PHARMACEUTICS I

Pharmaceutical Calculation

PHRM 210

DensityDensity (d) is mass per unit volume of a substance. It usually expressed as grams per cubic centimeter (g/cc).

Gram is defined as the mass of 1 cc of water at 4oC, so the density of water is 1 g/cc.

According to USP, 1 mL may be used as the equivalent of 1 cc. Therefore, The density of water may be expressed as 1 g/mL.

Density = MassVolume

The weight of 1 mL of mercury is 13.6 g, hence, its density is 13.6 g/mL

If 25.0g of olive oil are required for a prescription, what volume should be used? (Density of olive oil=0.910g/mL )

25.0g x 1 mL0.910g

= 27.5 mL

Problems:

1. Calculate the weight of 120 ml of oil whose density is

0.9624g/ml. What will be the weight of 5 liter of that oil?

2. What is the volume of 2.0 kg of glycerin if the density of

glycerin is 1.25g/ml?

3. Find out the volume in liter of an oil whose density is

0.75g/ml and the total weight of that oil is given 20.5 kg.

Specific gravity

Specific gravity is the ratio, expressed decimally, of the weight of a substance to the weight of an equal volume of a substance chosen as a standard.

Specific gravity =Weight of substance

Weight of equal volume of water

Water is used as the standard for the specific gravity of liquid and solids. The most useful standard for gases is hydrogen, although sometimes air is used.

Both substance should be at the same temperature or the temperature of each being definitely known. According to USP- the standard temperature for specific gravity is 250C.

At 25ºC and 1 atmosphere of pressure, one milliliter of distilled water weighs one gram. Therefore, the specific gravity of water is established as 1.

Specific gravity=

Weight of the substance

Weight of an equal volume of water

Because one milliliter of water weighs one gram:

Specific gravity=Number of grams of the substance

Number of milliliters of the substance

1. Known weight and volume:

Problem: If 54.96ml of an oil weight 52.78g, what is the specific gravity of the oil?

Answer:54.96 ml of an oil weight 52.78 gSpecific gravity of oil =52.78(g)/54.96(g) = 0.9603

Calculating the specific gravity of a liquid when its weight and volume are known.

Specific gravity of liquids

Problem: If a pint of a certain liquid weights 601g, what is the specific gravity of the liquid?

Answer:

1 pint = 473 ml

Specific gravity of liquid =

601(g)/ 473(g) = 1.27

2. Specific gravity bottle:

To calculate the specific gravity of a liquid by means of a specific gravity bottle, the container is filled and weighed first with water and then with liquid.

Problem: A specific gravity bottle weighs 23.66g. When filled with water, it weights 72.95g, when filled with another liquid, it weights 73.56g. What is the specific gravity of the liquid?

Answer: 73.56g – 23.66 g = 49.90 g of liquid72.95g – 23.66 g = 49.29 g of waterSpecific gravity of liquid = 49.90g / 49.29 g =1.012

3. Displacement or plummet method:

Calculating the specific gravity of a liquid determined by the displacement or plummet method is based on the Archimedes’ principle.

Problem: A glass plummet weighs 12.64g in air, 8.57g when immersed in water, and 9.12g when immersed in an oil. Calculate the specific gravity of the oil.

Archimedes’ principle: A body immersed in a liquid displaces an amount of the liquid equal to its own volume and suffers an apparent loss in weight equal to the weight of the displaced liquid.

Answer: 0.865

Specific gravity of solids

1. Solids heavier than and insoluble in water:

Problem: A piece of glass weighs 38.525g in air and 23.525g when immersed in water. What is its specific gravity?

Answer:

38.525g – 23.525g = 15g of displaced water (weight of an equal volume of water).

Specific gravity of glass = 38.525g / 15g

= 2.568

Weight of solid in air/weight of water that it displaces when immersed in it.

2. Solids heavier than and soluble in water:

Problem: A crystal of a chemical salt weighs 6.423g in air and 2.873g when immersed in an oil having a specific gravity of 0.858. What is the specific gravity of the salt?

Answer:

6.423g–2.873g = 3.550g of displaced oil.

Wt. of displaced oil/ Wt. of salt in air = Sp. gr. of oil / Sp. Gr. of salt

Sp. Gr. of salt= (0.858 x 6.423) / 3.550 = 1.55

The weights of equal volumes of any two substances are proportional to their specific gravities.

Specific volume

Specific volume is defined as an abstract number (has no unit) representing the ratio of the volume of a substance to the volume of an equal weight of another substance taken as a standard in same temperature.

