19

Pharmaceutical calculations

Arthur J. Winfield

Study Points

image Expressions of concentration

image Calculating quantities from master formulae

image Changing concentrations

image Small quantities (trituration)

image Solubility

image Calculations related to doses

image Reconstitution and rates of infusion

Introduction

Most pharmaceutical calculations are simple arithmetic. Calculating quantities in the dispensing process requires careful, methodical working which will minimize the risk of errors. Always try to relate the calculation to practice, visualize what you are doing and double-check everything.

How to minimize errors

As in all dispensing procedures, an organized, methodical approach is essential:

image Write out the calculation clearly – it is all too easy to end up reading from the wrong line

image If you are transferring data from a reference source, double-check what you have written down is correct

image Write down every step

image Do not take short cuts – you are more likely to make a mistake

image Try not to be totally dependent on your calculator – have an approximate idea of what the answer should be and then if you happen to hit the wrong button on the calculator you are more likely to be aware that an error has been made

image Finally, always double-check your calculation. There is frequently more than one way of doing a calculation, so if you get the same answer by two different methods the chances are that your answer will be correct. Alternatively, try working it in reverse and see if you get the starting numbers

Expressions of concentration

The metric system is the International System of Units (SI Units) for weight, volume and length. The basic unit for weight is the kilogram (kg), the basic unit for volume is the litre (L) and the basic unit of length is the metre (m). The prefix ‘milli’ indicates one-thousandth (10−3) and ‘micro’ one-millionth (10−6).

In some countries, the avoirdupois (or imperial) system (pounds and ounces) is still used in commerce and daily life. The imperial system of volume (pints and gallons) is still a common system for commerce and household measurement. Pharmacists need to know about these systems in order to avoid serious errors in interpretation of prescriptions. It is important to be able to change between the systems. Some conversion factors for the metric and avoirdupois systems are shown in Box 19.1 (see Examples 19.119.3).

Box 19.1   Weights and measures

image 1000 millilitres (mL) = 1 litre (L)

image 1000 micrograms (μL) = 1 milligram (mg)

image 1000 milligrams (mg) = 1 gram (g)

image 1000 grams (g) = 1 kilogram (kg)

image 1 kilogram (kg) = 2.2 pounds (lb)

image 1 teaspoonful (tsp) = 5 mL

image 1 tablespoonful = 15 mL (3 teaspoonfuls)

image 1 pint (pt) = 568 mL

image 1 fluid ounce (oz) = 29.57 mL (30 mL)

Example 19.1

A prescription is received for a dose of 125 mg of a drug. How many grams is the dose?

1 g = 1000 mg. Therefore:

125 mg = 125/1000 g = 0.125 g

Example 19.2

A 1-year-old child weighs 9 kg. What would this be in pounds (lb)?

1 kg = 2.2 lb. Therefore:

9 kg = 9 × 2.2 = 19.8 lb

Example 19.3

Gentamicin eye drops contain 200 drops in a 10 mL bottle. Calculate the volume of 1 drop.

200 drops = 10 mL. Therefore:

1 drop = 10/200 mL = 0.05 mL

Expressions of strength

Ratio is the relative magnitude of two like quantities. Thus:

image

If 1 g of sucrose is in 10 g of solution, the ratio is 1:10. Therefore, 10 g of sucrose is in 100 g of solution. This can be expressed as a percentage, so it is equivalent to a 10% w/w (weight in weight) solution.

Ratio strength is the expression of a concentration by means of a ratio, e.g. 1:10. Percentage strength is a ratio of parts per hundred, e.g. 10% (see Examples 19.419.6).

Example 19.4

Express 0.1% w/w as a ratio strength:

0.1 g/100 g = 1 part/y parts.

y = 100 × 1/0.1

y = 1000

Therefore, the ratio strength = 1:1000

Example 19.5

Express 1:2500 as a percentage strength:

1 part/2500 parts = y parts/100 parts.

Thus, y = 1 × 100/2500 = 0.04%

Example 19.6

Express 1 p.p.m. as a percentage strength:

1 p.p.m. = 1 part per million = 1:1 000 000

Let y be the percentage strength:

Thus, 1 part/1 000 000 = y parts/100 parts

y = 1 × 100/1 000 000 = 0.0001% = 1 × 10−4%

Percentage weight in weight (w/w)

Percentage weight in weight (w/w) is the number of grams of an ingredient in 100 grams (solid or liquid) (Example 19.7).

