Western Board 2020 Copy for

Part II
Pharmaceutical Sciences

Basic Pharmacy Calculations

Units

In metric system

Weight Volume
1 kilogram 1000 grams 1 pint 473 ml = 16 fluid
ounces
1 gram 1000 milligrams 1 fluid ounce 29.6 ml = 2
tablespoonfuls
1 milligram 1000 micrograms 1 fluidram 3.75 ml
1 microgram 0.001 milligrams 1 teaspoonful 5 ml
1 milligram 0.001 grams 1 tablespoonful 15 ml
1 microgram 10-6 grams 1 wine glassful 60 ml
1 nanogram 10-9 grams 1 tea cupful 120 ml
1 grain (1 gr) 65 milligrams 1 full glass 240 ml
1 kilogram 2.2 lbs 4 cups 1 pint
8 pints 1 gallon
1 gallon 3.8 liter
Official dropper 20 drops/ml

Body Surface Area

"BSA" stands for "body surface area." units are m2. BSA is sometimes used in dosing of medications
(more on this later). A universal equation for calculating BSA of both kids and adults is

BSA in m2 = (H0.3964) (W0.5378) (0.024265)

Where height is in centimeters and weight in kilograms.

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BSA in m2= VH x W/3600

Body Mass Index (BMI)

BMI = weight in Kg/height m2 or Weight in Kg/height m x height m

BMI = normal 18.9 to 24.9

BMI = overweight > 26 to 30

BMI = obese > 30

BMI = morbid obese > 35

Example:

1) A person body weight is 180 lbs and height is 1.5 m? What is right BMI?

A) normal

B) over-weight

C) obese

D) morbid obese

E) none

Tips: 180/2.2% = 82 kg

82/1.5 x 1.5 = 36 morbid obese

Density (specific gravity)

Density = Weight/volume (D = W/V)

Volume = Weight/Density

Weight = Volume x Density

Examples:

Water 1000 ml, its density is 1, what is weight water in grams?

so, 1000 ml = 1000 g of water or 1 Kg

Glycerin 1000 ml, its density is 1.2, what is weight in g of glycerin?

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Weight = Volume x Density

Weight = 1000 ml x 1.2 = 1200 g

A winter green oil density is 0.9, what is the weight of 1000 ml oil?

A topical preparation density is 1.1 and its volume is 720 ml. What is weight in kilograms.

A prescription order for glycerin 1200 ml, to prepare suppository in base. Glycerin density is 1.1. What is
weight of glycerin in suppository?

weight= volume x density = 1200 ml x 1.1 = 1320g

Fractions to decimal numbers

To convert fractions to decimal numbers  divide top number by bottom number

Example: ½ = 0.5

Example: 3/10 = 0.3

Example: 10/3 = 3.3

Mixed numbers to decimal number

To convert mixed numbers to decimal number

Example: 2 ¾ = 11/4 = 2.75 or 11 ¾ = 11.75

Step 1  multiple bottom number by whole number and add with top number.

Step 2  divide top number by bottom number

Decimal to mixed number

To convert decimal to mixed number

Example: 3.5 = 3 ½

Step 1  Write the decimal number over one, dividing with one = 3.5/1

Step 2  Move the decimal point in the top as many places as to right as necessary to form a whole
number.

Move the same in bottom. 35/10

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35/10 = 7/2 = 3 ½

Example: 1.25 = 125/100 = 5/4

Pounds and Kilograms or Kilograms (Kg) and Pounds (lbs)

To convert lbs into Kg = divide by 2.2

To convert Kg to lbs = multiply by 2.2

Example:

1) A newborn child weighs 10 lbs. What is weight in kg?

2) A child weighs 22 lbs, and doctor wants to give 1 mg/kg drug, how much you should give?

The child weight in kg? is 22 lbs/2.2 = 10 kg so give drug 10 mg

A patient weight 180 lbs has admitted to emergence for congestive heart failure and severe edema.
Patient was giving furosemide iv infusion for the past 24 hours. After discharge, the patient weight was
173 lbs. How many kg patient weight is lost?

180-173 = 7 Ibs

7/2.2 3.2 kg

Teaspoons (tsp) and Tablespoons (Tbsp)

The physician has prescribed 1 Tbsp of Riopan. The unit dose container available contains 30 ml of this
medication.

The correct dose is . . . . . . ml.

