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Improved and Novel Methods Of Pharmaceutical Calculations

R. Subramanya

Cheluvamba Hospital, Karri ataka, Mysore-570 001, India.

Abstract

It is a review and research article wherein general equations are derived for some problems and new rules are framed on
percentage calculations, ratio and proportions and on variation of surface area with respect to size of the particle. Percentage
calculations are projected from novel angles. Dimensions or units of quantities are retained up to the result. An easy method
for solution of ratio and proportions are suggested. Problems solved by alligation method in the books are solved by simple
algebraic method with little new rules.

Keywords: axis, decimal, percentage, serially, syntax,

1. Introduction

This article is review of books [1] & [2]. section-2, contains the review on basics of percentage calculations. A new general
equation is furnished at the end of this section to ease out solving of such problems. In most of the books same mistake is
committed as in [3] of [1]. Hence, proof of these general equations are given to confirm the mathematical statements.

In section 3, problems solved by dimensional analysis are solved by other methods that are easier than dimensional
analysis. [4]. In the second ratio of set of equivalent ratios of first method, different dimensions are changed into same kind
of dimensions i.e., ml, which has made the solution of problem easier and shorter. Similarly second method is also easier
and shorter.

A general equation for the total surface area with respect to the decrease in the particle size is derived. Accordingly, total
surface area is inversely proportional to the size of the particle. Because the equation belongs to magic series, increase in
surface area w.r.t the decrease in size of the particle is a wonder which is shown in example problems.

Section 5 gives a synoptic view of basic measurements.

Section 6 contains new ideas and laws on percentage calculations. Some of the problems are solved by using simple
mathematical operators like addition, subtraction, appropriate factor, and by simple division. In most of the books
proportional method is used for solving these problems. In this paper while solving alligation problems new methods are
used. And, algebraic solutions with multiple results are shown.

Section-7, gives a simple equation for conversion of percentage to milligram per ml and vice versa.

In section-8, new method i s furnished to solve the problems on dilution

In section-9, some problems are solved in different and easy methods than in books.

2.Review on Fundamentals of Percentage Calculations:

3
2. 1. Convert - to percent[3]

3
Given calculation in the book is -x 100 = 37.5%

The above equation does not satisfy the meaning of the symbol '=' (equals) used in between the expressions

(- x loo) and 37.5%.

3
: Actual value of - x 100 = 37.5

A„ rf 37.5 = 37.5x©

: = 3750%

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The correct method of solution is as following
: 3 -= 0.375

0.375 x^

= 37.5%

2.2. . A 1: 1000 solution has been ordered. .You have a 1% solution, a 0.5% solution, a 0.1% solution in
Will one oftliese work to fill the order. [4]
' 1 1

Solution: 1: 1000 = = = 0.1%

1000 10x100
In comparison with solution of [4]in [l],a few steps are reduced.
2.3. Express 0.02% as ratio strength.
2
0.02 Tnn 2 1

Solution: 0.02% = = J ^ L = = = 1: 5000

100 100 10000 5000

2.4. . Give the decimal fraction and percent equivalent of—

100

< Too

_2.22

"Too

2.5. General Equation:

100
Whence x is a real number
: Hence, x = 100x%

2.6. Proof of (1) and (2):

Let a be a real number,
Then, a = a x 1

.. 100

Also, a = ax —

100
_ 100a

_ 100

a = 100a%

And, a% = —

100

Further, by cross multiplication we have 100a% = a
100X — = a

100

Consequently, a = a

3. Another way for dimensional analysis:

3.1. How many fluidounces (fi.oz) are there in 2.5 liters (L)? [5]

Method- 1:

: Given, 1 fluidounce = 29.57 ml

2.5L _ 2500m/ _ 2500 _
: If I. oz ~ 29.57 m/ ~ 29.57 ~ 84 ' 55

Method -2:

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lfl.oz.= 29.57 ml
...Multiplying both sides by the factor — — that modifies 29.57 ml to required quantity, we g

29.57
: 84.55/1 oz.= 2.SL

4. General equation to find the total surface area of particles. (Imaginary and ideal case). [6].
4.1. A general equation is derived to find the total surface area of cubes when a cube is divided serially along the xy, yz and
zx -planes. When a cube is divided along the planes of axes x, y and z, it gets divided into eight equal cubes at every time i.e.
after every set of cutting. When a cube is cut through x, y and z plane it is called, 'a set of cutting'.

