OUR LADY OF FATIMA UNIVERSITY
College of Pharmacy
PHARMACEUTICAL
SEMINAR 1 (PSMA411)
Week 16-17 – Pharmaceutical Calculations and Techniques
Student Checklist:
 Read course and unit outline
 Read required learning materials
 Proactively participate in online class
 Participate in the Discussion Board
(Canvas)
 Answer and Submit course unit
tasks
UNIT OUTCOMES
At the end of the course, the students are expected to:
1) Demonstrate knowledge, understanding and skills to perform
basic mathematical operations and solve various problems
2) Perform laboratory techniques pertinent to the practice of
pharmacy with accuracy and precision;
3) Work independently and diligently manifesting the values of
analytical thinking, scientific discipline, honesty and patience.
UNIT OUTLINE
I.
International System of Units
II. Pharmaceutical Measurement
III. Reducing and Enlarging of Formulas
IV. Density, Specific Gravity
V. Calculation of Doses: Patient Parameters
VI. Percentage, Ratio Strength and other Expressions of
Concentrations
UNIT OUTLINE
VII. Dilution, Concentration and Alligation
VIII. Electrolyte solution: Milliequivalents, Millimoles, and
Milliosmoles
IX. Isotonic Solutions
X. HLB System
XI. Pharmacoeconomic Calculations
• International System of Units
• Also known as the Metric System
• is the internationally recognized decimal
system of weights and measures.
UNITS OF MEASUREMENT USED IN
PHARMACY
Metric System
• Is based on decimal system in which everything is measured in
multiples or fraction of 10.
• Major system of weights and measurement used in medicine.
• Uses standard measures:
• Meter (m) for length or distance
• Gram (g) for mass or weight
• Liter (L) for volume
MEASURE OF LENGTH
• meter (m) – primary unit of length
• Table of metric length:
1 kilometer (km) = 1,000,000 meters
1 hectometer (hm) = 100,000 meters
1 decameter (dam) = 10,000 meters
1 meter (m)
1 decimeter (dm) = 0.100 meter
1 centimeter (cm) = 0.010 meter
1 millimeter (mm) = 0.001 meter
1 micrometer (mm) = 0.000,001 meter
1 nanometer (nm) = 0.000,000,001 meter
1 meter = 0.001 kilometer
0.01 hectometer
0.1 decameter
10 decimeters
100 centimeters
1000 millimeters
1,000,000 micrometers
1,000,000,000 nanometers
MEASURE OF VOLUME
• Liter (L) – primary unit of volume
• Table of metric volume
1 kiloliter (kL) = 1000,000 liters
1 hectoliter (hL) = 100,000 liters
1 decaliter (daL) = 10,000 liters
1 liter (L)
1 deciliter (dL) = 0.100 liter
1 centiliter (cL) = 0.010 liter
1 milliliter (mL) = 0.001 liter
1 microliter (mL) = 0.000,001 liter
1 liter = 0.001 kiloliter
0.010 hectoliter
0.100 decaliter
10 deciliters
100 centiliters
1000 milliliters
1,000,000 microliters
MEASURE OF WEIGHT
• Gram (g) – primary unit of weight.
• Table of metric weight
1 kilogram (kg) = 1,000,000 grams
1 hectogram (hg) = 100,000 grams
1 dekagram (dag) = 10,000 grams
1 gram (g)
1 decigram (dg) = 0.100 gram
1 centigram (cg) = 0.010 gram
1 milligram (mg) = 0.001 gram
1 microgram (mg or mcg) = 0.000,001 gram
1 nanogram (ng) = 0.000,000,001 gram
1 picogram (pg) = 0.000,000,000,001 gram
1 femtogram (fg) = 0.000,000,000,000,001 gram
1 gram =
0.001 kilogram
0.010 hectogram
0.100 decagram
10 decigrams
100 centigrams
1000 milligrams
1,000,000 micrograms
1,000,000,000 nanograms
1,000,000,000,000 picograms
1,000,000,000,000,000 femtograms
Equivalents of length
1 inch = 2.54 cm
1 meter (m) = 39.37 in
Equivalents of volume
1 fluid ounce (fl. oz.) = 29.57 mL
Some useful
equivalents
1 pint (16 fl. oz.) = 473 mL
1 quart (32 fl. oz.) = 946 mL
1 gallon, US (128 fl. oz.) = 3785 mL
1 gallon, UK = 4545 mL
Equivalents of weight
1 pound (lb, avoirdupois) = 454 g
1 ounce (oz, avoirdupois) = 28.35 g
1 kilogram (kg) = 2.2 lb
POP QUIZ:
1. A low-strength aspirin tablet contains 81 mg of aspirin per
tablet. How many tablets may a manufacturer prepare from
0.5 kg of aspirin?
2. An intravenous solution contains 500 mg of a drug
substance in each milliliter. How many milligrams of the
drug would a patient receive from the intravenous infusion
of a liter of the solution?
Answer:
1. A low-strength aspirin tablet contains 81 mg of aspirin per tablet.
How many tablets may a manufacturer prepare from 0.5 kg of
aspirin?
81𝑚𝑔
1 𝑡𝑎𝑏𝑙𝑒𝑡
0.5 𝐾𝑔 𝑥
81𝑚𝑔
1 𝑡𝑎𝑏𝑙𝑒𝑡
=
0.5𝐾𝑔
𝑋
1000 𝑔 1000 𝑚𝑔
𝑥
1𝐾𝑔
1𝑔
=
= 500,000 𝑚𝑔
500,000 𝑚𝑔
𝑋
(81 𝑚𝑔)(𝑋) = (500,000 𝑚𝑔)(1 𝑡𝑎𝑏𝑙𝑒𝑡)
81 𝑚𝑔
81 𝑚𝑔
𝑋 = 6,172.83 𝑡𝑎𝑏𝑙𝑒𝑡𝑠
or 6,172 𝑡𝑎𝑏𝑙𝑒𝑡𝑠
Answer:
2. An intravenous solution contains 500 mg of a drug substance in
each milliliter. How many milligrams of the drug would a patient
receive from the intravenous infusion of a liter of the solution?
500𝑚𝑔
1 𝑚𝐿
1𝐿𝑥
=
1000 𝑚𝑙
1𝐿
500𝑚𝑔
1 𝑚𝐿
=
=
(500 𝑚𝑔)(1,000 𝑚𝐿)
=
1 𝑚𝐿
𝑋
1 𝐿𝑖𝑡𝑒𝑟
1,000 𝑚𝐿
𝑋
1,000 𝑚𝐿
(𝑋)(1 𝑚𝐿)
1 𝑚𝐿
𝑋 = 500,000 mg
PHARMACEUTICAL
MEASUREMENT
• employed in community and institutional pharmacies, in
pharmaceutical research, in the development and manufacture
of pharmaceuticals, in chemical and product analysis, and in
quality control.
Weights and measures
The measurement systems in place for pharmacy are:
Metric
Avoirdupois
Apothecary
• International System of Units (SI) – commonly referred to as the
metric system.
• Avoirdupois – the common system of commerce.
• Apothecaries – the traditional system of pharmacy.
• the traditional system of pharmacy
• Developed in England in 18th century
• components of this system are
occasionally found on prescriptions.
Apothecary
system
• First used by apothecaries/ early
pharmacist and moved from Europe
to colonial America.
• Household system evolved from the
apothecary system
• Older medications are still measured
in apothecary units
Apothecary System
Units of weight:
Units of volume:
• Grain
• Minims
• Dram
• Fluidounce
• Ounce
• Pint
• Pound
• Quart
• Gallon
Apothecary system
Measure of weight:
20 grains
= 1 scruple
3 scruple (60 grains)
= 1 drachm or dram
8 drachms (480 grains)
= 1 ounce
12 ounces (5760 grains)
= 1 pound
Apothecary System
Measure of Volume:
60 minims
= 1 fluidrachm
8 fluidrachm (480 minims)
= 1 fluidounce
16 fluidounces
= 1 pint
2 pints (32 fluidounces)
= 1 quart
4 quarts (8 pints)
= 1 gallon (gal)
• Originated in Europe
• Common system of
Commerce
AVOIRDUPOIS
SYSTEM
• Mainly used in measuring bulk
medication encountered in
manufacturing.
