Improved and Novel Methods Of Pharmaceutical Calculations. R. Subramanya.
Cheluvamba Hospital, Karnataka, Mysore-570 001, India. Abstract. It is a reviewΒ ...
IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
Improved and Novel Methods Of Pharmaceutical Calculations R. Subramanya Cheluvamba Hospital, Karnataka, Mysore-570 001, India.
Abstract It i s a review and r esearch article wherein general equations are derived for some problems and n ew 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 r atio 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 t he books same mistake is committed as in [3] of [1]. Hence, proof of t hese 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 m ethod i s 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 t he 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 l ike addition, subtraction, appropriate factor, and by simple division. In m ost of the books proportional m ethod i s 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: π. π. πΆπππ£πππ‘
3 π‘π πππππππ‘[3] 8
3 Γ 100 = 37.5% 8 The above equation does not satisfy the meaning of the symbol β=β (equals) used in between the expressions 3 Γ 100 and 37.5%. 8 3 βΆ π΄ππ‘π’ππ π£πππ’π ππ Γ 100 = 37.5 8 100 Aππ37.5 = 37.5 Γ 100 βΆ = 3750% Given calculation in the book is
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129 The correct method of solution is as following 3 βΆ = 0.375 8 100 βΆ = 0.375 Γ 37.5
100
βΆ = 100 βΆ = 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 of these work to fill the order. [4] 1 1 ππππ’π‘πππ: 1: 1000 = = = 0.1% 1000 10 Γ 100 In comparison with solution of [4]in [1],a few steps are reduced. 2.3. Express 0.02% as ratio strength. 2 0.02 100 2 1 ππππ’π‘πππ: 0.02% = = = = = 1: 5000 100 100 10000 5000
stock.
1
2.4. . Give the decimal fraction and percent equivalent of 45 βΆ
1 1 100 = Γ 45 45 100 2.22 100
βΆ
=
βΆ
= 2.22%
2.5. General Equation: π₯ βΆ π₯% = β β β β β β β β β β(1) 100 ππππππ π₯ ππ π ππππ ππ’ππππ : π»ππππ, π₯ = 100π₯% ---------------- (2) 2.6. Proof of (1) and (2): πΏππ‘ π ππ π ππππ ππ’ππππ, βΆ ππππ, π = π Γ 1 100 βΆ π΄ππ π, π = π Γ 100 100 π
βΆ = 100 βΆ π = 100π% π βΆ π΄ππ, π% = 100 :πΉπ’ππ‘πππ, ππ¦ ππππ π ππ’ππ‘πππππππ‘πππ π€π πππ£π 100π% = π π : 100 Γ 100 = π : πΆπππ πππ’πππ‘ππ¦, π = π βΆβ
3. Another way for dimensional analysis: 3.1. How many fluidounces (fi.oz) are there in 2.5 liters (L)? [5] .....Method-1: : πΊππ£ππ, 1 πππ’ππππ’πππ = 29.57 ππ 2.5πΏ 2500ππ 2500 βΆ = = = 84.55 1ππ. ππ§ 29.57 ππ 29.57 Method -2:
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129 1 ππ. ππ§. = 29.57 ππ 2500
.....Multiplying both sides by the factor29.57 that modifies 29.57 ml to required quantity, we get βΆ 1ππ. ππ§.Γ
2500 2500 = 29.57ππ Γ 29.57 29.57
βΆ 84.55ππ. ππ§. = 2.5πΏ
4. General equation to find the total surface area of particles. (Imaginary and ideal case).[6]. 4.1. A general equation i s derived to find the total surface area of cubes when a cube is divided serially along the xy, yz and zx βplanes. 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, βa set of cuttingβ. The sum of the surface area of cubes at βnthβ set of cutting is : π΄ = 2π +1 Γ 3π2 π ππ’πππ π’πππ‘π --(3) Where, βAβ denotes the sum of surface area of cubes and βnβ the ordinal value of set of cuttings from the first cube. Hence n is a set of whole number, i.e. π = 0,1,2,3 β¦ ; when cut along zx plane =
+
=
=
+
+
; when cut along yz plane
; when cut along xy plane
Fig.1. A typical view of a set of cuttings of the cubes across the zx, yz and xy planes. ππ π‘ππ 1π π‘ ππ’ππ, πππππππ ππ’ππππ ππ π ππ‘ ππ ππ’π‘π‘πππ ππ 0, π. π. , n = 0, and if he length is l units Total surface area of first cube is : π΄ = 2π+1 Γ 3 Γ π2 βΆ = 21 Γ 3 Γ π2 : = 6π2 : And if = 1 ; π΄ = 2π +1 Γ 3π2 βΆ
= 22 Γ 3 π2
:
= 12 π2
:And if π = 2 ; π΄ = 2π +1 Γ 3 Γ π2 βΆ
= 24π2
: And so on.
