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Percentage weight in weight (w/w)
Percentage weight in weight (w/w) is the number of
grams of an active ingredient in 100 grams (solid or
liquid) (Example 26.7).
Example 26.7
How many grams of a drug should be used to prepare
240 grams of a 5% w/w solution?
Let y be the weight of the drug needed:
Thus, y/240 = 5 g/100 g
y =5 240/100 = 12 g
Percentage weight in volume (w/v)
Percentage weight in volume (w/v) is the number of
grams of an active ingredient in 100 mL of liquid
(Example 26.8).
Example 26.8
If 5 g of iodine is in 250 mL of iodine tincture, calculate the
percentage of iodine in the tincture.
Let y be the percentage of iodine in the tincture:
y/100 mL = 5 g/250 mL
y =5 100/250 = 2% w/v
Percentage volume in volume (v/v)
Percentagevolume in volume (v/v) indicates the number of millilitres (mL) of an active ingredient in
100 mL of liquid (Example 26.9).
Example 26.9
If 15 mL of ethanol is mixed with water to make 60 mL of
solution, what is the percentage of ethanol in the
solution?
Let y be the percentage of ethanol in the solution:
y/100 mL = 15 mL/60 mL
y =15 100/60 = 25% v/v
Pharmaceutical calculations CHAPTER 26
Example 26.10
Express 30 g of dextrose in 600 mL of solution as a
percentage, indicating w/w, w/v or v/v.
Let y grams be the weight of dextrose in 100 mL:
y/100 mL = 30 g/600 mL
y =30 100/600 mL = 5% w/v
Example 26.11
What is the percentage of magnesium carbonate in the
following syrup?
Magnesium carbonate 15 g
Sucrose 820 g
Water, q.s. to 1000 mL
Percentage is the number of grams of magnesium carbonate in 100 mL of syrup.
y/100 mL = 15 g/1000 mL
y =15 100/1000 = 1.5% w/v (grams in 100 mL)
Example 26.12
Calculate the amount of drug in 5 mL of cough syrup if
100 mL contains 300 mg of drug.
By proportion, y mg/5 mL = 300 mg/100 mL
y =5 300/100 = 15 mg
Example 26.13
Compute the percentage of the ingredients in the
following ointment (to 2 decimal places):
Liquid paraffin 14 g
Soft paraffin 38 g
Hard paraffin 12 g
Total amount of ingredients
=14g+38g+12g=64g
To find the amounts of the ingredients in 100 g of
ointment, each figure will be multiplied by 100/64:
Liquid paraffin = (100/64) 14 = 21.88% w/w
Soft paraffin = (100/64) 38 = 59.38% w/w
Hard paraffin = (100/64) 12 = 18.75% w/w
Miscellaneous examples
(Examples 26.10–26.13 )
It is useful to double-check that these numbers add up to
100% (allowing for the rounding off to 2 decimal places).
289
SECTION FOUR Dispensing and related pharmaceutical practice activities
Moles and molarity
Concentrations can also be expressed in moles or
millimoles (see also Ch. 38). When a mixture contains the molecular weight of a drug in grams in 1 litre
of solution, the concentration is defined as a 1 molar
solution (1 mol). It has a molarity of 1. Thus, for
example, the molecular weight of potassium hydroxide (KOH) is the sum of the atomic weights of its
elements, i.e. KOH = 39 + 16 + 1 = 56. Therefore
a 1 molar solution (1 mol) of KOH contains 56 g of
KOH in 1 litre of solution.
A 1 millimole (mmol) solution of KOH contains
one-thousandth of a mole in 1 litre = 56 mg (Exam-
ples 26.14–26.16).
Example 26.14
Calculate the number of moles (molarity) of a solution if it
contains 117 g of sodium chloride (NaCl) in 1 L of solution
(atomic weights: Na = 23, Cl = 35.5).