• Commonly used to measure
weight
AVOIRDUPOIS SYSTEM
Units of measure:
• Grains
• Ounce
• Pounds
AVOIRDUPOIS SYSTEM
Measure of weight:
437.5 grains
= 1 ounce
16 ounces (7000 grains)
= 1 pound
INTERSYSTEM
CONVERSION
• To convert a given weight or
volume from units of one system
to equivalent units of another
system.
INTERSYSTEM CONVERSION
Measure of volume:
Measure of length:
1 mL
= 1 minims
1m
= 39.37 in
1 minims
= 0.06 mL
1 inch
= 2.54 cm
1 fl dram
= 3.69 mL
1 fl oz
= 29.57 mL
1 pt
= 473 mL
1 gal (US)
= 3785 mL
INTERSYSTEM CONVERSION
Measure of weight:
1g
= 14.432 gr
1 Kg
= 2.2 lb (avoir.)
1 gr
= 0.065 g (65 mg)
1 oz (avoir) = 28.35 g
1 oz (apoth)= 31.1 g
1 lb (avoir)
= 454 g
1 lb (apoth) = 373 g
Other equivalents:
1 oz (avoir)
= 437.5 gr
1 oz (apoth)
= 480 gr
1 gal (US)
= 128 fl oz.
Common household system
• 1 teaspoon = 5 mL
• 1 tablespoon = 15 mL
• 1 teaspoon = 60 drops
• 1 cup = 240 mL
• 2 tablespoon = 1 fluid ounce
• 8 fluid ounces = 1 cup
• 2 cups = 1 pint
• 4 quarts = 1 gallon
Other Household Measurement
POP QUIZ
1. If a child accidentally swallowed 2 fluidounces of FEOSOL
Elixir, containing 2/3 gr of ferrous sulfate per 5 mL, how
many milligrams of ferrous sulfate did the child ingest?
2. A formula for a cough syrup contains 1/8 gr of codeine
phosphate per teaspoonful (5 mL). How many grams of
codeine phosphate should be used in preparing 1 pint of
the cough syrup?
1. If a child accidentally swallowed 2 fluidounces of FEOSOL Elixir,
containing 2/3 gr of ferrous sulfate per 5 mL, how many milligrams of
ferrous sulfate did the child ingest?
2
𝑔𝑟
3
=
5 𝑚𝐿
2𝑓𝑙 𝑜𝑧 𝑥
29.57 𝑚𝐿
1𝑓𝑙 𝑜𝑧
𝑋 𝑚𝑔
2 𝑓𝑙 𝑜𝑧
=
59.14 𝑚𝐿
0.6667 𝑔𝑟
𝑋
=
5 𝑚𝐿
59.14 𝑚𝐿
(0.6667𝑔𝑟)(59.14𝑚𝐿)
= (𝑋)(5𝑚𝐿)
5 𝑚𝐿
5 𝑚𝐿
𝑋 = 7.8853 𝑔𝑟𝑎𝑖𝑛𝑠
7.8853 𝑔𝑟𝑎𝑖𝑛𝑠
65𝑚𝑔
𝑥 1𝑔𝑟𝑎𝑖𝑛
= 512.55 𝑚𝑔
2. A formula for a cough syrup contains 1/8 gr of codeine phosphate
per teaspoonful (5 mL). How many grams of codeine phosphate
should be used in preparing 1 pint of the cough syrup?
1
𝑔𝑟
8
=
5 𝑚𝐿
1𝑝𝑡 𝑥
473 𝑚𝐿
1𝑝𝑡
0.125 𝑔𝑟
5 𝑚𝐿
𝑋𝑔
1 𝑝𝑡
=
473 𝑚𝐿
𝑋
=
473 𝑚𝐿
(0.125𝑔𝑟)(473𝑚𝐿)
5 𝑚𝐿
= (𝑋)(5𝑚𝐿)
5 𝑚𝐿
𝑋 = 11.825 𝑔𝑟𝑎𝑖𝑛𝑠
11.825 𝑔𝑟𝑎𝑖𝑛𝑠
0.065𝑔
𝑥
1𝑔𝑟𝑎𝑖𝑛
= 0.77𝑔
ALIQUOT METHOD OF
WEIGHING AND MEASURING
ALIQUOT METHOD
A method by which small quantities of a
substance may be obtained within the
desired degree of accuracy by weighing a
larger-than-needed portion of the
substance, diluting it with an inert material,
and then weighing a portion (aliquot) of the
mixture calculated to contain the desired
amount of the needed substance.
ALIQUOT PROCEDURE
Preliminary Step.
Calculate the smallest quantity of a substance that can be weighed on the
balance with the desired precision.
Using the equation:
100% 𝑥 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡𝑠 𝑚𝑔
= Smallest Quantity (mg)
𝐴𝑐𝑐𝑒𝑝𝑡𝑎𝑏𝑙𝑒 𝑒𝑟𝑟𝑜𝑟 (%)
ALIQUOT METHOD
• PROCEDURE
1. Select a multiple of the desired quantity that can be weighed/measured
with the required precision.
2. Dilute the multiple quantity with an inert substance/diluent
3. Weigh/Measure the aliquot portion of the dilution that contains the
desired quantity.
EXAMPLE
On a balance with an SR of 6 mg, and with an acceptable error of no
greater than 5%,
100% 𝑥 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡
= Smallest quantity (mg)
𝐴𝑐𝑐𝑒𝑝𝑡𝑎𝑏𝑙𝑒 𝐸𝑟𝑟𝑜𝑟 (%)
100% 𝑥 6𝑚𝑔
= 120 mg
5%
 a quantity of not less than 120 mg must be weighed.
1. Select a multiple of the desired quantity that can be
weighed with the required precision
On a balance with an SR of 6 mg, and with an acceptable error
of no greater than 5%,
If 5mg required amount on a prescription
120 𝑚𝑔
= 24 or at le𝑎𝑠𝑡 25 (𝑡𝑕𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒)
5 𝑚𝑔
Required amount x multiple = Desired amount
5mg x 25 = 125 mg
2. Dilute the multiple quantity with an inert substance.
Smallest Quantity x Multiple selected = Total quantity of the drug-diluent
120𝑚𝑔 𝑥 25 = 3000 𝑚𝑔
Total quantity of drug-diluent - Amount of drug = Amount of diluent
3000 𝑚𝑔 − 125 𝑚𝑔 = 2875 𝑚𝑔
3. Weigh the aliquot portion of the dilution that contains
the desired quantity.
Acceptable error x Desired quantity of drug = Amount of drug
0.04 x 125 mg = 5 mg
Acceptable error x Desired quantity of diluent = Amount of diluent
0.04 x 2875 mg = 115 mg
Amount of Drug + Amount of Diluent = Aliquot Part
5 mg + 115 mg = 120 mg
PERCENTAGE
ERROR
PERCENTAGE OF ERROR
• As the maximum potential error multiplied by 100 and divided by the quantity
desired.
• The difference between an experimental and theoretical value, divided by the
theoretical value, multiplied by 100 to give a percent.
• Always expressed as a positive number.
• Equation is:
𝐸𝑟𝑟𝑜𝑟 𝑥 100
Percentage error =
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑑𝑒𝑠𝑖𝑟𝑒𝑑
Example:
Using a graduated cylinder, a pharmacist measured 30 mL of a liquid. On
subsequent examination, using a narrow-gauge burette, it was
determined that the pharmacist had actually measured 32 mL. What was
the percentage of error in the original measurement?
V𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟 = 32 𝑚𝐿 − 30 𝑚𝐿 = 2 𝑚𝐿
𝐸𝑟𝑟𝑜𝑟 𝑥 100
Percentage error =
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑑𝑒𝑠𝑖𝑟𝑒𝑑
2𝑚𝐿 𝑥 100%
= 6.7%
Percentage error =
30𝑚𝐿
Example:
A pharmacist attempts to weigh 0.375 g of morphine sulfate on a balance of
dubious accuracy. When checked on a highly accurate balance, the weight is
found to be 0.400 g. Calculate the percentage of error in the first weighing.