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129 4.2. Derivation of equation : π΄ = 2π+1 Γ 3π2 To start with, cube has six faces and if its length is l units the total surface area of the cube, when no cut is done along the axes is βΆ π0 = one cube Γ number of faces Γ area of one surface :i.e. π0 = 1 Γ 6 Γ π2 To make the above equation look like rest of the equations, it can be written as βΆ
2
π
π0 = 80 Γ 6 Γ
20
when this cube is cut across the Co-ordinate planes we get 8 cubes of dimensions equal to βΆ β΄ π1 = 8 Γ 6 Γ
π 2 2
2
: Further, π2 = 8 Γ 6 Γ
π
π 2
π’πππ‘π each.
2
22
:Finally, equation for ππ‘π term is π
2
βΆ ππ = 8π Γ 6 Γ π 2 :To simplify this equation, it can be written as = 23
: βΆ πΉπ’ππ‘πππ, ππ = 23π+1 Γ 3π2 2β2π :Replacing ππ by A, and simplifying further we get :π΄ = 3(2π +1 π2 ) βΆ = 2π +1 3π2 Thus derived.
π
Γ2Γ3Γ
π 2π
4.3. For example when π = 0 Surface area of the cube of length 1 cm is : 2π+1 3π2 ππ2 = 2 Γ 3ππ2 βΆ = 6ππ2 This is equal to the area of 2ππ Γ 3 ππ rectangle. Further π€πππ π = 5 ( 5 set of cuts) and π = 1ππ, βΆ ππππ, 2π+1 3π2 = 26 Γ 3 βΆ = 192 ππ2 This area is equal to the area of 12 ππ Γ 16 ππ rectangle. And when π = 20 πππ π = 1 ππ : 2π +1 3π2 = 221 Γ 3 : = 6291456ππ2 This is approximately equal to the area of 2.5π Γ 2.51π ππππππ πππ.
