Pharmaceutical Calculations PDF

Pharmaceutical calculation AIM : To understand the basic calculations required in Pharmaceutical practice  OBJECTIVES Appreciate a)Pharmaceutical formula b)Scaling up or down of a formula c) Interpretation and calculation of concentrations d) Expression of concentration in phar...

Pharmaceutical calculation AIM : To understand the basic calculations required in Pharmaceutical practice  OBJECTIVES Appreciate a)Pharmaceutical formula b)Scaling up or down of a formula c) Interpretation and calculation of concentrations d) Expression of concentration in pharmacy e) Dilution and concentration f) Alcohol dilutions  Pharmaceutical calculation OUTCOME : At the end of this section students should be able to a) interpret or write a pharmaceutical formula b) scale up or down of a pharmaceutical formula in order to obtain the required ingredients for the formulation c) interpret or calculate percentage concentration concentrations.  OUTCOME d) dilute adult preparation to children preparation or e) mixing of two similar preparations of different strengths and also dilute alcohol to the required concentration  Introduction The adoption of the metric system Pharmaceutical calculations involve simple arithmetic, however errors easily slip in, therefore, rules that help to prevent errors are strictly adopted. In general dispensing, a further check of the calculation by a second person is always recommended.  Introduction cont. When the ingredients in a formula adds up to a convenient figure, as is often the case, the quantities for any fraction of this can be added and should be the same fraction of the unreduced formula. For example, the ingredients for one-fifth of a formula totalling 1000 g should add up to 200 g.  Formula This is the list and quantities (either relative or specific) of ingredients in a medicinal preparation. These ingredients and their quantities are usually written vertically in the official books (e.g. B.N.F., the Pharmacopoeias). Look at the examples below:  Formula Example 1 Magnesium trisilicate powder 50 g Light magnesium carbonate 50 g Sodium bicarbonate 50 g Peppermint emulsion 25 ml Double strength chloroform water 500 ml Purified Water to 1000 ml  Formula Example 2 Coal tar 6g Zinc oxide, finely sifted 6g Starch 38 g Emulsifying wax 5g Yellow soft paraffin 45 g  Formula Add up when only weights are involved Standards – BP, BNF USP etc Scale factor (S.F.) and Formula writing Medicines are prepared according to amounts required, hence, formulae are either scale up or down. The amount required divided by the amount stated in the formula is termed the scale factor (S.F.). Scale factor (S.F.) and Formula writing Example, prepare 25 g of Zinc and coal tar paste B.N.F. Official Amounts amounts required ( x 0.25) Coal tar 6g 1.50 g Zinc oxide, finely sifted 6 g 1.50 g Starch 38 g 9.50 g Emulsifying wax 5g 1.25 g Yellow soft paraffin 45 g 11.25 g 100 g 25.00 g Scale factor (S.F.) and Formula writing Solution: Calculate the scale factor first Multiply every ingredient by the S.F. S.F. = Amount required = 25 g Final amount in formula 100 g = 0.25 Scale factor (S.F.) and Formula writing Note: The quantities have been arranged in orderly vertical columns. Errors caused by misplacing the decimal point are minimised by this procedure. The sum of the ‘Amounts required’ column is the correct fraction of the sum of the ‘Official amounts’ column  Relative quantities Formulas sometimes indicate relative quantities of ingredients, or proportional parts, to be used in obtaining any desired total amounts. Example 4 Salicylic acid 1 part Precipitated sulphur 8 parts Hydrophilic ointment 91 parts  Number of parts of an ingredient (as specified) Total number of parts in the formula (as specified) x Total quantity of formula (as desired) From the above formula, calculate the quantity of salicylic acid to use in preparing 30 g of ointment. Solution: 1 (salicylic acid as specified x 30 g (formula desired) 100 (total parts specified) = 0.3 g.  Summary Rules in pharmaceutical calculations are 1) further check of the calculation by a second person is always recommended, 2) quantities should be arranged in orderly vertical columns with the same number of figure(s) after the decimal points, 3) adhere strictly to the method of calculating the scale factor. 