Pharmaceutical Calculations
Pharmaceutical
Calculations
Pharmaceutical calculations are
directly tested on the NAPLEX®.
4
p. 165
2
Pharmaceutical Conversions
Pharmaceutical Conversions
King
Kilo-
Henry died
drinking chocolate
milk
Hecto- deca- U decicentimilli1000 x 100 x
10 x U
1/10 x
1/100 x
1/1000 x
Whenever units are expressed in either
the apothecary or the avoirdupois, it is
convenient to immediately convert the
units to metric units (SI units).
L 1 dL = 1/10 L = 0.1 L
1 Kg = 1000 g
g
1 mg =
1/1000 g
Micro: one millionth (1 x 10-6)
Nano: one billionth (1 x 10-9)
Mega: one million (1 x 106)
Giga: one billion (1 x 109)
Tera: one trillion (1 x 1012)
3
4
1
Weight Measurements
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
5
PG 166
6
PG 166
Roman Numerals
Length Measurements
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
1 inch = 2.54 centimeters (cm)
1 ft = 12 inches
PG 166
7
8
2
Roman Numerals
Rules for Roman Numerals
In writing prescriptions physicians or
other health care professionals may use
small or capital roman numerals.
z
z When the small letter i is used it should be
dotted to distinguish it from the letter l.
z
z Sometimes a j may be used for the final i in
a sequence (e.g. viij).
z
z Following the Latin custom, Roman
numerals are generally placed after the
symbol or term (e.g. capsules no. xxiv or
fluidounces xij)
z
z
„
„
Two or more letters express a quantity that is
the sum of their values if they are
successively equal or smaller in value.
z
vii = 7 XIII = 13
z e.g. ii = 2
„
„
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
IX = 9 CM = 900
z e.g. IV = 4
9
10
Examples
Drug Dosing
„
„
lxvi
= 66
Dosing based on body weight
„
„
xix
= 19
ƒ Two steps
„
„
xcix
= 99
„
„
cclxx = 270
„
„
cxl
1. Convert the pounds to kilograms
2. Multiply dose by the body weight
= 140
11
PG 166
12
3
Drug Dosing
Drug Dosing
Example A:
How many mg of Tobramycin are needed for
a patient weighing 160 lbs if the desired dose
is 0.8 mg/kg?
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?
1 kg = 2.2 lbs.
x kg
1 kg = 2.2 lbs.
x kg
160 lbs.
154 lbs.
1 kg
x = 72.7 kg
70 kg
1 tablet
0.8 mg = x mg
1 kg
72.7 kg
x tablets
x = 58 mg
PG 166
PG 167
= 50 mg
140 mg
x = 2.8 tablets or 3 tablets
(NOT: 2.8 x 10)
14
PG 167
Nomogram for Determining Surface Areas
Dosing Based on
Body Surface Area
m2
Surface area of patient (m 2 )
× Average adult dose
1.73
Patient' s BSA (m 2 ) =
x = 140 mg
For 10 days: 3 x 10 = 30 tablets
13
Child or adult dose =
x = 70 kg
2 mg = x mg
Patient' s height (cm) × Patient' s weight (kg)
3600
15
PG 168
16
4
Example C
Example D
The average child dose of a drug is
200 mg/m2. Estimate an appropriate
dose for a 4-year-old child who is 4
feet tall and weighs 80 lb.
The surface area of a 4-year-old
child is estimated to be 0.5 m2. How
many mg of a drug which has an
adult dose of 100 mg should be
administered?
m2 = 1.1
Child dose = surface area × adult dose
1.75
200 mg/m2 x 1.1 m2 = 220 mg
Dose = 0.5m2 × 100 mg
1.75
PG 168
17
Dosing for Children Based on Weight or Age
18
Dosing for Children Based on Weight or Age
Young’s Rule: Also intended for
children greater than 2 years of age
Clark’s Rule: a weight calculation
intended for children greater than 2
years of age
PG 169
Dose = 29 mg
19
20
5
Example
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.
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.
