Pharm Calculations for undergraduate students

Helpful for those who are willing to learn calculations

1. 1
Pharmaceutical
Calculations p. 165
4
2
Pharmaceutical calculations are
directly tested on the NAPLEX
®
.
Pharmaceutical Calculations
3
Whenever units are expressed in either
the apothecary or the avoirdupois, it is
convenient to immediately convert the
units to metric units (SI units).
Pharmaceutical Conversions
4
milk chocolate drinking died Henry King
Mega:one million (1 x 10
6
)
Giga:one billion (1 x 10
9
)
Tera:one trillion (1 x 10
12
)
Pharmaceutical Conversions
milli- centi- deci- U deca- Hecto- Kilo-
Micro:one millionth (1 x 10
-6
)
Nano:one billionth (1 x 10
-9
)
1/1000 x 1/100 x 1/10 x
U
10 x 100 x 1000 x
L 1 dL = 1/10 L = 0.1 L
1 Kg = 1000 gg1 mg =
1/1000 g

2. 2
5
Volume Measurements
1 teaspoonful (tsp) = 5 mL
1 tablespoonful (tbsp) = 15 mL
1 pint = 473 mL
1 quart = 946 mL (2 pints)
1 gallon = 3784 mL or 3785 mL (4 quarts)
1 mL = 20 drops
PG 166
6
Weight Measurements
PG 166
7
1 inch = 2.54 centimeters (cm)
1 ft = 12 inches
Length Measurements
PG 166
8
Roman Numerals Roman Numerals
z
ss = ½
z
I or i = 1
z
V or v = 5
z
X or x = 10
z
L or l = 50
z
C or c = 100
z
D or d = 500
z
M or m = 1,000
z
ss = ½
z
I or i = 1
z
V or v = 5
z
X or x = 10
z
L or l = 50
z
C or c = 100
z
D or d = 500
z
M or m = 1,000

3. 3
9
Roman Numerals Roman Numerals
z
In writing prescriptions physicians or
other health care professionals may use
small or capital roman numerals.
z
When the small letter iis used it should be
dotted to distinguish it from the letter
l
.
z
Sometimes a jmay be used for the final iin
a sequence (e.g. viij).
z
Following the Latin custom, Roman
numerals are generally placed after the
symbol or term (e.g. capsules no. xxiv or
fluidounces xij)
z
In writing prescriptions physicians or
other health care professionals may use
small or capital roman numerals.
z
When the small letter iis used it should be
dotted to distinguish it from the letter
l
.
z
Sometimes a jmay be used for the final iin
a sequence (e.g. viij).
z
Following the Latin custom, Roman
numerals are generally placed after the
symbol or term (e.g. capsules no. xxiv or
fluidounces xij)
10
Rules for Roman Numerals Rules for Roman Numerals
„
Two or more letters express a quantity that is
the sum of their values if they are
successively equal or smaller in value.
z
e.g. ii = 2 vii = 7 XIII = 13
„
Two or more letters express a quantity that is
the sum of the values remaining after the
value of each smaller letter has been
subtracted from that of a following greater
letter.
z
e.g. IV = 4 IX = 9 CM = 900
„
Two or more letters express a quantity that is
the sum of their values if they are
successively equal or smaller in value.
z
e.g. ii = 2 vii = 7 XIII = 13
„
Two or more letters express a quantity that is
the sum of the values remaining after the
value of each smaller letter has been
subtracted from that of a following greater
letter.
z
e.g. IV = 4 IX = 9 CM = 900
11
Examples Examples
„
lxvi =
„
xix =
„
xcix =
„
cclxx =
„
cxl =
„
lxvi =
„
xix =
„
xcix =
„
cclxx =
„
cxl =
66
19
99
270
140
66
19
99
270
140
12
Dosing based on body weight
ƒTwo steps
1. Convert the pounds to kilograms
2. Multiply dose by the body weight
Drug Dosing
PG 166

4. 4
13
Example A:
How many mg of Tobramycin are needed for
a patient weighing 160 lbs if the desired dose
is 0.8 mg/kg?
Drug Dosing
1 kg = 2.2 lbs.
x kg 160 lbs. x = 72.7 kg
0.8 mg = x mg
1 kg 72.7 kg x = 58 mg
PG 166
14
Example B:
The oral dosing regimen for Cytoxan is 2 mg/kg/day for 10 days.
How many 50-mg tablets should be dispensed by a pharmacist
to a 34-year-old patient weighing 154 pounds?
Drug Dosing
1 kg = 2.2 lbs.
x kg 154 lbs. x = 70 kg
2 mg = x mg
1 kg 70 kg x = 140 mg
1 tablet = 50 mg
x tablets 140 mg x = 2.8 tablets or 3 tablets
For 10 days: 3 x 10 = 30 tablets(NOT: 2.8 x 10)
PG 167
15
3600
(kg) weight s Patient' (cm) height s Patient'
) (m BSA s Patient'
2
×
=
Dosing Based on
Body Surface Area
PG 167
dose adult Average
73.1
) (m patient of area Surface
dose adult or Child
2
× =
16
Nomogram for Determining Surface Areas
m
2
PG 168

5. 5
17
Example C
The average child dose of a drug is
200 mg/m
2
. Estimate an appropriate
dose for a 4-year-old child who is 4
feet tall and weighs 80 lb.
m
2
= 1.1
200 mg/m
2
x 1.1 m
2
= 220 mg
PG 168
18
Example D
The surface area of a 4-year-old
child is estimated to be 0.5 m
2
. How
many mg of a drug which has an
adult dose of 100 mg should be
administered?
Child dose = surface area
×adult dose
1.75
Dose = 0.5m
2
×100 mg
Dose = 29 mg
1.75
19
Clark’s Rule:a weight calculation
intended for children greater than 2
years of age
Dosing for Children Based on Weight or Age
PG 169
20
Dosing for Children Based on Weight or Age
Young’s Rule:Also intended for
children greater than 2 years of age