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟 = 0.400𝑔 − 0.375𝑔 = 0.025 𝑔
𝐸𝑟𝑟𝑜𝑟 𝑥 100
Percentage error =
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑑𝑒𝑠𝑖𝑟𝑒𝑑
0.025𝑔 𝑥 100%
Percentage error =
0.375
= 6.67%
REDUCING AND ENLARGING
OF FORMULAS
• Determine the total weight or volume of
ingredients and convert, if necessary, to
the quantities desired. The quantities in
the original and new formulas will have
the same ratio.
REDUCING AND ENLARGING OF FORMULAS
• Reducing Calculating the amounts to be used in a
pharmaceutical formula to make a smaller amount than the
original formula.
• Enlarging Calculating the proper amounts to be used in a
pharmaceutical formula to make a larger amount than the
original formula.
METHODS TO
REDUCE OR
ENLARGE
FORMULA
Ratio and Proportion
Dimensional analysis
Factor Method
Factor Method
• Is based on the relative quantity of the total formula to be
prepared.
Equation used:
Factor
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑑𝑒𝑠𝑖𝑟𝑒𝑑
=
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑔𝑖𝑣𝑒𝑛
Steps in Factor Method
1. Using the following equation, determine the factor that defines the multiple
or the decimal fraction of the amount of formula to be prepared:
Factor =
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑑𝑒𝑠𝑖𝑟𝑒𝑑
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑔𝑖𝑣𝑒𝑛
2. Multiply the quantity of each ingredient in the formula by the factor to
determine the amount of each ingredient required in the reduced or enlarged
formula.
Example of factor method
• If a formula for 1000 mL contains 6 g of a drug, how many grams
of drug are needed to prepare 60 mL of the formula?
Step 1
Step 2
Factor =
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑑𝑒𝑠𝑖𝑟𝑒𝑑
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑔𝑖𝑣𝑒𝑛
Factor =
60 𝑚𝐿
1000 𝑚𝐿
Factor = 0.06
6 g x 0.06 = 0.36g
POP QUIZ
1. Calculate the quantity of each ingredient required to make 240 mL of
calamine lotion.
Calamine
80 g
Zinc oxide
80 g
Glycerin
20 mL
Bentonite magma
250 mL
Calcium hydroxide,
to make
1000 mL
2. Calculate the quantity of each ingredient required to prepare
a dozen 30-mL containers.
Polyvinyl alcohol
1.4 g
Povidone
0.6 g
Chlorobutanol
0.5 g
Sterile sodium chloride,
solution 0.9%, ad
100 mL
• Using the following methods:
• Factor method
• Ratio and proportion
• Dimensional analysis
Factor Method
Factor =
1. Calculate the quantity of each ingredient
required to make 240 mL of calamine
lotion.
Calamine
80 g
Zinc oxide
80 g
Glycerin
20 mL
Bentonite magma
Factor =
240 𝑚𝐿
1000 𝑚𝐿
Factor = 0.24
Calamine
80 g x 0.24 = 19.20 g
Zinc oxide
80 g x 0.24 = 19.20 g
Glycerin
20 mL x 0.24 = 4.80 mL
250 mL
Calcium hydroxide,
to make
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑑𝑒𝑠𝑖𝑟𝑒𝑑
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑔𝑖𝑣𝑒𝑛
1000 mL
Bentonite magma
250 mL x 0.24 = 60 mL
Calcium hydroxide,
to make
1000 mL x 0.24 = 240 mL
Ratio and proportion
Calamine
Calculate the quantity of each ingredient
required to make 240 mL of calamine lotion.
Calamine
80 g
Glycerin
20 g
Bentonite magma 250 mL
Zinc oxide
x = 19.20 g
80 𝑔
1000 𝑚𝐿
=
(80 g) (240 mL) =
(1000 mL)
Calcium hydroxide,
to make
=
1000 mL
𝑥
240 𝑚𝐿
(80 g) (240 mL) = (x) (1000 mL)
(1000 mL)
(1000 mL)
80 g
Zinc oxide
80 𝑔
1000 𝑚𝐿
𝑥
240 𝑚𝐿
(x) (1000 mL)
(1000 mL)
x = 19.20 g
Glycerin
20 𝑚𝐿
𝑥
=
1000 𝑚𝐿
240 𝑚𝐿
(x) (1000 mL)
(20 mL) (240 mL)
=
(1000 mL)
(1000 mL)
x = 4.80 mL
Bentonite magma
Ratio and proportion
Calculate the quantity of each ingredient
required to make 240 mL of calamine lotion.
Calamine
80 g
Zinc oxide
80 g
Glycerin
20 mL
Bentonite magma 250 mL
x = 60 mL
Calcium hydroxide
1000 𝑚𝐿
1000 𝑚𝐿
=
𝑥
240 𝑚𝐿
(x) (1000 mL)
(1000ml) (240 mL)
=
(1000 mL)
(1000 mL)
Calcium hydroxide,
to make
𝑥
250 𝑚𝐿
1000 𝑚𝐿 = 240 𝑚𝐿
(250 mL) (240 mL) = (x) (1000 mL)
(1000 mL)
(1000 mL)
1000 mL
x = 240 mL
Dimensional analysis
Calculate the quantity of each ingredient
required to make 240 mL of calamine lotion.
Calamine
80 g
Zinc oxide
80 g
Calamine
80 𝑔
𝑥 240 𝑚𝐿 = 19.20 𝑔
1000 𝑚𝐿
Zinc oxide
80 𝑔
𝑥 240 𝑚𝐿
1000 𝑚𝐿
= 19.20 𝑔
Glycerin
Glycerin
20 mL
Bentonite magma 250 mL
Calcium hydroxide,
to make
20 𝑚𝐿
𝑥 240 𝑚𝐿 = 4.80 mL
1000 𝑚𝐿
Bentonite magma
1000 mL
250 𝑚𝐿
𝑥 240 𝑚𝐿 = 60 mL
1000 𝑚𝐿
Calcium hydroxide
1000 𝑚𝐿
𝑥 240 𝑚𝐿 = 240 mL
1000 𝑚𝐿
2. Calculate the quantity of each ingredient
required to prepare a dozen
30-mL
containers.
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 = 12 X 30 mL = 360 mL
1.4 g
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑔𝑖𝑣𝑒𝑛 = 100 mL
Povidone
0.6 g
Factor = 100 𝑚𝐿
Chlorobutanol
0.5 g
Factor = 3.6
Polyvinyl alcohol
Sterile sodium chloride,
solution 0.9%, ad
Factor =
100 mL
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑑𝑒𝑠𝑖𝑟𝑒𝑑
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑔𝑖𝑣𝑒𝑛
360 𝑚𝐿
• Equation:
• 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
Density
𝑀𝑎𝑠𝑠
𝑉𝑜𝑙𝑢𝑚𝑒
• Mass = grams (g)
• Volume = cubic centimeter (cc) or
millimeter (mL)
Example:
• If 10 mL of sulfuric acid weighs 18 grams, what is its density?
𝑀𝑎𝑠𝑠
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑉𝑜𝑙𝑢𝑚𝑒
18 𝑔
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
10 𝑚𝐿
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 = 1.8 g/mL
Example:
• If 250 mL of alcohol weighs 203 g, what is its density?
𝑀𝑎𝑠𝑠
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑉𝑜𝑙𝑢𝑚𝑒
203 𝑔
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
250 𝑚𝐿
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 = 0.812 g/mL
Specific
Gravity
• - a ratio of the weight of a substance to the
weight of an equal volume of a substance
chosen as a standard, (both substances having
the same temperature or the temperature of
each being definitely known.)
• 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 =
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
• *USP standard for specific gravities is 25 º C
except for ALCOHOL 15.6º C
Specific gravity
• Substances that have sp. Gr. less than 1 are lighter
than water
• Substances that have sp. gr. Greater than 1 are
heavier than water
Example: (Using pycnometer)
A 50 mL pycnometer is found to weigh 120 g when empty, 171 g when filled with
water; and 160 g when filled with an unknown liquid. Calculate the specific gravity of
the unknown liquid.