Molecular weight of NaCl = 23 + 35.5 = 58.5 g
Therefore, 58.5 g of NaCl in 1 litre is equivalent to 1 mole
(1 mol) in solution.
Number of moles of NaCl = 117 g/58.5 g = 2 mol
Calculating quantities from a master formula
In extemporaneous dispensing, a list of the ingredients is provided on the prescription or is obtained
from a recognized reference source where the
quantities of each ingredient are indicated. It may
be that this ‘formula’ is for the quantity requested,
but more often the quantities provided by the master formula have to be scaled up or down, depending on the quantity of the product required. This
can be achieved using proportion or by deriving a
‘multiplying factor’. The latter is the ratio of the
required quantity divided by the formula quantity.
The following examples illustrate this process
(Examples 26.17 and 26.18).
In mos t formulae where a combination of
weights and volumes is required, the formula will
indicate that the preparation is to be made up to
the required weight or volume with the designated
vehicle. However, occasionally, as can be seen in
the next example, a combination of stated weights
and volumes is used and it is not possible to indicate what the exact final weight or volume of the
preparation will be. In these instances an excess
quantity is normally calculated for and the required
amount measured (Example 26.19).
Example 26.15
Calculate the number of milligrams of sodium hydroxide
(NaOH) to be dissolved in 1 L of water to give a
concentration of 10 mmol (atomic weights: H = 1, O = 16,
Na = 23).
Molecular weight of NaOH = 23 + 16 + 1 = 40
1 mmol = 40 mg in 1 L
Therefore, 10 mmol = 400 mg in 1 L
Example 26.16
Express 111 mg of calcium chloride (CaCl2)in1Lof
solution as millimoles (atomic weights: Ca = 40,
Cl = 35.5).
Molecular weight of CaCl
=40+(2 35.5) = 40 + 71 = 111 g
Therefore, 111 mg of CaCl
290
=Ca+(2 Cl)
2
= 1 mmol in 1 L
2
Example 26.17
Calculate the quantities to prepare the following
prescription:
50 g Compound Benzoic Acid Ointment BPC.
The master formula is for 100 g, the prescription is for
50 g, therefore the multiplying factor is 50/100, i.e. each
quantity in the master formula is multiplied by 50/
100 = 0.5 to give the scaled quantity.
Ingredient Master
formula
Benzoic acid
Salicylic acid
Emulsifying
ointment
Double-check: the quantities for the master formula add
up to 100 g and the scaled quantities add up to 50 g.
6 g 0.5
3 g 0.5 1.5 g
91 g 0.5 45.5 g
Multiplying
factor
Scaled
quantity
3g
Example 26.18
You are requested to dispense 200 mL of Ammonium
Chloride Mixture BPC. The formula can be found in a
variety of reference books such as Martindale. In this
example the master formula gives quantities sufficient for
10 mL. As the prescription is for 200 mL, the multiplying
factor is 200/10. Thus the quantity of each ingredient in
the master formula has to be multiplied by 20 to provide
the required amount.
Ingredient Master
formula
Ammonium chloride 1 g
Aromatic solution of
ammonia
Liquorice liquid extract 1 mL 20 mL
Water to 10 mL to 200 mL
Because this formula contains a mixture of volumes and
weights it is not possible to calculate the exact quantity of
water which is required. However, it is always good practice to have an idea of what the approximate quantity will
be. The liquid ingredients of the preparation, other than
the water, add up to 30 mL and there is 20 g of ammonium chloride. The volume of water required will therefore
be between 150 mL and 170 mL.
0.5 mL 10 mL
Scaled
quantity
20 g
Pharmaceutical calculations CHAPTER 26
Calculations involving parts
In the following example the quantities are expressed
as parts of the whole. The number of parts is added up
and the quantity of each ingredient calculated by proportion or multiplying factor, to provide the correct
amounts (Example 26.20).
There are some situations when extra care is nec-
essary in reading the prescription (Example 26.21).
Example 26.20
The quantity which is to be prepared of the following
formula is 60 g.
