   Solving the above equation, (Q2) = 1000 g.
   Because 10 g of the final mixture contains all of the drug and some diluent, (1000 g – 10 g) or 990 g of diluent is required to prepare the mixture at a concentration of 0.001 g of drug in 10 g of final mixture.
3. Calculate the percentage strength of a solution obtained by diluting 400 mL of a 5.0% solution to 800 mL.
   Let (Q1) = 400 mL, (C1) = 5%, and (Q2) = 800 mL.
   Using the equation for dilution, 400 mL × 5% = 800 mL × (C2)%.
   Solving the above equation, (C2) = 2.5% (w/v).

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.

1. One mg of Pancreatin contains not less than 25 USP Units of amylase activity, 2.0 USP Units of lipase activity, and 25 USP Units of protease activity. If the patient takes 0.1 g (100 mg) per day, what is the daily amylase activity ingested?
   1 mg of Pancreatin corresponds to 25 USP Units of amylase activity.
   100 mg of Pancreatin corresponds to 100 × (25 USP Units of amylase activity) = 2500 Units.
2. A dose of penicillin G benzathine for streptococcal infection is 1.2 million units intramuscularly. If a specific product contains 1180 units per mg, how many milligrams would be in the dose?
   1180 units of penicillin G benzathine are contained in 1 mg.
   1 unit is contained in 1/1180 mg.
   1,200,000 units are contained in (1,200,000 × 1)/1180 units = 1017 mg.

Frequently the base form of a drug is administered in an altered form such as an ester or salt for stability or other reasons such as taste or solubility. 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.

1. Four hundred milligrams of erythromycin ethylsuccinate (molecular weight, 862.1) is administered. Determine the amount of erythromycin (molecular weight, 733.9) in this dose.
   862.1 g of erythromycin ethylsuccinate corresponds to 733.9 g of erythromycin.
   1 g of erythromycin ethylsuccinate corresponds to (733.9/862.1) g of erythromycin.
   0.400 g erythromycin ethylsuccinate corresponds to (733.9/862.1) × 0.400 g or 0.3405 g of erythromycin.
2. The molecular weight of testosterone cypionate is 412.6 and that of testosterone is 288.4. What is the dose of testosterone cypionate that would be equivalent to 60.0 mg of testosterone?
   288.4 g of testosterone corresponds to 412.6 g of testosterone cypionate.
   1 g of testosterone corresponds to 412.6/288.4 g of testosterone cypionate.
   60.0 mg or 0.0600 g of testosterone corresponds to (412.6/288.4) × 0.0600 = 0.0858 g or 85.8 mg of testosterone cypionate.

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.

1. If the volume of 250 mg of ceftriaxone sodium is 0.1 mL, how much diluent should be added to 500 mg of ceftriaxone sodium powder to make a suspension having a concentration of 250 mg per mL?
   
2. Volume of 500 mg of ceftriaxone sodium =
   
3. Volume of the diluent required = (2 mL of suspension) (0.2 mL of Ceftriaxone Sodium) = 1.8 mL.
4. What is the volume of dry powder cefonicid, if 2.50 mL of diluent is added to 1 g of powder to make a solution having a concentration of 325 mg per mL?
   Volume of solution containing 1 g of the powder =

   
   Volume of dry powder cefonicid = 3.08 mL of solution 2.50 mL of diluent = 0.58 mL.

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.

EXAMPLES—

EXAMPLES—

See Concentrations in the General Notices.

EXAMPLES—

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.]