6. 6
21
Example
Determine the mg of an antibiotic
to be administered to an 8-year-old
child who weighs 84 lb. The usual
adult dose is 250 mg
.
Clark’s:Dose = 84 lb
×250 mg
150 lb
Dose = 140 mg
22
Example
Determine the mg of an antibiotic
to be administered to an 8-year-old
child who weighs 84 lb. The usual
adult dose is 250 mg
.
Young’s:
Dose = 8
×250mg
(8 + 12)
Dose = 100 mg
23
PG 169
Example
Determine the mg of an antibiotic
to be administered to an 8-year-old
child who weighs 84 lb. The usual
adult dose is 250 mg
.
Clark’s:
Young’s:
24
ƒPercent (%)
ƒParts per million (ppm)
ƒRatio strength (1 : something)
Concentration Expressions
PG 171

7. 7
25
The amount of active ingredient present
in a specified amount of final product.
A 10% w/vindicates that 10 gof
ingredient is present in every 100 mLof
formula.
A 10% w/wmeans a 10 gof drug are
contained in every 100 gof product
.
A 10% v/vmeans a 10 mLof drug are
contained in every 100 mLof product.Concentration Expressions
as Percentages
PG 171
26
1. Determine total weight or volume of the
formula or prescription.
2. Determine what percentage of a certain
ingredient is being requested for the final
product.
3. Calculate weight or volume of that ingredient.
4. Some cases calculate weight or volume of
solvent (vehicle) needed.
PG 172
The following thought process should be
followed to solve concentration problems:
27
Example A
Rx:
Procaine HCl 2% w/v
Pur. water qs—120 mL
1. Total volume 120 mL
2. Procaine HCl 2% w/v. = 2 g
100 mL
3. The amount of pure procaine will be:
2 g
= X g
100 mL 120 mL X = 2.4 g
28
Example B
Rx: Benzocaine 1% w/w
Petrolatum qs 100%
M & Ft Oint Disp—60 g
1. The prescription calls for 60 g of product.
2. Of the 60 g, 1% will be benzocaine. 1 g
= X g
X = 0.6 g
100 g 60 g
3. The amount of benzocaine will be 0.6 g
4. 60 g −0.6 g = 59.4 g of petrolatum.

8. 8
29
Concentration expressions based
on a denominator of 1 million
(1,000,000).
One part per million is: 1
1,000,000
Concentration Expressions –
Parts per Million
30
Example A
Express 0.025% w/v as ppm
.
Because 0.025% refers to 0.025 g per 100 mL of
solution
0.025 g = X
100 mL 1,000,000
X= 250 ppm
PG 173
31
Example B
Express 20 ppm as a percentage. We know 20ppm is 20
1,000,000
So:
20
= X
1,000,000 100 X = 0.002 = 0.002%
32
Example C
A water supply contains 12 ppm of calcium chloride
as an impurity. How many mg of calcium chloride are
present in every dL?
12 ppm indicates
12 parts (g) per 1,000,000 parts (mL)
12 g
= x g
1,000,000 mL 100 mL
x = 0.0012 g or 1.2 mg/100 mL

9. 9
33
Example D
The blood level of a drug is 8 µg/dL.
Express this concentration in terms of
parts per million.
34
Ratio Strength Ratio Strength
„
An expression of concentration using
ratios
„
Used when a low concentration is present
„
Always expressed as 1: something
„„
An expression of concentration using An expression of concentration using
ratios ratios
„„
Used when a low concentration is present Used when a low concentration is present
„„
Always expressed as 1: something Always expressed as 1: something
35
Express the resulting concentration of Drug Z as a
ratio strength if you were to dissolve 600 mg of
Drug Z in enough simple syrup to make 3000mL
Express the resulting concentration of Drug Z as a
ratio strength if you were to dissolve 600 mg of
Drug Z in enough simple syrup to make 3000mL 600 mg = 0.6 g
0.6 g
= 1
3000 mL X X = 5,000 mL Answer: 1 : 5,000
600 mg = 0.6 g
0.6 g
= 1
3000 mL X X = 5,000 mL Answer: 1 : 5,000
36
ƒConverting a liquid measurement to a weight quantity
ƒConverting a weight to a volume measurement
ƒDetermining the volume cost of a drug purchased
ƒDetermining the density of a solution Calculations Involving Specific Gravity
PG 174

10. 10
37
The specific gravity is a ratio of the
weight of a certain volume compared with
the weight of the same volume of water.
The density of water is: 1 g
= 1 g/mL
1 mL
The specific gravity of water = 1
38
Example A
An ointment requires 10 g of Coal Tar Solution,
which has a specific gravity of 0.84. What
volume of the solution should be measured?
SG = density = wt
vol
0.84 = 10 g
X mL
Vol = 10
= 11.9 mL
0.84
39
Example B
What weight of glycerin (SG = 1.25) must be
used to obtain 120 mL of glycerin?
SG = density = wt
vol
1.25 = wt
120 mL
Wt =120 x 1.25 = 150 g
40
1. Determine volume of 1 lb of elixir:
SG = density = wt
vol
0.92 = 454 g
x = 493 mL
x mL
2. Determine the cost per mL:
493 mL = $12.40
1 mL = xx = $0.025/mL
Example C
What is the cost per mL of an elixir (sp. gr. =0.92)
if the bulk cost is $12.40/lb?

11. 11
41
The pharmacist may have to adjust the
concentration of an existing preparation
to a new strength.
Problems involving adjustment of strength
may be classified into:
ƒDilution problems
ƒConcentration problems
ƒAlligation problems
Calculations Involving
Adjustments of Strengths
PG 175
DILUTION PROBLEMS DILUTION PROBLEMS
The pharmacist must be able to
state the new strength of a
solution if given the original
concentration and the amount of
diluent added.
The pharmacist must be able to
state the new strength of a
solution if given the original
concentration and the amount of
diluent added.
44
Step 1: Determine how much active ingredient is
present.
Step 2: Determine what the new total volume or weight
will be.
Step 3: State the strength by dividing Step 1 by Step 2.
Method1
PG 176