Ingredient Master formula Quantity
Zinc oxide 12.5 parts 7.5 g
Calamine 15 parts 9 g
Hydrous wool fat 25 parts 15 g
White soft paraffin
The total number of parts adds up to 100 so the proportions of each ingredient will be 12.5/100 of zinc oxide, 15/
100 of calamine and so on. The required quantity of each
ingredient can then be calculated. Zinc oxide 12.5/100 of
60 g, calamine 15/100 of 60 g, etc. as indicated above.
47.5 parts
for 60 g
28.5 g
Example 26.19
Calculate the quantities required to produce 300 mL
Turpentine Liniment BP 1988.
Ingredient Master formula
Soft soap 75 g
Camphor 50 g
Turpentine oil 650 mL
Water 225 mL
When the total number of units is added up for this formula it comes to 1000. However, because it is a combination of solids and liquids, it will not produce 1000 mL.
The prescription is for 300 mL and experience shows that
calculating for 340 units will provide slightly more than
300 mL. The required amount can then be measured.
Ingredient Master
formula
Soft soap 75 g 25.5 g
Camphor 50 g 17 g
Turpentine oil 650 mL 221 mL
Water 225 mL 76.5 mL
Scaled quantity
for 340 units
Example 26.21
Two products are to be dispensed:
Ò
Betnovate
Aqueous cream to 4 parts
Prepare 50 g
Haelan
White soft paraffin 4 parts
Prepare 50 g
At first glance these calculations look similar but the
quantities required for each are different. In the
Betnovate prescription the total number of parts is 4, i.e.
1 part of Betnovate and 3 parts of aqueous cream to
produce a total of 4 parts. However, in the Haelan
prescription the total number of parts is 5, i.e. 1 part of
Haelan ointment and 4 parts of white soft paraffin.
The quantities required for the prescriptions are as
follows:
Betnovate
Aqueous cream 37.5 g
Haelan
White soft paraffin 40 g
cream 1 part
Ò
ointment 1 part
Ò
cream 12.5 g
Ò
ointment 10 g
291
SECTION FOUR Dispensing and related pharmaceutical practice activities
Calculations involving
percentages
There are conventions which apply when dealing with
formulae which include percentages:
*
A solid in a formula where the final quantity is
stated as a weight is calculated as weight in weight
(w/w)
*
A solid in a formula where the final quantity is
stated as a volume is calculated as weight in volume
(w/v)
*
A liquid in a formula where the final quantity is
stated as a volume is calculated as volume in
volume (v/v)
*
A liquid in a formula where the final quantity is
stated as a weight is calculated as weight in weight
(w/w) (Example 26.22).
Example 26.22
Prepare 500 g of the following ointment
Ingredient Master
formula
Sulphur 2% 0.2 g 10 g
Salicylic acid 1% 0.1 g 5 g
White soft paraffin
to 10 g to 10 g 485 g (to 500 g)
The master formula is for a total of 10 g. To calculate the
quantities required for 500 g the multiplying factor for
each ingredient is 500/10 = 50. Remember do not multiply the percentage figure. This always remains the same
no matter how much is being prepared.
In the following example a liquid ingredient, the coal
tar solution, is stated as a percentage and a weight in
grams of final product is requested. The convention of
% w/w is applied (Example 26.23).
Quantity
for 500 g
When dealing with preparations where ingredients are
expressed as a percentage concentration it is important to check that the standard conventions apply
because there are some situations where they do not
apply. Two examples are given below:
1. Syrup BP is a liquid – a solution of sucrose and
water. If the normal convention applied it would
be w/v, i.e. a certain weight of sucrose in a final
volume of syrup. However, in the BP formula the
concentration of sucrose is quoted as w/w.
Therefore Syrup BP is:
Sucrose 66.7% w/w
Water to 100%
This means that when preparing Syrup BP the appropriate weight of sucrose is weighed out and water is
added to the required weight, not volume.
