Pharmaceutical Calculations Dosage PDF

# Pharmaceutical Calculations ## 1. O. Edafiogho and A. J. Winfield After studying this chapter you will be able to undertake pharmaceutical calculations dealing with: - Expressions of concentration - Master formulae to working quantities - Changing concentrations - Small quantities (trituration)...

# Pharmaceutical Calculations ## 1. O. Edafiogho and A. J. Winfield After studying this chapter you will be able to undertake pharmaceutical calculations dealing with: - Expressions of concentration - Master formulae to working quantities - Changing concentrations - Small quantities (trituration) - Solubility - Calculations related to doses - Reconstitution and rates of infusion There are some tutorial examples and answers provided to help you assess your progress. ## Introduction Many pharmacy students approach the calculations involved in dispensing and manufacturing with trepidation. There is no need. Most calculations are simple arithmetic. It is true that there are many steps in the dispensing process where things can go wrong and calculating quantities is one of them. However, careful, methodical working 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: - Write out the calculation clearly. It is all too easy to end up reading from the wrong line. - If you are transferring data from a reference source double check what you have written down. - Write down every step. - Do not take short cuts; you are more likely to make a mistake. - 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. - 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) while the basic unit for volume is the litre (L) and the basic unit of length is the metre (m). Appendix 5 gives information about the weights and measures commonly used in pharmacy. 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, avoirdupois and apothecary systems are shown in Box 8.1. | Weight and measures | | |---|---| | 1000 millilitres (ml) | 1 litre (L) | | 1000 micrograms | 1 milligram (mg) | | 1000 milligrams (mg) | 1 gram (g) | | 1000 grams (g) | 1 kilogram (kg) | | 1 kilogram (kg) | 2.2 pounds (lb) | | 1 teaspoonful (tsp) | 5 ml | | 1 tablespoonful | 15 mL (tea spoonfuls) | | 1 grain (Avoir. or Apoth.) | 64.8 mg | | 1 pint (pt) | 473 mL | | 1 gallon (gal) | 3785 ml | | 1 fluid ounce (oz) | 29.57 mL (30 mL) | | 1 fluid ounce (oz) | 480 minims | **Example 8.1:** Express 85 grains in the metric system. 64.8 x 85 / 1000 1 grain = 64.8 mg. Therefore 85 grains = 64.8 x 85 mg = 5.508 g = 5.51 g (to 2 decimal places) **Example 8.2:** If 60 minims make 1 fluid drachm, and 8 fluid drachms make 1 fluid ounce, what is the volume of 1 minim in the metric system? 60 minim = 1 fluid drachm 8 fluid drachms = 1 fluid ounce (30 ml) 60 x 8 minims = 30 mL; 480 minims = 30 mL Therefore 1 minim = 30 / 480 = 0.06 mL **Example 8.3:** Sulfacetamide 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: 1:10 = 1 part in 10 parts or 1 g in 10 g 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%. **Example 8.4:** Express 0.1% w/w as a ratio strength. 0.1 g = 1 part 100 g = y parts 100 x 1 / 0.1 = 1000 y = 1000 Therefore, the ratio strength = 1:1000 **Example 8.5:** Express 1:2500 as a percentage strength. 1 part = y parts 2500 parts = 100 parts 1 x 100 / 2500 = 0.04% **Example 8.6:** 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: 1 / 1.000 000 = y / 100 y = 1 x 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 active ingredient in 100 grams (solid or liquid). **Example 8.7:** 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 5 g / 100 g = y / 240 g y = 5 x 240 / 100 = 12g **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 8.8:** If 3 g of iodine is in 150 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 = 3 g / 150 mL y = 3 x 100 / 150 = 2% w/v **Percentage volume in volume (v/v)** Percentage volume in volume (v/v) indicates the number of millilitres (mL) of an active ingredient in 100mL of liquid. **Example 8.9:** If 20 mL of ethanol is mixed with water to make 40 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 = 20 mL / 40 mL y = 20 x 100 / 40 = 50% v/v ## Miscellaneous examples **Example 8.10:** Express 25 g of dextrose in 500 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 = 25 g / 500 mL y = 25 x 100 / 500 mL = 5% w/v **Example 8.11:** What is the percentage of magnesium carbonate in the following syrup? > Magnesium carbonate: 10g > Sucrose: 820 g > Water, q.s.: ad 1000 mL. Percentage is the number of grams of magnesium carbonate in 100 mL of syrup. y / 100 mL = 10 g / 1000 mL y = 10 x 100 / 1000 = 1% w/v (grams in mL) **Example 8.12:** Calculate the amount of drug in 5 mL of cough syrup if 100 mL contains 200 mg of drug. By proportion, y mg / 5 mL = 200 mg / 100 mL y = 5 x 200 / 100 = 10 mg **Example 8.13:** 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) x 14 = 21.88% w/w Soft paraffin = (100/64) x 38 = 59.38% w/w Hard paraffin = (100/64) X 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 also Ch. 25). 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) of KOH contains one-thousandth of a mole in 1 litre = 56 mg. **Example 8.14:** 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 8.15:** 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 mmol = 40 mg in 1 L; therefore 10 mmol = 400 mg in 1 L. **Example 8.16:** Express 111mg of calcium chloride (CaCl₂) in 1 L of solution as millimoles (atomic weights: Ca = 40, Cl = 35.5). Aolecular weight of CaCl₂ = Ca + (2 x Cl) = 40 + (2 x 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 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 are 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. **Example 8.17:** Calculate the quantities to prepare the following prescription: 50 g Compound Benzoic Acid Ointment BPC. |Ingredient|Master formula|Scaled quantity| |---|---|---| |Benzoic acid|6g|3g| |Salicylic acid|3g|1.5g| |Emulsifying ointment|91g|45.5g| |Double check: the quantities for the master formula add up to 100 g and the scaled quantities add up to 50 g.