12. 12
45
What is the new strength of a solution prepared by
diluting 120 mL of a 5% w/v solution with 380 mL
of water?
Step 1.120 mL x 5% = 6g (of active ingredient)
5 g
= X g
100 mL 120 mL X = 6 g
Step 2.120 mL + 380 mL = 500 mL (new volume)
Step 3.6 g
= X g
500 mL 100 mL
X = 1.2%
Example A
PG 176
46
A nurse dilutes 0.5 mL of a 1:1,000 epinephrine HCl
solution with 4.5 mL of sterile water for injection. What is the
concentration of this dilution expressed as a ratio strength?
Step 1.1g
= X g
1,000 mL 0.5 mL X = 0.0005 g
(active ingredient)
Step 2.4.5 mL + 0.5 mL = 5 mL (new volume)
Step 3.0.0005 g
= 1 g
5 mL x mL
0.005 x = 5
x = 10,000 or conc. is 1:10,000
Example B
PG 176
47
Q
1
C
1
= Q
2
C
2
Two guidelines must be followed:
1. The original and new quantities must be
expressed in identical units.
2. The original and new concentrations should be
expressed in identical terms (% w/v, % w/w, % v/v,
ppm, ratio strength, etc…)
Method2
PG 177
48
What is the new strength of a solution prepared by diluting 120 mL of a 5% w/v solution with
380 mL of water?
Q
1
C
1
= Q
2
C
2
(120 ml)(5% w/v) = (500 ml)(X% w/v)
600 = 500(X)
X= 1.2%
Example A

13. 13
49
A nurse dilutes 0.5 mL of a 1:1,000 epinephrine HCl solution
with 4.5 mL of sterile water for injection. What is the
concentration of this dilution expressed as a ratio strength?
Q
1
C
1
= Q
2
C
2
(0.5 mL)(1/1,000) = (5 mL)(1/x)
0.5
=5
1,000 x
0.5 x = 5,000
x = 5,000/0.5 = 10,000 or conc. is 1:10,000
Example B
The pharmacist must be able to
determine the amount of diluent
that should be added to a given
solution to obtain a desired new
concentration.
The pharmacist must be able to
determine the amount of diluent
that should be added to a given
solution to obtain a desired new
concentration.
51
PG 177
EXAMPLE C
A pharmacist has 120 mL of a 4% aluminum chloride
solution.How many mL of water must be added to
obtain a 0.24% solution?
Q
1
C
1
= Q
2
C
2
(120 mL)(4% w/v) = (x mL)(0.24% w/v)
480 = 0.24X
X = 2,000 mL
The question asks for the amount of diluent needed,
2,000 mL - 120 mL = 1,880 mL
The pharmacist must be able to
determine the amount of solution
of a given strength that may be
prepared from a second solution
of another strength.
The pharmacist must be able to
determine the amount of solution
of a given strength that may be
prepared from a second solution
of another strength.

14. 14
53
PG 178
EXAMPLE D
Determine the amount of solution of a 6% w/v strength
that may be prepared from 1 pint of 25% w/v strength?
Q
1
C
1
= Q
2
C
2
(473 mL)(25% w/v) = (x mL)(6% w/v)
11,825 = 6x
x = 1,970 mL
54
EXAMPLE E
How much of a 1/1,500 strength solution can be prepared by
the dilution of 60 mL of a 1/120 strength solution?
Q
1
C
1
= Q
2
C
2
(60 mL)(1/120) = (xmL)(1/1,500)
60
= x
120 1,500
120x = 90,000
x = 750 mL
The pharmacist must be able to
determine the amount of solution
of a given strength needed to
prepare a specified amount of a
second solution.
The pharmacist must be able to
determine the amount of solution
of a given strength needed to
prepare a specified amount of a
second solution.
56
EXAMPLE F
A pharmacist needs 16 mL of a 0.2% w/v vanillin in alcohol
solution. How many mL of a 5% w/v solution should be
diluted with alcohol to obtain the desired solution?
Q
1
C
1
= Q
2
C
2
(x mL)(5% w/v) = (16 mL)(0.2% w/v)
5x = 3.2x = 0.64 mL
or
0.2 g
= x g
100 16 mL x= 0.032 g
5 g
= 0.032 g
100 xmLx= 0.64 mL

15. 15
57
EXAMPLE A
A company has prepared 5 liters of a crude herbal extract,
which assays at 0.05% active drug. How much alcohol
menstruum must be evaporated to obtain a 1% w/v
concentration?
Q
1
C
1
= Q
2
C
2
(5,000 mL)(0.05%) = (x mL)(1%)
x = 250 mL of final product may be
made; therefore,
5,000 mL - 250 mL = 4,750 mL must be evaporated
Concentration Problems
PG 179
58
EXAMPLE A
A company has prepared 5 liters of a crude herbal
extract, which assays at 0.05% active drug. How
much alcohol menstruum must be evaporated to
obtain a 1% w/v concentration?
0.05 g
= X g
100 mL 5000 mL X = 2.5 g
1 g
= 2.5 g
100 mL X mL X = 250 mL
5,000 mL −250 mL = 4,750 mLmust be evaporated
59
Adjustment of Strength Using Mixtures
–
Alligation Alternate Method
Parts of high-concentration ingredient [D] = [C] −[B]
Parts of low-concentration ingredient [E] = [A] −[C]
60
Example A
How many mL of a 20% w/v solution of aluminum
chloride must be mixed with a 5% w/v solution to
prepare 120 mL of a 12% strength?
Step #1
PG 180
Step #2
7 parts
8 parts
Total: 15 parts
120 mL
x mL

16. 16
61
Step# 3
20% solution *7
parts
=XmL
15
total parts
120 mL X= 56 mL
5% solution
8
parts
= XmL
15
total parts
120 mL X= 64 mL
Mixing 56 mL of 20% solution and 64 mL of 5% solution will
result in 120 mL of a 12% concentration.
62
Alligation Alligation
„
A second situation is when one of the
ingredients is available in a limited supply
(the amount of one of the ingredients is
provided, and the amount of the other
ingredient is to be calculated).
„
A second situation is when one of the
ingredients is available in a limited supply
(the amount of one of the ingredients is
provided, and the amount of the other
ingredient is to be calculated).
63
Example B
A pharmacist wishes to prepare a 5% ichthammol
ointment by using 20% w/w ichthammol ointment and
200 g of 2% w/w ichthammol ointment in stock. How
many grams of the 20% ointment are needed?
Step #1
20% [A] parts of 20% [D]
5% [C]
2% [B] parts of 2% [E]
Step #2Parts of high-conc. [D] = [C] – [B] = 5 - 2 = 3 parts
Part of low-conc. [E] = [A] - [C] = 20 – 5 = 15 parts
*
PG 181
64
Step #2Parts of high-conc. [D] = [C] – [B] = 5 - 2 = 3 parts (of 20% oint)
Part of low-conc. [E] = [A] - [C] = 20 – 5 = 15 parts (of 2% oint) The amount of the 2% ointment is 200 g:
3 parts of 20% oint = x g
15 parts of 2% oint 200 g
x = 3 x 200 = 40 g of 20% ointment
15