Example 19.7

How many grams of a drug should be used to prepare 240 g 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 ingredient in 100 mL of liquid (Example 19.8).

Example 19.8

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)

Percentage volume in volume (v/v) indicates the number of millilitres (mL) of an ingredient in 100 mL of liquid (Example 19.9).

Example 19.9

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

Miscellaneous examples

See Examples 19.1019.13.

Example 19.10

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 19.11

What is the percentage of magnesium carbonate in the following syrup? Percentage is the number of grams of magnesium carbonate in 100 mL of syrup.

Magnesium carbonate 15 g
Sucrose 820 g
Water, q.s. to 1000 mL

y/100 mL = 15 g/1000 mL

y = 15 × 100/1000 = 1.5% w/v (grams in 100 mL)

Example 19.12

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 19.13

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

= 14 g + 38 g + 12 g = 64 g

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

It is useful to double-check that these numbers add up to 100% (allowing for the rounding off to 2 decimal places).

Moles and molarity

Concentrations can also be expressed in moles or millimoles (see Ch. 41). 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 (see Examples 19.1419.16).

Example 19.14

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

Example 19.15

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 19.16

Express 111 mg of calcium chloride (CaCl2) in 1 L of solution as millimoles (atomic weights: Ca = 40, Cl = 35.5):

Molecular weight of CaCl2 = Ca + (2 × Cl)

= 40 + (2 × 35.5) = 40 + 71 = 111 g

Therefore, 111 mg of CaCl2 = 1 mmol in 1 L

Calculating quantities from a master formula

In extemporaneous dispensing, a list of the ingredients is provided on the prescription or is obtained from a reference source. 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 19.17 and 19.18).

Example 19.17

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.

image

Double-check: the quantities for the master formula add up to 100 g and the scaled quantities add up to 50 g.

Example 19.18

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 Scaled quantity
Ammonium chloride 1 g 20 g
Aromatic solution of ammonia 0.5 mL 10 mL
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.

In most 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 19.19)

Example 19.19

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 Scaled quantity for 340 units
Soft soap 75 g 25.5 g
Camphor 50 g 17 g
Turpentine oil 650 mL 221 mL
Water 225 mL 76.5 mL

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 19.20).

Example 19.20

The quantity which is to be prepared of the following formula is 60 g.

Ingredient Master formula Quantity for 60 g
Zinc oxide 12.5 parts 7.5 g
Calamine 15 parts 9 g
Hydrous wool fat 25 parts 15 g
White soft paraffin 47.5 parts 28.5 g

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.

There are some situations when extra care is necessary in reading the prescription (see Example 19.21).

Example 19.21

Two products are to be dispensed:

Betnovate® cream 1 part
Aqueous cream to 4 parts

Prepare 50 g

Haelan® ointment 1 part
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® cream 12.5 g
Aqueous cream 37.5 g
Haelan® ointment 10 g
White soft paraffin 40 g

Calculations involving percentages

There are conventions which apply when dealing with formulae which include percentages:

image A solid in a formula where the final quantity is stated as a weight is calculated as weight in weight (w/w)

image A solid in a formula where the final quantity is stated as a volume is calculated as weight in volume (w/v)

image A liquid in a formula where the final quantity is stated as a volume is calculated as volume in volume (v/v)

image A liquid in a formula where the final quantity is stated as a weight is calculated as weight in weight (w/w) (see Example 19.22).

Example 19.22

Prepare 500 g of the following ointment

Ingredient Master formula Quantity for 500 g
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 19.23).

Example 19.23

The quantity to be made is 30 g.

Ingredient Master formula Quantity for 30 g
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

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:

image

where C1 and V1 are the initial concentration and initial volume respectively; and C2 and V2 are the 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 19.2419.27).

Example 19.24

What is the final concentration if 60 mL of a 12% w/v chlorhexidine solution is diluted to 120 mL with water?

C1 = 12%, V1 = 60 mL, C2 = y%, V2 = 120 mL

12 × 60 = 120 × y

Therefore

y = 12 × 60/120 = 6% w/v

Example 19.25

What concentration is produced when 400 mL of a 2.5% w/v solution is diluted to 1500 mL (answer to 2 decimal places)?

C1 = 2.5%, V1 = 400 mL, C2 = y, V2 = 1500 mL

2.5%×400 mL = y × 1500 mL

Therefore

y = 2.5 × 400/1500 = 0.67% w/v

Example 19.26

What volume of 1% w/v solution can be made from 75 mL of 5% w/v solution?