40 𝑔
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 =
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Weight of pycnometer = 120 g
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 =
51 𝑔
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 0.78
Weight of unknown liquid = Weight of pycnometer with unknown liquid – weight of pycnometer
Weight of unknown liquid = 160 g – 120 g
Weight of unknown liquid = 40 g
Weight of water = Weight of pycnometer with water – weight of pycnometer
Weight of water = 171 g – 120 g
Weight of water = 51 g
Example: (Using pycnometer)
A specific gravity bottle weighs 23.66 g. When filled with water, it weighs 72.95 g; when
filled with another liquid, it weighs 73.56 g. What is the specific gravity of the liquid?
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 =
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Weight of pycnometer = 23.66 g
49.90 𝑔
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 =
49.29 𝑔
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 1.01
Weight of liquid = Weight of pycnometer with liquid – weight of pycnometer
Weight of unknown liquid = 73.56 g – 23.66 g
Weight of unknown liquid = 49.90 g
Weight of water = Weight of pycnometer with water – weight of pycnometer
Weight of water = 72.95 g – 23.66 g
Weight of water = 49.29 g
Density is a concrete number (1.8 g/mL)
Difference
between
Density and
Specific
Gravity
Density of water can be expressed as
1g/mL, 1000 g/L, 455 gr/floz, 62 ½ lb /cu ft
Specific gravity is an abstract number
(dimensionless)
Specific gravity of water is always 1.
Use of
Specific
Gravity in
Calculations
of Weight and
Volume
• Specific gravity a factor that expresses
how much heavier or lighter a substance
is than water, the standard with a
specific gravity of 1.0.
• For example,
• A liquid with a specific gravity o 1.25 is
1.25 times as heavy as water, and a
liquid with a specif c gravity of 0.85 is
0.85 times as heavy as water.
Calculating Weight, Knowing the Volume
and Specific Gravity
• Equation:
• Weight of substance = volume of substance x Specific gravity
• Grams = Milliliters × Specific gravity
• Grams other liquid = Grams (of equal volume of water) x Specific gravity (of other
liquid)
Example:
• What is the weight, in grams, of 3620 mL of alcohol with a specific
gravity of 0.82?
Weight of substance = volume of substance x Specific gravity
Volume of substance = 3620 mL alcohol  volume of water = 3620 mL = 3620 g
Specific gravity = 0.82
Weight of substance = 3620 g x
0.82
Weight of substance = 2968.40 g
Example:
• What is the weight, in grams, of 2 fl. oz. of a liquid having a specific
gravity of 1.118?
2 fl.oz. x
29.57 𝑚𝐿
1 𝑓𝑙.𝑜𝑧.
= 59.14 mL
Weight of substance = volume of substance x Specific gravity
Volume of substance = 59.14 mL liquid  volume of water = 59.14 mL = 59.14 g
Specific gravity = 1.118
Weight of substance = 59.14 g x 1.118
Weight of substance = 2968.40 g
Calculating Volume, Knowing the Weight
and Specific Gravity
• Equation:
Milliliters =
𝐺𝑟𝑎𝑚𝑠
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
Example:
• What is the volume, in milliliters, of 492 g of a liquid with a specific
gravity of 1.40?
Volume of substance =
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 (𝑔)
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
492 𝑔
Volume of substance =
1.40
Volume of substance= 351.43 g
Volume of substance = 351.43 mL
Volume of substance =
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 (𝑔)
𝑔
)
𝑚𝐿
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 (
Volume of substance = 𝑚𝐿
Example:
• What is the volume, in milliliters, of 1 lb. of a liquid with a specific
gravity of 1.185? 1 lb. = 454 g
Volume of substance =
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 (𝑔)
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
454𝑔
Volume of substance =
1.185
Volume of substance= 383.12 g
Volume of substance = 383.12 mL
Volume of substance =
𝑊𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 (𝑔)
𝑔
)
𝑚𝐿
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 (
Volume of substance = 𝑚𝐿
is the branch of medicine that deals with disease in
children rom birth through adolescence.
Pediatric
Patients
Groups:
Neonate/newborn
= Birth to 1 month
Infant
= 1 month to 1 year
Early childhood
= 1 year to 5 years
Late childhood
= 6 years to 12 years
Adolescence
= 13 years to 17 years
• Doses should be based on accepted clinical studies as
reported in the literature.
• Doses should be age appropriate and generally based on
body weight or body surface area.
Special
Considerations
in Dose
Determinations
for Pediatric
Patients
• Pediatric patients should be weighed as closely as
possible to the time of admittance to a health care facility
and that weight recorded in kilograms.
• As available, pediatric formulations rather than those
intended for adults should be administered.
• All calculations of dose should be double-checked by a
second health professional.
• All caregivers should be properly advised with regard to
dosage, dose administration, and important clinical signs
to observe.
• Calibrated oral syringes should be used to measure and
administer oral liquids.
Geriatric Patients
• Also known as elderly.
• Medical care for older adults, an age group that is not easy to
define precisely.
• Therapy is often initiated with a lower-than-usual adult
dose.
Special
Considerations
in Dose
Determinations
for Elderly
Patients
• Dose adjustment may be required based on the
therapeutic response.
• The patient’s physical condition may determine the
drug dose and the route of administration used.
• The dose may be determined, in part, on the patient’s
weight, body sur ace area, health and disease status,
and pharmacokinetic actors.
• Concomitant drug therapy may affect drug/dose
effectiveness.
• A drug’s dose may produce undesired adverse effects
and may affect patient adherence.
• Complex dosage regimens of multiple drug therapy
may affect patient adherence.
Rules for Approximate Doses for
Infants and Children
Based on Age
• Young’s Rule (for children 2 years old and older)
• Cowling’s Rule
• Fried’s Rule
Based on Weight
• Clark’s Rule
Drug Dosage Based on Age
Young’s Rule:
Dose for a Child =
𝐴𝑔𝑒 (𝑦𝑒𝑎𝑟𝑠)
x
𝐴𝑔𝑒 𝑦𝑒𝑎𝑟𝑠 :12
Adult Dose
Example:
Determine the dose for a 2–year old child if the
usual adult dose is 250 mg. (use young’s rule)
Dose for a Child =
𝐴𝑔𝑒 (𝑦𝑒𝑎𝑟𝑠)
x
𝐴𝑔𝑒 𝑦𝑒𝑎𝑟𝑠 :12
Dose for a Child =
2 𝑦𝑒𝑎𝑟𝑠
x
2 𝑦𝑒𝑎𝑟𝑠 :12
Dose for a Child = 35.71 mg
Adult Dose
250 mg
Drug Dosage Based on Age
Cowling’s Rule:
Dose for a Child =
𝐴𝑔𝑒 𝑛𝑒𝑥𝑡 𝑏𝑖𝑟𝑡𝑕𝑑𝑎𝑦 ( 𝑖𝑛 𝑦𝑒𝑎𝑟𝑠)
x
24
Adult Dose
Example:
Determine the dose for a 1-year old child if the
adult dose is 125 mg (use cowling’s rule).
Dose for a Child =
𝐴𝑔𝑒 𝑛𝑒𝑥𝑡 𝑏𝑖𝑟𝑡𝑕𝑑𝑎𝑦 ( 𝑖𝑛 𝑦𝑒𝑎𝑟𝑠)
x
24
Dose for a Child =
2
x
24
125 mg
Dose for a Child = 10.42 mg
Adult Dose
Drug Dosage Based on Age
Fried’s Rule:
Dose for a Child =
𝐴𝑔𝑒 𝑖𝑛 𝑚𝑜𝑛𝑡𝑕𝑠 𝑥 𝐴𝑑𝑢𝑙𝑡 𝐷𝑜𝑠𝑒
150
Example:
Determine the dose for a 3-months old child if the
usual adult dose is 375mg (use fried’s rule).
Dose for a Child =
𝐴𝑔𝑒 𝑖𝑛 𝑚𝑜𝑛𝑡𝑕𝑠 𝑥 𝐴𝑑𝑢𝑙𝑡 𝐷𝑜𝑠𝑒
150
Dose for a Child =
3 𝑚𝑜𝑛𝑡𝑕𝑠 𝑥 375 𝑚𝑔
150
Dose for a Child = 7.5 𝑚𝑔
Drug Dosage Based on Weight
Clark’s Rule:
Dose for a Child =
𝑊𝑒𝑖𝑔𝑕𝑡 (𝑙𝑏) 𝑥 𝐴𝑑𝑢𝑙𝑡 𝐷𝑜𝑠𝑒
150
Example:
Determine the dose for a 2-year old child weighing
40 lb if the usual adult dose is 250 mg (use clark’s
rule).