The sum of the surface area of cubes at 'n th ' set of cutting is : A = 2 n+1 X 3l 2 square units- (3)

Where, 'A' denotes the sum of surface area of cubes and 'n' the ordinal value of set of cuttings from the first cube. Hence n is
a set of whole number, i.e. n = {0,1,2,3 ... }

ar

; when cut along zx plane

ff f f

; when cut along yz plane

; when cut along xy plane

r r
rr rr

Fig.l. A typical view of a set of cuttings of the
cubes across the zx, yz and xy planes.

To the 1 st cube, ordinal number of set of cutting is 0, i. e. , n = 0, and if he length is I u
Total surface area of first cube is : A = 2 n+1 x3x( 2

: And if = 1 ; A =

: = 12 I 2

:Andifn = 2; A = 2 n+1 x3xl 2

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<&) 2

4.2. Derivation of equation : A = 2 n+1 X 3l 2

To start with, cube has six faces and if its length is I units the total surface area of the cube, when no cut is done along the
axes is

: a = one cube x number of faces x area of one surface

:i.e. a = 1 x 6 x I 2

To make the above equation look like rest of the equations, it can be written as

: a =8»x6x(i) 2

when this cube is cut across the Co-ordinate planes we get 8 cubes of dimensions equal to - units each.
: .% a x = 8 X 6 X (0
Further, a 2 = 8 2 x 6 x (^)
Finally, equation for n th term is

a n = 8 n x 6 x (-^)
To simplify this equation, it can be written as

= (2 3 )" :
Further, a n = 2 3n+1 x 3/ 2 (2" 2n )
Replacing a n by A, and simplifying further we get
A = 3(2 n+1 Z 2 )
= (2 n+1 )3l 2
Thus derived.

4.3. For example when n =

Surface area of the cube of length 1 cm is
: (2 n+1 )3l 2 cm 2 = 2 x 3cm 2
: = 6cm 2

This is equal to the area of 2cm X 3 cm rectangle.

Further when n = 5 ( 5 set of cuts) and / = 1cm,
: Then, (2 n+1 )3Z 2 = 2 6 x 3
: = 192 cm 2

This area is equal to the area of 12 cm X 16 cm rectangle.

And when n = 20 and I = 1 cm

: (2 n+1 )3Z 2 = 2 21 X 3

: = 6291456cm 2

This is approximately equal to the area of 2.5m X 2.51m dimension.

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5. MEASURMENT

Table- 1:. Measurement Of Length, Mass And Volume (SI Units)

Name of the unit

Expressed in terms
of m/L/g

Values expressed in
scientific notation

Fractional notation in
m/L/g

1 Kilo-meter/liter/gram. (km/kL/kg)

1000 m/L/g

10 3 m/L/g

1 hecto-meter/liter/gram. (hm/hL/hg)

100 m/L/g

10 2 m/L/g

1 deka-meter/liter/gram. dam/daL/dag

10 m/L/g

10 1 m/L/g

1 meter/liter/gram. m/L/g

1 m/L/g

10° m/L/g

1 deci-meter/liter/gram. dm/dL/dg

O.lm/L/g

10" 1 m/L/g

QV/<7

1 centi-meter/liter/gram. cm/cL/g

0.01 m/L/g

lO" 2 m/L/<7

[wS m/L/9

1 milli-meter/liter/gram (mm/mL/mg)

0.001 m/L/g

10" 3 m/L/g

(ib^o) m/L/5

1 micro-meter/liter/gram
(mcm/mcL/mcg)

0.000001 m/L/g

lO" 6 m/L/g

/ 1 \ th

VlOOOOOO/

m/L/g

1 nano-meter/liter/gram. (nm/nL/ng)

0.000000001
m/L/g

lO" 9 m/L/g

/ 1 f

Vioooooooooy

m/L/g

Note: all measurements are basically compared with m/L/g in this table
5.2. Add 0.5 kg, 50 mg, 2.5 dg, reduce the result to grams. [7]
: =0.5kg+50mg+2.5dg

: = 0.5 x (1 kg ) + 50 x {Img) + 2.5 x (ldg) (4)

Conversion factors are

: lkg=1000g, lmg=^0, ldg=^g

: Substituting the above values in (4), we get

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: = 0.5 x (1000 5 ) + 50 x(^) + 2.5 x(^)
: = 0.5 X (1000 g ) + 50 X (j^g) + 2.5 X (±g)
■■ = SOOg + O.OSg + 0.2S g
: = 500.3 g