The client has been taking 20 ml of a medication in the hospital. in preparing him to administer this tsp
of this drug. medication in the home setting, you would teach him to take . . . . . .tsp of this drug.

Milligrams and Grains

1. You are to administer phenobarbital gr ¾ . There are scored 60 mg tablets available in floor
stock.

You would administer . . . . . . mg or . . . . . . tablet(s).

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Morphine sulfate gr 1/8 IM is to be prepared from a solution containing 10 mg/1 mt. The volume to
be given is . . . . . . ml.

Milliliters and Ounces

The physician has ordered ss ounce of a laxative. The correct dose is . . . . . . ml.

You are to administer 5 ml of ferrous gluconate that must be diluted in water to protect the client's
teeth and gastrointestinal track. The directions in your drug reference state that each mi of this
medication must be diluted in 20 ml of water.

Once diluted, the total volume to be administered is . . . . . . ml.

Calculations involving percentage, ratios, proportions

Percentage: To convert percent to fractions

25% = 25/100 = ¼

50% = 50/100 = ½

200% = 200/100 = 2

To convert fraction to percentage: multiply by 100

¾ * 100 = 75%

Ratios: Relation between two quantities

W/W 1 gram in 100 grams =1 g:100

W/V 1 gram in 100 ml = 1 g:100 ml

V/W 1 ml in 100 grams 1 ml:100 g

V/V 1 ml in 100 ml = 1 mL:100 mL

Ratio Strength

For solids in liquids (w/v) For liquids in liquids (v/v) For solids in solids (w/w)
Grams /1000 ml of mixture ml/1000 ml of liquid Grams/1000 grams of Mixture
or Grains/1000 grains of
mixture
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Example

1) Express 0.02% as ratio strength

Solution: 0.02  100 ml

2  10 000 than 1  5000

2) Dissolving 4.8 g of NaCI in water can make how many milliliters of a 1:1500 solution?

Solution:

We have 1 gram for every 1500 ml of solution

So for 4.8 we will give x ml of solution

X ml = 4.8 * 1500 /1 = 7200 ml

3-Syrup is an 85% w/v solution of sucrose in water. It has a density of 1.313 g/ml. How many
milliliters of water should be used to make 125 ml of syrup? (D = W/V)

Working:

Density of solution =mass / volume

Density of solution = Weight of solution / 125

Therefore, the weight of the solution will be 164.125 g

Now we have 85 grams of sucrose in 100 ml of solution

Therefore for 125 ml of solution we need: (125 * 85)/100 = 106.25 grams the weight of water in this
solution = 164.125 -106.25 = 57.875

Ratio percent: W/W %

Example: 5 g of glucose in 50 mL water, what is ratio percent  10% W/N

Examples: 50 g of glucose in a 1 kg glycerin, what is ratio percent  5% W/W

Example: If 2% of glucose in water, what is ratio? 2 g/100 ml = 1/50 = 1:50

12.5 g glucose in 100 ml Water, what is ratio percent? = 12.5% w/v

Example: 2.5% glucose in 80 ml of water? Glucose?

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2.5 g …………100 ml

?....................80 ml

= (80 ml x 2.5 g)/100 ml = 2 g of glucose is present in 80 ml

Example: if 10% of dextrose in water, what is ratio? 1g :10ml W/V

Example: if 50g/L glucose, what is ratio percent? 5%W/V

Example: 100g/1L water, what is ratio percent? 10%

= 5g/50 mL 5:50 or 1:10

Example: 50 grams of glucose in 1 L?

= 50g/1L = 50:1000 = 5:100 = 1:20

Example: Convert 5 grams of glucose in 100 mL into percent?

5g/100mL. 100 = 5 W/V%

5 grams of glucose in 1L water, what is ratio percent?

5/1000.100 = 0.5 W/V%

50 gram of glucose in 500 ml of water, what is ratio percent?

50/500 .100 = 10 W/V%

5 grams of glucose in S00 mL water, what is ratio percent? 5/500 .100 = 1 W/V%

Convert 10 grams of glucose in 10 ml into percent? 10 g/10 mL = 100/1 = 100 w/v%

A drug x is 1 mg/10 ml present. What %WW/V? 0.01% W/N

Two ratios with the same value are equivalent.