8
× 250mg
(8 + 12)
Young’s: Dose =
Clark’s: Dose = 84 lb × 250 mg
150 lb
Dose = 100 mg
Dose = 140 mg
21
Example
22
Concentration Expressions
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.
ƒ Percent (%)
ƒ Parts per million (ppm)
ƒ Ratio strength (1 : something)
Clark’s:
Young’s:
PG 169
23
PG 171
24
6
Concentration Expressions
as Percentages
The following thought process should be
followed to solve concentration problems:
The amount of active ingredient present
in a specified amount of final product.
1. Determine total weight or volume of the
formula or prescription.
A 10% w/v indicates that 10 g of
ingredient is present in every 100 mL of
formula.
2. Determine what percentage of a certain
ingredient is being requested for the final
product.
A 10% w/w means a 10 g of drug are
contained in every 100 g of product.
3. Calculate weight or volume of that ingredient.
4. Some cases calculate weight or volume of
solvent (vehicle) needed.
A 10% v/v means a 10 mL of drug are
contained in every 100 mL of product.
25
PG 171
26
PG 172
Example A
Example B
Rx:
Rx:
Procaine HCl 2% w/v
Benzocaine 1% w/w
Pur. water qs—120 mL
Petrolatum qs 100%
1. Total volume 120 mL
2. Procaine HCl 2% w/v. =
M & Ft Oint Disp—60 g
2g
100 mL
1. The prescription calls for 60 g of product.
3. The amount of pure procaine will be:
2g
100 mL
=
2. Of the 60 g, 1% will be benzocaine. 1 g = X g
100 g 60 g
Xg
120 mL
X = 2.4 g
27
3.
The amount of benzocaine will be 0.6 g
4.
60 g − 0.6 g = 59.4 g of petrolatum.
X = 0.6 g
28
7
Concentration Expressions –
Parts per Million
Example A
Express 0.025% w/v as ppm.
Because 0.025% refers to 0.025 g per 100 mL of
solution
Concentration expressions based
on a denominator of 1 million
(1,000,000).
0.025 g
100 mL
= X
1,000,000
X= 250 ppm
One part per million is:
1
1,000,000
29
30
PG 173
Example C
Example B
Express 20 ppm as a percentage.
A water supply contains 12 ppm of calcium chloride
as an impurity. How many mg of calcium chloride are
present in every dL?
We know 20ppm is
12 ppm indicates
20
1,000,000
12 parts (g) per 1,000,000 parts (mL)
So:
= X
20
1,000,000 100
= xg
12 g
1,000,000 mL 100 mL
X = 0.002 = 0.002%
x = 0.0012 g or 1.2 mg/100 mL
31
32
8
Example D
Ratio Strength
The blood level of a drug is 8 µg/dL.
Express this concentration in terms of
parts per million.
„
„
An expression of concentration using
ratios
„
„
Used when a low concentration is present
„
„
Always expressed as 1: something
33
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
34
Calculations Involving Specific Gravity
ƒ Converting a liquid measurement to a weight quantity
ƒ Converting a weight to a volume measurement
600 mg = 0.6 g
ƒ Determining the volume cost of a drug purchased
ƒ Determining the density of a solution
0.6 g = 1
3000 mL X
Answer:
X = 5,000 mL
1 : 5,000
35
PG 174
36
9
Example A
The specific gravity is a ratio of the
weight of a certain volume compared with
the weight of the same volume of water.
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
The density of water is: 1 g = 1 g/mL
1 mL
Vol = 10 = 11.9 mL
0.84
The specific gravity of water = 1
37
Example C
Example B
What is the cost per mL of an elixir (sp. gr. =0.92)
if the bulk cost is $12.40/lb?
What weight of glycerin (SG = 1.25) must be
used to obtain 120 mL of glycerin?
1. Determine volume of 1 lb of elixir:
SG = density = wt
vol
1.25 =
38
SG = density = wt
vol
0.92 = 454 g
x mL
2. Determine the cost per mL:
493 mL = $12.40
1 mL = x
wt
120 mL
Wt =120 x 1.25 = 150 g
39
x = 493 mL
x = $0.025/mL
40
10
Calculations Involving
Adjustments of Strengths
The pharmacist may have to adjust the
concentration of an existing preparation
to a new strength.