17. 17
65
Alligation Alligation
„
Alligation may also be used when either pure
chemical (100%) or pure diluent (0% active
ingredient) is mixed with a certain
concentration to obtain a new strength.
„
Alligation is used to solve problems in which
concentration of a solid or semi-solid by the
addition of drug or active ingredient is desired
(diluent cannot be evaporated).
„
Alligation may also be used when either pure
chemical (100%) or pure diluent (0% active
ingredient) is mixed with a certain
concentration to obtain a new strength.
„
Alligation is used to solve problems in which
concentration of a solid or semi-solid by the
addition of drug or active ingredient is desired
(diluent cannot be evaporated).
PG 182
66
Example D
How many grams of hydrocortisone powder must be
mixed with 1 lb of 2% hydrocortisone ointment to obtain
a 5% w/w ointment?
Step #1 100% [A] parts of HC [D]
5% [C]
2% [B] parts of 2% [E]
Step #2Parts of high-conc. [D] = [C] – [B] = 5 – 2 = 3 parts
Part of low-conc. [E] = [A] - [C] = 100 – 5 = 95 parts
PG 182
67
Step# 3
Hydrocortisone 3
parts of Hydrocortisone
= X g
95
parts of 2% ointment
454 g X= 14.33 g
To confirm your answer, determine the amount of
hydrocortisone in the new ointment and express it as
a % concentration:
Amount of HC in 2% ointment:
454 x 2% = 9.08 g
Total amount of HC in new ointment:
9.08 + 14.33 = 23.41 g
Weight of the new ointment:
454 + 14.33 = 468.33 g
% of hydrocortisone in new ointment
23.41 g
= X g
468.33 g 100 g X= 5%
PG 183
68
Based on Sodium Chloride Equivalents
Isotonicity Calculations
“E” is Key
What is “E”
“E” = NaCl equivalent
NaCl equivalentis the amount of NaCl
represented by another ingredient.
The E-value of a substance is the amount of
NaCl equivalent to 1 gram of that substance.
PG 185

18. 18
69
Based on Sodium Chloride Equivalents
Isotonicity Calculations
Isotonicity is based on
0.9% NaCl.
70
Based on Sodium Chloride Equivalents
Isotonicity Calculations
Steps to solve isotonicity problems:
1. Determine the weight in mg of all chemicals present.
2. Multiply each weight by the listed E value of the
chemical.
3. Add these weights together.
4. Determine the theoretical amount (in mg) of sodium
chloride that would be necessary if no other chemical
were present.
5. Subtract Step 3 value from Step 4 value.
PG 185
71
Based on Sodium Chloride Equivalents
Isotonicity Calculations
If another substance (e.g. boric acid) will be used
instead of sodium chloride to make the solution
isotonic, then we need to add an extra step:
PG 183
6. Calculate the amount of the substance to be
used by dividing the amount of sodium chloride
calculated in step 5 by the E-value of this
substance.
72
Based on Sodium Chloride Equivalents
Isotonicity Calculations
Steps to solve:
PG 183
1. Determine the weightof each item.
2. Multiply each weight by the listed $/lb
(convert the item to $).
3. Addthese amounts in $.
4. Determine the amount of $ on the gift
card.
5. Subtract Step 3 value from Step 4
valueto determine change to be
returned.
Gift Card
$

19. 19
73
Based on Sodium Chloride Equivalents
Isotonicity Calculations
Steps to solve:
e.g.
PG 185
6. If store will not return cash, and you are
forced to buy gum with the amount
remaining on the gift card, then you
dividethe $ amount remaining by the
price of a pack of gum to find how many
packs of gum you should get.
If you are owed $ 2.00, then you should
get:
$2.00 ÷ $0.50 = 4 packs of gum
$0.50/pack
74
Example A
How many mg of sodium chloride are needed to render 60 mL
of the following solution isotonic? Rx
(E Value)
Zinc chloride 0.2% w/v 0.62
Phenacaine HCl 1.0% 0.17
Pur. water qs 60 mL
Step#1 Zinc chloride 0.2 g
= x g
100 mL 60 mL x = 0.12 g = 120 mg
Phenacaine HCl 1 g
= x g
100 mL 60 mL x = 0.6 g = 600 mg
PG 186
75
Example A
How many mg of sodium chloride are needed to render 60 mL
of the following solution isotonic? Rx
(E value)
Zinc chloride 0.2% w/v 0.62
Phenacaine HCl 1.0% 0.17
Pur. water qs 60 mL
Step #2 Zinc chloride 120 mg x 0.62 = 74.4 mg
Phenacaine HCl 600 mg x 0.17 = 102 mg
76
Example A
How many mg of sodium chloride are needed to render 60 mL
of the following solution isotonic? Rx
(E value)
Zinc chloride 0.2% w/v 0.62
Phenacaine HCl 1.0% 0.17
Pur. water qs 60 mL
Zinc chloride 74.4 mg
Phenacaine HCl 102 mg
Step#3
176.4 mg

20. 20
77
Example A
Step#4
0.9 g NaCl = x g
100 mL 60 mL x = 0.54 g
or 540 mg of NaCl
(if no other chemical present)
Step#5 540 mg – 176 mg = 364 mg sodium chloride needed in every
60 mL of formula
78
Example B
How many mg of boric acid could be used in
place of sodium chloride in the last example?
364 mg sodium chloride needed in every 60 mL of formula
1 gram of boric acid is equivalent to 0.5 grams of sodium chloride
there for boric acid “E” value = 0.5
mg boric acid = 364
0.5
mg of boric acid = 728 mg of boric acid
79
Example C
How much sodium chloride is needed to make the following
solution isotonic? Assume that the phenylephrine solution is the
commercial isotonic solution, and the E value of phenylephrine HCl
is 0.184.
Rx
Phenylephrine 0.5% ophthalmic sol. – 30 mL
Sodium chloride qs
Pur. Water qs – 60 mL
In this example, the 30 mL of phenylephrine sol. Is already
isotonic. Therefore, it is necessary to add only sufficient sodium
chloride to adjust the remaining 30 mL of vehicle.
0.9 g x g x = 30 x 0.9
= 0.27 g or 270 mg
100 mL 30 mL 100
80
Based on Freezing Point Depression
Isotonicity Calculations
“D” is Key
What is “D”
“D” = FP depression caused by a
1% solution of the ingredient
PG 187

