IOSR Journal of Pharmacy f\/\
Vol. 2, Issue 1, Jan-Feb.2012, pp. 1 13-129 fc?NJ
IOSR
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%
ISSN: 2250-3013 www.iosrphr.org 113IPage
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
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:
ISSN: 2250-3013 www.iosrphr.org
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
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
ISSN: 2250-3013 www.iosrphr.org
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
IOSR
<&) 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.
ISSN: 2250-3013 www.iosrphr.org 116IPage
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
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
ISSN: 2250-3013
www.iosrphr.org
IOSR Journal of Pharmacy f\/\
Vol. 2, Issue 1, Jan-Feb.2012, pp. 1 13-129 fc?NJ
IOSR
: = 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.
ISSN: 2250-3013 www.iosrphr.org 118IPage
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
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
ISSN: 2250-3013
www.iosrphr.org
119IPage
IOSR Journal of Pharmacy f\/\
Vol. 2, Issue 1, Jan-Feb.2012, pp. 1 13-129 t^NJ
IOSR
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,
ISSN: 2250-3013 www.iosrphr.org 120 I P a g e
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
•■+£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
ISSN: 2250-3013 www.iosrphr.org
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
IOSR
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)
ISSN: 2250-3013 www.iosrphr.org
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
: 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
ISSN: 2250-3013 www.iosrphr.org
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
: 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
ISSN: 2250-3013 www.iosrphr.org
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
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
ISSN: 2250-3013 www.iosrphr.org
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
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)
ISSN: 2250-3013 www.iosrphr.org
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
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
ISSN: 2250-3013 www.iosrphr.org 127 I P a g e
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\
'(%)
*(£)♦<(£)
ISSN: 2250-3013 www.iosrphr.org 128 I P a g e
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
ISSN: 2250-3013 www.iosrphr.org 129 I P a g e