Water is standard for liquids and solids.

Problem: If 25g of glycerin measure 20ml and 25g of water measure 25ml under the same conditions. What is the specific volume of glycerin?

Answer:

The specific volume of the glycerin=

vol. of glycerin / vol. of water

= 20ml /25ml = 0.8

Problem: Calculate Specific volume of a syrup, 91.0ml of which weigh 107.16g.

Answer: 0.85

Percentage Calculations

The term percent and its corresponding sign (%) mean “by the hundred” or “in a hundred”, and percentage means “rate per hundred”.

Percent weight-in-volume (W/V):

W/V or Weight/Volume percent is defined as the number of grams in 100 milliliters of solution or liquid preparation.

Example: A 10 percent (w/v) potassium chloride (KCl) elixir would contain 10 grams of potassium chloride in every 100 milliliters of KCl elixir.

Percent volume-in-volume (V/V)

V/V percent or Volume/Volume percent is defined as the number of milliliters in every 100 ml of solution or liquid preparation.

Example: A 70% (v/v) alcoholic solution would contain 70 milliliters of alcohol in every 100 ml of solution.

Percent weight-in-weight (W/W)

W/W percent or Weight/Weight percent is defined as the number of grams in 100 grams of a solid preparation.

Example: A 5 percent (w/w) boric acid ointment would contain 5 grams of boric acid in each 100 grams of boric acid ointment.

Drug Concentrations Expressed As Percent(%)

Weight/Weight =

Weight/Volume =

grams of solute100 g of prep.

1 % =

1 % =

1 % =

1g of solute 100 g of solution

Example:

1 mL of solute 100 mL

1 g of solute 100 mL

Volume/Volume =mL of solute 100 mL

grams of solute 100 mL

Dilution and concentration

Pharmacy personnel will often go to a stock solution to obtain the amount of active ingredient that is needed to make a preparation. This is especially true if the amount required is so small that it cannot be accurately weighed on a torsion balance. It is easier to measure an amount of stock solution than to set up a balance, weigh the ingredients, and compound the entire product.

The use of stock preparations is an important aspect of pharmacy.

c. Formula: V C = V1 C1

(1) V = Volume of stock preparation

(2) C = Concentration of stock preparation

(3) V1 = Volume of desired preparation

(4) C1 = Concentration of desired preparation

d. Formula: W C = W1 C1

(1) W = Weight of stock preparation

(2) C = Concentration of stock preparation

(3) W1 = Weight of desired preparation

(4) C1 = Concentration of desired preparation

Formulas:

a. Volumes and weights must be expressed in the same units.

b. Concentrations must be expressed in the same units.

Answer:

Let the volume of syrup 100ml.

100ml syrup evaporated to 85% of its volume, so we will have 85ml

85ml/100ml=65%/X%

X%= 76.47% or 76% (answer)

Problem: If a syrup containing 65% (w/v) of sucrose is evaporated to 85% of its volume, what percent (w/v) of sucrose will it contain?

Problem: How many milliliters of water should be added to a pint of a 5% (w/v) solution of boric acid to make a 2% (w/v) solution?

Answer:

1pint = 473 ml

Therefore,

2%/5% = 473ml/ Xml X= 1182.5 ml

So water should be added (1182.5-473)ml= 709.5ml (answer)

Problem: How many milliliters of a 1:200 silver nitrate solution would be needed to make 2000 ml of a 1:4000 solution?

Answer:

V C = V1 C1

(X) ( 1/200 ) = (2000 ml) ( 1/4000 )

(X) (4000/200)=(2000ml) (4000/4000)

20 X = 2000 ml

X = 100 ml (answer)

Problem: How many milliliters of a 2% stock solution of potassium permanganate (KMn04) would be needed to

compound 120 ml 0.2% solution of potassium permanganate ?

Answer

V C = V1 C1

(X) (2%) = (120 ml) (.02%)

2 X = 2.4

X = 1.2 ml (answer)

Problem: How many grams of 14% zinc oxide ointment can be made from one pound of 20% zinc oxide ointment?

Answer:

1 pound = 454 gm

W C = W1 C1

(X) (14%) = (454 g) (20%)

14 X = 9080

X = 648.57grams (answer)

Answer:

V C = V1 C1

(X) (10%) = (4 L) ( 1/2000 )

Change 4 liters to milliliters

Change 10% to a ratio by placing the 10 over 100.