1% × V1 = 5% × 75 mL

V1 = 5/1 × 75 = 375 mL

Example 19.27

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 (C1) is 100% w/w. The initial weight of the atropine powder (W1) = 200 mg = 0.2 g. Therefore, we can modify the dilution equation to read:

C1W1 = C2 W2

where C2 and W2 are the final concentration and final weight respectively, of the diluted drug. The diluting medium is the lactose.

Thus, 100% × 0.2 g = C2 × 50 g

Therefore C2 = 100 × 0.2/50 = 0.4% w/w

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 19.28 and 19.29).

Example 19.28

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.

image

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

Example 19.29

Promazine oral syrup is available as 25 mg/5 mL and 50 mg/5 mL. Calculate the quantities used 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:

image

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.

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 quantity has to be diluted with an inert diluent. The process is called ‘trituration’ (see Ch. 38).

Small quantities in powders

The method for preparing divided powders is described in Chapter 38 (see Example 19.30).

Example 19.30

Calculate the quantities required to make 10 powders each containing 200 micrograms (μg) 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 mg =  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.

Trituration A

Digoxin 100 mg
Lactose 900 mg
Total 1000 mg

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 1000 mg

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)

Mixture B 200 mg (= 2 mg digoxin)
Lactose (1200–200) = 1000 mg
Total 1200 mg

Each 120 mg of this mixture (C) will contain 200 μg (0.2 mg) of digoxin.

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 19.31)

Example 19.31

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.

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. 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:

image

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 solubility can help in extemporaneous dispensing can be found in Chapter 30.

Calculations involving doses

A simple calculation which pharmacists sometimes have to make while dispensing, is to calculate the number of tablets or capsules or volume of a liquid medicine to be dispensed (Examples 19.32 and 19.33).

Example 19.32

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 8 hours, means 3 doses per day.

Therefore, the total number of tablets required is 2 × 3 × 28 = 168 tablets.

Example 19.33

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.

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 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 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 ‘×milligrams daily, in divided doses’. An explanation of the terminology will usually be found in the introduction to the book (Example 19.34).

Example 19.34

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, given 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 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’s rule for infants

Age (month) × adult dose/150 = dose for infant

Clark’s rule

Weight (in kg) × adult dose/75 = dose for child

Body surface area (BSA) method

image

Calculation of doses by weight and surface area

For some drugs the 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.

image

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. Those for injection are sterile powders and are dissolved in sterile water aseptically (see Ch. 40). There are a number of calculations which may be required around the reconstitution processes (Example 19.35).

Example 19.35

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:

Wt1/Wt2 = Vol1/Vol2

5 g = 5000 mg.

Substituting we get:

5000 mg/y mg = 200 mL/5 mL

y = 125 mg.

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 19.36).

Example 19.36

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.

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 may not be correct and we need to be able to calculate exactly how much water to add (Examples 19.37 and 19.38).

Example 19.37

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.

Example 19.38

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 = 7500 mg = 7.5 g

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 = 172 mL

Volume occupied by powder: 150 mL–111 mL = 39 mL

Therefore, volume to be added:

172 mL–39 mL = 133 mL.

Drugs for injection solutions do not normally contain 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 19.39).

Example 19.39

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.

Calculation of infusion rates

Drugs may be given to patients intravenously by adding them to an intravenous (IV) infusion (see Ch. 41). 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 (Example 19.40).

Example 19.40

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 administration is first calculated in terms of millilitres per minute (Example 19.41).

Example 19.41

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 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 19.42).

Example 19.42

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

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/mL = 21 drops/min.

A variation on this is when the doctor wishes a drug solution to be added to the infusion (Example 19.43).

Example 19.43

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:

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.

Key Points

image Always work methodically and write down calculations clearly

image Check calculations, using a different method where possible

image Estimate the answer before you start

image Try to visualize the quantities you are using in the calculation

image Look carefully to see if a formula gives the quantities of all ingredients or uses ‘to’ for the vehicle

image Always check the units being used and be careful not to mix them during a calculation

image Be careful to read the wording; small changes in terminology can alter the calculation

image Triturates with solids and liquids normally use a 1 in 10 dilution per step

image Be very careful in checking doses, particularly with ‘in divided doses’ and ‘mg/kg’ statements in the reference books

image On completion of a calculation, ask yourself whether the answer is ‘reasonable’ given the numbers you are using