6. SYNTAX OF PERCENTAGE (%) AND PARTS PER n (PPn)CALCULATIONS

Percent means 'for every hundred'. Symbolically it is written as %. And mathematically it is written as the ratio
x/ 100 or — . The bar between two numbers represents the phrase 'for every'. It can also be read as 'Parts per cent' the
word cent is Latin derivative which means, hundred. And this 'Parts per cent' can be abbreviated as PPc. I suggest to
write this as PPh which is the abbreviation of 'Parts Per Hundred'. This kind of abbreviation helps to extend the same
idea to any quantity. For example PPt for parts per ten,, i.e., PPh for 'Parts Per Hundred, PPth, parts per thousand, PPtth
'Parts Per Ten Thousand', PPhth 'Parts Per Hundred Thousand', Parts Per Million 'PPm'and so on. Alpha numerically it
can also be written as PP10 , PP10, PP10 2 ,PP10 3 •■•PP10 n . Usually, comparison is done in multiple of 10.

Although general expression Pails Per n can be written as PPn,where nis a positive intiger.

If x is the number of parts then general expression is ■■ xPPn.

Mathematically xPPn can be written as following in the fractional form

The staked fractional notation of xPPn is -

The skewed fractional notation is x /n

The linear fractional notation is x/n

And the decimal notation of - is a , and it is a real number

Also, y units of xPPn = —^ L

let,z = y§,

Then, y units of xPPn = -

And henceforth [y] (xPPn) stands for "y units of xPPn"

Accordingly, [y](xPPn) = (5)

Here the numerator indicates active ingredient and the denominator indicates solution or drug.

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Table-2: x Parts Per '10 n '

X

10"

Verbal form of the
expression

Abbreviation

Alphanumeric
notation

Prevailing
notation

Decimal
form

X

10

x Parts per ten

xPPt

xPPIO

0.x

X
TO 1

x Parts per hundred

Or

Parts per cent

Or

PERCENT

xPPh

or

PPc

x PP10 2

x%

O.Ox

X
TO 1

x Parts per
Thousand

xPPth

x PP10 3

x%

O.OOx

10*

x Parts per Ten
thousand

x PPtth

x PP10 4

O.OOOx

105

x Parts per hundred
thousand

x PPhth

x PP10 5

0.0000X

106

x Parts per million

xPPm

x PP10 6

0.00000X

In addition to this, some more basic factors are intertwined in nature, with the x PPn. Those are volume, mass and weight,
x/n may be understood in terms of : x/n{v/v )orx/n (w/w) or x/n (v/w)or x/n (w/v).
If the volume does not change after mixing and if ml and g are the units of volume and mass ,
Then,-{-) = — — .because n = (y + x)

n W (y+x)

Similarly , x/n (w/w) = xg , x/n(v/w) =

(y+x)g '

<y+i

.Of X

Only in case of, x/n(w/v) = , here, volume of the active ingredient is not consideredbecause

increase in volume of solution after dissolution of active ingredient will be negligible.

Example:. Peppermint spirit contains 10 % v/v peppermint oil. Wln.il is the volume of the solvent when volume of spirit is

75 ml ?[8]

:10%(v/v) = ^^ L

v ' J 100 ml

Factor that is required to make the numerator to 75 is — , multiplying both the numerator and denominator by this factor
we get

lOmZx^ 7 sml

100ml

"10

75 750 ml

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This shows that volume of the solvent of 10% peppermint sprit is 675 ml.

6.1. Summing of x (PPn)
Case-1

If x 1 PPn 1 ,x 2 PPn 1 ,x 3 PPn 1 ■■■ ,x a PPn x are the solutions of different strengths and of same PPn.
( Here a and n are natural numbers)

: then, — + — + — + •■• + — = (6)

n x n x n x n x an x

: to simplify(6')let,x 1 + x 2 + x 3 H \- x a = y

: 77ien(6), = ^ (7)

:Andif,( y / a ) = z

: Then (6) = f-

•■= zln x ( — or-or — or—)

•■= zPPn x ( — or-or — or — )

If n x = 100, the result will be = z%

Example: What is the percentage strength of the solution when equal quantities of 10%, 20%, 30% solutions are mixed
together.