1/3 x 2/2 = 2/6 = 1/3

If two ratios are equal, then their reciprocal are equal:

If 1/3 = 2/6 then 3/1 = 6/2

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Proportions: (two ratios) relation of two ratios is the proportion

Example: How many mg (?) of glucose in 500 ml if this is a equal to 1g in 1000 mi of water.

Proportion and Percent Calculations

Weight/weight w/w % Weight/Volume w/v % Volume / Volume v/v %

Number of grams of substance Number of grams of Number of milliliters of a
in 100 grams of solvent or constituent per 100 ml of substance in 100 ml of the
mixture solvent solvent

1-How much drug should be added to 30 ml of water to make 10% w/w solution?

Solution:

The mass of the final solution in this case in unknown, we only have the mass of solvent (water) as
30 ml will weight 30 grams.

So, know we need a solution which contains 10 grams (10%) of drug in 100 grams of the solution
(solvent + drug)

In this case the solvent will represent 100-10 = 90%

So if 90%  30 grams, how much is 10 %?

The weight of the drug is = 10 * 30/90 = 3.33

2-What is the percentage strength of an injection that contains 50 mg of pentobarbital sodium in

each milliliter of solution?

Solution:

[(50 mg/1 mL) x1 g /1,000 mg] x 100 = 5 % w/v

3-If an injection contains 0.5% w/v of diltiazem hydrochloride; calculate the number of milligrams of
the drug in 25 ml of injection.

Solution: [(0.5 g/100 mL) x (1,000 mg/ 1 g)] x 25 mL = 125 mg

4-How many grams of potassium permanganate should be used in compounding the following
prescription?

Rx
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Potassium permanganate 0.02% w/v

Purified water ad 250 mL

Solution:

(0.02 g/ 100 mL) x 250 mL = 0.05g = 50 mg = 50, 000 µg

Parts Per Million (ppm)

Occasionally, you will see a number followed by the term "ppm." This stands for "parts per million"
and is most often used to indicate the amount of trace substances in water. The standard dilution
for fluoride added to a municipal water source, for example, is 1ppm, In every 1,000,000 ml of
water, therefore, there is 1 g of fluoride.

1) Express 5 ppm of iron in water as ratio strength and in percentage strength

Solution: 5 ppm 5 parts in 1,000,000 parts

Ratio strength: 5: 1,000,000 = 1: 200,000

Percent strength: (5/1,000,000) x 100 = 0.0005%

2) Express 10 ppm, in percent strength? 0.001%

10:1,000,000 = 0.001%

3) Express 0.0001% in how many ppm?

1ppm

4) Express 0.0005% of iron in water as parts per million (ppm)?

(0.0005 x 1000000)/ 100 = 5:1000, 000 = 5 ppm

5) Express 0.023% of iron in water as ppm?

[0.023 x 1,000, 000]/100 = 230 ppm

1-A parenteral solutions used in pharmacy. If 100 g of parenteral solution dissolved in 900 mL of
petrolatum (Density of petrolatum is 0.9 g/mL), the concentration of parenteral solution is: D = W/V

900 mL x 0.9 g/mL 810

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g 810g+100g = 910g

100g/910g .100 = 10.9%

2) If 500 mL of a 15% v/v solution is diluted to 1,500 mL, what is the resultant percentage
strength?

Q1 (quantity) x C, (concentration) = Q2 (quantity) x C2 (concentration)

500 (mL) x 15 (%) = 1,500 (mL) x? (%)

1,500 x = 7,500

X = 7,500/1,500

x = 5% v/v

3-If syrup containing 65% w/v of sucrose is evaporated to 85% of its volume, what percent of
sucrose Will it contain?

Note any convenient volume of syrup may be selected, say 100 ml. Then, 85% x 100 ml = 85 mL (Q2)

Q1 x C1 = Q2 x C2

100 (mL) x 65% = 85 (mL) x X (%)

85x = 6,500

X = 76.47% w/v

4-If 1 gallon of a 30% w/v solution is evaporated so that the solution has strength of 50% w/v, what
is its volume in milliliters?

Solution: Q, x C, = Q2 X C2

3,785 (mL) x 30 (%) = x (mL) x 50(%)

x = 2,271 mL

4) A pharmacist mixed 100 ml of 37% w/w concentrated hydrochloric acid (specific gravity, 1.20)
with enough purified water to make 360 mL of diluted acid. Calculate the percentage strength
(w/v) of the diluted acid.