21. 21
81
Based on Freezing Point Depression
Isotonicity Calculations
Isotonicity is based on the premise that an
aqueous solution that has a total freezing
point depression of 0.52°C is isotonic.
82
Based on Freezing Point Depression
Isotonicity Calculations
Steps to solve isotonicity problems:
1. Convert the weight in mg of all chemicals present to
% concentration.
2. Multiply each % concentration by the listed D value of
the chemical.
3. Add these FP depressions together.
4. Subtract Step 3 from 0.52 (FP depression of isotonic
solution).
5. Determine how much sodium chloride needs to be
added per 100 mL of solution (knowing that a 0.9%
NaCl solution causes a FP depression of 0.52°C).
83
Example D
How many mg of sodium chloride are needed to render the following solution isotonic? Rx
(D Value)
Pilocarpine HCl 1 % 0.138
Benzyl alcohol 2.0 % 0.09
Pur. water qs 100 mL
Step#1 Not needed (already % concentration)
PG 187
84
Example D
How many mg of sodium chloride are needed to
render the following solution isotonic? Rx
(D Value)
Pilocarpine HCl 1 % 0.138
Benzyl alcohol 2.0 % 0.09
Pur. water qs 100 mL
Step #2
Pilocarpine HCl 1 x 0.138 = 0.138
Benzyl alcohol 2 x 0.09 = 0.180

22. 22
85
Example D
How many mg of sodium chloride are needed to
render the following solution isotonic? Rx
(D Value)
Pilocarpine HCl 1 % 0.138
Benzyl alcohol 2.0 % 0.09
Pur. water qs 100 mL
Pilocarpine HCl 0.138
Benzyl alcohol 0.180
Step#30.318
86
Step#4
0.52°C – 0.318°C = 0.20°C further depression needed
Step#5
0.9% NaCl = x% NaCl
0.52°C 0.20°C
x = 0.35% or 350 mg NaClper 100 mL, Ans.
Note:
Ifthe final volume was ordered as 60 mLinstead of 100 mL,
0.35 g = x g
100 mL 60 mL x = 0.21 g or 210 mg
87
Adjustments of Products to Isotonicity
An isotonic solution has approximately 300 mOsm/L.
How many mL of water should be added to 8 ounces of
Comply Liquid (410 mOsm/L) to obtain an approximate
isoosmotic solution (300 mOsm/L)?
Q
1
C
1
= Q
2
C
2
(240 mL)(410 mOsm/L) = (x mL)(300 mOsm/L)
x = 328 mL of total solution
Therefore, 328 mL – 240 mL = 88 mL
88
Calculations Involving
Milliequivalents
Milliequivalents⎯the amount, in milligrams, of the
solute equal to 1/1,000 of its gram equivalent
weight
Alternatively, a milliequivalent (mEq) is the
equivalent weight expressed in mg.
PG 188

23. 23
89
Calculations Involving
Milliequivalents
Knowledge of valences of common ions used in medicine.
PG 188
90
Method 1 is the standard method usually presented in
chemistry.
1. Determine the atomic, molecular, or formula weight of the
ion or molecule.
2. Determine the equivalent weight by dividing the above
weight by the valence of either the anion or cation.
3. Express this equivalent weight in mg to obtain the
milliequivalent weight.
Method 1
91
Example
A 20-mL vial contains 20 mEq of potassium chloride. What weight of
chemical is present? (K = 39; Cl = 35.5)
Determine the atomic, molecular, or formula weight of the ion or molecule.
39 + 35.5 = 74.5
Determine the equivalent weight by dividing the above weight by the
valence of either the anion or cation
Eq. wt = 74.5 = 74.5
1
Express this equivalent weight in mg to obtain the milliequivalent weight.
1 mEq = 74.5 mg
20 mEq x mg
x = 20 ×74.5 = 1490 mg, Ans.
92
PG 189
Example
A 20-mL vial contains 20 mEq of potassium chloride. What weight of
chemical is present? (K = 39; Cl = 35.5)
1,490mg of KCl present
How many mg of KCl are present in each mL?
1,490 mg
= X mg
20 mL 1 mL
X = 74.5 mg/mL

24. 24
93
Determine the concentration of potassium ionas mEq
in a solution containing 0.3 g of KCl per 100 mL. (Mol. wt
KCl = 74.5; K = 39; Cl = 35.5)
Method I:
Step 1. Molecular weight of KCl = 74.5
Step 2. 1 Eq wt of KCl = 74.5
Step 3. 1 mEq of KCl
= 74.5 mg
x mEq 300 mg x = 4 mEq, Ans.
Note that 1 mEq of KCl provides 1 mEq of K
+
and 1 mEq of Cl
-
In the book, K
+
content of KCl is calculated, not necessary (same
answer).
94
Milliequivalents Milliequivalents
Note that:
1 mEq of KCl provides 1 mEq of K
+
and 1 mEq of Cl
-
1 mEq of CaCl
2
provides
1 mEq of Ca
++
and 1 mEq of Cl
-
1 mEq of NaCl provides
1 mEq of Na
+
and 1 mEq of Cl
-
1 mEq of Na
2
SO
4
provides
1 mEq of Na
+
and 1 mEq of SO
4
--
Note that:
1 mEq of KCl provides 1 mEq of K
+
and 1 mEq of Cl
-
1 mEq of CaCl
2
provides
1 mEq of Ca
++
and 1 mEq of Cl
-
1 mEq of NaCl provides
1 mEq of Na
+
and 1 mEq of Cl
-
1 mEq of Na
2
SO
4
provides
1 mEq of Na
+
and 1 mEq of SO
4
--
95
Method 2
96
Method 2
Determine the concentration of potassium ion as mEq in a
solution containing 0.3 g of KCl per 100 mL. (mol. wt KCl =
74.5; K = 39; Cl = 35.5)
mg KCl = (mEq) (molecular wt)
valence
300 mg = x mEq (74.5)
1
x = 4 mEq