Dose for a Child =
𝑊𝑒𝑖𝑔𝑕𝑡 𝑙𝑏 𝑥 𝐴𝑑𝑢𝑙𝑡 𝐷𝑜𝑠𝑒
150
Dose for a Child =
40 𝑙𝑏 𝑥 250 𝑚𝑔
150
Dose for a Child = 66.67 𝑚𝑔
Drug Dosage Based on Body Weight
Patient’s Dose (mg) = Patient’s weight (kg) x
𝐷𝑟𝑢𝑔 𝐷𝑜𝑠𝑒 (𝑚𝑔)
1 (𝐾𝑔)
Example:
The dose of gentamicin for premature and full-term neonates is 2.5 mg/kg
administered every 12 hours. What would be the daily dose for a newborn
weighing 5.6 lb?
𝐷𝑟𝑢𝑔 𝐷𝑜𝑠𝑒 (𝑚𝑔)
1 (𝐾𝑔)
Patient’s Dose (mg) = Patient’s weight (kg) x
1 𝑘𝑔
Patient’s Weight (kg) = 5.6 lb x 2.2 𝑙𝑏 = 2.55 kg
Drug Dose =
2.5 𝑚𝑔
1 𝐾𝑔
Patient’s Dose (mg) = 2.55 kg x
2.5𝑚𝑔
1 𝐾𝑔
Patient’s Dose (mg) = 6.38 mg/12 hours x
24 𝑕𝑟𝑠
1 𝑑𝑎𝑦
= 12.76 mg
Based on Body Surface Area
The Square Meter Surface Area Method:
Patient’s Dose (mg) =
𝑃𝑎𝑡𝑖𝑒𝑛𝑡 ′ 𝑠 𝐵𝑆𝐴(𝑚2)
x
1.73𝑚2
Drug Dose (mg)
Determining Patient’s BSA:
Nomogram
BSA Equation (Mosteller’s Formula)
Nomogram
Mosteller’s Formula;
BSA
(m2)
=
𝐻𝑡 𝑐𝑚 𝑥 𝑊𝑡 (𝑘𝑔)
3600
Example:
If the adult dose of a drug is 75 mg, what would be the dose
for a child weighing 40lb and measuring 32 inches in height
using the BSA Nomogram?
From the nomogram, the BSA = 0.60 m2
Patient’s Dose (mg) =
0.60 𝑚2
1.73𝑚2
x 75 mg
Patient’s Dose (mg) = 26. 01 mg or 26 mg
Example:
The daily dose for a child 4 years of age, 39 inches in
height, and weighing 32 lb for a drug with an adult dose
of 100 mg (Use the BSA equation)
Patient’s Dose (mg) =
BSA (m2) =
𝑃𝑎𝑡𝑖𝑒𝑛𝑡 ′ 𝑠 𝐵𝑆𝐴(𝑚2)
x
1.73𝑚2
Drug Dose (mg)
𝐻𝑡 𝑐𝑚 𝑥 𝑊𝑡 (𝑘𝑔)
3600
BSA (m2) =
Ht (cm) = 39 in x
2.54 𝑐𝑚
1 𝑖𝑛
Wt (kg) = 32 lb x
1 𝑘𝑔
2.2 𝑙𝑏
99.06 𝑐𝑚 𝑥 14.5454 𝑘𝑔
3600
= 0.63 m2
= 99.06 cm
= 14.5454 kg
Patient’s Dose (mg) =
0.63𝑚2
x
1.73𝑚2
100 mg
Patient’s Dose (mg) = 36.42 mg
PERCENTAGE, RATIO STRENGTH
AND OTHER EXPRESSIONS OF
CONCENTRATIONS
PERCENT
• Corresponding (%) sign
• By the hundred
• Parts in a hundred
• An essential component of pharmaceutical calculations
• Uses:
a. To express the strength of a component in a pharmaceutical preparation
b. To determine the quantity of a component to use when a percent strength
is desired
Percent Preparations
• For solutions or suspensions of solids in liquids, percent weight in
volume
• For solutions of liquids in liquids, percent volume in volume
• For mixtures of solids or semisolids, percent weight in weight
• For solutions of gases in liquids, percent weight in volume
PERCENT CONCENTRATIONS
•Percent weight-in-volume (w/v)
•Percent volume-in-volume (v/v)
•Percent weight-in-weight (w/w)
Percent weight-in-volume (w/v)
• expresses the number of grams of a constituent in 100 mL of
solution or liquid preparation and is used regardless of whether
water or another liquid is the solvent or vehicle.
• Expressed as: % w/v.
• Equation
Percent weight-in-volume(%w/v) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑔)
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑚𝐿)
𝑥 100
Percent weight-in-volume (w/v)
Equation
Percent weight-in-volume(%w/v) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑔)
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑚𝐿)
𝑥 100
If the amount of solute (g) is unknown
Amount of solute (g) = %w/v expressed in decimal x amount of solution (mL)
%w/v expressed in decimal =
%
100
If the amount of solution is unknown
Amount of solution (mL) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑔)
𝑤
𝑣
% 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑖𝑛 𝑑𝑒𝑐𝑖𝑚𝑎𝑙
Example:
How many grams of dextrose are required to prepare 4000 mL of 5%
solution?
Amount of solute (g) = %w/v expressed in decimal x amount of solution (mL)
%w/v expressed in decimal =
%
100
5%
%w/v expressed in decimal = 100 = 0.05 (g/mL)
Amount of solute (g) = 0.05
𝑔
(𝑚𝐿)
Amount of solute (g) = 200 g
x 4000 mL
Example:
Rx
Antipyrine
5%
Glycerin ad
60 mL
How many grams of antipyrine should be used in preparing the prescription?
Amount of solute (g) = %w/v expressed in decimal x amount of solution (mL)
%
%w/v expressed in decimal = 100
5%
%w/v expressed in decimal = 100 = 0.05 (g/mL)
Amount of solute (g) = 0.05
Amount of solute (g) = 3 g
𝑔
(𝑚𝐿)
x 60 mL
POP QUIZ
1. Rx
Ofloxacin ophthalmic solution 0.3%
Disp. 10 mL
How many milligrams of ofloxacin are contained in each milliliter of the dispensed prescription?
Answer: 3 mg of ofloxacin
2. A formula or an antifungal shampoo contains 2% w/v ketoconazole. How many grams of
ketoconazole would be needed to prepare 240 mL of the shampoo?
Answer: 4.8 g of ketoconazole
Percent volume-in-volume (v/v)
• expresses the number of milliliters of a constituent in 100 mL of
solution or liquid preparation.
• Expressed as: % v/v.
• Equation
Percent volume-in-volume(%v/v) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝐿)
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑚𝐿)
𝑥 100
Percent volume-in-volume (v/v)
Equation
Percent volume-in-volume(%v/v) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝐿)
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑚𝐿)
𝑥 100
If the amount of solute (mL) is unknown
Amount of solute (mL) = %v/v expressed in decimal x amount of solution
(mL)
%v/v expressed in decimal =
%
100
If the amount of solution is unknown
Amount of solution (mL) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝐿)
𝑣
%𝑣 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑖𝑛 𝑑𝑒𝑐𝑖𝑚𝑎𝑙
Example:
How many liters of a mouthwash can be prepared from 100 mL of cinnamon
flavor if its concentration is to be 0.5% (v/v)?
Amount of solution (mL) =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝐿)
𝑣
𝑣
% 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑖𝑛 𝑑𝑒𝑐𝑖𝑚𝑎𝑙
%
%v/v expressed in decimal = 100
%v/v expressed in decimal =
Amount of solution (mL) =
0.5 %
100
= 0.005 (mL/mL)
100 𝑚𝐿
𝑚𝐿
0.005
𝑚𝐿
1𝐿
Amount of solution (mL) = 20,000 mL x 1000 𝑚𝐿 = 20 L of mouthwash
Example:
The formula for 1 liter of an elixir contains 0.25mL of a flavoring oil.
What is the percentage (v/v) of the flavoring oil in the elixir?