100 ml x 1.20 (specific gravity) = 120 g of concentrated acid

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120 g x 37% w/w = 44.4 g HCI

(44.4g/360 ml) x 100 = 12.33% w/v

Drug Strength Expressions

Units (International units): Insulin, Penicillin G, Vitamin K, Vitamin E, Vitamin A, Vitamin D, Heparin,
LMWH (Fragmin), interferon alpha, Nystatin, Polymixin, and Bacitracin.

Example: U-100 insulin contains 100 units/mL

Insulin 40units

For 60 days and bid.

How many ml of insulin you dispense?

Solution: (1 mL /100 units) x 40 units = 0.4 mL

For 60 days and two times a day = 48 ml

2-How many milliliters of a heparin sodium injection containing 200,000 heparin units in 10 mL
should be used to obtain 5,000 heparin units?

Solution: (10 mL/200,000 units) x 5,000 units 0.25 ml

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Dilution and Concentrations

Stock solution: Solutions of known concentration that are prepared in the most concentrated form.
Sometimes a stock solution will be pure drug in powder or crystalline form. At other times it will be
a liquid or a solid paste or cream.

Stock solutions and additives: Here are some common IV stock solutions. A pharmacist or pharmacy
technician will withdraw a calculated number of milliliters from the vial and place it into a bag of
fluid, thereby diluting the original concentration of the stock solution. These stock solutions are also
called "additives" because they are added to another IV solution.

Most commonly dilution and concentration can be solved by inverse proportion method and by
determination of percentage or ratio strength. The following formula can be used to calculate
dilutions and concentrations:

Q1 (quantity) X C1 (concentration) = Q2 (quantity) X C2 (concentration)

Or We can use the equation = C1 * V1 = C2 * V2

Examples:

1) A prescription for hydrocortisone cream 0.1%. Pharmacy has 0.25% available in 30 g tube. How
many grams diluents base (vanishing cream) should be added?

A) 30 g

B) 45 g

C) 50 g

D) 75 g

E) 25 g

C1Q1 = C2Q2

so 0.25% x 30 g = 0.1% x g?

g? = 0.25% x 30 g/0.1% 75 g

but the added tube contains 30 g

so the used base = 75 g -30 g 45 g

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2) Dexamethasone is available as 4 mg/ml preparation; in infant is to receive 0.35 mg. Prepare a

dilution so that the final concentration is 1 mg/ml. How much diluent will you need if the original
product is in a 1ml vial and you use the full vial?

A) 4 ml B) 3ml C) 1ml D) 0.35ml

Ans: B

Step1: determine the volume of final product. Since dexamethasone is 4mg/mL, a 1 ml vial have
4mg of drug

X ml/ 4 mg = 1 ml/1 mg = 4 mL

Step2: subtract the volume of concentrate from the total volume to determine the amount of
diluents needed.

4 ml-1ml = 3 mL

3-lf a 600 ml of a 15% (v/v) solution of methyl salicylates in alcohol are diluted to 1500 ml what will
be the percentage strength.

Q1 * C1 = Q2 * C2

C2 = Q2/Q1*C1 1

500 mL/ 600 mL = 15 % / X % = 6%

4-If a potassium chloride elixir contained 20 mEq of potassium ion in each 15 mL of elixir, how many
milliliters will provide 25 mEq of potassium ion to the patient?

Solution/Answer:

20 mEq / 15 ml = 25 mEq / X ml

X = 15 x 25 / 20

x = 18.75 mL

5-How many grams of dextrose are required to prepare 4,000 mL of a 5% w/v solution? Equivalent
factor: a 5% w/v solution = 5 g in 100 mL of solution.

Solution/Answer

(5 g /100 mL) x 4,000 mL = 200 g

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Calculations involving dilution and concentration of Stock solutions

A solution of known concentration that is prepared in the most concentrated form is referred as
stock solutions. Sometimes a stock solution will be pure drug in powder or crystalline form. At other
times it will be a liquid or a solid paste or cream.

1-How many mL of a 1:500 (w/v) stock solution should be used to make 4 liters of 1:2000 (w/v)
solution?

Solution/Answer:

1:500 = 0.2%

4 liters = 4000 ml

1:2000 = 0.05%

0.2% / 0.05% = 4000 mL / x mL = 1000 mL

2- A parenteral solutions used in hospital pharmacy. If 250 g of parenteral solution dissolved in 1000
ml of glycerin (density of glycerin is 1.25 g/mL), the concentration of parenteral solution is?