Here,x 1 = 10, x 2 = 20, x 3 = 30, and n x can be = 100, thenusing (6), we have

10 20 30 10 + 20 + 30

100 100 100 300
_ 20
100

: = 20%

Case-2: To find the strength of the solution to a specified base when different strength, different quantity,
PPn solutions are mixed together.

Let [b]PPn v [c]PPn 2 , [d]PPn 3 , — ,[f]PPn a be the solutions to be mixed,

and let n y be the specified base i. e. , PPn y and let zPPn y be the final solution,

■ then, sum of them is = [b] (—) + [c] (— ) + [d] (—) + ••■+[/] (— )

»m <<® o /©

= — ^+ — -+— T-+-+— r-

b c d f

bx x cx 2 dx 3 fx a

(b + c + d + - + f)
The required and specified base of strength is n y , then (8) will be,

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•■+£0

(b + c + d + ••• + /)

+£0

z, then (9)wiM foe

: Then, (9)can fee written as,zPPn y , if n y = 100, the result will be z%

Exmple-l:Wza? is the percentage strength (v/v) of alcohol in a mixture of 3000 ml of 40 % v/v alcohol, 1000
% v/v alcolwl, and 1000 ml of 70% v/v alcohol? Assume no contraction of volume after mixing. [9]
:Heren y = 100 and x x = 40, x 2 = 60, x 3 = 70, Also, b = 3000, c = 1000, d = 1000
and n x = n 2 = n 3 = 100,
putting these values £n(9) we get

3000 X 40 + 1000 X 60 + 1000 X 70

(3000+1000+1000)
: Percentage strength = — —

Note: since n 1 =n 2 =n- i = 100, and n y = 100 , they get cancelled by one another.

Example-2: When the same problem is solved for n y = 1000, i.e., for c

f 3000 X 40 + 1000 X 60 + 1000 X 70 >t

1UUU l 100 )_

(3000+1000+1000)

' x t ~ 1000

500

~ 1000

Example-3: Same problem, when n x = 1000, n 2 = 2000, n 3 = 5000 and n y
^3000 x 40 x 10 + 1000 x 60 x 5 + 1000 x 70 x 2\

loop

10000
: xPPc =

5000

100

: = 3.28%

This problem can also be solved by another method as following

40 3000 (lMo) 120

: 3000 ml of — — strength solution = ^„„„ = --—

; 1000 w 3000 3000

60 1000 (2§Sn) 30

: 1000 ml of —— strength solution = ^ " uu/ = — —

' 2000 w 1000 1000

70 1000 (5nno) 14

: 1000 ml of —— strength solution = ^ " uu/ = — —

; 5000 w 1000 1000

Sum of these solutions are
120 30 14

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Example-4: A pharmacist adds 10 ml of a 20% (w/v) solution of a drug to 500 ml of DSW for parenteral infusion.

What is the percentage strength of the drug in the infusion solution? [10]

Data:x 1 = 20, x 2 = 0, since the active ingredient dextrose in D S W is not considered in this case,

b = 10, c = 500 andn = 100, ^ = n 2 = 100

100 (-

1 0x20 500 x
100 + IPC

_ 0.392
100

= 0.392%

This is also solvable in another way as following:

10 ml of 20% solution is added to 500 ml of D5W.Here D5W is a dummy drug with respect to this problem. Hence

this statement can be written mathematically as following.

n ?no/„

: xPPh(w/v) =

500ml (D5W) 10ml
To convert 20% into the quantity of active ingredient it is multiplied by 10 which is the quantity added.

500 ml (D5W) 10ml

2g

500 ml (DSW) 10ml

By direct addition, we get.

2g
510 ml (D5W)

2g

5.10 x 100m/ (D5W)

0-39 g
100ml (D5W)
= 0.39% (w/v)

Example-5-.Prepare 250 ml of dextrose 7.5 % w/v using dextrose 5% (D S W) w/v and dextrose 50 % (D 50 W)
How many milliliters of each will be needed?[ll]

-- 100, n 2 = 100. n v

Method- 1: using (9)

Data: x

= 7.5,n = 100

,x x = 5,x 2

"•[(&) +

(™)1

b + c

100

100

7.5

5b + 50c

100

100(6 + c)

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: 500ft + 5000c = 750ft + 750c

: 250ft = 4250c

ft _ 17
c ~T

: ft = 17 and c = 1

Check: Putting values of b and c in (9) we get.