4) check if ‘to’ is in the formula, 5) check if the sum of the ‘Amounts required’ column is the correct fraction of the sum of the ‘Official amounts’ column. Percentage Concentration Expressions Percentage weight in volume (% w/v) - the number of grams of solute per 100 ml of solution. It is given by the formula – concentration % w/v = weight of solute in gram x 100 volume of solution in ml Percentage Concentration Expressions conti. Percentage volume in volume (% v/v) - Volume in ml (cm3) of solute in 100 ml of solution. This is usually used for liquid dissolved in another. Formula is % v/v = volume of liquid in ml x 100 volume of solution in ml Percentage Concentration Expressions conti. 3. Percentage weight in weight - % w/w - Number of grams of solute per 100 g of solution. Use in gas in liquid solutions of gases Solutions expressed in % w/w are not affected by temp. and offers precise description of concentration. Percentage Concentration Expressions conti. 4. Percentage volume in weight - % v/w - Volume in ml (cm3) of solute (usually liquid) in 100 g of solution product).  Example Calculate the quantity of sodium chloride required for 500ml of a 0.9 per cent solution. 0.9%w/v solution means 0.9g of sodium chloride in 100ml solution 500ml solution = (500/100) x 0.9g = 4.5g Percentage Concentration Expressions conti. Note: The 4.5 g would not be added to 500 ml of solvent because this would result in a volume greater than the 500 ml due to the displacement caused by solution of the medicament. The correct procedure is to dissolve the solid in part of the solvent and then adjust the volume to 500 ml.  Other concentration expressions Parts The Pharmacopoeias also express some concentrations in terms of the number of ‘parts’ of solute dissolved in a stated number of ‘parts’ of solution. Use of this method to describe the strength of a solution of a solid in a liquid infers that a given number of parts by volume (ml) of solution contain a certain number of parts by weight (g) of solid  Other concentration expressions Parts An expression such as “1 in 10” means that 1 part by volume of a liquid is to be diluted with or 1 part by weight of a solid to be dissolved in sufficient of the diluent or solvent to make the volume of the finished solution 10 parts by volume. To avoid unnecessary bulk of solutions Concentration expressed in parts can be converted to percentage concentrations  Other concentration expressions Parts Example 5 Convert i) 1 in 400 to % concentrations 1 in 400 = 100 x 1 per cent = 0.25 % `400 2. Molarity M,c - Number of moles of solute in 1 liter of solution (mol l-1) 3. Molality m – Moles of solute in 1 kg (1000 g) of solvent. This offers a more precise description of concentration  Other concentration expressions Normality N Number of gram equivalent weight of solute in I litre of solution. Gram equivalent weight is the weight in grams of an atom or radical divided by the valence of the atom or radical. E.g. the gram equivalent weight of potassium is 39 g/1 = 39 Example 6 Convert 5 % w/v sulphuric acid soln to normality; Molecular wt. of sulphuric acid is 98 g.  Other concentration expressions Normality N H2SO4 has two displaceable hydrogen atoms or valence of 2. Hence the gram equivalent weight = 98 g/2 = 49 g Concentration of solution = 5 g in 100 ml solution Therefore number of gram eq. in the solution = 5 g/49 g = 0.102 Normality N = 0.102 x 1000 ml = 1.02 100 ml  Other concentration expressions Mole fraction X or N Ratio of the moles of one constituent e.g. the solute of a solution to the total moles of all constituent (solute and solvent) i.e. mole fraction of solute X1 = n1 and solvent X2 = n2 n1+n2 n1+n2 Where n1 and n2 are the numbers of moles of solute and solvent respectively. The sum of mole fractions of any solution is always 1.  Other concentration expressions Mole and milligram % Mole percent – Mole fraction x 100. Milligram % - Number of milligrams of solute in 100 mls of solution.  