2. A gas in a solution is always calculated as w/w,
unless specified otherwise. Formaldehyde
Solution BP is a solution of 34–38% w/w
formaldehyde in water.
Changing concentrations
Sometimes it is necessary to increase or decrease the
concentration of a medicine by the addition of more
drug or a diluent. On other occasions, instructions
have to be provided to prepare a dilution for use.
These problems can be solved by the dilution equation:
C1V1¼ C2V
where C1and V1are the initial concentration and
initial volume respectively; and C
final concentration and final quantity of the mixture respectively.
When three terms of the equation are known, the
fourth term can be made the subject of the formula,
and solved (Examples 26.24–26.27).
2
and V2are the
2
Example 26.23
The quantity to be made is 30 g.
Ingredient Master
formula
Coal tar
solution 3% 3 g 0.9 g
Zinc oxide 5 g 5 g 1.5 g
Yellow soft
paraffin to 100 g 92 g 27.6 g
292
Quantity
for 30 g
Example 26.24
What is the final concentration if 60 mL of a 12% w/v
chlorhexidine solution is diluted to 120 mL with water?
= 12%, V1= 60 mL, C2= y%, V2= 120 mL
C
1
12 60 = 120 y, therefore
y =12 60/120 = 6% w/v
Pharmaceutical calculations CHAPTER 26
Example 26.25
What concentration is produced when 400 mL of a 2.5%
w/v solution is diluted to 1500 mL (answer to 2 decimal
places)?
= 2.5%, V1= 400 mL, C2= y,V2= 1500 mL
C
1
2.5% 400 mL = y 1500 mL, therefore
y = 2.5 400/1500 = 0.67% w/v
Example 26.26
What volume of 1% w/v solution can be made from
75 mL of 5% w/v solution?
1% V
V
=5% 75 mL
1
= 5/1 75 = 375 mL
1
Example 26.27
What percentage of atropine is produced when 200 mg of
atropine powder is made up to 50 g with lactose as a
diluent?
The atropine powder is a pure drug, so its concentration
) is 100% w/w. The initial weight of the atropine powder
(C
1
) = 200 mg = 0.2 g. Therefore, we can modify the
(W
1
dilution equation to read:
Alligation
Alligation is a method for solving the number of parts
of two or more components of known concentration
to be mixed when the final desired concentration is
known. When the relative amounts of components
must be calculated for making a mixture of a desired
concentration, the problem is most easily solved by
alligation (Examples 26.28 and 26.29).
Calculations where quantity of ingredients is too small to weigh or measure accurately
When preparing medicines by extemporaneous dispensing, the quantity of active ingredient required
may be too small to weigh or measure with the equipment available. In these situations a measurable
¼ C2W
C
1W1
where C2and W2are the final concentration and final
weight respectively, of the diluted drug. The diluting medium is the lactose.
Thus, 100% 0.2 g = C
Therefore C
= 100 0.2/50 = 0.4% w/w
2
50 g
2
2
Example 26.28
Calculate the amounts of a 2% w/w metronidazole cream
and of metronidazole powder required to produce 150 g of
6% w/w metronidazole cream (to 2 decimal places).
In alligation, the two starting material concentrations are
placed above each other on the left hand side of the
calculation. The target concentration is placed in the
centre. The arithmetic difference between the starting
material and the target is calculated and the answer
recorded on the right hand end of the diagonal. The
proportions of the two starting materials are then given by
reading horizontally across the diagram.
As shown above, the difference between the concentration
of the pure drug powder (100%, recorded top left) and the
desired concentration (6%) is 94 (recorded bottom right).
This is equivalent to the number of parts of 2% cream
required (read horizontally across the bottom). Similarly,
the difference between the concentration of 2% cream
(recorded bottom left) and the desired concentration (6%)
is 4 (recorded top right). This is equivalent to the number of
parts of 100% drug (metronidazole powder) needed for the
mixture (read horizontally across the top).