(X) (10/100) = (4000 ml) ( 1/2000 )

(X)(20,000/100)=(4000 ml)(2000/2000)

200 X = 4000 ml

X = 20 ml (answer)

Problem: How many milliliters of 10% povidone-iodine solution would be needed to make 4 liters of a 1:2000 povidone-iodine solution?

Solve the Problems:

1. How many milliliters of a 3% hydrogen peroxide solution would be needed to make 120 ml of 1% hydrogen peroxide solution? How many milliliters of water should be added? [40ml & 80ml]

2. The strength of real acetic acid is 33% (W/V). Using this, prepare 200ml of a solution of acetic acid containing 4% (W/V) of real acetic acid. [24.2ml]

3. Prepare 200 ml of a solution of ammonia containing 4% by weight of ammonia. The strong solution of ammonia contains 32.5% of ammonia W/W. [24.62gm]

4. Prepare 400ml of a 5% solution and label with direction for preparing 2 liter quantities of a 1 in 2000 solution. [20gm & 20ml up to 2 liter]

Alcohol dilution:

Problems: Prepare 400 ml of 45% alcohol from 95% alcohol.

Volume of stronger alcohol to be used=

400x45

95

= 189.47 ml

= 190 ml

190 ml of 95% alcohol is diluted with water to produce 400ml. The strength of dilute alcohol will be 45%.

Solve the Problems:

1. Prepare 600 ml of 60% alcohol from 95% alcohol. [378.95ml]

2. Prepare 500 ml of 40% alcohol from 95% alcohol. [210.52]

3. How much water should be mixed with 5000 ml of 85.5% (V/V) alcohol to make 50% ( V/V) alcohol? [3550ml of water]

4. How many milliliters of water must be added to 5 gallons of 100% isopropyl alcohol to make a 70% dilute Isopropyl alcohol? (1 gallon= 3785 ml)

5. How many milliliters of distilled water must be added to one gallon of 100% Isopropyl Alcohol to make a 70% alcohol?

6. A 25% solution of ethyl alcohol may be used to bathe a small child for the purpose of cooling and reducing fever. How many milliliters of ethyl alcohol (95%) and how many milliliters of distilled water must be combined to make two quarts of the 25% ethyl alcohol solution? (1 quart = 946 ml)

Problem: How many grams of zinc oxide are needed to make 240 grams of a 4% (w/w) zinc oxide ointment?

IF 4 g ZnO=

100 g Oint

100 X = 960 X = 9.6 g of zinc oxide

THEN X g ZnO

240 g Oint

Answer:

(answer)

Compounding of same formula

Problem: How many milliliters of a 5% (w/v) boric acid solution can be made from 20 grams of boric acid?

Answer:

(answer)

Problem: How many milliliters of paraldehyde are needed to make 120 ml of a 10% (v/v) paraldehyde solution?

Answer: 12ml

Problem: Prepare 500 ml of a 2 in 4000 solution of potassium permanganate.

Answer: 50mg

One percent Method:

Problem: How many grams of ephedrine sulfate are needed to make 120 ml of a 2% (w/v) ephedrine sulfate solution?

Answer:1gm dissolved in 100ml = 1% w/v solution

2gm dissolved in 100ml = 2% w/v solution

For 120ml solution the quantity of ephedrine sulfate required-

2x120

100= 2.4 gm of ephedrine sulfate needed.

(answer)

How many grams of potassium permanganate (KMnO4) are needed to compound the above prescription?

Distilled water qsad 240

Rx

KMnO4 0.02%

mlProblem:

Answer: 0.048 grams

Problem: Prepare 150ml of 4% (W/V) potassium permanganate and label the directions for preparing 500ml of 1 in 2500 solution.

Solve the Problems:

► If 5 g of a chemical is dissolved in enough water to make the preparation measure one liter, what is the percentage strength of the solution?

► How many milliliters of a 0.02% W/V solution can be made from 2.5 g of a chemical?

► Normal saline solution contains 0.9% W/V NaCl. How many grams of sodium chloride should be used to make 1.5 liters of normal saline?

► How many grams of strong silver protein (SSP) are required to make 250 ml of a 0.25% (w/v) solution?

Problem: Beximco uses raw material of paracetamol in which

purity of paracetamol is 94% for manufacturing Napa tablet (500mg paracetamol). The batch size of Napa is 50,000 Tablets. Calculate the amount of Raw Materials of Paracetamol to produce a batch of NAPA.

If Beximco uses 3% overage then calculate the amount of RM of paracetamol to manufacture a batch of NAPA.