A) 8% B) 20% C) 24% D) 16% E) 32%

To find out concentration: %C = [Solute/Solute + Solvent] x 100

D = M/V

M = 1.25 x 1000 mL = 1250 g

Then 250 g + 1250 g = 1500 g (total volume)

(250g / 1500 g) x 100 = 16%

Allegation Method
Allegation medial: a method for calculating the average concentration of a mixture of two or more
substances.

1) What is the final percentage of ZnO ointment made by mixing ZnO ointment of the following
strengths?

200 g of 10% + 50 g of 20 % +100 g of 5%

Solution:

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200 * 10% = 20

50 * 20% =10

100 * 5% = 5

Therefore, we have 350 g of the ointment which contain 35 g of ZnO

Therefore, the percentage is 35/350=10%

Ointment mixture of 15 g of 70%, 5 g of 90%, 10 g of 40%, 5 g of 10%, what is final concentration of

this mixture?

Ans: 15 g + 5 g + 10 g + 5 g = 35 g

(15 x 70) /100 + (5x 90) /100 + (10 x 40) /100 + (5x10) / 100 =

10.5 g + 4.5 g + 4 g + 0.5 g = 19.5 g

(19.5 g/35 g) × 100 = 55%

ALLEGATION ALTERNATE:
A METHOD OF CALCULATION OF THE NUMBER OF PARTS OF TWO OR MORE COMPONENTS OF
KNOWN CONCENTRATION TO BE MIXED WHEN THE FINAL DESIRED CONCENTRATION IS KNOWN,

2) Which proportion of 95% alcohol and 50% alcohol should be used to make a solution of 500 mL
of 70% alcohol?

Solution:

95% 20 parts of 95 %

70 %

50% 25 parts of 50 %

the final proportion is 20:25 = total 45 parts

Take 95%:

(20/45) x 500 ml 223 ml

Take 50%

(25/45) x 500 ml = 277 ml

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Calculation involving electrolyte solutions

Mole: molecular weight in grams

Reference atomic weights: Na = 23, C = 12, O = 16, K 39, Cl = 35, Ca = 40

Eq. wt Molecular Weight / Valence

Converting between milligrams (mg) and milliequivalent (mEq)

Number of mEq = Weight of substance in mg / mEq weight %

One mole of NaCl = 58.5 g

One mole of KCl = 75.5 g

One mole of HCl = 36 g

One mole of Na2CO3 = 106 g

One mole of CaCl2 = 111 g

Molarity
Molarity is the expression of the number of moles of solute is dissolved in liter of solution. Molarity
can be calculated by diving the moles of solute by the volume of solution in liters. One mole
dissolved in 1liter solution is 1M.

1M HCI= 36.5g of HCl dissolved in 1L

1M NaCl 58.5g of NaCl dissolved in 1L

One mole of substance dissolved in a liter solution

Molarity = molecular weight in grams/ Liters

1molar NaCI = 58 grams in liter

1) To prepare 100 ml of 1M NaCl, how many grams of NaCl (Mol. Wt. of NaCl = 58.5g) needed?

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Solution: 58.5/1000 x 100 = 5.85g

2- To prepare 28 mL of 0.5M NaCl, how many grams of NaCl needed? (M. Wt of Na = 58.5)

Solution: 58.5/2 = 29 g

29/1000 x 28 mL = 0.8 g

1- How many mEq of magnesium sulphate are represented in 1 g anhydrous magnesium sulfate?

(M. wt. of MgS0, =120)

A) 120 mEq B) 32 mEq C) 16.6 mEq D) 33.2 mEq E) 66.6

Ans: C

Tips:

1g = 1000mg mEq

mEq = [(weight) (valence)] / Molecular weight

= [(1000 mg) (2)] / 120= 16.6 mEq

2- A solution contain 10 mg% of Ca2+, describe this concentration in mEq/L. (Atomic weight = 40
and valence = 2

10 mg% is = 10 mg/100 ml

10 mg% for 1L is 100 mg/L

mEq/L [(mg/L) (valence)]/atomic weight

[(100 mg/L) (2)] / 40 mg = 5 mEq/L

3- What is the concentration in g per ml of a solution containing 4mEq of calcium chloride (CaCl, x
2H2O) M. wt. = 147

Mg / L = [(4 x 1470)] / 2 = 294 mg = 0.294g / L = 0.002 g/mL

Millimole

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Millimole (mmol)/L = Molecular Wt in milligrams/Liters

1mmol of NaCl solution contain how many milligrams of sodium chloride? 58.5 mg /L

1 mmol of 100 mL NaCl contain how many milligrams of NaCl? 5. 8 mg

0.5 mmol of NaCl contain how many milligrams of NaCl? 29 mg

Magnesium has an atomic weight of 24, what is weight of 1 mmol?