The total amount (4 parts + 94 parts = 98 parts) is 150 g.
Thus, 1 part = 150/98 g.
Therefore, the amount of 2% cream required
= 94 parts 150/98 g = 143.88 g.
The amount of pure metronidazole (100%) required
= 4 parts 150/98 = 6.12 g.
293
SECTION FOUR Dispensing and related pharmaceutical practice activities
Example 26.29
Promazine oral syrup is available as 25 mg/5 mL and
50 mg/5 mL. Calculate the quantities to use to prepare
150 mL of 40 mg/5 mL of the oral syrup.
Convert all the concentrations to percentages.
Therefore, 25 mg/5 mL is equivalent to 0.025 g in
5 mL = 0.500 g in 100 mL = 0.5% w/v.
Similarly, 50 mg/5 mL = 1% w/v; and 40 mg/5 mL = 0.8%
w/v.
Using the alligation method:
Total number of parts
= 0.3 part + 0.2 parts = 0.5 parts = 150 mL.
Amount of 50 mg/5 mL (1.0% w/v) oral syrup needed
= 0.3/0.5 150 mL = 90 mL.
Amount of 25 mg/5 mL (0.5% w/v) oral syrup needed
= 0.2/0.5 150 mL = 60 mL.
quantity has to be diluted with an inert diluent. The
process is called ‘trituration’ (see Ch. 35).
Small quantities in powders
The method for preparing divided powders is described in Chapter 35 (Example 26.30).
Example 26.30
Calculate the quantities required to make 10 powders each
containing 200 micrograms of digoxin.
Assume that the balance available has a minimum
weighable quantity of 100 mg. An inert diluent, in this case
lactose, will be used for the trituration. The convenient
weight of each divided powder is 120 mg.
The total weight of powder mixture required will be
10 120 = 1200 milligrams = 1.2 g.
Quantities for 10 powders:
Digoxin 2 mg
Lactose 1198 mg
Total 1200 mg
The weight of digoxin is too small to weigh. The minimum
weighable quantity of 100 mg is weighed and used in the
triturate. A 1 in 10 dilution is produced.
Small quantities in liquids
If the quantity of a solid to be incorporated into a
solution is too small to weigh, again dilutions are
used. In this case a solution is prepared, so the
solubility of the substance needs to be considered.
Normally a 1 in 10 or 1 in 100 dilution is used
(Example 26.31).
Each 100 mg of this mixture (A) contains 10 mg of digoxin.
Trituration B
Mixture A 100 mg (= 10 mg digoxin)
Lactose 900 mg
Total
Each 100 mg of this mixture (B) contains 1 mg of digoxin.
This amount of digoxin is less than the required amount, so
mixture B can be used to give the required quantity.
200 mg of mixture B provides the 2 mg digoxin required.
Final trituration (C)
1000 mg
Trituration A
Digoxin 100 mg
Lactose 900 mg
Total 1000 mg
294
Mixture B 200 mg (= 2 mg digoxin)
Lactose
Total
Each 120 mg of this mixture (C) will contain 200 micrograms (0.2 mg) of digoxin.
(1200–200) = 1000 mg
1200 mg
Example 26.31
Calculate the quantities required to prepare 100 mL of a
solution containing 2.5 mg morphine hydrochloride/5 mL.
Quantities for 100 mL:
Morphine hydrochloride 50 mg
Chloroform water to 100 mL
The solubility of morphine hydrochloride is 1 in 24 of water.
Pharmaceutical calculations CHAPTER 26
The minimum quantity of 100 mg of morphine
hydrochloride is weighed and made up to 10 mL with
chloroform water (this weight of morphine hydrochloride
will dissolve in 2.4 mL).
5 mL of this solution (A) provides the 50 mg of morphine
hydrochloride required.Take 5 mL of solution A and make
up to 100 mL with chloroform water.