/50 x IV

. 17 + 1

100
_ 7.5
' ~100

: = 7.5%

Now, to prepare 250 ml, -
are required

Example-6: In what proportion should alcohols of 95% and 50% strengths be mixed to make 70% alcohol? [12]
Method- l:This can be solve by using (9) as usual.

Method-2: Here determinant is used to solve the problem.

Let x and y be the factors that modify the given ratios of percentage to —,

95x 50y _ 70
'100x + lOOy '

Then, 95x + 50y = 70 (10)

And,W0x + 100y= 100

i.e.,x + y= 1 (11)

Using (10)&(ll)tfte detrminant A can be written as following,

■H7 Tl

: =45

Let A x & A y fte the remaining determinants formed w. r. t RHS,
170 501

: =20

= ***,- 17 Tl

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: NOW, X =

andy =

A

: i.e.,x =

20
= 45

andy

25
"45

■■ X =

4

= 9'

indy =

5
9

x

4

y

5

■■ Then,x =

= 4 and

y = 5

Examplc-7: A hospital pharmacist wants to use lots of zinc oxide ointment containing, respectively, 50 %,
20%, and 5% of zinc oxide. In what proportion should they be mixed to prepare a 10% zinc oxide
ointment?[l3]

Method- 1:

Data: n y = 100, x x = 50, x 2 = 20, x 3 = 5,^ = n 2 = n 3 = 100, and required strength is 10%
Using (9), we

luu Uoo + ioo + iooJ

10 _ (b + c + d)

100 100

10 _ 50fc + 20c + 5d
' 100 ~ lOOfc + 100c + lOOd

: 10006 + 1000c + lOOOd = 5000b + 2000c + 500d

: Id = 8b + 2c

■■ When, d = 1, above equation will be, 8b + 2c = 1

1
: Now, by trial and error method we can find b = — a

1 1
" 16'4'

By multiplying RHS of the above equation by 16, we have,
: =1:4:16

Then, the proportion of b:c:d =

Assigning different values to d we can get different proportions. For example when d=10, consequently b=l and c=l
which is the result obtained in [14]. Also, we can get the result by giving convenient values to b and c such that

satisfies the equation.

Given, 50 %, 20 %, and 5% zinc oxide ointment and the required percentage is 10 % . Keeping 50 % and 20
% ointments as fixed. And considering 5 % ointment as the variable.

Then, let the variable base (solvent)t = 100

Then, 5 = 0.05 t

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i.e,50% + 20% + 0.05t% =

■■ 10%

50 20
100 + 100

0.05t
t

10

(12)

100

70 + 0.05t

10

200 + t

100

100(70 + 0.05t) =

10(200 +

t)

7000 + 5t :

= 2000+ lOt

t = 1000

50

20

50

10

= Too

: Proportion of denominator of LHS is 100: 100: 1000
: which is equal to 1: 1: 10

Without using (9) answer can be found directly as following

Let all the three percentages be variables. And let t 1# t 2 , t 3 be the variable quantities

0.5t! 0.2t 2 0.05to 10

: Then, - + - + = — —

h t 2 t 3 100

: 50^ + 20t 2 + 5t 3 = 10^ + 10t 2 + 10t 3

: 8^ + 2t 2 = t 3

As usual answers can be found by giving different values to t 3 or to t t &. t 2

Note: Since problem has three unknown and only two rows, square matrix cannot be formed. Hence it is not possible

to draw solution by determinant method. Although, by modifying some of the rules of determinants or matrix,

answer can be obtained.

Example-8: If 50 ml of a 1:20 w/v solution are diluted to 1000 ml, what is the ratio strength (w/vj?[14]

Method-1: Using (9).

Here two ratio strength solutions are found that are 1) 1:20 solution and 2) the diluent which can be written as 0:20.
Here base of the diluent is arbitrary hence convenient base is chosen in this case. Also, n y is also a random number which
can be chosen according to our convenience.

Data: x 1 = 1, x 2 = , b = 50, c = 950, n y = 1000

fSOxl , 950 x 0\

1000 g

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In this way problems on percentage and PPn of different parameters can be solved by using (9).