Percent volume-in-volume(%v/v) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝐿)
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑚𝐿)
0.25 𝑚𝐿
Percent volume-in-volume(%v/v) =
1000 𝑚𝐿
𝑥 100
Percent volume-in-volume(%v/v) = 0.025 % v/v of flavoring oil
𝑥 100
POP QUIZ
1. A dermatologic lotion contains 1.25 mL of liquefied phenol in 500
mL. Calculate the percent strength of liquefied phenol in the lotion.
Answer: 0.25% liquefied phenol
2. What is the percent strength (v/v) if 225 g of a liquid having a
specific gravity of 0.8 are added to enough water to make 1.5 L of
the solution?
Answer: 18.75% v/v
Percent weight-in-weight (w/w)
• expresses the number of grams of a constituent in 100 g of
solution or preparation.
• Expressed as: % w/w
• Equation
Percent weight-in-weight(%w/w) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑔)
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑔)
𝑥 100
Percent weight-in-weight (w/w)
Equation
Percent weight-in-weight(%w/w) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑔)
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑔)
𝑥 100
If the amount of solute (g) is unknown
Amount of solute (g) = %w/w expressed in decimal x amount of solution (g)
%
%w/w expressed in decimal = 100
If the amount of solution is unknown
Amount of solution (g) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑔)
𝑔
𝑔
% 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑖𝑛 𝑑𝑒𝑐𝑖𝑚𝑎𝑙
Example:
Antibacterial gel contains 0.05% w/w of drug. Calculate the quantity of
this agent (in grams) in each 60 g tube of the product.
%
100
%w/w expressed in decimal =
%w/w expressed in decimal =
0.05 %
100
Amount of solute (g) = 0.0005
Amount of solute (g) = 0.03 g
𝑔
(𝑔 )
= 0.0005 (g/g)
x 60 g
Example:
How many grams of azelaic acid are contained in 30 grams tubes of
the 15% w/w ointment?
%w/w expressed in decimal =
%w/w expressed in decimal =
Amount of solute (g) = 0.15
15 %
100
𝑔
(𝑔 )
%
100
= 0.15 (g/g)
x 30 g
Amount of solute (g) = 4.5 g of azelaic acid
POP QUIZ
1. What is the percentage strength (w/w) of a solution made by
dissolving 62.5 g of potassium chloride in 187.5 mL of water?
Answer: 25% potassium chloride
2. How many grams of hydrocortisone should be used in preparing
120 suppositories, each weighing 2 g and containing 1% of
hydrocortisone?
Answer: 2.4 g of hydrocortisonw
RATIO STRENGTH
• Expresses the concentration of weak solution.
• For example,
5% means 5 parts per 100 or 5:100.
Although 5 parts per 100 designates a ratio strength, it is customary to
translate this designation into a ratio, the first figure of which is 1; thus,
5:100 = 1:20.
Ratio Strength
For solids in liquids =1 g of solute or constituent in 1000 mL of
solution or liquid preparation.
For liquids in liquids = 1 mL of constituent in 1000 mL of
solution or liquid preparation.
For solids in solids = 1 g of constituent in 1000 g of mixture.
Example:
Express 0.02% as a ratio strength.
0.02 (%)
100 (%)
=
1 (𝑝𝑎𝑟𝑡)
𝑥 (𝑝𝑎𝑟𝑡𝑠)
x (0.02%) = (1 part) (100%)
0.02%
0.02%
= 5000 parts
Ratio strength = 1:5000
Example:
Express 1:4000 as a percentage strength.
4000 𝑝𝑎𝑟𝑡𝑠
1 𝑝𝑎𝑟𝑡
=
100 %
𝑥 (%)
4000 parts (x) = (1 part) (100%)
4000 parts
4000 parts
= 0.025 %
Percent strength = 0.025 %
PARTS PER MILLION (ppm) AND PARTS PER
BILLION (ppb)
• Strengths of very dilute solutions
• Expressed in terms of ppm and ppb
• Example:
5 ppm = 5 parts in 1,000,000 parts
2 ppb = 2 parts in 1,000,000,000 parts
Example:
The concentration of a drug additive in animal feed is 12.5 ppm. How many
milligrams of the drug should be used in preparing 5.2 kg of feed?
12.5 ppm = 12.5 g (drug) in 1,000,000 g (feed)
5.2 kg x
1000 𝑔
1 𝑘𝑔
1,000,000 𝑔
12.5 𝑔
=
= 5,200 g
5,200 𝑔
𝑥
(1,000,000 g) (x) = (12.5 g) (5,200 g)
1,000,000 g
1,000,000 g
1000 𝑚𝑔
x = 0.065 g x
1𝑔
= 65 mg
Milligrams Percent
- Express the number of milligrams of substance in 100 mL of
liquid.
- Denotes the concentration of a drug or natural substance in
biologic fluid, as in blood.
- Example: non-protein nitrogen in blood is 30 mg% means that
each 100 mL of blood contains 30 mg of non-protein blood.
mg/mL
• Measurement of a solutions concentration.
Example:
Convert 4% (w/v) to mg/mL
4% w/v =
4000 𝑚𝑔
100 𝑚𝑙
4𝑔
100 𝑚𝐿
4gx
= 40 𝑚𝑔/𝑚𝐿
1000 𝑚𝑔
1𝑔
= 4000 mg
PROOF STRENGTH
• Twice the percentage strength of alcohol
• Example: 50% is 100 proof
Proof Spirit
• Is an aqueous solution containing 50 % (v/v) of
alcohol.
Proof gallon
Proof gallons =
𝑊𝑖𝑛𝑒 𝑔𝑎𝑙𝑙𝑜𝑛 𝑥 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
50%
Proof gallons =
𝑊𝑖𝑛𝑒 𝑔𝑎𝑙𝑙𝑜𝑛 𝑥 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑝𝑟𝑜𝑜𝑓)
100 𝑝𝑟𝑜𝑜𝑓
Example:
How many proof gallons are contained in 5 wine gallons of
75% v/v alcohol?
Proof gallons =
𝑊𝑖𝑛𝑒 𝑔𝑎𝑙𝑙𝑜𝑛 𝑥 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
50%
Proof gallons =
5 𝑊𝑖𝑛𝑒 𝑔𝑎𝑙𝑙𝑜𝑛 𝑥 75%
50%
Proof gallons = 7.5 𝑝𝑟𝑜𝑜𝑓 𝑔𝑎𝑙𝑙𝑜𝑛𝑠
Altering product strength
• The percentage or ratio strength (concentration) of a
component in a pharmaceutical preparation is based on its
quantity relative to the total quantity of the preparation.
• If the quantity of the component remains constant, any
change in the total quantity of the preparation, through
dilution or concentration, changes the concentration of the
component in the preparation inversely.
In pharmacy practice
• The reduction in the strength of a commercially available
pharmaceutical product may be desired to treat a particular
patient, based on the patient’s age (e.g., pediatric or elderly)
or medical status, or to assess a patient’s initial response to a
new medication.
• The strengthening of a product may be desired to meet the
specific medication needs of an individual patient
Altering product strength
Problems may be solved by any of the following
methods:
1. Inverse proportion
2. The equation: (1st quantity) × (1st concentration) = (2nd quantity)
× (2nd concentration or Q1 x C1 = Q2 x C2
3. Traditional calculations, by determining the quantity of active
ingredient present and relating that amount to the quantity of the
total preparation
Example:
If 500 mL of a 15% v/v solution are diluted to
1500 mL, what is the percent strength (v/v) of the
dilution?
• By inverse proportion
1500 mL
_____________
15%
=
_________
500 mL
x
(1500 mL) (X) = (500 mL) (15%)
1500 mL
1500 mL
X = 5% v/v
Example:
If 500 mL of a 15% v/v solution are diluted to 1500
mL, what is the percent strength (v/v) of the
dilution?
• By traditional calculation
Amount of solution X Concentration expressed in decimal = Amount of solute
𝑚𝐿
Amount of solute (g) = 0.15 (𝑚𝐿) x 75 mL
Percent volume-in-volume(%v/v) =
Percent volume-in-volume(%v/v) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝐿)
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑚𝐿)
75 𝑚𝐿
𝑥
100
1500 𝑚𝐿
Percent volume-in-volume(%v/v) = 5% v/v
𝑥 100
Example:
If 500 mL of a 15% v/v solution are diluted to
1500 mL, what is the percent strength (v/v) of the
dilution?