25. 25
97
*Given the following prescription for calcium carbonate,
how many Milliequivalents of Calcium will the patient
consume each day. (The atomic weights of atoms in
CaCO
3
are Ca = 40, C = 12 O = 16)
Rx:
Calcium Carbonate 500mg tablets
# 180
Sig: 2 tablet tid
Daily dose = 2 x 500 mg x 3 = 3,000 mg
mg CaCO
3
= mEq (molecular wt)
valence
3,000 mg = X mEq (100)
2
X = 60 mEq
98
Example B Example B
„
A 10-mL vial is labeled Potassium Chloride (2
mEq/mL). How many grams of potassium
chloride are present? (Mol. Wt.: potassium
chloride = 74.5)
10 mL . 2 mEq/mL = 20 mEq KCl total
1 mEq = 74.5 mg of KCl
20 mEq x mg of KCl
x = 1,490 mg or 1.49 g
„
A 10-mL vial is labeled Potassium Chloride (2
mEq/mL). How many grams of potassium
chloride are present? (Mol. Wt.: potassium
chloride = 74.5)
10 mL . 2 mEq/mL = 20 mEq KCl total
1 mEq = 74.5 mg of KCl
20 mEq x mg of KCl
x = 1,490 mg or 1.49 g
99
Example C Example C
„
A pharmacist has a 1-liter bottle containing 24.5 g of hydrated
calcium chloride. How many mEq per mL of calcium chloride are
present? (Anhydrous calcium chloride = 111; hydrated calcium
chloride = 147)
1 mEq = 147 mg of calcium chloride
2
1 mEq = 73.5 mg of calcium chloride
x mEq 24,500 mg of calcium chloride
x = 333 mEq in 1 Liter
333 mEq = x mEq
1000 mL 1 mL
x = 0.333 mEq/mL
„
A pharmacist has a 1-liter bottle containing 24.5 g of hydrated
calcium chloride. How many mEq per mL of calcium chloride are
present? (Anhydrous calcium chloride = 111; hydrated calcium
chloride = 147)
1 mEq = 147 mg of calcium chloride
2
1 mEq = 73.5 mg of calcium chloride
x mEq 24,500 mg of calcium chloride
x = 333 mEq in 1 Liter
333 mEq = x mEq
1000 mL 1 mL
x = 0.333 mEq/mL
100
Example D Example D
„
A solution contains 10 mg% of potassium ions.
Express this concentration as mEq/L.
(atomic weight: K = 39; Cl = 35.5)
10 mg% means 10 mg per 100 mL or 100 mg/L
100 mg = (x mEq)(39)
1
x = 2.6 mEq/L
„
A solution contains 10 mg% of potassium ions.
Express this concentration as mEq/L.
(atomic weight: K = 39; Cl = 35.5)
10 mg% means 10 mg per 100 mL or 100 mg/L
100 mg = (x mEq)(39)
1
x = 2.6 mEq/L

26. 26
101
Example E Example E
„
Potassium gluconate elixir contains 20 mEq of
potassium gluconate per tablespoon. How many grams
of potassium gluconate are present in every 100 mL?
(K = 39; potassium gluconate = 234)
x mg = (20 mEq)(234)
= 4,680 mg in 15 mL
1
4,680 mg
=x mg
15 mL 100 mL
x = 31,200 mg or 31.2 g
„
Potassium gluconate elixir contains 20 mEq of
potassium gluconate per tablespoon. How many grams
of potassium gluconate are present in every 100 mL?
(K = 39; potassium gluconate = 234)
x mg = (20 mEq)(234)
= 4,680 mg in 15 mL
1
4,680 mg
=x mg
15 mL 100 mL
x = 31,200 mg or 31.2 g
102
Example F Example F
„
How many mg of anhydrous aluminum chloride are needed to
prepare 200 mL of a solution that will contain 40 mEq of
aluminum in 1 liter? (Mol. Wt: aluminum chloride = 133, Al = 27)
1 mEq = 133 mg of Aluminum Chloride
3
1 mEq = 44.33 mg of Aluminum Chloride
40 mEq x mg of Aluminum Chloride
x = 1,777 mg in 1 Liter
1,777 mg = x mg
1,000 mL 200 mL
x = 355 mg
„
How many mg of anhydrous aluminum chloride are needed to
prepare 200 mL of a solution that will contain 40 mEq of
aluminum in 1 liter? (Mol. Wt: aluminum chloride = 133, Al = 27)
1 mEq = 133 mg of Aluminum Chloride
3
1 mEq = 44.33 mg of Aluminum Chloride
40 mEq x mg of Aluminum Chloride
x = 1,777 mg in 1 Liter
1,777 mg = x mg
1,000 mL 200 mL
x = 355 mg
103
Example G Example G
„
How many mEq of sodium are present in the following
admixture order? (Na = 23, Cl = 35.5)
“Add sodium chloride (2.5 mEq/mL) 20 mL to 1 liter D5W/1/2NS
and infuse over 8 hours”
mEq of NaCl in 20 mL = 2.5 x 20 = 50 mEq
Amount of NaCl in 1 liter of ½ NS:
0.45 g = x g x = 4.5 g or 4,500 mg
100 mL 1000 mL
1 mEq of Na
+
= 58.5 mg of NaCl
x mEq of Na
+
4,500 mg of NaCl
x = 76.9 mEq
Total mEq of Na
+
= 50 + 76.9 = 126.9 or 127 mEq
„
How many mEq of sodium are present in the following
admixture order? (Na = 23, Cl = 35.5)
“Add sodium chloride (2.5 mEq/mL) 20 mL to 1 liter D5W/1/2NS
and infuse over 8 hours”
mEq of NaCl in 20 mL = 2.5 x 20 = 50 mEq
Amount of NaCl in 1 liter of ½ NS:
0.45 g = x g x = 4.5 g or 4,500 mg
100 mL 1000 mL
1 mEq of Na
+
= 58.5 mg of NaCl
x mEq of Na
+
4,500 mg of NaCl
x = 76.9 mEq
Total mEq of Na
+
= 50 + 76.9 = 126.9 or 127 mEq
104
The milliosmole (mOsm) is a measurement that is used for
parenteral solutions.
Determination of mOsm involves two simple steps.
1. Determine the number of millimoles present.
Weight of drug in grams = moles ×1000 = millimoles
molecular weight
2. Multiply this value by the theoretical number of particles or
ions present (assuming complete disassociation).
mOsm = Wt. of a substance in g ×1000×# of species
molecular weight
Osmolarity Calculations
PG 192