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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
1 Kilo-meter/liter/gram. (km/kL/kg)
1000 m/L/g
103 m/L/g
1 hecto-meter/liter/gram. (hm/hL/hg)
100 m/L/g
102 m/L/g
1 deka-meter/liter/gram. dam/daL/dag
10 m/L/g
101 m/L/g
1 meter/liter/gram. m/L/g
1 m/L/g
100 m/L/g
1 deci-meter/liter/gram. dm/dL/dg
0.1m/L/g
10β1 m/L/g
1 centi-meter/liter/gram. cm/cL/g
0.01 m/L/g
10β2 π/πΏ/π
1 100
1 milli-meter/liter/gram (mm/mL/mg)
0.001 m/L/g
10β3 π/πΏ/π
1 1000
1 micro-meter/liter/gram 0.000001 m/L/g
10β6 m/L/g
(πππ/πππΏ/πππ)
Fractional notation in m/L/g
π‘π
1 10
π/πΏ/π π‘π
π/πΏ/π π‘π
π/πΏ/π
1 1000000
π‘π
m/L/g
1 nano-meter/liter/gram. (nm/nL/ng)
0.000000001 m/L/g
10β9 m/L/g
1 1000000000
π‘π
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] : βΆ
=0.5kg+50mg+2.5dg = 0.5 Γ 1 ππ + 50 Γ 1ππ + 2.5 Γ (1ππ) -----------(4)
Conversion factors are 1
1
: 1kg=1000g, 1mg=1000 π, 1dg=10 g : Substituting the above values in (4), we get
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: = 0.5 Γ 1000 π + 50 Γ
1
1
βΆ = 0.5 Γ 1000 π + 50 Γ
1
1000
1000
π + 2.5 Γ (10 π) 1
π + 2.5 Γ (10 π)
βΆ = 500π + 0.05π + 0.25 π βΆ = 500.3 π
6. SYNTAX OF PERCENTAGE (%) AND PARTS PER n (PPn)CALCULATIONS Percent means βfor every hundredβ. Symbolically it is written as %. And mathematically it is written as the ratio π₯ π₯ 100 ππ . The bar between two numbers represents the phrase βfor everyβ. It can also be read as βParts per centβ the 100 word cent is Latin derivative which means, hundred. And this βParts per centβ can be abbreviated as PPc. I suggest to write this as PPh which is the abbreviation of βParts Per Hundredβ. This kind of abbreviation helps to extend the same idea to any quantity. For example PPt for parts per ten,. i.e., PPh for βParts Per Hundred, PPth, parts per thousand, PPtth βParts Per Ten Thousandβ, PPhth βParts Per Hundred Thousandβ, Parts Per Million βPPmβand so on. Alpha numerically it can also be written as ππ100 , PP10, ππ102 , ππ103 β― ππ10π . Usually, comparison is done in multiple of 10. Although general expression Parts Per n can be written as πππ, π€ππππ π ππ π πππ ππ‘ππ£π πππ‘ππππ. πΌπ π₯ ππ π‘ππ ππ’ππππ ππ ππππ‘π π‘πππ πππππππ ππ₯ππππ π πππ ππ βΆ π₯πππ. Mathematically xPPn can be written as following in the fractional form π₯ The staked fractional notation of xPPn is π The skewed fractional notation is π₯ π The linear fractional notation is π₯ π π₯ And the decimal notation of π ππ π , πππ ππ‘ ππ π ππππ ππ’ππππ π₯ π¦ π π΄ππ π, π¦ π’πππ‘π ππ π₯πππ = π¦ π₯ πππ‘, π§ = π¦ π , π§
ππππ, π¦ π’πππ‘π ππ π₯πππ = π¦ And henceforth [y](xPPn) stands for "y units of xPPn" z π΄ππππππππππ¦, y xPPn = β β β β β β(5) y Here the numerator indicates active ingredient and the denominator indicates solution or drug.
