PHARMACEUTICAL CALCULATIONS.

Pharmaceutical Calculations Density: • Def.: It is a mass per unit volume of substance (1.8 g/ml). • Calculated by dividing mass by volume. Example: If 10 ml of sulfuric acid weight 18 g, Density = 18 (g) ...

Pharmaceutical Calculations Density: • Def.: It is a mass per unit volume of substance (1.8 g/ml). • Calculated by dividing mass by volume. Example: If 10 ml of sulfuric acid weight 18 g, Density = 18 (g) 10 (ml) = 1.8 g per ml (g/ml) Specific gravity: (sp.gr.) • Def.: Weight of substance / weight of equal volume of standard substances have the same temperature. • For specific gravity of liquids and solids, water used as standard. • It is a ratio between like quantities (no dimension). Example: in the above example, calculate the specific gravity? 10 ml of H2O (with same conditions) = 10 g (weight) Specific gravity of acid: wt of 10 ml of acid wt of 10 ml of water  18 (g)  1.8 10 (g) • It is a constant value for each substance. Specific Gravity of Liquids • Calculate specific gravity by three methods (liquids): 1- Weight and volume known. 2- Specific gravity bottle (pycnometer). 3- Displacement or plummet method. 1- Specific gravity by known weight and volume of liquids: Ex: If 54.96 ml of oil weight 52.8 g Specific gravity of oil = wt of 54.96 ml of oil wt of 54.96 ml of water = 52.78  0.9603 54.96 N.B.: 54.96 ml of water weight 54.96 gm (under similar conditions). 2- Specific gravity of liquids by pycnometer (specific gravity bottle): 1- Wt of container (specific gravity bottle) (empty)  (1). 2- Filled the container with water and weight  (2). 3- Filled the container with liquid and weight  (3). 4- Wt of water = (2) – (1)  (4) 5- Wt of liquid = (3) – (1)  (5) Specific gravity = wt of liquid  (5) Example wt of water (4) • Specific gravity bottle weights = 23.66 g (1) • Filled with water weights = 72.95 g (2) • Filled with liquid weights = 73.56 g (3) Specific gravity ? • Wt of water = 72.95 – 23.66 = (2) – (1) = 49.29 gm • Wt of liquid = 73.56 – 23.66 = (3) – (1) = 49.90 gm Specific gravity = 49.90  1.012 49.29 Specific Volume • Specific volume of substance = volume of substance volume of equal wt of water • Specific gravity: comparison Wts of equal volumes • Specific volume: comparison Vs of equal weights • Are reciprocals of each other (specific gravity = 1/ specific volume) (Specific volume= 1/specific gravity) • If multiplied together = 1 Specific volume calculation (liquid): 1- Given volume of specified Wt. 2- Given its specific gravity 3- Calculate specific gravity by having specific volume. Example: 1- Determine specific volume of syrup, 91.0 ml (its wt = 107.16 g) Specific volume = volume 91.0 of standard (water)  91.0 107.16  0.850 107.16 g of water = 107.16 g ml (volume) 2- What is specific volume of phosphoric acid having sp. gr. 1.71? sp. volume = 1  0.585 1.71 3- What is sp. gr. of liquid has sp. volume 1.396 sp. gr. = 1  0.716 1.396 Calculations of weight (liquid): • Calculation the weight of liquid when given - Volume - specific gravity sp. gr. water sp. gr. liquid  wt of equal volume x of water x = weight of liquid N.B: sp. gr. of water = 1  Wt of liquid = wt of equal volume of water x sp. gr. liquid Example: What is the wt of 3620 ml of alcohol having sp. gr. of 0.820? • Sp. gr. of liquid x wt of equal volume of water = wt of liquid • 3620 ml of water = wt of 3620 g • Wt of liquid = 3620 (g) of water x 0.820 = 2968 g. Calculations of volume (liquid): • Calculate the volume of liquid: - Wt - sp. gr. sp. gr. liquid sp. gr. water  volume of equal wt water x x = volume of liquid V liquid = volume of equal wt water sp. gr. liquid Example: What is the volume of 492 g nitric acid with sp. gr. 1.40? 492 g of water measure 492 ml  Volume = 492  351 ml 1.4 Percentages Def.: Rate for hundred, no. and percent (%). Ex: 25% = 25 parts of 100 parts = 25 / 100 = 0.25 Uses in Pharmaceutics: 1- Express the conc. of solute in solution. 2- Amount of active ingredient (drug or preparation). 