DILUTION PROBLEMS
Problems involving adjustment of strength
may be classified into:
ƒ Dilution problems
ƒ Concentration problems
ƒ Alligation problems
PG 175
The pharmacist must be able to
state the new strength of a
solution if given the original
concentration and the amount of
diluent added.
41
Method 1
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.
PG 176
44
11
Example A
Example B
What is the new strength of a solution prepared by
diluting 120 mL of a 5% w/v solution with 380 mL
of water?
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. 120 mL x 5% = 6g (of active ingredient)
Step 1.
5g
= Xg
100 mL 120 mL
X=6g
Step 3.
= Xg
6g
500 mL
100 mL
0.0005 g
5 mL
=
1g
x mL
0.005 x = 5
x = 10,000 or conc. is 1:10,000
X = 1.2%
45
PG 176
X = 0.0005 g
(active ingredient)
Step 2. 4.5 mL + 0.5 mL = 5 mL (new volume)
Step 2. 120 mL + 380 mL = 500 mL (new volume)
Step 3.
=
Xg
1g
1,000 mL 0.5 mL
46
PG 176
Method 2
Example A
Q1C1 = Q2C2
What is the new strength of a solution prepared
by diluting 120 mL of a 5% w/v solution with
380 mL of water?
Two guidelines must be followed:
Q1C1 = Q2C2
1. The original and new quantities must be
expressed in identical units.
(120 ml)(5% w/v) = (500 ml)(X% w/v)
2. The original and new concentrations should be
expressed in identical terms (% w/v, % w/w, % v/v,
ppm, ratio strength, etc…)
PG 177
47
600 = 500(X)
X = 1.2%
48
12
Example B
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?
Q1C1 = Q2C2
(0.5 mL)(1/1,000) = (5 mL)(1/x)
0.5 = 5
1,000 x
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.
0.5 x = 5,000
x = 5,000/0.5 = 10,000 or conc. is 1:10,000
49
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?
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.
Q1C1 = Q2C2
(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
PG 177
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13
EXAMPLE E
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?
How much of a 1/1,500 strength solution can be prepared by
the dilution of 60 mL of a 1/120 strength solution?
Q1C1 = Q2C2
(473 mL)(25% w/v) = (x mL)(6% w/v)
(60 mL)(1/120) = (x mL)(1/1,500)
60 = x
120 1,500
11,825 = 6x
120x = 90,000
Q1C1 = Q2C2
x = 1,970 mL
PG 178
x = 750 mL
53
54
EXAMPLE F
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.
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?
Q1C1 = Q2C2
(x mL)(5% w/v) = (16 mL)(0.2% w/v)
5x = 3.2
x = 0.64 mL
or
0.2 g = x g
100 16 mL
5 g = 0.032 g
100
x mL
x = 0.032 g
x = 0.64 mL
56
14
EXAMPLE A
Concentration Problems
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?
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?
Xg
0.05 g =
100 mL
5000 mL
1 g = 2.5 g
100 mL X mL
Q1C1 = Q2C2
(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
X = 2.5 g
X = 250 mL
5,000 mL − 250 mL = 4,750 mL must be evaporated
57
PG 179
Adjustment of Strength Using Mixtures –
Alligation Alternate Method
58
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
Step #2
7 parts
x mL
8 parts
Parts of high-concentration ingredient [D] = [C] − [B]
Parts of low-concentration ingredient [E] = [A] − [C]
Total: 15 parts
59
PG 180
120 mL
60
15
Step# 3
Alligation
20% solution
*7parts
= XmL
„
„
15total parts 120 mL X= 56 mL
5% solution
8 parts = XmL
15total parts 120 mL X= 64 mL
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).
Mixing 56 mL of 20% solution and 64 mL of 5% solution will
result in 120 mL of a 12% concentration.