Other concentration expressions Milliequivalents The concentrations of solutes in body fluids and in solutions used as replacements for these fluids are usually expressed in terms of the number of millimoles (1 millimole – one thousandth of a mole) in a litre of solution. In electrolyte replacement therapy, solutions containing various electrolytes are injected into the body to correct serious electrolyte imbalances  Other concentration expressions Milliequivalents The concentrations are usually expressed as equivalents per liter or milliequivalents per liter. A milliequivalent (mEq) of an ion is 1/1000th of the gram equivalent of the ion and is therefore the same figure expressed in milligrams. 1 mEq = ionic wt in mg Valence  Other concentration expressions Milliequivalents Milliequivalent best expresses the electrical activity of ions in replacement fluids, which in this instance is important. The gram equivalent wt. of potassium (K) is 39 g. The mEq of K is gram equivalent wt divided by 1000. Thus, 39 g/1000 = 0.039 g = 39 mg, One mEq of K+ (ion) combines with 1 mEq Cl- to give 1 mEq of KCl. The mEq of KCl is 74.5 mg, (1 mEq K + is 39 mg + mEq Cl- is 35.5 mg)  Other concentration expressions Milliequivalents Water of hydration contributes to the molecular weight of a compound but not to the valence, and the total weight is used to calculate mEq. Example 8 How many mEq of Mg are there in 10 ml of 50 % solution of Magnesium sulphate injection? The mol.wt of MgSO4.7H2O is 246. 50 % soln means 5 g MgSO4.7H2O in 10 ml.  Other concentration expressions Milliequivalents Eq wt = gram mol. Wt = 246 = 123 g Valence 2 mEq wt = 123 g = 0.123 g 1000 Total mEq = 5 = 40.7 mEq in 10 ml. 123 mg/mEq  Dilution and Concentration Stock solutions Addition of inert substances Example How much of diluent must be added to 50 g of a 10% ointment to make it a 5 % ointment ?  Mixing Different Strengths - Proportion Calculation and Alligation The sum of the products obtained by multiplying a series of quantities by their respective concentrations equals the product obtained by multiplying a concentration by the sum of the quantities. This rule is applied in mixing of two similar preparations, but of different strengths, to produce a preparation of intermediate strength. Proportion Calculation and Alligation Example 100 g of a powder containing 8 % of sulphur are to be made from powder containing 10 % of sulphur and powder containing 2 % of sulphur. Let n be the number of g of the 10 % powder req’d Then 100 – n g will be the wt of 2 % powder req’d Therefore, 10 x n + 2 x (100 – n) = 8 x 100 100 100 100 Proportion Calculation and Alligation 10n + 200 – 2n = 800 8n = 600 n = 75 That is 75 g of the 10 % powder will be mixed with 25 g of the 2 % powder.  Alligation method This type of calculation can also be done by an alternate, alligation 10 2 8 6 2 Subtract 8 from 10 = 2 (10 – 8 = 2) (required percentage from stronger percentage) Subtract 2 from 8 = 6 (8 – 2 = 6) (weaker percentage from required percentage)  Alligation method Then 6 parts of the 10 % powder plus 2 parts of the 2 % powder will produce 8 parts of 8 % powder. 100 g are required; Therefore, quantity of 10 % powder required 100 x 6 = 75 g 8 and quantity of 2% powder required 100 x 2 = 25 g OR 100 g – 75 g = 25 g 8  Alligation method Although this method produces correct answer, it is not recommended except as a method of checking since its logic is not immediately obvious, and if a slip in writing a figure is made it may not readily be apparent.  Parts Per Million (ppm). ❖ Occasionally used in compounding prescription orders of very dilute preparations. ❖A 0.1 % solution may be expressed as 0.1 part/100 or 1 part/1000. A 1 ppm solution contains 1 part of solute/1 million parts of solution; 5 ppm is 5 parts of solute/1 million parts of solution.  Parts Per Million (ppm). Example Rx Sodium fluoride, q.s. Purified water, q.s. 60 ml Make solution such that when 5 ml is diluted to 240 ml of water a 2 ppm soln results. The amount of NaF needed is not known  Parts Per Million (ppm). The concentration of NaF in the 240 ml is 2 ppm The NaF solution diluted into 240 ml was 5 ml Now to calculate; 240 ml contains 2 ppm NaF = 2 g x240 ml = 0. 