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Table-2: x Parts Per β10π β π₯ 10π π₯ 10 π₯ 102 π₯ 103 π₯ 104 π₯ 105 π₯ 106
Verbal form of the expression
Abbreviation
Alphanumeric notation
Prevailing notation
Decimal form
x Parts per ten
x PPt
x PP10
-
x Parts per hundred Or Parts per cent Or PERCENT
x PPh or PPc
x PP102
π₯%
0.0π₯
x Parts Thousand
per
x PPth
x PP103
π₯%0
0.00π₯
x Parts per Ten thousand
x PPtth
x PP104
-
0.000π₯
x Parts per hundred thousand
x PPhth
x PP105
-
0.0000π₯
x Parts per million
x PPm
x PP106
-
0.00000π₯
0. π₯
In addition to this, some more basic factors are intertwined in nature, with the x PPn. Those are volume, mass and weight. π₯ π πππ¦ ππ π’πππππ π‘πππ ππ π‘ππππ ππ : π₯ π π£ π£ ππ π₯ π π€ π€ ππ π₯ π π£ π€ ππ π₯ π π€ π£ . πΌπ π‘ππ π£πππ’ππ ππππ πππ‘ ππππππ πππ‘ππ πππ₯πππ πππ ππ ππ πππ π πππ π‘ππ π’πππ‘π ππ π£πππ’ππ πππ πππ π , π₯ π£ π₯ ππππ, π π£ = (π¦+π₯) , πππππ’π π π = (π¦ + π₯) πππππππππ¦ , π₯ π π€ π€ =
π₯π π¦+π₯ π π₯π
,π₯ π π£ π€ =
π₯ ππ π¦+π€π‘ .ππ π₯ ππ π
Only in case of, π₯ π π€ π£ = π ππ , here, volume of the active ingredient is not considered because increase in volume of solution after dissolution of active ingredient will be negligible. Example:. Peppermint spirit contains 10 % v/v peppermint oil. What is the volume of the solvent when volume of spirit is 75 ml ?[8] 10 ππ
:10%(π£ π£) = 100 ππ Factor that is required to make the numerator to 75 is we 75 10 ππ Γ 10 75 ππ βΆ = 75 750 ππ 100ππ Γ 10 75 ππ βΆ = (675 + 75)ππ
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75 10
, multiplying both the numerator and denominator by this factor get
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129 This shows that volume of the solvent of 10% peppermint sprit is 675 ml. 6.1. Summing of x (PPn) Case-1 If π₯1 πππ1 , π₯2 πππ1 , π₯3 πππ1 β― , π₯π πππ1 πππ π‘ππ π πππ’π‘ππππ ππ πππππππππ‘ π π‘πππππ‘ππ πππ ππ π πππ πππ. ( π»πππ π πππ π πππ πππ‘π’πππ ππ’πππππ ) π₯1 π₯2 π₯3 π₯π π₯1 + π₯2 + π₯3 + β― + π₯π βΆ π‘πππ, + + + β― + = β β β β β β(6) π1 π1 π1 π1 ππ1 βΆ π‘π π πππππππ¦ 6 πππ‘, π₯1 + π₯2 + π₯3 + β― + π₯π = π¦ π¦
π π1
βΆ ππππ 6 , = βΆ π΄ππ ππ,
π¦
β β β β β β(7)
π =π§ π§
βΆ ππππ (6) = π βΆ= π§ π1 βΆ= π§πππ1
π€ π€
1
π£
π€
π£
ππ π£ ππ π£ ππ π€ π€ π€
π£
π€
π£
ππ π£ ππ π£ ππ π€
πΌπ π1 = 100, π‘ππ πππ π’ππ‘ π€πππ ππ = π§% Example: What is the percentage strength of the solution when equal quantities of 10%, 20%, 30% solutions are mixed together. π»πππ, π₯1 = 10, π₯2 = 20, π₯3 = 30, πππ π1 πππ ππ = 100, π‘πππ π’π πππ 6 , π€π πππ£π βΆ
10 20 30 10 + 20 + 30 + + = 100 100 100 300 20 100
βΆ
=
βΆ
= 20%
Case-2: To find the strength of the solution to a specified base when different strength, different quantity, PPn solutions are mixed together.