Solubilities
When preparing pharmaceutical products, the solubility of any solid ingredients should be checked. This
will give useful information on how the product
should be prepared. Examples of the calculations
are given in Chapters 25 and 30. The objective of this
section is to clarify the terminology used when solubilities are stated.
The solubility of a drug can be found in reference
sources such as the drug monograph in Martindale.
The method of stating solubilities is as follows:
Sodium chloride is soluble 1 in 2:8 of water;
1 in 250 of alcohol and 1 in 10 of glycerol
This means that 1 g of sodium chloride requires
2.8 mL of water, 250 mL of alcohol or 10 mL of
glycerol to dissolve it. An example of how knowledge of a substance’s solubili ty can help in extemporaneous dispensing can be found in Chapter 25.
Some examples of calculating quantities of liquids
required to dissolve solids are found in the selfassessment section (questions 6.1–6.4) at the end
of this chapter.
Calculations involving doses
A simple calculation which pharmacists sometimes
have to make while dispensing is to calculate the number of tabletsor capsules or volumeof a liquidmedicine
to be dispensed (Examples 26.32 and 26.33).
Calculating doses
An overdose of a drug, if given to a patient, can have
very serious consequences and may be fatal. It is the
responsibility of everyone involved in supplying or
administering drugs to ensure that the accuracy and
suitability of the dose are checked. The following are
some examples of areas where errors can occur.
The standard way to check whether a drug dose is
appropriate is to consult a recognized reference book.
One of the commonest used for this purpose is the
British National Formulary (BNF). When first using
any reference source it is important to be aware of the
terminology used, to avoid misinterpreting the
entries, especially where doses are quoted as ‘x milligrams daily, in divided doses’. An explanation of the
terminology will usually be given in the introduction
to the book (Example 26.34).
Example 26.32
The doctor prescribes orphenadrine tablets, 100 mg to
be taken every 8 hours for 28 days. Orphenadrine is
available as 50 mg tablets. How many tablets should be
supplied?
For each dose, 2 tablets are required. Every eight hours
means 3 doses per day.
Therefore, the total number of tablets required is
2 3 28 = 168 tablets.
Example 26.33
The following prescription is received:
Sodium valproate oral solution:
100 mg to be given twice daily for 2 weeks.
Sodium valproate oral solution contains sodium
valproate 200 mg/5 mL.
This prescription is therefore translated as:
2.5 mL to be given twice daily for 2 weeks.
The quantity to be dispensed will be:
2.5 2 14 = 70 mL.
295
SECTION FOUR Dispensing and related pharmaceutical practice activities
Example 26.34
The following prescription is received:
Verapamil tablets 160 milligrams
Send 56
Take two tablets twice daily
There are a variety of doses quoted for verapamil in the
BNF depending on the condition being treated. They are as
follows for oral administration:
Supraventricular arrhythmias, 40–120 mg three times daily
Angina, 80–120 mg three times daily
Hypertension, 240–480 mg daily in 2–3 divided doses.
The dose given for hypertension is stated in a
significantly different way. Whereas the other doses can
be given three times daily, indicating a maximum of
360 mg in any one day, the hypertension dose is the
total to be given in any one day and is divided up and
given at the stated frequencies, i.e. a maximum of
240 mg, g iven twice daily or a maximum of 160 mg,
given three times daily.
The prescription is for a dose higher than recommended,
so consultation with the prescriber would be required. Be
alert – variation in terminology and a lack of awareness
could have very serious consequences.
Calculations of children’s doses
Children often require different doses from those of
adults. Ideally these should be arrived at as a result of
extensive clinical studies, although this is often not
possible. When this is the case an estimate of the dose
has to be made. This is best carried out using body
weight (see next section), but where this is not available, there are three formulas which relate the child’s
dose to the adult dose.