61
62
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 #2 Parts 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 =
Step #1
20% [A]
parts of 20% [D]
15 parts of 2% oint
x g
200 g
5% [C]
2% [B]
x = 3 x 200 = 40 g of 20% ointment
parts of 2% [E]
15
Step #2 Parts of high-conc. [D] = [C] – [B] = 5 - 2 = 3 parts
Part of low-conc. [E] = [A] - [C] = 20 – 5 = 15 parts
PG 181
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64
16
Example D
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.
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]
„
„
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).
65
PG 182
Step #2 Parts of high-conc. [D] = [C] – [B] = 5 – 2 = 3 parts
Part of low-conc. [E] = [A] - [C] = 100 – 5 = 95 parts
Isotonicity Calculations
Hydrocortisone
Based on Sodium Chloride Equivalents
“E” is Key
What is “E”
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
66
PG 182
Step# 3
3 parts of Hydrocortisone = X g
95 parts of 2% ointment
454 g
parts of 2% [E]
67
“E” = NaCl equivalent
NaCl equivalent is 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
68
17
Isotonicity Calculations
Isotonicity Calculations
Based on Sodium Chloride Equivalents
Based on Sodium Chloride Equivalents
Steps to solve isotonicity problems:
1. Determine the weight in mg of all chemicals present.
Isotonicity is based on
2. Multiply each weight by the listed E value of the
chemical.
0.9% NaCl.
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.
69
70
PG 185
Isotonicity Calculations
Isotonicity Calculations
Based on Sodium Chloride Equivalents
Based on Sodium Chloride Equivalents
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:
Steps to solve:
2. Multiply each weight by the listed $/lb
(convert the item to $).
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.
3. Add these amounts in $.
4. Determine the amount of $ on the gift
card.
Gift Card
$
PG 183
1. Determine the weight of each item.
71
PG 183
5. Subtract Step 3 value from Step 4
value to determine change to be
returned.
72
18
Isotonicity Calculations
Example A
Based on Sodium Chloride Equivalents
Steps to solve: 6. If store will not return cash, and you are
forced to buy gum with the amount
remaining on the gift card, then you
divide the $ 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:
e.g.
$0.50/pack
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
$2.00 ÷ $0.50 = 4 packs of gum
0.2 g = x g
100 mL
60 mL
x = 0.12 g = 120 mg
1g = xg
100 mL 60 mL
x = 0.6 g = 600 mg
Phenacaine HCl
73
PG 185
74
PG 186
Example A
Example A
How many mg of sodium chloride are needed to render 60 mL
of the following solution isotonic?
How many mg of sodium chloride are needed to render 60 mL
of the following solution isotonic?
Rx
Rx
(E value)
(E value)
Zinc chloride
0.2% w/v
0.62
Zinc chloride
0.2% w/v
0.62
Phenacaine HCl
1.0%
0.17
Phenacaine HCl
1.0%
0.17
Pur. water qs
60 mL
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
75
Zinc chloride
74.4 mg
Phenacaine HCl
102 mg
Step#3
176.4 mg
76
19
Example A
Example B
How many mg of boric acid could be used in
place of sodium chloride in the last example?
Step#4
0.9 g NaCl =
xg
100 mL
60 mL
364 mg sodium chloride needed in every 60 mL of formula
x = 0.54 g
or 540 mg of NaCl
(if no other chemical present)
1 gram of boric acid is equivalent to 0.5 grams of sodium chloride
there for boric acid “E” value = 0.5
Step#5
mg boric acid = 364
0.5
540 mg – 176 mg = 364 mg sodium chloride needed in every
60 mL of formula
mg of boric acid = 728 mg of boric acid
77
78
Example C
Isotonicity Calculations
Based on Freezing Point Depression
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.
“D” is Key
What is “D”
Rx
Phenylephrine 0.5% ophthalmic sol. – 30 mL
Sodium chloride qs
Pur. Water qs – 60 mL
“D” = FP depression caused by a
1% solution of the ingredient
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
100 mL
xg
30 mL
x = 30 x 0.9 = 0.27 g or 270 mg
100
79
PG 187
80
20
Isotonicity Calculations
Isotonicity Calculations
Based on Freezing Point Depression
Based on Freezing Point Depression
Steps to solve isotonicity problems:
1. Convert the weight in mg of all chemicals present to
% concentration.