| **Example 8.18:** 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|1g|20g| |Aromatic solution|0.5ml|10ml| |Liquorice liquid extract|1ml|20ml| |Water|to 10ml|to 200ml| 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 in the region of 150 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 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 8.19:** Calculate the quantities required to produce 300 mL Turpentine Liniment BP 1988. |Ingredient|Master formula| |---|---| |Soft soap|75g| |Camphor|50g| |Turpentine oil|650ml| |Water|225ml| 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 so calculate for 340 units, and this 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 8.20:** The quantity which is to be prepared of the following formula is 30 g. |Ingredient|Master formula|Quantily for 30 g| |---|---|---| |Zinc oxide|12.5 parts|3.75 g| |Calamine|15 parts|4.5 g| |Hydrous wool fat|25 parts|7.5 g| |White soft paraffin|47.5 parts|14.25 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 30 g, calamine = 15/100 of 30 g etc., as indicated above. There are some situations when extra care is necessary in reading the prescription. **Example 8.21:** 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: to 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. ## 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 8.22:** Prepare 250 g of the following ointment |Ingredient|Master formula|Master formula|Quantity for 250 g| |---|---|---|---| |Sulphur|2%|0.2g|5g| |Salicylic acid|1%|0.1g|2.5g| |White soft paraffin|to 10 g|to 10 g|242.5g| The master formula is for a total of 10 g. To calculate the quantities required for 250 g multiply the weight of each ingredient required to prepare 10 g by 25. 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 8.23:** The quantity to be made is 60 g. |Ingredient|Master formula|Master formula|Quantity for 60 g| |---|---|---|---| |Coal tar solution|3%|3g|1.8g| |Zinc oxide|5g|5g|3g| |Yellow soft paraffin|to 100 g|92g|55.2g| 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: **C₁V₁=C₂V₂** where C₁ and V₁ are the initial concentration and initial quantity 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. **Example 8.24:** What is the final concentration if 120 mL of a 12% w/v chlorhexidine solution is diluted to 240 mL with water? C₁ = 12%, V₁ = 120 mL, C₂ = y%, V₂ = 240 mL 12 x 120 = 240 x y, therefore y = 12 x 120 / 240 = 6% w/v **Example 8.25:** What concentration is produced when 200 mL of a 2.5% w/v solution is diluted to 750 mL (answer to 2 decimal places)? C₁ 2.5%, V₁ = 200 mL, C₂ = y%, V₂ = 750 mL 2.5% x 200 mL = y x 750 mL, therefore y = 2.5 x 200 / 750 = 0.67% w/v **Example 8.26:** What volume of 1% w/v solution can be made from 75 mL of 5% w/v solution? 1% × V₁ = 5% x 75 mL V₁ = 5 / 1 x 75 = 375 mL **Example 8.27:** What percentage of atropine is produced when 150 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) = 150 mg = 0.15 g Therefore, we can modify the dilution equation to read: C₁W₁ = C₂W₂ where C₂ and W₂ are the final concentration and final weight respectively, of the diluted drug. The diluting medium is the lactose. Thus, 100% x 0.15 g = C₂ x 50 g C₂ = 100 x 0.15 / 50 = 0.3% 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. **Example 8.28:** 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. **Using the alligation method:** 2 % w/v (100 mg/5 mL) 0.8% w/v (40 mg/5 mL) 0.5% w/v (25 mg/5 mL) Total number of parts = 0.3 parts + 1.2 parts = 1.5 parts = 100 mL Amount of 100 mg/5 mL (2% w/v) suspension needed = 0.3 / 1.5 x 100 mL = 20 mL. Amount of 25 mg/5 mL (0.5% w/v) suspension needed = 1.2 / 1.5 x 100 mL = 80 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'. ## Small quantities in powders **Example 8.30:** 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 x 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 produces. **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 micrograms (0.2 mg) of digoxin. The method for preparing divided powders is described in Chapter 22. ## 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 8.31:** 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. Examples of the calculations are given in Chapters 7 and 17. 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' where it is in the section dealing with physical properties. 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 solubility can help in extemporaneous dispensing can be found in Chapter 7. Some examples of calculating quantities of liquids required to dissolve solids are found in the tutorial section at the end of this chapter. ## Calculations Involving Doses A simple calculation which pharmacists sometimes have to make whilst dispensing is to calculate the number of tablets or capsules or volume of a liquid medicine to be dispensed. **Example 8.32:** The doctor prescribes levodopa capsules, 1000 mg to be taken every 8 hours for 28 days. Levodopa is available as 500 mg capsules. How many capsules should be supplied? For each dose, 2 capsules are required. Every eight hours means 3 doses per day. Therefore, the total number of capsules required is 2 x 3 x 28 = 168 capsules. **Example 8.33:** 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 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 8.34:** 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 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's rule for infants: >age (months) × adult dose / 150 = dose for infant. - Clark's rule: >weight (in kg) × adult dose / 75 = dose for child. - Body surface area method (BSA): >BSA of child (m²) × adult dose / 1.73 m² (average adult BSA) = approximate child's dose. The BNF also gives a percentage method for calculating paediatric doses of drugs which have a wide therapeutic window, i.e., where accuracy is less critical. ## 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, where 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. 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 (m²) = Weight (kg)^0.425 × Height (cm)^0.725 × 0.007184 ## 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