Fried’sruleforinfants
Age ðmonthÞadult dose=150 ¼ dose for infant
Clark’srule
Weight ðin kgÞadult dose=75 ¼ dose for child
Body surface area(BSA) method
BSA of child ðm2Þadult dose=1:73 m
ðaverage adult BSAÞ¼approximate child0s dose
2
Calculation of doses by weight
and surface area
Forsome drugsthe amount of drug has to be calculated
accurately for the particular patient. This is normally
carried out using either body weight or body surface
area. When body weight is being used, the dose will be
expressed as mg/kg. In countries which still use
pounds, it will be necessary to convert the patient’s
weight in pounds into kilograms by dividing by 2.2.
The total dose required is then obtained by multiplying
the weight of the patient by the dose per kilogram.
Body surface area is a more accurate method for
calculating doses and is used where extreme accuracy
is required. This is necessary where there is a very
narrow range of plasma concentration between the
desired therapeutic effect and severe toxicity, such
as with the drugs used to treat cancer. The body
surface area can be calculated from body weight and
height using the equation given below, but it is more
usual to use a nomogram for its determination. The
actual nomogram is published in many reference
sources.
Body surface area ðm2Þ¼weight ðkgÞ
height ðcmÞ
0:37
0:024265
0:5378
Reconstitution and infusion
Some drugs are not chemically stable in solution and
so are supplied as dry powders for reconstitution just
before use. Many of these are antibiotics, but there is
also a range of chemotherapeutic agents used in cancer treatment. The antibiotics may be for oral use or
for injection. An oral antibiotic for reconstitution
comes as a powder in a bottle with sufficient space
to add the water. The powder itself will remain stable
for up to 2 years when dry. When reconstituted, a
shelf life of 10–14 days is normal, depending on
whether it is refrigerated or not. Those for injection
are equally stable when dry, but are intended to be
used within hours of reconstitution. Because they are
for injection they are sterile powders and are dissolved
in sterile water aseptically (see Ch. 29). There are a
number of calculations which may be required around
the reconstitution processes (Example 26.35).
296
Pharmaceutical calculations CHAPTER 26
Example 26.35
What dose of antibiotic will be contained in a 5 mL
spoonful when a bottle containing 5 g of penicillin V is
reconstituted to give 200 mL of syrup?
For this type of calculation, the simple proportion
equation can be used:
/Wt2= Vol1/Vol
Wt
1
5 g = 5000 mg.
Substituting we get:
5000 mg/y mg = 200 mL/5 mL
y = 125 mg.
2
Example 26.36
We have an ampicillin product for reconstitution. It
contains 2.5 g of ampicillin to be made up to 100 mL. To
what volume should it be made to give 100 mg per 5 mL
dose?
The normal mixture will give a dose of:
2500 mg/y mg = 100 mL/5 mL
y = 125 mg per 5 mL
To calculate the amount of water to add, the same
equation is used:
2500 mg/100 mg = y mL/5 mL
y = 125 mL.
Sometimes, the doctor may request a more or less
concentrated syrup to be produced which requires
altering the amount of water added from that indicated by the manufacturer (Example 26.36).
However, this type of oral mixture is likely to have
other ingredients – thickeners, colours, flavours, etc. –
which will occupy some of the final volume. So this
Example 26.37
The label on an ampicillin bottle indicates that 78 mL of
water must be added to produce 100 mL of final syrup.
How much water must be added to give the 125 mL final
volume?
Thus, the volume of powder in the final syrup is:
100 mL 78 mL = 22 mL.
Therefore, the volume to add to give 125 mL is:
125 mL 22 mL = 103 mL.
may not be correct and we need to be able to calculate
exactly how much water to add (Examples 26.37 and
26.38).
Drugs for injection solutions do not normally con-
tain ingredients other than the drug (or they make an
insignificant contribution to the final volume). However, they are usually packed as a quantity of drug
with the final volume left to be calculated by the
pharmacist (Example 26.39).
Calculation of infusion rates
Drugs may be given to patients intravenously by adding them to an intravenous (IV) infusion (see Ch. 38).