3- Amount of active ingredient in dosage form. Percentage calculation (concentration): I- Percent (w/v) % No. grams of substance in 100 ml (solution liquid). Ex: water or another. II- Percent (v/v) % No. of milliliters of substances in 100 ml of solution III- Percent (w/w) % No. of grams in 100 gm of solution or liquid. N.B.: • mg. % - Number of milligrams of substances in 100 ml of liquid - Used to determine conc. of drug or natural substances in biological fluid (blood). • ppm (part per million): - Used for very dilute solutions (determine conc.) - For test limits. I) Percentage weight-in-volume 1- Calculate (wt) of substance of known % (w/v) in specific volume. 2- Calculate the % (w/v) of solution by knowing: • wt. solute or • volume of solution. 3- Calculate the volume of solution by knowing: • (w/v) % or • (wt) of solute gr. solute = vol. in ml x % (w/v) Examples: 1- How many grams of dextrose required to prepare 4000 ml of 5% solution? 5 g. solute = 4000 x 100 = 200 g 5g.  100 ml , X  4000 ml X = 4000 x 5  200 g 100 2- What is the % strength (w/v) of solution of urea, if 80 ml contain 12 g.? % = g. solute of constituen t x100  volume in ml % = 12 x100  15% 80 3- How many milliliters of 3% solution can be made from 27 g. ephedrine sulfate? Volume in ml = g. solute %  27 0.03  900 ml II) Percentage volume-in-volume (v/v) %: 1- Calculate (v) of active ingredient by knowing: • (V) Liquid or • % (v/v) 2- Calculate % (v/v) by knowing: • (V) Active ingredient or • volume of solution 3- Calculate (V) of solution by knowing: • % (v/v) or • (V) active ingredient. Examples: 1- How many ml. liquefied phenol should be used in: R/ Liquefied phenol 2.5% Calamine lotion to 240.0 ml (V) Active ingredient = (V) in ml. x % ml 240 ml x 0.025 = 6 ml. or 2.5 – 100 ml x – 240.0 ml  x = 240 x 2.5  6 ml 100 2- Calculate the % (v/v) of solution given the volume of active ingredient and volume of solution? ml. % = of active ingredient x100 volume in ml. Ex: - 250 ml of lotion - used 4 ml liquefied phenol Calculate the % (v/v) liquefied phenol in lotion? 4 % = 250 x 100  1.6% (V/V) 3- Calculate volume of solution given the volume of active ingredient and its % (v). Example: Peppermint spirit contains 10% (v/v) of peppermint oil. What volume of spirit contain 75 ml of active ingredient? Volume in ml = 75  750 ml 0.1 III) Percentage weight-in-weight (w/w) %: 1- Calculate the (wt) g drug by knowing: • % or • (wt) solution. 2- Calculate the (wt) of solution by knowing: • % or • (wt) of active ingredient 3- Calculate % by knowing: • (wt) of solution or • (wt) of solution Examples: 1- How many grams of phenol should be used to prepare 240 g. of 5% (w/w) solute in water? g. of solute = wt solution (g.) x % Weight of phenol = 240 x 0.05 = 12 g 2- How many g. of drug to make 120 ml of 20% (w/w) solute in having sp. gr. of 1.15? 1- wt of solution (120 ml) = 120 x 1.15 = 138 g. 2- wt of solute = 138 x 0.2 = 27.6 g. 3- If 1500 g. of solution contains 75 g. of drug substance, what is the % (w/w) of solution? % = 75 1500 x100  5% 3- Calculate wt. of solution either by knowing % or wt. of active ingredient. Ex: What wt. of 5% (w/w) solution can be prepared from 2 g. active ingredient? g. of solute %  2 0.05  40 g. Ratio Strength • For expression the conc. of weak solution. • 5% = 5 parts per 100 parts 5: 100 1: 20 (ratio strength). Example: 1- Express 0.02% as ratio strength? 0.02  100  1: 5000 2  100 100 1/50  100 Ratio strength = 1: 5000 2- Express 1: 4000 as percentage strength? 1  4000 x  100 x = 100  0.025% 4000 Dilution and Concentration of Liquids A- Calculate the % or ratio strength of solution made by diluting or concentrating a solute, giving • strength or • quantity (solution) Example: If 500 ml of 15% (v/v) solution of methyl salicylate in alcohol are diluted to 1500 ml. What % (v/v) ? * 1500  15%  x = 5% 500 x Quantity x conc. (Known) = quantity x conc. (Unknown) * 500 x 15% = 1500 x (X %)  X = 5% B- Calculate the amount of solution with known strength by either diluting or concentrating a specific quantity of solution of given strength Example: How many grams (g.) of 10% (w/w) ammonia water can be made from 1800 g. of 28% (w/w) ammonia water? * 10% 28%  1800 x x = 5040 g. Quantity x conc. (Known) = quantity x conc. (Unknown) Or * 1800 (g.) x 28% = X (g.) x 10% x = 5040 g. Stock Solutions Definition: • Are solutions of known concentration. • Are strong solutions from which weaker ones are made. • Are prepared in (w/v). • Their concentrations expressed by ratio strength ( ). Example: How many milliliters of 1:400 (w/v) stock solutions should be used to make 4 liters of a 1:2000 (w/v) solution? 