00048 1,000,000  Parts Per Million (ppm). This concn was in the 5 ml dose diluted to the 240 ml. Therefore , amount of NaF needed for the 60 ml prepn = 0.00048 g x 60 ml = 0.00576 g = 5.76 mg 5 ml.  Parts Per Million (ppm). If the prescriber had requested the concentration in terms of fluoride ion (F-), then the atomic weight of F- and molecular weight of NaF are used in the calculation Atomic weight of F- is 19, and mwt of NaF is 42 Therefore, 5.76 mg NaF = 5.76 mg x 19 g 42 g = 2.606 mg.  Alcohol Dilutions The Official dilute alcohols are made from 95 % alcohol - 95 parts by volume of ethyl alcohol (C2H5OH) and 5 parts by volume of water and is a volume in volume percentage solution. On mixing alcohol with water, contraction of volume and rise in temperature occur and, consequently, it is necessary to cool the mixture to about 20oC before diluting to the final volume.  Alcohol Dilutions The turbidity that occurs at first is caused by minute air bubbles evolved from the alcohol on dilution as air is less soluble in both solvent. The method of calculating alcohol dilutions is as follows; Example Prepare 600 ml of 60 % alcohol from 95 % alcohol. Let n = number of ml of 95 % alcohol required  Alcohol Dilutions These n ml will contain the same amount of alcohol as 600 ml of 60 % alcohol 95 x n = 60 x 600 100 100 Therefore, n = 60 x 600 = 379 ml 95 It follows that to calculate the amount of a stronger alcohol required to make a weaker alcohol the procedure is  Alcohol Dilutions It follows that to calculate the amount of a stronger alcohol required to make a weaker alcohol the procedure is Multiply the number of ml of diluted alcohol required by the percentage strength required and divide by the percentage of alcohol used. Vol. req’d x % req’d = Volume of stronger % used alcohol to be used  Proof Spirit In U.S. the strength of pure alcohol or absolute alcohol (C2H5OH) is calculated by means proof degrees. 100 proof spirit contains 50 % (by volume) or 42.5 % (by weight) of C2H5OH, and its specific gravity is 0.93426 at 15.56oC. Thus, 2 proof degrees equals 1 % (by volume) of C2H5OH. 100 ml proof spirit contains 50 ml of C2H5OH and water added to 100 ml.  Proof Spirit The term 10 degree under proof (𝟏𝟎° 𝒖. 𝒑) signifies that 100 volumes of the spirit contains 90 volumes of proof spirit plus 10 volumes of water, and 30 degree over proof 𝟑𝟎° 𝒐. 𝒑 indicates that 100 volumes diluted with water yields 130 volumes of proof spirit. To prepare proof spirit (proof strength), 50 volumes of 𝑪𝟐 𝑯𝟓 𝑶𝑯 are mixed with 53.71 volumes of water to allow for the contraction which occurs to yield 100 volumes of product.  Use of impure powders for preparations In pharmaceutical preparations it is usual to use slightly impure active ingredients. The purity of pharmaceutical ingredients are always included in the label of the container, e.g. 97 % 𝒘Τ𝒘. In the use of these ingredient, it is always assumed that the impurity is inert To weigh the quantity required for formulation, use this formula; 𝟏𝟎𝟎 × amount required = amount to be taken 𝒑𝒖𝒓𝒊𝒕𝒚  Formulation of liquid preparations from tablets Liquid preparation are the right dosage form for the infant, children and the very elderly who cannot swallow tablets. When a pure powders are unavailable to serve a prescription order, the pharmacists requires to prepare this from tablets.  Formulation of liquid preparations from tablets Example 17 Prepare 𝟏𝟎𝟎𝒎𝒍 of paracetamol syrup for an infant such that each 𝟓 𝒎𝒍 syrup contain 𝟏𝟐𝟎 𝒎𝒈 paracetamol. Available is 𝟔𝟎𝟎 𝒎𝒈 paracetamol tablets each containing 𝟓𝟎𝟎 𝒎𝒈 paracetamol.  Formulation of liquid preparations from tablets Volume of paracetamol syrup required = 𝟏𝟎𝟎 𝒎𝒍 𝟏𝟎𝟎 𝒎𝒍 Quantity of paracetamol in 𝟏𝟎𝟎𝒎𝒍 syrup = × 𝟓𝒎𝒍 𝟏𝟐𝟎 𝒎𝒈 = 𝟐𝟒𝟎𝟎 𝒎𝒈 Weight of paracetamol in tablet = 𝟓𝟎𝟎 𝒎𝒈 𝟐𝟒𝟎𝟎 𝒎𝒈 Number of paracetamol tablets required = = 𝟓𝟎𝟎 𝒎𝒈 𝟒. 𝟖 tablets 4.8 tablets are required, however, 4.8 tablets cannot be taken, hence, 5 tablets are selected and equivalent excess volume is prepared and 𝟏𝟎𝟎 𝒎𝒍 is then measured.  Formulation of liquid preparations from tablets Thus 1𝟎𝟎 𝒎𝒍 contains 𝟐𝟒𝟎𝟎 𝒎𝒈. 5 tablets = 𝟓 × 𝟓𝟎𝟎 𝒎𝒈 = 𝟐𝟓𝟎𝟎 𝒎𝒈 𝟐𝟓𝟎𝟎 𝒎𝒈 Volume to be prepared = 𝟐𝟒𝟎𝟎 𝒎𝒈 × 𝟏𝟎𝟎 𝒎𝒍 = `𝟏𝟎𝟒. 𝟏𝟔 𝒎𝒍 = 𝟏𝟎𝟒 𝒎𝒍. Take 5 tablets, prepare 𝟏𝟎𝟒 𝒎𝒍 syrup and measure 𝟏𝟎𝟎 𝒎𝒍 for the prescription.