Calculations involve working out how much drug solution should be added, working out how fast, in
terms of mL/min, the infusion should be administered, and calculating what this means in terms of
‘drops per minute’ through the giving set. When an
infusion pump is used, this can be set to deliver a
specified number of mL/min. The final stage of
drops/min is only required for traditional IV ‘ drips’
(see Ch. 38)(Example 26.40).
Example 26.38
A child weighing 60 lb requires a dose of 8 mg/kg of
ampicillin. Given that a 5 mL dose is to be given, what
volume of water must be added when the powder is
reconstituted? Instructions on the label indicate that
dilution to 150 mL (by adding 111 mL) gives 250 mg
ampicillin per 5 mL.
Conversion of weight to kg: 60/2.2 = 27.27 kg
Calculation of amount of ampicillin required:
27.27 8 = 218 mg.
Calculation of amount of ampicillin in container:
250 mg/y mg = 5 mL/150 mL, therefore
y =7500mg=7.5g
Calculation of amount of water (a) to add to give 218 mg per
5 mL: 218 mg/7500 mg = 5 mL/ a mL, therefore
a =172mL
Volume occupied by powder: 150 mL 111 mL = 39 mL
Therefore, volume to be added:
172 mL 39 mL = 133 mL.
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SECTION FOUR Dispensing and related pharmaceutical practice activities
Example 26.39
Calculate the amount of sterile water to be added to a vial
containing 200 000 units of penicillin G in order to
produce a solution containing 40 000 units per millilitre.
Again, simple proportion is used:
40 000 units/200 000 units = 1 mL/y mL, therefore
y = 5 mL.
Example 26.40
An ampoule of flucloxacillin contains 250 mg of powder
with instructions to dissolve it in 5 mL of water for
injections. What volume of this solution should be added
to 500 mL of saline infusion to provide a dose of 175 mg?
250 mg in 5 mL = 50 mg per mL
Therefore, we require: 175 mg/50 mg/mL = 3.5 mL.
When administering intravenous infusions, the rate
of addition is first calculated in terms of millilitres per
minute (Example 26.41).
Example 26.41
100 mg of phenylephrine hydrochloride are added to
500 mL of saline infusion. What should be the rate of
infusion to give a dose of 1 mg per minute? How long will
the infusion take?
Using simple proportion: 100 mg/1 mg = 500 mL/y mL
y = 5 mL and contains the required amount of drug
The infusion rate should be 5 mL per minute.
The total volume is 500 mL, therefore the time taken at
5 mL/min is:
500 mL/5 mL/min = 100 min.
Most infusions are administered using a giving set
with a dropping device on the tube (called venoclysis
set; see Ch. 38). Partial clamping of the tube can be
used to adjust the rate of dropping. Depending on the
drop size – that is the number of drops per millilitre –
it is then possible to convert a rate of millilitres per
minute into drops per minute which the nurse can
adjust (Example 26.42).
A variation on this is when the doctor wishes a
drug solution to be added to the infusion (Example
26.43).
Example 26.42
A doctor requires an infusion of 1000 mL of 5% dextrose to
be administered over an 8 hour period. Using an IV giving
set which delivers 10 drops/mL, how many drops per
minute should be delivered to the patient?
First convert the time into minutes:
8 hour = 8 60 min = 480 min
Example 26.43
20 mL of a drug solution is added to a 500 mL infusion
solution. It has to be administered to the patient over a
5 hour period. Using a set giving 15 drops per millilitre, how
many drops per minute are required?
The total volume of infusion is:
20 mL + 500 mL = 520 mL
Then calculate the number of drops which will be
administered in total:
298
Next calculate how many mL/min are required:
1000 mL/480 min = 2.1 mL/min
Then calculate the number of drops this requires:
2.1 mL/min 10 drops/min = 21 drops/min.
520 mL 15 drops = 7800 drops
The duration of the infusion is to be:
5 (hours) 60 = 300 min
Calculate how many drops are required per minute:
7800 drops/300 min = 26 drops per min.