1mM = 24g/1000 = 0.024 g = 24 mg

Normality
A method of dealing with acids, bases, and electrolytes which involves the use of equivalents.

One equivalent of acid is the quantity of that acid that supplies or donates of mole of H ions.

One equivalent of base is quantity that gives off one mole of OH ions.

One equivalent of acid (H") reacts with one equivalent of base (OH'). Equivalent can be calculated
for atoms or molecules.

Equivalent wt. dissolved in 1L

Equivalent Wt. = M. wt. in g/ Valence

1N HCI = 36.0 g of HCl dissolved in 1L

The salts with valence 1 have the same molarity and normality. The valence in salts is referring to
metal ions.

The salts with valence 1: NaCl, HCI, KCI, Li2CO3, Na2CO3, NaHCO3

The salts with valence 2: CaCO3, CaCl2, MgCl2, Mg (OH)2, ZnCl2

The salts with valence 3: Al, citrate

One mole of NaCl contain one equivalent of Na+ (Na mol. weight 23 g)

One mole of NaCl contain one equivalent of CI- (Cl mol. Weight 35 g)

0.9% NaCl contain 0.9g of NaCl in every 100 ml

0.9% NaCl contain 9 g of NaCl in every 1liter

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Milliequivalent: The amount, in mg, of a solute equal to 1/1000 of its gram equivalent weight per
unit volume.

Valence

Na, K, Li = 1

Ca, Mg = 2

Al = 3

Important Formula

mEq = (mg x valence) / Mol. Wt.

mg = (mEq x Mol. Wt.) / valence

Converting milliequivalents per unit volume to weight per unit volume.

1) Sodium atomic weight is 23 and valence is 1. How many milliequivalents are in 92 mg of Na?

Number of mEq = weight of substance in mg/ mEq weight

92/23 = 4 mEq

2) What is the concentration in mg/ml of a solution containing 2 mEq of KCI per milliliters (KCI
M.Wt. = 74.5)?

wt. Of KCI = 74.5 g

Eq. wt. of KCI= 74.5 g

1mEq. of KCI = (1/1000) * 74.5 = 0.0745 g= 74.5 mg

2mEq. Of KCI= 74.5 mg * 2 = 149 mg/ml

OR

mg/ml = 2 (mEq/ml) * 74.5 = 149 mg/ml

2) What is the concentration, in grams per milliliter of a solution containing 4 mEq. Of CaCl2.2H20
per milliliter?

(M.weight of CaCl2.2H20 = 147)

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Eq. wt. Of CaCI2.2H20 = 147/2 = 73.5

1 mEq. CaCl2.2H20 = (1/1000) * 73.5 g = .0735 g

4 mEq. Of CaC 2.2H20 = 0.0735 g * 4 = 0.294 g/ml

OR

by mg/ml = [4 x 147]/ 2 = 294 mg/ml = 0.294 g/ml

Converting Milligram % to mEq/L

1- A solution contains 10 mg% of K+ ions. Express the conc. In terms of mEq/L?

Solution:

Atomic weight of K = 39

Equivalent weight of K = 39

1 mEq. Of K = 1/1000 * 39 g = 0.039 g = 39 mg

10 mg% of K = 10 mg K per 100 ml

= 100 mg per liter

100 mg * 39 = 2.56 mEq/L

Or

MEq/L = [100 mg/L*1] /39 = 2.56 mEq/L

2- A solution contains 10 mg % of Ca2+ ions. Express this concentration in terms of mEg/Liter

mEq/Liter [100 mg /L *2]/40 (atomic weight of Ca2+ )

Converting weight to milliequivalents

1-How many mEq. Of KCL are represented in a 15 ml dose of 10 % w/v KCL elixir? Mol, Weight KCI =
74.5 g

Solution:

Equivalent weight = 74.5 g

1 mEq of KCL = 0.0745 grams =74.5 mg

15 ml dose of 10 % w/v elixir = 1.5 g or 1500mg KCL

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74.5 mg  1 mEq

1800 mg  x x = 20.1 mEq

Milliosmole
It is the unit of measuring the osmotic concentration. Osmotic pressure is directly proportional to
the number of particles in the solution.