different
πΏππ‘ π πππ1 , π πππ2 , π πππ3 , β― , π ππππ ππ π‘ππ π πππ’π‘ππππ π‘π ππ πππ₯ππ, πππ πππ‘ ππ¦ ππ π‘ππ π ππππππππ πππ π π. π. , ππππ¦ πππ πππ‘ π§ππππ¦ ππ π‘ππ πππππ π πππ’π‘πππ, βΆ π‘πππ, π π’π ππ π‘πππ ππ = π
π₯1 + π π1
π₯2 + π π2
π₯3 + β―+ π π3
π₯π ππ
π₯ π₯ π₯1 π₯ π 3 π π π 2 π3 ππ π1 π2 βΆ = + + +β―+ π π π π ππ₯1 ππ₯2 ππ₯3 ππ₯π π + π2 + π3 + β― + ππ βΆ = 1 ββββββ 8 π +π +π +β―+π The required and specified base of strength is ππ¦ , then (8) will be, π
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129 ππ₯π ππ₯1 ππ₯2 ππ₯3 π1 + π2 + π3 + β― + ππ π + π + π + β―+ π = ββββββ 9 ππ¦ ππ₯1 ππ₯2 ππ₯3 ππ₯ + + +β―+ π π1 π2 π3 ππ = π§, π‘πππ 9 π€πππ ππ (π + π + π + β― + π) ππ¦
βΆ ππ¦ πΏππ‘,
π§ ππ¦ βΆ ππππ, 9 πππ ππ π€πππ‘π‘ππ ππ , π§ππππ¦ , ππ ππ¦ = 100, π‘ππ πππ π’ππ‘ π€πππ ππ π§% Exmple-1:What is the percentage strength (v/v) of alcohol in a mixture of 3000 ml of 40 % v/v alcohol, 1000 % v/v alcohol, and 1000 ml of 70% v/v alcohol? Assume no contraction of volume after mixing.[9] : π»πππ ππ¦ = 100 πππ π₯1 = 40, π₯2 = 60, π₯3 = 70, π΄ππ π, π = 3000, π = 1000, π = 1000 πππ π1 = π2 = π3 = 100, ππ’π‘π‘πππ π‘πππ π π£πππ’ππ ππ 9 π€π πππ‘ βΆ=
ml of 60
3000 Γ 40 + 1000 Γ 60 + 1000 Γ 70 3000 + 1000 + 1000 βΆ πππππππ‘πππ π π‘πππππ‘π = 100 Note: since π1 = π2 = π3 = 100, πππ ππ¦ = 100 , π‘πππ¦ πππ‘ πππππππππ ππ¦ πππ ππππ‘πππ. 50
βΆ= 100 βΆ= 50% π₯
Example-2: When the same problem is solved for ππ¦ = 1000, i.e., for 1000 ππ πππ‘π 3000 Γ 40 + 1000 Γ 60 + 1000 Γ 70 1000 100 3000 + 1000 + 1000 : π₯πππ‘π = 1000 500 : = 1000 Example-3: Same problem, when π1 = 1000, π2 = 2000, π3 = 5000 πππ ππ¦ = 100 3000 Γ 40 Γ 10 + 1000 Γ 60 Γ 5 + 1000 Γ 70 Γ 2 100 10000 5000 : π₯πππ = 100 : = 3.28% This problem can also be solved by another method as following 40 3000 40 1000 = 120 : 3000 ππ ππ π π‘πππππ‘π π πππ’π‘πππ = 1000 3000 3000 60 1000 60 2000 = 30 : 1000 ππ ππ π π‘πππππ‘π π πππ’π‘πππ = 2000 1000 1000 70 1000 70 5000 = 14 : 1000 ππ ππ π π‘πππππ‘π π πππ’π‘πππ = 5000 1000 1000 Sum of these solutions are 120 30 14 + + 3000 1000 1000 164 := 5000 164 := 50 Γ 100 : = 3.28%
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129 Example-4: A pharmacist adds 10 ml of a 20 % (w/v) solution of a drug to 500 ml of D5W for parenteral What is the percentage strength of the drug in the infusion solution?[10]
infusion.