• By equation
Q1 x C1 = Q2 x C2
(500 mL) (15%) = (1500 mL) (x)
1500 mL
1500 mL
X = 5%
Stock Solutions
• Stock solutions are concentrated solutions of
active (e.g., drug) or inactive (e.g., colorant)
substances and are used by pharmacists as a
convenience to prepare solutions of lesser
concentration.
Q1 x C1 = Q2 x C2
Example
How many milliliters of a 1% w/v stock solution of a certified red dye
should be used in preparing 4000 mL of a mouthwash that is to contain
1:20,000 w/v of the certified red dye as a coloring agent?
1:20,000 = 0.005%w/v
Q1 x C1 = Q2 x C2
(4000 mL) (0.005%) = (x) (1%)
1%
1%
X = 20 mL
POP QUIZ
• If 250 mL of 1:800 (v/v) solution were diluted in 1000 mL,
what would be the ratio strength (v/v)?
• If a pharmacist added 12 grams of azelaic acid to 50 g of an
ointment containing 15% azelaic acid , what would be the
final concentration of the azelaic acid in the ointment?
Electrolyte Preparations
• are used in the treatment of disturbances of the electrolyte
and fluid balance in the body.
• They are provided by the pharmacy as oral solutions, syrups,
tablets, capsules, and, when necessary, intravenous
infusions.
Electrolyte Solutions
• Electrolyte ions in the blood plasma include the
cations Na+, K+, Ca2+, and Mg2+ and the anions Cl−,
HCO3−, HPO42−, SO42−, organic acids, and protein.
• Electrolytes in body fluids play an important role in
maintaining the acid–base balance.
• They also play a part in controlling body water
volumes and help regulate metabolism.
Milliequivalents (mEq)
• is used almost exclusively in the United States by clinicians,
physicians, pharmacists, and manufacturers to express the
concentration of electrolytes in solution.
• This unit of measure is related to the total number of ionic charges
in solution, and it takes note of the valence of the ions
• a unit of measurement of the amount of chemical activity of an
electrolyte.
Milliequivalent
____MW____
Valence
1 mEq = _________________
1000
Sample Problems
1. A physician prescribes 10 mEq of potassium chloride for a patient. How
many illigrams of KCl would provide the prescribed quantity?
2. A physician prescribes 3 mEq/kg of NaCl to be administered to a 165-lb
patient. How many milliliters of a half–normal saline solution (0.45% NaCl)
should be administered?
3.
What is the concentration, in milligrams per milliliter, of a solution
containing 2 mEq of potassium chloride (KCl) per milliliter?
4. What is the concentration, in grams per milliliter, of a solution containing 4
mEq of calcium chloride (CaCl2 · 2H2O) per milliliter?
Millimoles
•A mole is the molecular weight of a substance in
grams.
•A millimole is one-thousandth of a mole and is,
therefore, the molecular weight of a substance in
milligrams.
Millimoles
MW
1 mmol = __________
1000
Sample problem
1. How many millimoles of monobasic sodium phosphate
monohydrate (m.w. 138) are present in 100 g of the substance?
2. What is the weight, in milligrams, of 5 mmol of potassium
phosphate dibasic (m.w. 174)?
3.
If lactated Ringer’s injection contains 20 mg of calcium chloride
dihydrate (CaCl2 · 2H2O) m.w 147 in each 100 mL, calculate the
millimoles of calcium present in 1 L of lactated Ringer’s injection.
Osmolarity
•Osmotic pressure is proportional to the total number of
particles in solution. The unit used to measure osmotic
concentration is the milliosmole (mOsmol).
•For dextrose, a nonelectrolyte, 1 mmol (1 ormula
weight in milligrams) represents 1 mOsmol.
Milliosmoles
Wt. of substance(g/L)
mOsmol = ___________________ x No. of Species x 1000
Molecular weight (g)
1.
2.
3.
A solution contains 10% of anhydrous dextrose in water for
injection. How many milliosmoles per liter are represented by this
concentration? Molecular weight of anhydrous dextrose = 180
Calculate the osmolarity, in milliosmoles per liter, of a parenteral
solution containing 2 mEq/mL of potassium acetate (KC
2H3O2—m.w. 98).
What is the osmolarity of an 8.4% w/v solution of sodium
bicarbonate( m.w. 84)?
ISOTONIC SOLUTION
What is Isotonicity
•Two solutions that have the same osmotic pressure are
termed isosmotic. Many solutions intended to be mixed with
body fluids are designed to have the same osmotic pressure
for greater patient comfort, efficacy, and safety.
• A solution having the same osmotic pressure as a specific
body fluid is termed isotonic(meaning of equal tone) with that
specific body fluid.
• When a solvent passes through a semipermeable membrane from a dilute solution into
a more concentrated one, the concentration
become equalized and this is called as
osmosis.
154
• The pressure responsible for this phenomenon
is termed as osmotic pressure and varies with
the nature of the solute.
155
• If the solute is non-electrolyte, its solution contains
only molecules and the osmotic pressure varies with
the concentration of the solute. If the solute is an
electrolyte, its solution contains ions and the osmotic
pressure varies with the both the concentration of the
solute and its degree of dissociation.
156
• Isotonic – a solution having the same osmotic pressure as the
specific body fluid.
• Hypotonic – solutions of lower osmotic pressure with the body
fluid.
• Hypertonic – solutions having higher osmotic pressure with the
body fluid.
157
Example: Calculation of the Factor
1. Ferrous Sulfate is a 2 ion electrolyte, dissociating 60% in a
certain concentration. Calculate its dissociation (i) factor.
158
2. Sodium Carbonate dissociates in a certain concentration at
70%. Calculate its dissociation (i) factor.
159
• Most medicinal salts approximate the dissociation of sodium chloride in weak
solutions. If the number of ions is known, we may use the following values:
Non-electrolytes and subs of slight dissociation: 1.0
Subs. that dissociate in to 2 ions: 1.8
3 ions: 2.6
4 ions: 3.4
5 ions: 4.2
160
Calculations of the NaCl equivalent:
Formula:
MW of NaCl
i factor of NaCl
X i factor of Subs = NaCl equivalent
MW of Subs
161
Examples:
1. Ephedrine hydrochloride (mw= 202) is a 2 ion electrolyte,
dissociating 75% in a given concentration. Calculate the
sodium chloride content.
162
2. Morphine sulfate (mw=759) is a 3 ion electrolyte,
dissociating 60% in a concentration. Calculate the sodium
chloride equivalent.
163
Procedure in the calculation of isotonic solutions with NaCl equivalents:
1. Calculate the amount in grams of sodium chloride represented by
the ingredients in the prescription. Multiply the amount in grams of
each substance by its sodium chloride equivalent.
2. Calculate the amount in grams of sodium chloride alone, that would
be contained in an isotonic solution of the volume specified in the
prescription, namely, the amount of sodium chloride in a 0.9%
solution of the specified volume.
164
3. Subtract the amount of sodium chloride represented by the
ingredients in the prescription (Step 1) from the amount of
sodium chloride, alone, that would be represented in the
specific volume of an isotonic solution (Step 2). The answer
represents the amount (in grams) of sodium chloride to be
added to make the isotonic solution.
165
4. If an agent other than sodium chloride, such as boric acid,
dextrose or potassium nitrate is to be used to make a solution
isotonic, divide the amount of sodium chloride (Step 3) by the
sodium chloride equivalent of the other substance.
166
Example Calculations of tonicic agent required:
1. Rx
Cocaine hydrochloride
0.6
Eucatropine hydrochloride
Chlorobutanol
Sodium chloride
0.6
0.1
q.s.
Purified water ad
Make isoton. Sol.
30
Sig. For the eye.
How many grams of sodium chloride should be used in compounding the
prescription?
167
2. Rx
Tetracaine hydrochloride
0.1
Zinc sulfate
0.05
Boric acid
q.s.
Purified water ad
30
Make isoton. Sol.
Sig. Drop in eye
How many grams of boric acid should be used in compounding the
prescription?