Isotonicity is based on the premise that an
aqueous solution that has a total freezing
point depression of 0.52°C is isotonic.
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).
81
82
Example D
Example D
How many mg of sodium chloride are needed to
render the following solution isotonic?
How many mg of sodium chloride are needed to
render the following solution isotonic?
Rx
Rx
(D Value)
Pilocarpine HCl
1%
0.138
Pilocarpine HCl
1%
0.138
Benzyl alcohol
2.0 %
0.09
Benzyl alcohol
2.0 %
0.09
Pur. water qs
100 mL
Pur. water qs
100 mL
Step #2
Pilocarpine HCl
Benzyl alcohol
Step#1
Not needed (already % concentration)
PG 187
(D Value)
83
1 x 0.138 = 0.138
2 x 0.09 = 0.180
84
21
Step#4
Example D
0.52°C – 0.318°C = 0.20°C further depression needed
How many mg of sodium chloride are needed to
render the following solution isotonic?
Rx
Step#5
(D Value)
Pilocarpine HCl
1%
0.138
Benzyl alcohol
2.0 %
0.09
Pur. water qs
100 mL
Pilocarpine HCl
Benzyl alcohol
0.138
0.180
Step#3
0.318
0.9% NaCl = x% NaCl
0.52°C
0.20°C
x = 0.35% or 350 mg NaCl per 100 mL, Ans.
Note:
If the final volume was ordered as 60 mL instead of 100 mL,
0.35 g = x g
100 mL 60 mL
x = 0.21 g or 210 mg
85
86
Calculations Involving
Milliequivalents
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)?
Milliequivalents ⎯ the amount, in milligrams, of the
solute equal to 1/1,000 of its gram equivalent
weight
Q1C1 = Q2C2
(240 mL)(410 mOsm/L) = (x mL)(300 mOsm/L)
x = 328 mL of total solution
Alternatively, a milliequivalent (mEq) is the
equivalent weight expressed in mg.
Therefore, 328 mL – 240 mL = 88 mL
87
PG 188
88
22
Calculations Involving
Milliequivalents
Method 1
Method 1 is the standard method usually presented in
chemistry.
Knowledge of valences of common ions used in medicine.
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.
89
PG 188
90
Example
Example
A 20-mL vial contains 20 mEq of potassium chloride. What weight of
chemical is present? (K = 39; Cl = 35.5)
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.
1,490mg of KCl present
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
How many mg of KCl are present in each mL?
= 74.5
1,490 mg = X mg
20 mL
1 mL
1
Express this equivalent weight in mg to obtain the milliequivalent weight.
1 mEq
20 mEq
X = 74.5 mg/mL
= 74.5 mg
x mg
x = 20 × 74.5 = 1490 mg, Ans.
91
PG 189
92
23
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)
Milliequivalents
Note that:
Method I:
1 mEq of KCl provides 1 mEq of K++ and 1 mEq of Cl--
Step 1. Molecular weight of KCl = 74.5
1 mEq of CaCl22 provides
Step 2. 1 Eq wt of KCl = 74.5
Step 3. 1 mEq of KCl = 74.5 mg
x mEq
300 mg
++ and 1 mEq of Cl-1 mEq of Ca++
x = 4 mEq, Ans.
1 mEq of NaCl provides
1 mEq of Na++ and 1 mEq of Cl--
Note that 1 mEq of KCl provides 1 mEq of K+
and 1 mEq of ClIn the book,
answer).