1. Potassium (K+) has a gram-atomic weight of 39.10. The valence of K+ is 1+. Calculate its milliequivalent weight (mEq wt).
   Eq wt = 39.10 g/1 = 39.10 g
   mEq wt = 39.10 g/1000 = 0.03910 g = 39.10 mg
2. Calcium (Ca2+) has a gram-atomic weight of 40.08. Calculate its milliequivalent weight (mEq wt).
   Eq wt = 40.08 g/2 = 20.04 g
   mEq wt. = 20.04 g/1000 = 0.02004 g = 20.04 mg
   NOTE—The equivalent weight of a compound may be determined by dividing the molecular weight in g by the product of the valence of either relevant ion and the number of times this ion occurs in one molecule of the compound.
3. How many milliequivalents of potassium ion (K+) are there in a 250-mg Penicillin V Potassium Tablet? [NOTE—Molecular weight of penicillin V potassium is 388.48 g per mol; there is one potassium atom in the molecule; and the valence of K+ is 1.]
   Eq wt = 388.48 g/[1(valence) × 1(number of charges)] = 388.48 g.
   mEq wt = 388.48 g/1000 = 0.38848 g = 388.48 mg.
   (250 mg per Tablet)/(388.48 mg per mEq) = 0.644 mEq of K+ per Tablet.
4. How many equivalents of magnesium ion and sulfate ion are contained in 2 mL of a 50% Magnesium Sulfate Injection? (Molecular weight of MgSO4·7H2O is 246.48 g per mol.)
   Amount of magnesium sulfate in 2 mL of 50% Magnesium Sulfate Injection
   
   Eq wt of MgSO4.7H2O = MW (g)/(valence of specified ion × number of specified ions in one mole of salt).
   For the magnesium ion:
   The number of equivalents is calculated as follows:
   246.48/[2(valence) × 1 (number of ions in the compound)] = 123.24 g/Eq of magnesium ion.
   The number of equivalents in 1 g is 1g/ 123.24 g/Eq = 0.008114 Eq.
   The number of mEq may be calculated as follows.
   The mEq wt = Eq wt (g)/1000 = (123.24 g/Eq)/1000 = 0.12324 g.
   The number of milliequivalents of magnesium ion in 1 g is 1g/0.12324 g/mEq = 8.114 mEq.
   For the sulfate ion:
   The number of equivalents is calculated as follows:
   246.48/[2(valence) × 1 (number of ions in the compound)] = 123.24 g/Eq of sulfate ion.
   The number of equivalents in 1 g is 1g/123.24 g/Eq = 0.008114 Eq.
   The number of mEq may be calculated as follows.
   The mEq wt = Eq wt (g)/1000 = (123.24 g/Eq)/1000 = 0.12324 g.
   The number of milliequivalents of sulfate ion in 1 g is 1g/0.12324 g/mEq = 8.114 mEq.
5. A vial of Sodium Chloride Injection contains 3 mEq of sodium chloride per mL. What is the percentage strength of this solution? (Molecular weight of sodium chloride is 58.44 g per mol.)
   1 mEq = 1 Eq/1000 = 58.44 g/1000 = 0.05844 g = 58.44 mg.
   Amount of sodium chloride in 3 mEq per mL = 58.44 mg per mEq × 3 mEq per mL = 175.32 mg per mL.
   

EXAMPLES—

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 which 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 Excipent

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

1. Sodium Heparin 8,000 units in 250 mL Sodium Chloride Injection 0.9% solution are to be infused over 4 hours. The administration set delivers 20 drops per mL.
   What is the flow rate in mL per hour?
   In 4 hours, 250 mL are to be delivered.
   In 1 hour, 250 mL/4 = 62.5 mL are delivered.
   What is the flow rate in drops per minute?
   In 60 minutes, 62.5 mL are delivered.
   In 1 minute, 62.5 mL/60 = 1.04 mL are delivered.
   1 mL = 20 drops.
   1.04 mL = 1.04 × 20 drops = 20.8 drops.
   Thus in 1 minute, 20.8 or 21 drops are administered.
2. A 14.5 kg patient is to receive 50 mg of Sodium Nitroprusside in 250 mL of dextrose 5% in water (D5W) at the rate of 1.3 µg per kg per minute. The set delivers 50 drops per mL.
   Calculate the flow rate in mL per hour.
   The dose for 1 kg is 1.3 µg per minute.
   The 14.5 kg patient should receive 14.5 × 1.3 µg = 18.85 µg per minute.
   50 mg or 50000 µg of drug are contained in 250 mL of D5W.
   18.85 µg are contained in 250 mL × 18.85/50000 = 0.09425 mL D5W, which is administered every minute.
   In 1 minute, 0.09425 mL are administered.
   In 1 hour or 60 minutes, 60 × 0.09425 mL = 5.655 or 5.7 mL are administered.
   Calculate the flow rate in drops per minute.
   1 mL corresponds to 50 drops per minute.
   0.09425 mL corresponds to 0.09425 × 50 = 4.712 or 4.7 drops per minute.