Osmotic pressure ------ Number of Particles in solution

Osmol/L = (wt. Of substance in g/L / Molecular weight in g) X number of species

mOsmol/L = Wt. Of substance in g/L = (Wt. of substance in g/L / Molecular weight in g) x number of
species x 1000

Solutes, which dissociate exert osmotic pressure based on the number of particles present in the
solution after they have dissociated.

Some Example salts:

For NaCl, the number of species =2

Nacl  Na+ + Cl-

For CaCl2, the number of species = 3

CaCl2  Ca2+ + 2CI-

For Li2CO3 the number of species = 3

Li2CO3  2Li+ + CO3 2-

For MgSO4 the number of species = 2

For Solutes which do not dissociate, the milliosmole = millimole

1) Solution contains 5% anhydrous dextrose in water injection. How many milliOsmoles per liter
are present in this concentration? (M. wt = 180 g)

Solution:

mOsmol/L = (50 g /180) x 1 x 1000 = 278 mOsmol/L

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Note: dextrose, gluconate are examples of substances that do not dissociates, however salts dissociates
into ions.

2) How many milliosmoles per liter are present in 0.9% NaCl solution?

Solution: 0.9% Nacl is 0.9g in 100ml, however, solution in one liters, thereby NaCl concentration is 9 g

(9 x 2 x1000)/58.5 = 307 mOsmol/L

Isotonic solutions preparations

Isotonic → "Normal saline" and is 0.9% NaCl concentra on

Hypotonic  Less than 0.9% NaCl concentration

Hypertonic  More than 0.9% NaCl concentration

Tonicity is affected by number of particles in solution. Substances that dissociate have greater tonic
effect than non-dissociated substances. Greater the dissociation greater the osmotic pressure and
greater the tonic effect.

The dissociation factor is the measure of the number of particles resulted in when a substance is placed
in aqueous solution.

Non-electrolyte substances have low dissociation factor. Dissociation factor for non-electrolytes
substances are assigned a value of 1.

Substance dissociate into two ions dissociation factor (i) = 1.8

For three ions (i) = 2.6

For four ions (i) = 3.4

For five ions (i) = 4.2

Salts that dissociate into two ions: NaCl, KCI, LiCI, NaHCO3

Salts that dissociate into three ions: Li2CO3, Na2CO3, ZnCl2, CaCl2, Mg (OH)2

Sodium chloride equivalent

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1-Calculate the sodium chloride equivalent for fluorescein sodium, which dissociates into three ions and
has a molecular weight of 376.

i factor for sodium chloride = 1.8

i factor for fluorescein sodium 2.6

(Mol. wt of sodium chloride / i factor of sodium chloride) x (i factor of substance / Mol. Wt of substance)
= sodium chloride equivalent

(58.5 / 1.8) x (2.6 / 376) = 0.22

NaCl equivalent (E) = 0.22

2-How much NaCl must be added to the following Rx to make it isotonic? NaCl equivalent of ZnSO4 =
0.16, NaCI Equivalent of phenylephrine = 0.29.

Rx

ZnSO4 ------------1/4%

Phenylephrine--1/8 %

NaCl----------------Q.S.

Aq. Distilled ad to 30 mL

you must follow the steps as follows

First step:

Calculate the amount of each ingredient in the prescription

So, if we look at the prescription you will find that the final volume is 30 ml

So, for ZnSO4, we need 0.25 for each 100 ml so for 30 ml we need 30* 0.25/100 = 0.075 grams.

now for phenylephrine do the same so we will need 30 * 0.125/100 =0.0375 grams.

Second step:

By using the Nacl equivalent of each of them calculate the contribution of these salts to the isotonicity
of the solution.

now for ZnSO4: 1 gram of ZnSO4 is equivalent to 0.16 grams of NaCl, but in our prescription we have
only 0.075 so this makes the solution as if it contains

I gram ------> 0.16 NaCl

0.075 gram -----> x NaCI

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x = 0.075 * 0.16/1= 0.012 NaCl

Applying the same for phenylephrine: 0.0375 *0.29/1=0.0108 NaCl

therefore, total contribution of the salts = 0.012 + 0.0108 = 0.022875 NaCl

Third step:

Find the amount of NaCl needed to make 30 ml - which is the volume of the final solution

isotonic.