K+
1 mEq of Na22SO44 provides
1 mEq of Na++ and 1 mEq of SO44----
content of KCl is calculated, not necessary (same
93
94
Method 2
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
95
96
24
*Given the following prescription for calcium carbonate,
how many Milliequivalents of Calcium will the patient
consume each day. (The atomic weights of atoms in
CaCO3 are Ca = 40, C = 12 O = 16)
Rx:
Example B
„
„
Calcium Carbonate 500mg tablets
# 180
Sig: 2 tablet tid
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)
Daily dose = 2 x 500 mg x 3 = 3,000 mg
10 mL . 2 mEq/mL = 20 mEq KCl total
mg CaCO3 = mEq (molecular wt)
valence
1 mEq = 74.5 mg of KCl
20 mEq
x mg of KCl
3,000 mg = X mEq (100)
2
x = 1,490 mg or 1.49 g
X = 60 mEq
97
98
Example C
„
„
Example D
A
A pharmacist
pharmacist has
has aa 1-liter
1-liter bottle
bottle containing
containing 24.5
24.5 gg of
of hydrated
hydrated
calcium
calcium chloride.
chloride. How
How many
many mEq
mEq per
per mL
mL of
of calcium
calcium chloride
chloride are
are
present?
present? (Anhydrous
(Anhydrous calcium
calcium chloride
chloride == 111;
111; hydrated
hydrated calcium
calcium
chloride
chloride == 147)
147)
11 mEq
mEq == 147
147 mg
mg of
of calcium
calcium chloride
chloride
22
11 mEq
mEq == 73.5
73.5 mg
mg of
of calcium
calcium chloride
chloride
xx mEq
mEq 24,500
24,500 mg
mg of
of calcium
calcium chloride
chloride
„
„
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
xx == 333
333 mEq
mEq in
in 11 Liter
Liter
100 mg = (x mEq)(39)
1
x = 2.6 mEq/L
333
333 mEq
mEq == xx mEq
mEq
1000
1000 mL
mL 11 mL
mL
xx == 0.333
0.333 mEq/mL
mEq/mL
99
100
25
Example E
„
„
Example F
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)
„
„
How
How many
many mg
mg of
of anhydrous
anhydrous aluminum
aluminum chloride
chloride are
are needed
needed to
to
prepare
prepare 200
200 mL
mL of
of aa solution
solution that
that will
will contain
contain 40
40 mEq
mEq of
of
aluminum
aluminum in
in 11 liter?
liter? (Mol.
(Mol. Wt:
Wt: aluminum
aluminum chloride
chloride == 133,
133, Al
Al == 27)
27)
11 mEq
mEq == 133
133 mg
mg of
of Aluminum
Aluminum Chloride
Chloride
33
11 mEq
mEq == 44.33
44.33 mg
mg of
of Aluminum
Aluminum Chloride
Chloride
40
xx mg
40 mEq
mEq
mg of
of Aluminum
Aluminum Chloride
Chloride
x mg = (20 mEq)(234) = 4,680 mg in 15 mL
1
4,680 mg = x mg
15 mL
100 mL
xx == 1,777
1,777 mg
mg in
in 11 Liter
Liter
1,777
1,777 mg
mg == xx mg
mg
1,000
1,000 mL
mL 200
200 mL
mL
x = 31,200 mg or 31.2 g
xx == 355
355 mg
mg
101
102
Example G
„
„
Osmolarity Calculations
How
How many
many mEq
mEq of
of sodium
sodium are
are present
present in
in the
the following
following
admixture
admixture order?
order? (Na
(Na == 23,
23, Cl
Cl == 35.5)
35.5)
“Add
“Add sodium
sodium chloride
chloride (2.5
(2.5 mEq/mL)
mEq/mL) 20
20 mL
mL to
to 11 liter
liter D5W/1/2NS
D5W/1/2NS
and
and infuse
infuse over
over 88 hours”
hours”
The milliosmole (mOsm) is a measurement that is used for
parenteral solutions.
mEq
mEq of
of NaCl
NaCl in
in 20
20 mL
mL == 2.5
2.5 xx 20
20 == 50
50 mEq
mEq
1. Determine the number of millimoles present.
Determination of mOsm involves two simple steps.