1) 4 liter = 4000 ml 100 1:400 (ratio strength) = 400 1:2000 = 0.05% = 0.25%  0.25(%)  4000 (ml) x = 800 ml 0.05(%) x (ml) 2)  4000 (ml) x (ml) x = 800 ml 800 ml taken from stock solution and completed to 4000 ml. Dilution of Alcohol Examples: 1- How much water should be mixed with 5000 ml of 85% (v/v) alcohol to make 50% (v/v) alcohol? 50 (%) 85 (%)  5000 (ml) x (ml) x = 8500 ml i.e. 5000 ml (85%) alcohol  add water to 8500 ml. 2- How much water should be added to 4000 g. of 90% (w/w) alcohol to make 40% (w/w) alcohol? 40 (%)  4000 (g.) x = 9000 g. 90 (%) x (g.) i.e. 4000 g. (90%) alcohol  add water to 9000 ml. Alligation 1- Alligation medical 2- Alligation alternate 1- Alligation Medical: Def.: A method by which the weighted average percentage strength of mixture of two or more substances whose quantities and concentrations known  quickly calculated. Determine percentage strength of a mixture. Calculate the (%) of mixture of two or more components of known (%). Examples: 1- What is the percentage (v/v) of alcohol in a mixture of 3000 ml of 40% (v/v) alcohol, 1000 ml of 60% (v/v) alcohol and 1000 ml of 70% (v/v) of alcohol? 40 x 3000 = 120,000 60 x 1000 = 60,000 70 x 1000 = 70,000 Total 5000 = 250,000 % of alcohol in mixture = 250000  50% 5000 2- What is the % of zinc oxide in ointment prepared by mixing 200 g. of 10% ointment, 50 g. of 20% ointment, and 100 g. of 5% ointment? 10 x 200 = 2000 20 x 50 = 1000 5 x 100 = 500 = 3500  10% 350 Total 350 = 3500 2- Alligation Alternate: Def. Calculate the relative amounts of solutions or substances of different strengths that used to make a mixture of required strength. Example: In what proportion should 20% benzocaine ointment be mixed with ointment base (without drug) to produce 2.5% benzocaine ointment? 20% minus 2.5 parts of 20% (drug + base) gives 0% drug (base) 17.5 parts of oint. base Relative amounts = 2.5: 17.5 or 1:7 Check: 20 x 1 = 20 0 x 7 = 0 Total 8 = 20 Percentage of oint. = 20 / 8 = 2.5% 2- Calculate the quantity of a solution or mixture of given strength that should be mixed with a specified quantity of another solution or mixture of given strength to make a solution or mixture of desired strength. Example: How many grams of 2.5% hydrocortisone cream should be mixed with 360.0 g. of 0.25% cream to make a 1% hydrocortisone cream? 2.5% 0.75 parts of 2.5% cream 0.25% 1.5 parts of 0.25% Relative amounts = 0.75 : 1.5 or 1:2 Check: Homework 2 (parts)  360.0 (g.) 1 (part)  x (g.) x = 360  180 g 2 180 : 360 = 1:2 • 180 (g.) of 2.5% hydrocortisone cream + 360 (g.) of 0.25% cream  1% hydrocortisone cream. Specific Gravity of Mixtures I- To calculate specific gravity of a mixture given the specific gravity of its ingredients. Examples: What is sp. gr. of mixture of 1- 1000 ml of syrup sp. gr. = 1.3 2- 400 ml glycerin sp. gr. = 1.25 3- 1000 ml alixir sp. gr. = 0.950 1.3 x 1000 = 1300 1.25 x 400 = 500 0.95 x 1000 = 950 Total 2400 = 2750 Sp. gr. of mixture = 2750  1.146 2400 II-Calculate the relative or specific amounts of ingredients of given sp. gr. required to make mixture of desired sp. gr.? Examples: 1- In what proportion must glycerin with sp. gr. of 1.25 and water be mixed to give a liquid having sp. gr. 1.10 ? Gly. 1.25 0.10 parts of glycerin 1.10 water 1.00 0.15 parts of water Relative amounts = 0.10 : 0.15 or 2:3 (gly. : water) 2- How many milliliters of each of two liquids with sp. gr. 0.950 and 0.875 should be used to prepare 1500 ml of liquid having sp. gr. of 0.925%. (1) 0.950 0.050 parts of liquid (1) (2) 0.875 0.025 parts (2) Relative amounts = 0.050 : 0.025 = 2 : 1 Total = 3 parts 3 parts 2 parts  1500 ml x ml x = 1000 ml of liquid with sp. gr. 0.950 3 parts  1500 ml y = 500 ml of liquid with sp. gr. 0.875 1 part y ml 1000 + 500 = 1500