OVERAGE:

Overage means excess. Usually companies use certain percent (within the specification limit) excess of active ingredient in their preparation.

"PARTS-PER" NOTATION

The parts-per notation is used in some areas of science and engineering because it does not require conversion from weights or volumes to more chemically relevant units such as normality or molarity. It describes the amount of one substance in another.

It is the ratio of the amount of the substance of interest to the amount of that substance plus the amount of the substance it is in.

Parts per hundred denotes the amount of a given substance in a total amount of 100 regardless of the units of measure as long as they are the same. e.g. 1 gram per 100 gram. 1 part in 102.

Parts per thousand denotes the amount of a given substance in a total amount of 1000 regardless of the units of measure as long as they are the same. e.g. 1 milligram per gram, or 1 gram per kilogram. 1 part in 103.

Parts per million ('ppm') denotes the amount of a given substance in a total amount of 1,000,000 regardless of the units of measure used as long as they are the same. e.g. 1 milligram per kilogram. 1 part in 106.

Problem: Calculate the quantity of Sodium Fluoride (NaF) required for the preparation of 500ml of 2 ppm NaF aqueous solution.

Answer- 0.001 gm

CALCULATION OF DOSES

Posology:

The word posology derived from Greek word ‘posos’, meaning how much and ‘logos’, means science. That means it is a branch of medical science which deals with doses or quantity of drugs which can be administered to produce the required pharmacological actions.

The term dose refers to the amount of medication that a patient must take at one time to produce the optimum therapeutic effect.

Calculation of child dose:

I. According to age:

a) Young’s formula:

Child's dose =Child's age in years Child's age in years + 12 years X Adult dose

The above formula is used for calculating the doses for children under 12 years of age.

Problem: The adult maintenance dose of a drug is 325 mg. Calculate the dosage for a 3-year-old child.

Answer: 65mg

b) Dilling’s formula:

Because of easy and quicker calculations, Dilling’s formula is considered better.

Problem: If the adult dose is 60mg and the age of the child is 6 years, what will be child’s dose?

Answer: 18mg

Child's dose =Child's age in years

20 X Adult dose

c) Fried's rule:

Child's dose =Child's age in months

150 months X Adult dose

Problem: The adult dose of an antihistamine is 50 mg. Calculate the dosage for a 2½ year-old child.

Answer: 10mgUsually used for infant’s dose calculation.

II. According to body weight:

Clark's Rule:

Problem: The adult dose of Doxycycline is 100 mg. Calculate the dosage for a child weighing 50 pounds.

Answer: 33.3 mg

Child's dose =Child's weight in lbs.

150 lbsX Adult dose

III. According to body surface area:

Child's dose =Surface area of childSurface area of adult

X Adult dose

The average body surface area for an adult = 1.73m2. Hence,

Problem: Calculate the dosage for a child that has a body surface area of 0.57 m2, when the adult dose of a drug is 50 mg.

Answer: 16.5 mg

Child's dose =Surface area of child

1.73m2X Adult dose

In chemistry, normality is a measure of concentration, it is equal to the number of gram equivalents of a solute per liter of solution.

It is denoted by N.

Normality =Gram equivalents

Liters solution

Normality:

Gram equivalents =Atomic weight or molecular weight

Valence

Problem: 5.30 gm of Na2CO3 was dissolved in water and the volume made to 100ml. Calculate the normality of the solution.

Answer:

Molecular weight of Na2CO3= 106The equivalent weight of Na2CO3 is half the molecule

weight, i.e., 106/2= 53

100 ml solution contains 5.30 gm Na2CO3 Therefore,1000 ml solution contains (5.30x10) gm

= 53 gm

Normality of the solution is 1

It defined as the presence of number of moles of solute in 1000ml or 1 litre of solution.

It is denoted by M.

Molarity =Moles soluteLiters solution

Molarity:

Problem: 58.44 grams of NaCl dissolved in exactly 2 L of solution. What would be the molarity of the solution?

Answer:

Molecular weight of NaCl is 58.44 grams/mol

58.44 grams 58.44 grams/mol =1mol

Then, dividing 1.00 mol by 2.00 L gives 0.500 mol/L (or 0.500 M).

Answer: 0.28 M

Problem: Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL. Molecular weight of KBr is 119 grams/mol.

Molality denotes the number of moles of a given substance per kilogram of solvent. (not solution)

It is denoted by m.

Molarity =Moles soluteKg Solvent

Molality: