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)
= 1.8 g per ml (g/ml)
10 (ml)
Specific gravity:
 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: 10 ml of H2O (with same conditions) = 10 g (weight)
Specific gravity of acid:
wt of 10 ml of acid
18 (g)

 1.8
wt of 10 ml of water 10 (g)
 It is a constant value for each substance.
Pharmaceutical calculations:
I- Specific gravity of liquids
II- Specific gravity of solids
I- 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.
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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
52.78
 0.9603
=
wt of 54.96 ml of water
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)

wt of water (4)
Example
 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
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II- Specific Gravity of Solids
A- Heavier than water.
1- Insoluble
2- Soluble
B- Lighter than water and insoluble
C- Granulated solids (heavier and insoluble in water).
A1) Specific gravity of solids heavier than and insoluble in
water:
Specific gravity =
wt of solid in air
wt of displaced in water
N.B.: Wt of displaced water = apparent loss of wt in water.
Examples:
 Wt of glass in air = 38.525 gm
 Wt of glass when immersed in water = 23.525 g
Specific gravity of glass
Wt of displaced water
?
= apparent loss of wt in water
= 38.525 – 23.525 = 15.000 g
Specific gravity of glass =
38.525
 2.268
15.000
A2) Specific gravity of solids heavier than and soluble in water:
Example:
Wt of equal volumes of two substances and their specific gravities.
 Crystal of chemical salt weight
= 6.423 g (air)
 When immersed in oil weight
= 2.873 g (oil)
 Oil specific gravity
= 0.858
Specific gravity of salt
?
Wt of displaced oil = 6.423 – 2.873 = 3.550 g
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3.550 (g. oil)
0.858 (sp. gr. oil)

6.423 (g. salt in air)
x (sp. gr. salt) ?
x (specific gravity of salt) = 1.55
B) Specific gravity of solid lighter than and insoluble in water:
- Use sinker to prevent solid from floating.
 Wt of sinker when immersed in water alone.
 Wt of solid in air.
 Combined wt of sinker (water) and solid in air (before immersion).
 Wt of water displaced by solid
= combined wt (sinker + solid) before immersion – combined wt
(sinker + solid) after immersion.
Specific gravity solid =
wt of solid in air
wt of displaced water
Example:
 Wt of empty bottle = 80.0 g
 Wt of filled water bottle = 96.8 g
 Wt of granulated metal = 28.8 g
 Wt of bottle (total weight) (metal filled water) = 118.4 g
Specific gravity of metal
??
1- Wt of water filling bottle = 96.8 – 80.0 = 46.8 g
2- Combined wt of water and metal = 46.8 + 28.8 = 75.6 g
3- Combined wt of water and metal in bottle = 118.4 – 80.0 = 68.4 g
4- Wt of water displaced by metal = 75.6 – 68.4 = 7.2 g
(wt of equal volume of water)
Specific gravity of metal =
wt of solid
28.8

 4.0
wt of an equal volume of water
7.2
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Specific Volume
 Specific volume of substance =
volume of substance
volume of equal wt of water
(liquids and solids) standard under the same conditions
 Specific gravity: comparison Wts of equal volumes
 Specific volume: comparison Vs of equal weights
 Are reciprocals of each other
 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 =
91.0
91.0

 0.850
volume of standard (water) 107.16
107.16 g of water = 107.16 g ml (volume)
2- What is specific volume of phosphoric acid having sp. gr. 1.71 ?
1
 0.585
sp. volume =
1.71
3- What is sp. gr. of liquid has sp. volume 1.396
1
 0.716
sp. gr. =
1.396
Calculations of weight (liquid):
 Calculation the weight of liquid when given
- Volume
- specific gravity
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sp. gr. water wt of equal volume of water

sp. gr. liquid
x
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
volume of equal wt water

sp. gr. 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).
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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 (w/v) %:
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:
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1- How many grams of dextrose required to prepare 4000 ml of 8%
solution ?
g. solute = 4000 x
S
= 200 g
100
S g.  100 ml
X=
,
X  4000 ml
4000 x S
 200 g
100
2- What is the % strength (w/v) of solution of urea, if 80 ml contain
12 g. ?
%=
g. solute or constituen t
W
x100 
volume in ml
V
%=
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

 900 ml
%
0.03
II) Percentage volume-in-volume (v/v) %:
1- Calculate (v) of active ingredient by knowing:
 (V) liquid
 % (v/v)
or
2- Calculate % (v/v) by knowing:
 (V) active ingredient
 volume of solution
or
3- Calculate (V) of solution by knowing:
 % (v/v)
 (V) active ingredient.
or
Examples:
1- How many ml. liquefied phenol should be used in:
R/ Liquefied phenol
Calamine lotion ad
2.5%
240.0 ml
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(V) active ingredient = (V) in ml. x % …….. ml
240 ml x 0.025 = 6 ml.
2.5 – 100 ml
or
x=
x – 240.0 ml
240 x 2.5
 6 ml
100
2- Calculate the % (v/v) of solution given the volume of active ingredient
and volume of solution ?
%=
Ex:
ml. of active ingredient
V
x100 
volume in ml.
V
- 250 ml of lotion
- used 4 ml liquefied phenol
Calculate the % (v/v) liquefied phenol in lotion ?
%=
4
x 100  1.6% (V/V)
250
3- Calculate volume of solution given the volume of active ingredient and
its % (v).
Volume in ml =
ml of active ingredient
%
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
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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
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.
2- If 1500 g. of solution contain 75 g. of drug substance, what is the %
(w/w) of solution ?
%=
g. of solute
x 100
wt. solution g.
%=
75
x100  5%
1500
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 ?
Wt. of solution (g.) =
g. of solute
2

 40 g.
%
0.05
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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:
1- If 500 ml of 15% (v/v) solution of methyl salicylate in alcohol are
diluted to 1500 ml. What % (v/v) ?
1500 15%

500
x
 x = 5%
2- Quantity x conc. = quantity x conc.
500 x 15% = 1500 x (x%)
 x = 5%
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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.) 10% (w/w) ammonia water can be made from
1800 g. of 28% (w/w) ammonia water ?

10% 1800

28%
x
x = 5040 g.
or  1800 (g.) x 28% = x (g.) x 10%
x = 5040 g.
Stock Solutions
Definition:
 Are solution of known concentration.
 Are strong solution from which weaker ones are made.
 Are prepared in (w/v).
 Their concentration expressed by ratio strength (--- : ---).
Example:
- How many milliliters of 1:400 (w/v) stock solution should be used to
make 4 liters of a 1:2000 (w/v) solution ?
1) 4 liter = 4000 ml
1:400 (ratio strength) =
100
= 0.25%
400
1:2000 = 0.05%

0.25(%) 4000 (ml)

0.05(%)
x (ml)
1
4000 (ml)
2) 1 400 
x (ml)
x = 800 ml
x = 800 ml
2000
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(quantity x conc. = quantity x conc. )
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 (%) 5000 (ml)

85 (%)
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.)

90 (%)
x (g.)
x = 9000 g.
Triturations
 Powders as dosage forms
Definition:
 Are dilutions of potent medicinal substances.
 Are prepared by mixing them with suitable diluent (lactose) in a
definite proportions by wt.
 Dilute one part by (wt) of drug + g. parts (wt) of finely powdered
lactose.
- Calculate the quantity of trituration required to obtain a given amount
of medicinal substance.
Example:
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- How many grams of 1:10 trituration atropine sulfate are required to
obtain 25 mg of atropine sulfate ?
10 g. of trituration contain 1 g. atropine sulfate
1 g.  10 g.
25 mg  x
0.025 g  x
25 mg. = 0.025 g.
x=
10 (g.) x 0.025 (g.)
1 (g.)
x = 0.25 g trituration
0.25 g. trituration contain 25 mg atropine
25 mg : 0.25 g
0.025 g. : 0.25 g.
1:10
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Alligation
1- Alligation medial
2- Alligation alternate
1- Alligation Medial:
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.
Example: 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
% of zinc oxide =
20 x 50 = 1000
5 x 100 = 800
Total 350 = 3500
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3500
 10%
380
2- Alligation Alternate:
1- 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 ?
Ointment ?
20%
2.8 parts of 20% (drug + base)
2.5%
0% (base + drug)
17.5 parts of oint. base
Relative amounts = 2.5 : 17.5 or 1:7
Check:
20 x 1 = 20
0x7=0
Total
8 = 20
Percentage of mixture 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
1%
0.25%
1.5 parts of 0.25%
Relative amounts = 0.75 (2.5%) : 1.5 (0.85%) or 1:2
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2 (parts)  360.0 (g.)
1 (part)  x (g.)
x=
360
 180 g
2
180 : 360 = 1:2
 180 (g.) of 8.5% hydrocortisone cream + 360 (g.) of 0.25% cream 
1% hydrocortisone cream.
3- Calculate amount of active ingredient that must be added to
increase the strength of a mixture of given amount and strength.
Example:
How many grams of coal tar should be added to 8200 g. of 5% coal tar
ointment to prepare an ointment containing 20% of coal tar ?
Coal tar (active ingredient) = 100%
100%
15 parts of 100% coal tar
20%
5%
80 parts of 5% ointment
Relative amounts = 15:80 or 3:16
16 (parts) 3200 (g.)

2 (parts)
x (g.)
x= 600 g. from 100% coal tar
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
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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)
0.025 parts
(2)
0.925
(2) 0.875
Relative amounts = 0.050 : 0.025 = 2 : 1
Total = 3 parts
3 parts 1500 ml

2 parts
x ml
x = 1000 ml of liquid with sp. gr. 0.950
3 parts 1500 ml

1 part
y ml
y = 500 ml of liquid with sp. gr. 0.875
1000 + 500 = 1500
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