π·ππ‘π: π₯1 = 20 , π₯2 = 0, π ππππ π‘ππ πππ‘ππ£π ππππππππππ‘ πππ₯π‘πππ π ππ π·5 π ππ πππ‘ ππππ ππππππ ππ π‘πππ πππ π, π = 10, π = 500 πππ ππ¦ = 100, π1 = π2 = 100 100 π₯πππ = =
10 Γ 20 500 Γ 0 100 + 100 10 + 500 100
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. this statement can be written mathematically as following. βΆ π₯πππ(π€ π£) =
Hence
0 20% + 500 ππ π·5π 10ππ
To convert 20% into the quantity of active ingredient it is multiplied by 10 which is the quantity added. 20 Γ 10 0 + 100 500 ππ π·5π 10ππ 0 2π = + 500 ππ π·5π 10ππ
βΆ=
By direct addition, we get. 2π 510 ππ π·5π 2π = 5.10 Γ 100ππ π·5π 0.39 π = 100ππ π·5π = 0.39% π€ π£ =
Example-5:Prepare 250 ml of dextrose 7.5 % w/v using dextrose 5% (D5W) w/v and dextrose 50 % (D50W) How many milliliters of each will be needed?[11]
w/v.
Method-1: using (9) Data: π₯ = 7.5, π = 100, π₯1 = 5, π₯2 = 50, π1 = 100, π2 = 100. ππ¦ = 100. 5π 50π 100 100 + 100 7.5 π+π βΆ = 100 100 βΆ
7.5 5π + 50π = 100 100 π + π
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129 βΆ 500π + 5000π = 750π + 750π βΆ 250π = 4250π π 17 = π 1
βΆ
βΆ π = 17 πππ π = 1 Check: Putting values of b and c in (9) we get. 100 βΆ= ;=
5 Γ 17 50 Γ 1 + 100 100 17 + 1 100
7.5 100
βΆ = 7.5% Now, to prepare 250 ml,
250 1+17
= 13.89 ππ ππ 50% π·ππ₯π‘ππ π πππ 250 β 13.89 = 236.11 ππ ππ 5% π·ππ₯π‘πππ π
are required Example-6: In what proportion should alcohols of 95% and 50% strengths be mixed to make 70% alcohol?[12] Method-1:This can be solve by using (9) as usual. Method-2: Here determinant is used to solve the problem. πΏππ‘ π₯ πππ π¦ ππ π‘ππ ππππ‘πππ π‘πππ‘ ππππππ¦ π‘ππ πππ£ππ πππ‘πππ ππ πππππππ‘πππ π‘π π‘πππ πππ ππ π€πππ‘π‘ππ ππ ,
70 100
,
95π₯ 50π¦ 70 + = 100π₯ 100π¦ 100
ππππ, 95π₯ + 50π¦ = 70 β β β β β β 10 π΄ππ, 100π₯ + 100π¦ = 100 π. π. , π₯ + π¦ = 1 β β β β β β(11) ππ πππ 10 & 11 π‘ππ πππ‘πππππππ‘ β πππ ππ π€πππ‘π‘ππ ππ ππππππ€πππ, βΆ β=
95 50 1 1
: =45 πΏππ‘ βπ₯ & βπ¦ ππ π‘ππ πππππππππ πππ‘ππππππππ‘π ππππππ π€. π. π‘ π
π»π, βΆ ππππ, βπ₯ = βΆ
= 20
: π΄ππ, βπ¦ = :
70 50 1 1
95 70 1 1
= 25
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βΆ πππ€, π₯ = βΆ
π. π. , π₯ =
βΆ
π₯=
βπ₯ β
πππ π¦ =
βπ¦ β
20 25 πππ π¦ = 45 45 4 5 πππ π¦ = 9 9 π₯ 4 = π¦ 5
βΆ βΆ
ππππ, π₯ = 4 πππ π¦ = 5
Example-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?[13] Method-1: Data: ππ¦ = 100, π₯1 = 50, π₯2 = 20, π₯3 = 5, π1 = π2 = π3 = 100, πππ ππππ’ππππ π π‘πππππ‘π ππ 10% Using
(9),
we
have,
50π 20π 5π 100 100 + 100 + 100 10 π+π+π βΆ = 100 100 βΆ
10 50π + 20π + 5π = 100 100π + 100π + 100π