168
Using an isotonic Sodium chloride to prepare other isotonic solutions:
A 0.9% w/v sodium chloride solution may be used to compound
isotonic solutions of other substances as follows:
Step 1: Calculate the quantity of the drug substance needed to fill
the prescription or medication order.
Step 2: Use the following equation to calculate the volume of water
needed to render a solution of the drug substance isotonic.
169
g of drug
x drug’s E value
0.009
= mL of water needed to
to make an isoton.sol’n
Step 3: Calculate the volume 0.9% w/v sodium chloride
solution to complete the required volume of the prescription
or medication order.
170
Example:
1. Rx
Phenylephrine hydrochloride
Chlorobutanol
1%
0.5%
Sodium chloride
q.s.
Purified water ad
15
Make isoton. Sol.
Sig: Use as directed
How many mL of a 0.9% solution of sodium chloride should be used in
compounding the prescription?
171
Freezing Point Data in Isotonicity Calculations
Freezing point data (∆Tƒ) can be used in isotonicity
calculations when the agent has a tonicic effect and does not
penetrate the biologic membranes in question. The freezing
point of both blood and lacrimal fluid is -0.52°C. A
pharmaceutical solution that has a freezing pt same with the
blood is said to be isotonic.
172
1. How many milligrams of Naphazoline hydrochloride and
sodium chloride are needed to produce 75 mL of 1% isotonic
solution with lacrimal fluid?
173
2. How many mg of Atropine sulfate and boric acid are required
to prepare 15 mL of 1% of atropine sulfate isotonic with
tears?
174
3. How many milligrams of dextrose and sodium chloride are
needed to produce 1L of 1% dextrose isotonic with blood?
175
How is HLB Values computed?
• The systematic choice of emulsifying agents in the
formulation of many emulsion systems depends on their HLB
(hydrophile-Lipophile-Balance) values.
• These values form the basis of the so-called HLB system,
which was developed by Griffin.
The system presupposes a scale of HLB
numbers and is based on the facts
1. that every surfactant or emulsifier molecule in part
hydrophilic and in part lipophilic
2. that a certain balance between these two parts is necessary
for various types of surfactant functions. In this scheme,
each surfactant or emulsifying agent is assigned a number
that varies from 1 to 20.
• The lower values are assigned to substances that are
predominantly lipophilic (oil loving) and have a tendency to
form water-in-oil (w/o) emulsions.
• The higher values are given to those materials that show
hydrophilic (water-loving) characteristics and favor the
formation of oil-in-water (o/w) emulsions.
• Consequently, the HLB number of an emulsifying agent is an
index of the type of emulsion that has the greatest tendency
to form.
• When two or more emulsifiers are combined, the HLB of the
combination is determined arithmetically by adding the
contribution that each makes to the HLB total of the mixture
SAMPLE PROBLEM
• What is the HLB of a mixture of 40% of Span 60 and 60% of Tween 60?
HLB of Span 60 = 4.7
HLB of Tween 60 = 14.9
HLB
% of mixture
Span 60
4.7
x
40%
= 1.9
Tween 60
14.9
x
60%
= 8.9
HLB of mixture
= 10.8
SAMPLE PROBLEM
• In what proportion should Tween 80 and Span 80 be blended to obtain a required HLB of 12?
HLB of Tween 80 = 15
HLB of Span 80
15
= 4.3
7.7 parts of tween 80
12
4.3
3.0 of Span 80
What is pharmacoeconomics?
• The term pharmacoeconomics encompasses the economic
aspects of drug, from the costs associated with drug
discovery and development to the costs of drug therapy
analyzed against therapeutic outcomes.
• Drug therapy and other means of treatment are intended to
serve the health care interests the patient while being cost
effective. In prescribing drug therapy, clinical as well as
economic factors are important considerations in the
selection of the drug substance and drug product.
• For example, if an expensive drug reduces morbidity and
hospitalization time, it is considered both therapeutically
advantageous and cost effective. If, however, a less
expensive drug would provide therapeutic benefit comparable
to the more expensive drug, the less costly drug is likely to be
selected for use.
Cost differential between drugs
Example:
An antihypertensive drug is available from various
manufacturers at prices per 100 tablets ranging from P6.26 to
P25.50, with a mean price of P10.75. If a patient presents a
prescription for a 6 month supply of the drug calling for two
tablets daily, calculate the differentials in the cost of the drug to
the pharmacy between the highest, mean and lowest cost
products.
Solution
6 month supply = approximately 180 days
2 tablets a day x 180 days = 360 tablets
Lowest price:
P6.25/100 tablets = P0.0625/tablets x 360 tablets = P 22.5
Mean price
P10.75/100 tablets = P0.1075/tablets x 360 tablets = P 38.70
Highest price:
P25.50/100 tablets = P0.2550/tablets x 360 tablets = P 91.80
Differentials:
Highest price to lowest price:
P91.80 – P22.50 = P69.30
Highest price to mean price:
P91.80 – P38.70 = P53.10
Mean price to lowest price:
P38.70 – P22.50 = P16.20
Depending on which product is dispensed, these differentials would be reflected in the
prescription price charges to the patient.
Discounts
• Discounts provided by suppliers may be based on quantity buying
and/or payment of invoices within a specified time period.
• In addition, for nonprescription products, discounts may be
available for certain seasonal or other promotional products,
bonuses in terms of free merchandise, and advertising and display
allowances.
• These discounts provide the pharmacy with a means of increasing
the gross profit on selected merchandise.
Net cost given list price and allowable discount
• Example:
The list price of an antihistamine elixir is P6.50 per pint, less 40%.
What is the nest cost per pint of elixir?
List Price
Discount
Net Cost
100%
-
40%
=
60%
P6.50
x
0.60
=
P3.90, answer
Markup
• The term mark up, sometimes used interchangeably with the
term margin of profit (gross profit), refers to the difference
between the cost of merchandise and its selling price.
•
Calculating the selling price of merchandise to yield a
given percent of gross profit on the cost involves the
following.
Example
The cost of 100 antacid tablets is P2.10. What would be the selling price
per tablet to yield a 66 2/3% gross profit on cost?
Cost x % of gross profit = gross profit
P2.10 x 66 2/3% = P1.40
Cost + Gross profit = Selling price
P2.10 + P1.40
= P3.50
Example
The cost of 100 antacid tablets is $2.10. What should be the selling price per 100 tablets to yield a 40% gross profit on
the selling price?
Selling price = 100%
Selling price - Gross profit = Cost
100% - 40% = 60%
60 (%)
($) 2.10
----------- = ----------100 (%)
($) x
x = $3.50, answer.
What are the Common Methods of Prescription
Pricing?
1. Percent Mark up. In this common method, the desired
percent markup is taken of the cost of the ingredients and
added to the cost of the ingredients to obtain the
prescription price.
Prescription Price = Cost of ingredients + (cost of ingredient x % markup)
Example
If the cost of the quantity of a drug product to be dispensed is
P4.00 and the pharmacist applies an 80% markup on cost,
what would be the prescription price?
P4.00 + (P4.00 x 80%) = P4.00 + P3.20 = P7.20
2. Percent Markup plus a minimum professional fee. In this
method, both a percent markup and a minimum professional
fee are added to the cost of the ingredient.
Prescription price= Cost of ingredients + (cost of ingredients x % markup) + minimum professional fee
Example
If the cost of a drug product to be dispensed in P4.00 and
pharmacist applies a 40% markup on cost plus a professional
fee of P2.25, what would be the prescription price?
P4.00 + (P4.00 x 80%) + P2.25 = P4.00 + P3.20 + P2.25 = P7.85
3. Professional Fee. This method involves addition of a
specified professional fee to the cost of ingredients used in
filling a prescription.
Cost of the ingredients + professional fee = prescription price
Example
If the cost of the quantity of a drug product to be dispensed is
P4.00 and the pharmacist applies a professional fee of P4.25,
what would be the prescription price?
P 4.00 + P 4.25 = P 8.25
END OF
DISCUSSION
REFERENCE
• Ansel, H. C., Pharmaceutical Calculations, 15th ed., Wolters Kluwer,
2017.
• Remington’s Pharmaceutical Sciences
Thanks!