So, we need 0.9 grams NaCI for every 100 ml ...so for 30 ml we need

0.9 g--- 100 ml

Xg -------- 30 x = 30* 0.9/100 = 0.27g NaCl

therefore, the amount of NaCl needed = 0.27-0.022875 =0.247 grams = 247 mg NaCI

2) You are given ZnCl2 0.7%, phenylephrine 0.1% and boric acid 1.1% with E values 0.16, 0.32 and 0.5
respectively. This solution will be:

A) Hypotonic B) Hypertonic C) isotonic DJ Non isotonic

Solution:

[(0.7/100) x 0.16] + [(0.1/100) x 0.32] + [ (1.1/100) x 0.5)]

= 0.00112 + 0.00032 + 0.0055 = 0.00694

Nacl = 0.9/100 = 0.009 > 0.007 = Hypotonic

How much NaCl required making isotonic solution? 0.009 - 0.007 = 0.002 or 2 mg

Dissociation factors
1-Zinc sulfate is a two-ion electrolyte, dissociating 70% in weak solutions. Calculate its dissociation
factor.

ZnSO4  Zn + SO4+ ZnSO4

Solution:

Based on 70% dissociation, 100 particles of zinc sulphate (ZnSO4) yield:

70 zinc ions

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70 sulphate ions

30 undissociated particles

170 total particles

Because 170 particles represent 1.7 times as many particles as were present before dissociation, the
dissociation factor is 1.7.

Calculations involving Balance sensitivity

1) What is the minimum quantity that can be weight on a balance with sensitivity requirements of
15 mg of a 5% error is permissible?

Sensitivity Requirement = Weight x Error

15 = weight x 5/100 = 300 mg

2) What is the sensitivity of a balance that can weight 120 mg of a substance and has a permissible
error of 5%?

SR = Weight x Error

SR = 120 mg x 5/100 = 6 mg

Error = sensitivity requirement/weight

Weight = sensitivity requirement/error

What is the sensitivity of a balance that can weight 120 of substance and has accuracy of 98%?

Sensitivity requirement = weight x error

120 X 2/100 = 2.4

Tips Practice format: Electrolyte Solutions

Molarity = molecular weight in grams / Liters

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Millimole (mmol)/L = molecular wt in milligrams/ Liters

Converting milliequivalents per unit volume to weight per unit volume

mEq = [mg x valence] / (Mol.wt)

mg = [mEq x (mol.wt)]/valence

mOsmol/L= = (Wt. Of substance in g/L / Molecular weight in g) x number of species x 1000

Sensitivity Requirement = Weight x Error

Error = sensitivity requirement/weight

Weight = sensitivity requirement /error

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Dosage Calculations
One of the most common calculations in pharmacy practice is that of dosages. The available supply is
usually labeled as a ratio of an active ingredient to a solution:

active ingredient (available)/ solution (available)

The prescription gives the amount of the active ingredient to be administered. The unknown quantity to
be calculated is the amount of solution needed in order to achieve the desired dosage of the active
ingredient. This yields another ratio:

active ingredient (to be administered)/solution (needed)

The amount of solution needed can be determined by setting the two ratios equal:

[active drug / Solution available] = (active drug (to be administered)/ solution (needed)]

When solving medication-dosing problems, use ratios to describe the amount of drug in a dosage form
(tablet, capsule, or volume of solution). It is important to remember that the numerators and
denominators of both fractions must be in the same units - for example, mg/ml, 5 mg/ml or mg/tablets
= mg/tablets.

CHILDREN DOSES:

1. Young's Rule: Child dose = Age /Age +12 x Adult's dose

2. Cowlings Rule: Infant dose = [Age (at next birthday) x Adult's dose]/ 24

3. Frieds Rule: Infant dose = (Age (in months) x Adult's dose]/ 150

4. Clarks Rule: Child dose [ weight (in pounds) x Adult's dose)]/150

Tips: Rarely applied now, generally dose for children is depending on weight of pt. & the technician
should ask the agent or caregiver about the weight of pt. so that the pharmacist can check the dose.

Example If the adult dose of X, is 5 mg. What is the dose for child of 8 years?

Since we know according to

Young’s rule: Child dose = Age x Adult’s dose

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