Weight of drug in grams = moles × 1000 = millimoles
Amount
Amount of
of NaCl
NaCl in
in 11 liter
liter of
of ½
½ NS:
NS:
0.45
xx == 4.5
0.45 gg == xx gg
4.5 gg or
or 4,500
4,500 mg
mg
100
100 mL
mL 1000
1000 mL
mL
molecular weight
2. Multiply this value by the theoretical number of particles or
ions present (assuming complete disassociation).
11 mEq
58.5 mg
mg of
of NaCl
NaCl
mEq of
of Na
Na++ == 58.5
xx mEq
4,500 mg
mg of
of NaCl
NaCl
mEq of
of Na
Na++ 4,500
mOsm = Wt. of a substance in g × 1000 × # of species
molecular weight
xx == 76.9
76.9 mEq
mEq
Total
Total mEq
mEq of
of Na
Na++ == 50
50 ++ 76.9
76.9 == 126.9
126.9 or 127
127 mEq
mEq
103
PG 192
104
26
Osmolarity Calculations
Example A
# of species:
Sodium chloride
NaCl
1 Na+ + 1 Cl- = 2
Calcium chloride
CaCl2
1 Ca++ + 2 Cl- = 3
K+
Cl-
Potassium chloride
KCl
1
Sodium sulfate
Na2SO4
2 Na+ + 1 SO4-- = 3
Magnesium sulfate
MgSO4
1 Mg++ + 1 SO4-- = 2
Zinc sulfate
ZnSO4
1 Zn++ + 1 SO4-- = 2
Sodium acetate
NaAc
1 Na+ + 1 Ac- = 2
Glucose
Glucose
1 Glucose = 1
+1
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
=2
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
105
106
Example B
Example C
How many mOsm are present in 1 liter of D5W?
(Mol. Wt. of dextrose = 180)
5g
= X g
100 mL 1000 mL
Determine the mOsm/L concentration of calcium chloride
(mol. wt = 147) when 132 mg is dissolved in 100 mL of water.
0.132 g
100 mL
X = 50 g of dextrose in 1 liter
Step 1.
=
X g
1000 mL
X = 1.32 g of CaCl2 in 1 liter
Step 1.
millimoles = Wt. of a substance in g × 1000
molecular weight
millimoles = Wt. of a substance in g × 1000
molecular weight
millimoles = 50 g × 1000 = 278 millimoles
180
millimoles = 1.32 g × 1000 = 9 millimoles
147
Step 2.
Step 2.
mOsm = millimoles x # of species
mOsm= 278 x 1 = 278 mOsm/L
mOsm = millimoles x # of species
mOsm= 9 x 3 = 27 mOsm/L
107
108
27
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
Some Terms
Hypotonic solution⎯
0.468 g X 1,000 = 8 millimoles × 2 = 16mOsm/500mL
58.5
12 + 16 = 28 mOsm = X mOsm
500 mL
1000 mL
Hypertonic solution⎯
X = 56 mOsm/L
109
Some Terms
110
Some Terms
Hypotonic solution ⎯
having a lesser
osmotic pressure, cells would swell
Hypotonic solution⎯having a lesser
Hypertonic solution ⎯
Hypertonic solution⎯having a greater
osmotic pressure, cells would swell
osmotic pressure, cells would shrink
111
112
28
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.
Miscellaneous Pharmaceutical
Calculations
DOUBLE DILUTION
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.
Q1C1 = Q2C2
Q1C1 = Q2C2
x mL (17%) = 240 mL (2%)
10 mL (1/x) = 1000 mL (1/5000)
x = 28.2 mL
x = 50
Conc. is 1: 50 or 2%
114
Miscellaneous Pharmaceutical
Calculations
Miscellaneous Pharmaceutical
Calculations
Dilution of Acids
Dilution of Acids
Example:
How many milliliters of 85.7% w/w lactic acid
(sp. gr. = 1.19) are needed to make 120 mL of
10% w/v lactic acid?