27. 27
105
# of species:
Sodium chloride
Calcium chloride
Potassium chloride
Sodium sulfate
Magnesium sulfate
Zinc sulfate
Sodium acetate
Glucose
Osmolarity Calculations
NaCl1Na
+
+1Cl
-
=2
CaCl
2
1Ca
++
+2 Cl
-
=3
KCl1K
+
+1Cl
-
=2
Na
2
SO
4
2Na
+
+1SO
4
--
=3
MgSO
4
1Mg
++
+1SO
4
--
=2
ZnSO
4
1Zn
++
+1SO
4
--
=2
NaAc1Na
+
+1Ac
-
=2
Glucose1Glucose =1
106
Example A
How many mOsm are present in 1 liter of sodium chloride
injection? (Mol. wt: sodium chloride = 58.5)
0.9 g
= X g
100 mL 1000 mL X = 9 g of NaCl in 1 liter
Step 1.
millimoles = Wt. of a substance in g
×1000
molecular weight
millimoles = 9 g
×1000 = 154 millimoles
58.5
Step 2.
mOsm = millimoles x # of species
mOsm= 154 x 2 =308 mOsm
107
Example B
How many mOsm are present in 1 liter of D5W?
(Mol. Wt. of dextrose = 180)
5 g
= X g
100 mL 1000 mL X = 50 g of dextrose in 1 liter
Step 1.
millimoles = Wt. of a substance in g
×1000
molecular weight
millimoles = 50 g
×1000 = 278 millimoles
180
Step 2.
mOsm = millimoles x # of species
mOsm= 278 x 1 =278 mOsm/L
108
Example C
Determine the mOsm/L concentration of calcium chloride
(mol. wt = 147) when 132 mg is dissolved in 100 mL of water.
0.132 g
= X g
100 mL 1000 mL X = 1.32 g of CaCl
2
in 1 liter
Step 1.
millimoles = Wt. of a substance in g
×1000
molecular weight
millimoles = 1.32 g
×1000 = 9 millimoles
147
Step 2.
mOsm = millimoles x # of species
mOsm= 9 x 3 =27 mOsm/L

28. 28
109
A solution contains 448mg of KCl (MW=74.5) and 468mg
of NaCl (MW = 58.5) in 500mL. What is the osmolar
concentration of this solution?
0.448 g
X 1,000 = 6 millimoles ×2 = 12 mOsm/500mL
74.5
0.468 g
X 1,000 = 8 millimoles ×2 = 16mOsm/500mL
58.5
12 + 16 = 28 mOsm
= X mOsm
500 mL 1000 mL X = 56 mOsm/L
A solution contains 448mg of KCl (MW=74.5) and 468mg
of NaCl (MW = 58.5) in 500mL. What is the osmolar
concentration of this solution?
0.448 g
X 1,000 = 6 millimoles ×2 = 12 mOsm/500mL
74.5
0.468 g
X 1,000 = 8 millimoles ×2 = 16mOsm/500mL
58.5
12 + 16 = 28 mOsm
= X mOsm
500 mL 1000 mLX = 56 mOsm/L
110
Some Terms
Hypotonic solution⎯
Hypertonic solution⎯
111
Some Terms
Hypotonic solution
⎯
having a lesser
osmotic pressure, cells would swell
Hypertonic solution
⎯
112
Some Terms
Hypotonic solution⎯
having a lesser
osmotic pressure, cells would swell
Hypertonic solution⎯
having a greater
osmotic pressure, cells would shrink

29. 29
DOUBLE DILUTION
Miscellaneous Pharmaceutical
Calculations
How many mL of 17% benzalkonium chloride
should be used to make 240 mL of a solution of
benzalkonium chloride such that 10 mL diluted to
a liter equals a 1:5000 solution.
114
How many mL of 17% benzalkonium chloride should be used to make 240 mL of a solution of
benzalkonium chloride such that 10 mL diluted to
one liter equals a 1:5000 solution.
Q
1
C
1
= Q
2
C
2
x mL (17%) = 240 mL (2%)
x = 28.2 mL
Q
1
C
1
= Q
2
C
2
10 mL (1/x) = 1000 mL (1/5000)
x = 50
Conc. is 1: 50 or 2%
115
Dilution of Acids
• Concentrated acids are expressed as w/w
• Diluted acids are expressed as w/v
• Conversion:
w/w x density = w/v
Miscellaneous Pharmaceutical
Calculations
PG 194
116
Dilution of Acids
Example: Example:
How many milliliters of 85.7% How many milliliters of 85.7% w/ww/wlactic acid lactic acid
(sp. gr. = 1.19) are needed to make 120 (sp. gr. = 1.19) are needed to make 120 mLmLofof
10% 10% w/vw/vlactic acid? lactic acid?
Miscellaneous Pharmaceutical
Calculations
PG 194
Q
1
C
1
= Q
2
C
2
x mL (85.7% x 1.19) = 120 mL (10%)
x = 11.8 mL

30. 30
117
Special Concentration Expressions for Alcohol
Alcohol USP consists of 95% v/v ethanol.
Whereas Alcohol USP is used for formulation work,
commercial labels indicate a product’s concentration
in terms of absolute alcohol. That is, a label stating
40% alcohol contains the equivalent of 40 mL of pure
alcohol in each 100 mL despite the fact that Alcohol
USP was used during manufacturing.
Miscellaneous Pharmaceutical
Calculations
PG 194
118
How many mL of alcohol USP are needed to
prepare 1 gallon of cough elixir to contain 18% v/v
ethanol? (
1 gal = 3,784
)
Q
1
C
1
= Q
2
C
2
3784 mL (18%) = X mL(95%)
X = 717mL
119
Alcohol concentrations can also be expressed
in terms of “proof” or “proof strength.”
Simply remember that the proof strength is
double that of the actual percent concentration.
Thus, Alcohol USP is 190 proof because
95% ×2 = 190.
Volumes can be expressed as “proof gallons”
.
A proof gallon is the equivalent of 1 gallon of
50% v/v alcohol.
120
How many proof gallons are present in 4 gallons of alcohol USP?
Q
1
C
1
= Q
2
C
2
x gal (50%) = 4 gal (95%)
X = 7.6 gal