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.

1. Convert 77 F to Celsius degrees.
   
   9(C) = 5(F) – 160
   C = [5(F) – 160]/9 = [(5 × 77) – 160]/9 = 25 C
2. Convert 30 C to Fahrenheit degrees.
   
   9(C) = 5(F) – 160
   F = [9(C) + 160]/5 = [(9 × 30) + 160]/5 = 86 F

K = C + 273.1,

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.

1. The means of the highest and lowest temperatures for 52 weeks are 25 C each. Calculate the MKT.
   n = 52
   DH/R = 10,000 K
   T1, T2, ...,Tn = 25 C = 273.1 + 25 = 298.1 K
   R = 0.0083144 kJ K–1mol–1
   DH = 83.144 kJ per mol
   
   
   
   
   The calculated MKT is 25.0 C. Therefore the controlled room temperature requirement is met by this pharmacy. [NOTE—If the averages of the highest and lowest weekly temperatures differed from each other and were in the allowed range of 15 C to 30 C (see Controlled Room Temperature under Preservation, Packaging, Storage, and Labeling in the General Notices), then each average would be substituted individually into the equation. The remaining two examples illustrate such calculations, except that the monthly averages are used]
2. A pharmacy recorded a yearly MKT on a monthly basis, starting in January and ending in December. Each month, the pharmacy recorded the monthly highest temperature and the monthly lowest temperature, and the average of the two was calculated and recorded for the MKT calculation at the end of the year.
   Table 2. Data for Calculation of MKT

   n	Month	Lowest Temperature (in C)	Highest Temperature (in C)	Average Temperature (in C)	Average Temperature (in K)	DH/RT	e–DH/RT
   1	Jan.	15	27	21	294.1	34.002	1.710 × 10–15
   2	Feb.	20	25	22.5	295.6	33.830	2.033 × 10–15
   3	Mar.	17	25	21	294.1	34.002	1.710 × 10–15
   4	Apr.	20	25	22.5	295.6	33.830	2.033 × 10–15
   5	May	22	27	24.5	297.6	33.602	2.551 × 10–15
   6	June	15	25	20	293.1	34.118	1.523 × 10–15
   7	July	20	26	23	296.1	33.772	2.152 × 10–15
   8	Aug.	22	26	24	297.1	33.659	2.411 × 10–15
   9	Sept.	23	27	25	298.1	33.546	2.699 × 10–15
   10	Oct.	20	28	24	297.1	33.659	2.411 × 10–15
   11	Nov.	20	24	22	295.1	33.887	1.919 × 10–15
   12	Dec.	22	28	25	298.1	33.546	2.699 × 10–15

   From these data the MKT may be estimated or it may be calculated. If more than half of the observed temperatures are lower than 25 C and a mean lower than 23 C is obtained, the MKT may be estimated without performing the actual calculation.
   To estimate the MKT, the recorded temperatures are evaluated and the average is calculated. In this case, the calculated arithmetic mean is 22.9 C. Therefore, the above requirements are met and it can be concluded that the mean kinetic temperature is lower than 25 C. Therefore, the controlled room temperature requirement is met.
   The second approach is to perform the actual calculation.
   n = 12
   
   
   
   
   The calculated MKT is 23.0 C, so the controlled room temperature requirement is met. [NOTE—These data and calculations are used only as an example.]
3. An article was stored for one year in a pharmacy where the observed monthly average of the highest and lowest temperatures was 25 C (298.1 K), except for one month with an average of 28 C (301.1 K). Calculate the MKT of the pharmacy.
   
   

   n = 12
   
   
   
   
   
   
   
   The controlled room temperature requirement is not met because the calculated MKT exceeds 25 C. (See Note in Example 2 above.)