7. Conversion of percentage to mg/ml and vice versa:

Let x% (g/n

»*<»** -i£s

_ lOOOx mg

100 ml

_ 10 x mg

lml

: x%(g/ml) = 10 x mg/ml (10)

Dividing both sides by 10 we get,

:^% = x mg/ml (11)

Example: A certain injectable contains 2mg of a drug per milliliter of solution. What is the ratio Strength (w/v) of
the solution? [15]

Using (11), we have.
: 2 mg /ml = — %

:. MEANING OF 0%

= OPPn
Proof: -

Given,

multiply it by nPPn

■-©

: = OPPn

: Decimal notation of this is =

Thus proved

PPn or 0/n or - is used to denote pure base, solvent or vehicle.

Example- 1:

/100\

:0 = 0x (ioo)

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Decimal notation of 0% Sis

Example-2: If 50 ml of a 1:20 w/v solution are diluted to 1000 ml, what is the ratio strength (w/v)?

50 ml of 1: 20 strength solution is diluted to 1000 ml. This statement can mathematically be written a

. 50 (Jo-). o

50 950

Adding numerator and denominator directly (6), we have

8.1. MEANING OF X/0 OR J

This kind of notation may leads to confusion. Addition of ingredient is in one way simple and in another way

complex because of factors like solubility and saturation point of a solution. Hence this kind of notation is not necessary in
pharmacy.

This kind of notation can be used to indicate that the drug is not in the form of solution or in any form of mixture.

And it may also imply that the solvent is completely removed.

8.2. Removal of solvent by any means. (Usually by evaporation)

Example: If a syrup containing 65% w/v of sucrose is evaporated to 85% of its volume, what percentage (w/v) of
sucrose will it contain? [16]

Method -1:

65 g

■■65% w/v --

lOOni

This syrup is evaporated to 85% of its volume means, 15% of solvent is evaporated or 15 ml of it is evaporated if
volume of solution is 100ml.

65g 0__

' l ' e "100ni 15ni

By direct subtraction of numerator and denominator we have the above expression equal to

= 65g_
&5ni

■■ Furt her ,to find the

. 100 woni
< ~85

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IOSR Journal of Pharmacy

Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129

To find the percentage strength
„ 700

76,-/7
100

9. Solution to Miscellaneous problems.

Problem No.-l:In acute hypersensitivity reactions, 0.5mL of a 1:1000 (w/v) solution of epinephrine may be

administered suhcutaneously or intramuscularly, calculate the milligrams of epinephrine <>Jven.[17]
Solution:

-o/*o =

Iff

Hence 0.5 mg of epinephrine is given

ProblemNo.2: How many grams of 10 % w/w ammonia solution can be made from 1800g of 28% strong
ammonia solution ?[17]

Solution: Let a be the reqidred tvlwne of 10% ammonia solution ,t hen

(10\ (28\

■ <iJ = 1800 (lod)

: a = 5040

Problem.No.3: How many grams of a substance should be added to 240 ml of water to make
solution'! '[18]

Solution:

4

: 4%{W W) =

y ' } 96 + 4
To make 96 equal to 240 multiply all the elements of RHS by the factor — .
, (240\

'(%)
*(£)♦<(£)

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IOSR Journal of Pharmacy

Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129

log

Conclusion:

Aim of this article is to give more scientific and algebraic touch to the pharmaceutical calculations. Also, an effort is made to
make the steps of the solutions more pragmatic and easily understandable. And, an attempt is made to draw solutions from
basic datum or from data. A good relation is maintained between every steps of the solutions.

References
Books:

[I] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health I Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010)

[2] Don A. Ballington, Tova Wiegand Green, Pharmacy Calculations (EMC Corporation, USA, 2007)

PageNo. and Section No.

[3] Howard C. Ansel,, Pharmaceutical Calculations (Wolters Kluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.3
[4] Don A. Ballington, Tova Wiegand Green, Pharmacy Calculations (EMC Corporation, USA,

2007),sec.2.2.3
[5] Howard C. Ansel,, Pharmaceutical Calculations (Wolters Kluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.9
[6] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.22
[7] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.30, Pr. No. 1
[8] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.85
[9] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.264
[10] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.95, pr.No.20

[II] Don A. Ballington, Tova Wiegand Green, Pharmacy Calculations (EMC Corporation, USA,
2007),sec.8.3.1

[12] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.265
[13] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.266
[14] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.253
[15] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p. 90
[16] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.253
[17] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),98. pr.No.65
[18] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health I Lippincott Williams & Wilkins

530 Walnut Street,Philadelphia Pa 19106, 2010),p.86

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