31. 31
121
How much water should be added to 4 gallons
of alcohol USP to prepare 40% alcohol?
Q
1
C
1
= Q
2
C
2
x gal (40%) = 4 gal (95%)
X = 9.5 gal
The exact amount of water to be added cannot
be calculated.
Add enough water to make 9.5 gallons.
122
Magnesium Sulfate
Magnesium Sulfate is an interesting pharmaceutical
chemical.
There are two forms available.
1. Anhydrous magnesium sulfate (MgSO
4
)
2. Hydrous magnesium sulfate (MgSO
4
•7H
2
O)
The different forms of MgSO
4
have the following molecular
weights:
• Anhydrous magnesium sulfate = 120
• Hydrous magnesium sulfate = 246
• Atomic wt magnesium = 24
123
Magnesium Sulfate
Anhydrous Hydrous
magnesium sulfate magnesium sulfate
1 molar
120 g/L 246 g/L
1 molal
120 g + 1,000 g of water 246 g + 1,000 g of water
1 normal
60 g/L 123 g/L
124
Conversion Between the Two
Forms of Magnesium
How many grams of anhydrous magnesium sulfate
are needed to obtain 40 g of hydrous magnesium
sulfate?
PG 196

32. 32
125
PG 196
Example
How many grams of hydrous magnesium sulfate are
needed to obtain 16 mEq of magnesium ion?
1 mEq = 246 mg of magnesium sulfate
2
1 mEq = 123 mg of magnesium sulfate
16 mEq x mg of magnesium sulfate
x = 1,968 mg
or 1.97 g of hydrous magnesium sulfate
126
Statistics
Use of Statistics and Graphs
Median.The median is the middle value in a series.
Themedian is determined by eliminating the highest
value against the lowest and repeating the process until
only one value remains.
PG 195
Themean is calculated by totaling all of the values and
dividing by the N
127
Example:An experiment reported the following
values: 45, 55, 20, 38, 52. Calculate the mean and the
median.
Calculation of
Mean and Median
Mean = (45 + 55 + 20 +38 +52)
= 42
5
Themedian is determined by eliminating the highest
value against the lowest and repeating the process until
only one value remains.
20, 38, 45, 52, 55 Median =45
128
Mean Deviation Standard Deviation
Candidates for the NAPLEX® will not be
expected to calculate these values but should
understand what a standard deviation
represents. It is the square root of variability.

33. 33
129
Probability (p)indicates chances that something will happen by accident or
will be outside a certain range. For example, a p value of 0.05 means that an
accidental or an erratic value will occur only 5% of the time
Bias or systematic errordescribes the tendency for measuring something
other than that intended; for example, showing a high incidence of hospital
drug-related deaths by using cancer ward patients
Precision (reproducibility)refers to close agreement in the values obtained
Accuracyis closeness of values to the correct value
t Test of significance (Student’s t test) and chi-square test of
significanceare mathematic methods of comparing sets of data to see if
they are significantly different. Must use prepared tables to evaluate
comparisons.
PG 198
130
Standard Deviation Standard Deviation
„
Indication of the spread of the data.
„
A small standard deviation is an indication of
a narrow spread of the data.
„
A large standard deviation is an indication of
a wide spread of the data.
„
Indication of the spread of the data.
„
A small standard deviation is an indication of
a narrow spread of the data.
„
A large standard deviation is an indication of
a wide spread of the data.
131
Gaussian Distribution Gaussian Distribution
„
For a Gaussian distribution:
z
68% of data are within ± 1 SD
z
95.5% of data are within ± 2 SD
z
99.7% of data are within ± 3 SD
„
For a Gaussian distribution:
z
68% of data are within ± 1 SD
z
95.5% of data are within ± 2 SD
z
99.7% of data are within ± 3 SD
132
Standard Deviation Standard Deviation
„
Example
If 100 capsules had an average weight of
500 mg ± 20 mg, how many weigh between:
480 and 520 mg?
68 capsules
„
Example
If 100 capsules had an average weight of
500 mg ± 20 mg, how many weigh between:
480 and 520 mg?
68 capsules

34. 34
133
Standard Deviation Standard Deviation
„
Example
If 100 capsules had an average weight of
500 mg ± 20 mg, how many weigh between:
460 and 540 mg?
95 capsules
„
Example
If 100 capsules had an average weight of
500 mg ± 20 mg, how many weigh between:
460 and 540 mg?
95 capsules
134
Standard Deviation Standard Deviation
„
Example
If 100 capsules had an average weight of
500 mg ± 20 mg, how many weigh between:
460 and 500 mg?
47 capsules
(95.5 ÷ 2)
„
Example
If 100 capsules had an average weight of
500 mg ± 20 mg, how many weigh between:
460 and 500 mg?
47 capsules
(95.5 ÷ 2)
135
Standard Deviation Standard Deviation
„
Example
If 100 capsules had an average weight of
500 mg ± 20 mg, how many weigh between:
500 and 520 mg?
34 capsules
(68 ÷ 2)
„
Example
If 100 capsules had an average weight of
500 mg ± 20 mg, how many weigh between:
500 and 520 mg?
34 capsules
(68 ÷ 2)
136
From Figure 1, one may conclude that:
I. Sodium lauryl sulfate appears to increase the rate of dissolution.
II. At 10 min, approximately 40 mg of sodium lauryl sulfate has dissolved.
III. The rate of dissoluti on of tablets containing magnesium stearate is greater than
the control.
Evaluation of Graphic Data
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
PG 199

35. 35
137
From Figure 1, one may conclude that:
I. Sodium lauryl sulfate appears to increase the rate of dissolution.
II. At 10 min, approximately 40 mg of sodium lauryl sulfate has dissolved.
III. The rate of dissoluti on of tablets containing magnesium stearate is greater than
the control.
Evaluation of Graphic Data
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
PG 199
138
From Figure 2, one may conclude that:
I. The presence of starch decreases the dissolution rate of salicylic acid tablets.
II. The dissolution rate for tablets with 10% starch follows first-order kinetics.
III. Inclusion of 20% starch increases th e dissolution of the tablets by more than
double the rate of 10% starch.
Evaluation of Graphic Data
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
PG 200
139
From Figure 2, one may conclude that:
I. The presence of starch decreases the dissolution rate of salicylic acid tablets.
II. The dissolution rate for tablets with 10% starch follows first-order kinetics.
III. Inclusion of 20% starch increases th e dissolution of the tablets by more than
double the rate of 10% starch.
Evaluation of Graphic Data
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
PG 200