Pharmaceutical compounding calculations
ان تحضير معظم وصفات التركيبات يحتاج الى إلمام بالحسابات الصيدلانية
و حتى في اكثر العمليات الحسابية بساطة فإن اهمال الارقام العشرية بعد الفاصلة او وضع قيم تقريبية .. ممكن ان يكون لها آثار خطيرة و قد تكون قاتلة .
فعندما نريد تحويل وحدة قياس الى وحدة قياس اخرى يجب استخدام القيمة المطابقة و ليست التقريبية ..
Ex.
The correct metric *****alent value for 1fl oz is 29.57 ml, not 30ml or even 29.6 ml.
Measurement systems:
The metric system is the official system of measurement used in pharmacy and medicine.
Least measurable quantity of weight:
A pharmacy’s tortion balance has a sensitivity reading of 5 mg. what is the smallest quantity that can be accurately weighed with a maximum error of 5%?.
a//// 100% X {maximum potential error (sensitivity reading) / permissible error (in percent)} =least measurable quantity.////a
a//// x= 100% X (5mg / 5%) = 100mg. ////a
Density factors in weighing /measuring:
A pharmacist receives a prescription for 120 ml of a 3% (w/v) hydrochloric acid (Hcl) solution. The density of concentrated hydrochloric acid (37 % w/w) is 1.18 g/ml. how many milliliters of the concentrated acid would be required for the prescription?
Step 1: calculate the weight of the required quantity of the 3% Hcl solution:
a//// 3 %( w/v) =0.03 g/ml. ////a
a//// 0.03 g/ml x 120 ml =3.6 g. ////a
Step 2: calculate the volume of the required quantity of the 3% Hcl solution:
a//// 3.6 g/1.18 g/ml = 3.05 ml. ////a
Step 3: calculate the volume of the 37% Hcl required for the prescription:
a//// 37%= 0.37. ////a
a//// 3.05 ml / 0.37= 8.24 ml of 37% Hcl, answer. ////a
Dilution aliquots:
A pharmacist needs 0.015 ml of flavoring oil to prepare an oral liquid. Using alcohol as a solvent and a pipet accurate to 0.01 ml, how can this volume of flavoring oil be obtained?
Step 1: Select a multiple of the desired quantity that can be accurately measured with the pipet. For this example, use 10 as the multiple and measure 0.1 ml of flavoring oil (10 x 0.01ml).place the oil in a suitable graduated cylinder.
Step 2: dilute the quantity of flavoring oil calculated in step 1 with alcohol to a quantity of solution that is divisible by the selected multiple, for this ex. add sufficient alcohol to make 50ml of solution, mix well.
Step 3: calculate the aliquot of the sol that contains the desired quantity of flavoring oil a//// (0.015 ml) as follows: ////a
a//// 0.1ml/50ml=0.015ml/x. ////a
a//// X=7.5 ml aliquot, answer. . ////a
Step 4: remove 7.5 ml of the sol that contains the required 0.015 ml of the flavoring oil.
Solutions:
Solubility ratios in the literature and in the United States Pharmacopoeia 25/National Formulary 20 are given as “x:y” and “x in y” (e.g., 1:4 and 1 in 4). These expressions are stated as one part of solute plus four parts of solvent. The resulting product is described as a 1:5 solution; that is, the product contains one part of solute in 5 parts of solution.
Another ex.: a substance with a solubility of 1:3 contains 1 g of solute in 3 ml of solvent. The resulting product, a 1:4 solution, contains 1 part of solute in 4 parts of solution. In contrast, a 1:3 solution
contains 1 g of solute in 3 ml of solution (i.e., sufficient solvent is added to make 3 ml of solution).
adwia_adwia82@yahoo.com
ان تحضير معظم وصفات التركيبات يحتاج الى إلمام بالحسابات الصيدلانية
و حتى في اكثر العمليات الحسابية بساطة فإن اهمال الارقام العشرية بعد الفاصلة او وضع قيم تقريبية .. ممكن ان يكون لها آثار خطيرة و قد تكون قاتلة .
فعندما نريد تحويل وحدة قياس الى وحدة قياس اخرى يجب استخدام القيمة المطابقة و ليست التقريبية ..
Ex.
The correct metric *****alent value for 1fl oz is 29.57 ml, not 30ml or even 29.6 ml.
Measurement systems:
The metric system is the official system of measurement used in pharmacy and medicine.
Least measurable quantity of weight:
A pharmacy’s tortion balance has a sensitivity reading of 5 mg. what is the smallest quantity that can be accurately weighed with a maximum error of 5%?.
a//// 100% X {maximum potential error (sensitivity reading) / permissible error (in percent)} =least measurable quantity.////a
a//// x= 100% X (5mg / 5%) = 100mg. ////a
Density factors in weighing /measuring:
A pharmacist receives a prescription for 120 ml of a 3% (w/v) hydrochloric acid (Hcl) solution. The density of concentrated hydrochloric acid (37 % w/w) is 1.18 g/ml. how many milliliters of the concentrated acid would be required for the prescription?
Step 1: calculate the weight of the required quantity of the 3% Hcl solution:
a//// 3 %( w/v) =0.03 g/ml. ////a
a//// 0.03 g/ml x 120 ml =3.6 g. ////a
Step 2: calculate the volume of the required quantity of the 3% Hcl solution:
a//// 3.6 g/1.18 g/ml = 3.05 ml. ////a
Step 3: calculate the volume of the 37% Hcl required for the prescription:
a//// 37%= 0.37. ////a
a//// 3.05 ml / 0.37= 8.24 ml of 37% Hcl, answer. ////a
Dilution aliquots:
A pharmacist needs 0.015 ml of flavoring oil to prepare an oral liquid. Using alcohol as a solvent and a pipet accurate to 0.01 ml, how can this volume of flavoring oil be obtained?
Step 1: Select a multiple of the desired quantity that can be accurately measured with the pipet. For this example, use 10 as the multiple and measure 0.1 ml of flavoring oil (10 x 0.01ml).place the oil in a suitable graduated cylinder.
Step 2: dilute the quantity of flavoring oil calculated in step 1 with alcohol to a quantity of solution that is divisible by the selected multiple, for this ex. add sufficient alcohol to make 50ml of solution, mix well.
Step 3: calculate the aliquot of the sol that contains the desired quantity of flavoring oil a//// (0.015 ml) as follows: ////a
a//// 0.1ml/50ml=0.015ml/x. ////a
a//// X=7.5 ml aliquot, answer. . ////a
Step 4: remove 7.5 ml of the sol that contains the required 0.015 ml of the flavoring oil.
Solutions:
Solubility ratios in the literature and in the United States Pharmacopoeia 25/National Formulary 20 are given as “x:y” and “x in y” (e.g., 1:4 and 1 in 4). These expressions are stated as one part of solute plus four parts of solvent. The resulting product is described as a 1:5 solution; that is, the product contains one part of solute in 5 parts of solution.
Another ex.: a substance with a solubility of 1:3 contains 1 g of solute in 3 ml of solvent. The resulting product, a 1:4 solution, contains 1 part of solute in 4 parts of solution. In contrast, a 1:3 solution
contains 1 g of solute in 3 ml of solution (i.e., sufficient solvent is added to make 3 ml of solution).
adwia_adwia82@yahoo.com
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