• Concentrated acids are expressed as w/w
• Diluted acids are expressed as w/v
Q1C1 = Q2C2
• Conversion:
x mL (85.7% x 1.19) = 120 mL (10%)
w/w x density = w/v
PG 194
x = 11.8 mL
115
PG 194
116
29
Miscellaneous Pharmaceutical
Calculations
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)
Special Concentration Expressions for Alcohol
Q1C1 = Q2C2
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.
PG 194
3784 mL (18%) = X mL(95%)
X = 717mL
117
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.
118
How many proof gallons are present in 4
gallons of alcohol USP?
Q1C1 = Q2C2
x gal (50%) = 4 gal (95%)
X = 7.6 gal
Volumes can be expressed as “proof gallons”.
A proof gallon is the equivalent of 1 gallon of
50% v/v alcohol.
119
120
30
Magnesium Sulfate
How much water should be added to 4 gallons
of alcohol USP to prepare 40% alcohol?
Magnesium Sulfate is an interesting pharmaceutical
chemical.
There are two forms available.
Q1C1 = Q2C2
1. Anhydrous magnesium sulfate (MgSO4)
x gal (40%) = 4 gal (95%)
2. Hydrous magnesium sulfate (MgSO4 • 7H2O)
X = 9.5 gal
The exact amount of water to be added cannot
be calculated.
The different forms of MgSO4 have the following molecular
weights:
• Anhydrous magnesium sulfate = 120
• Hydrous magnesium sulfate = 246
Add enough water to make 9.5 gallons.
• Atomic wt magnesium = 24
121
122
Conversion Between the Two
Forms of Magnesium
Magnesium Sulfate
Anhydrous
magnesium sulfate
Hydrous
magnesium sulfate
1 molar
120 g/L
246 g/L
How many grams of anhydrous magnesium sulfate
are needed to obtain 40 g of hydrous magnesium
sulfate?
1 molal
120 g + 1,000 g of water 246 g + 1,000 g of water
1 normal
60 g/L
123 g/L
123
PG 196
124
31
Use of Statistics and Graphs
Example
How many grams of hydrous magnesium sulfate are
needed to obtain 16 mEq of magnesium ion?
Statistics
The mean is calculated by totaling all of the values and
dividing by the N
1 mEq = 246 mg of magnesium sulfate
2
1 mEq = 123 mg of magnesium sulfate
16 mEq
x mg of magnesium sulfate
Median. The median is the middle value in a series.
x = 1,968 mg
or 1.97 g of hydrous magnesium sulfate
The median is determined by eliminating the highest
value against the lowest and repeating the process until
only one value remains.
125
PG 196
Calculation of
Mean and Median
PG 195
Mean Deviation
126
Standard Deviation
Example: An experiment reported the following
values: 45, 55, 20, 38, 52. Calculate the mean and the
median.
Mean = (45 + 55 + 20 +38 +52) = 42
5
The median 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
127
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.
128
32
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 error describes 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
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.
Precision (reproducibility) refers to close agreement in the values obtained
Accuracy is closeness of values to the correct value
t Test of significance (Student’s t test) and chi-square test of
significance are mathematic methods of comparing sets of data to see if
they are significantly different. Must use prepared tables to evaluate
comparisons.
129
PG 198
130
Gaussian Distribution
Standard Deviation
„
„
„
„
For a Gaussian distribution:
z
z 68% of data are within ± 1 SD
z
z 95.5% of data are within ± 2 SD
z
z 99.7% of data are within ± 3 SD
Example
If 100 capsules had an average weight of
500 mg ± 20 mg, how many weigh between:
480 and 520 mg?
68 capsules
131
132
33
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?
„
„
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)
95 capsules
133
134
Evaluation of Graphic Data
Standard Deviation
„
„
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
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)
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 dissolution of tablets containing magnesium stearate is greater than
the control.
135
PG 199
136
34
Evaluation of Graphic Data
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
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
From Figure 2, one may conclude that:
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 dissolution of tablets containing magnesium stearate is greater than
the control.
137
PG 199
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 the dissolution of the tablets by more than
double the rate of 10% starch.
PG 200
138
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
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 the dissolution of the tablets by more than
double the rate of 10% starch.
PG 200
139
35