βΆ 1000π + 1000π + 1000π = 5000π + 2000π + 500π βΆ 1π = 8π + 2π βΆ ππππ, π = 1, ππππ£π πππ’ππ‘πππ π€πππ ππ, 8π + 2π = 1 βΆ πππ€, ππ¦ π‘ππππ πππ πππππ πππ‘πππ π€π πππ ππππ π = ππππ, π‘ππ πππππππ‘πππ ππ π: π: π =
1 1 πππ π = 16 4
1 1 : :1 16 4
By multiplying RHS of the above equation by 16, we have, βΆ
= 1: 4: 16
Assigning different values to d we can get different proportions. For example when d=10, consequently b=1 and c=1 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. Method-2: 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)π‘ = 100 Then, 5 = 0.05 π‘
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it
IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129 βΆ π. π, 50% + 20% + 0.05π‘ % = 10% βΆ
50 20 0.05π‘ 10 + + = β β β β(12) 100 100 π‘ 100
βΆ
70 + 0.05π‘ 10 = 200 + π‘ 100
βΆ 100 70 + 0.05π‘ = 10(200 + π‘) βΆ 7000 + 5π‘ = 2000 + 10π‘ βΆ π‘ = 1000 βΆ πΆπππ πππ’πππ‘ππ¦ 12 ππ ,
50 20 50 10 + + = 100 100 1000 100
βΆ πππππππ‘πππ ππ πππππππππ‘ππ ππ πΏπ»π ππ 100: 100: 1000 βΆ π€ππππ ππ πππ’ππ π‘π 1: 1: 10 Without using (9) answer can be found directly as following Let all the three percentages be variables. And let π‘1 , π‘2 , π‘3 ππ π‘ππ π£πππππππ ππ’πππ‘ππ‘πππ βΆ ππππ,
0.5π‘1 0.2π‘2 0.05π‘3 10 + + = π‘1 π‘2 π‘3 100
βΆ 50π‘1 + 20π‘2 + 5π‘3 = 10π‘1 + 10π‘2 + 10π‘3 βΆ 8π‘1 + 2π‘2 = π‘3 As usual answers can be found by giving different values to π‘3 ππ π‘π π‘1 & π‘2 Note: Since problem has three unknown and only two rows, square matrix cannot be formed. Hence it is not to draw solution by determinant m ethod. Although, by modifying some of t he rules of determinants or answer can be obtained.
possible m atrix,
Example-8: If 50 ml of a 1:20 w/v solution are diluted to 1000 ml, what is the ratio strength (w/v)?[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, ππ¦ is also a random number which can be chosen according t o our convenience. π·ππ‘π: π₯1 = 1, π₯2 = 0 , π = 50, π = 950, ππ¦ = 1000 1000 βΆ π
ππ‘ππ π π‘πππππ‘π = βΆ
=
2.5 1000
βΆ
=
1 400
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50 Γ 1 950 Γ 0 20 + 20 50 + 950 1000
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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:
ππ πΏππ‘ π₯% π ππ ππ π‘ππ πππππππ‘πππ π‘π ππ ππππ£πππ‘ππ π‘π , π‘πππ ππ π₯π : π₯% π ππ = 100ππ 1000π₯ ππ : = 100 ππ 10 π₯ ππ : = 1 ππ βΆ π₯% π ππ = 10 π₯ ππ ππ β β β β β β( 10) π·ππ£πππππ πππ‘π π ππππ ππ¦ 10 π€π πππ‘, π₯ : % = π₯ ππ ππ β β β β β β(11) 10 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
: 2 mg /ml = 10 % βΆ = 0.2% =
2 = 1: 500 1000
8. MEANING OF 0% AND 0 PPn 0 = 0πππ Proof:πΊππ£ππ, 0 ππ’ππ‘ππππ¦ ππ‘ ππ¦ ππππ βΆ =0
π π
βΆ =
0Γπ π
βΆ =
0 π
βΆ = 0πππ βΆ π·ππππππ πππ‘ππ‘πππ ππ π‘πππ ππ = 0 Thus proved 0 PPn or 0 π ππ
0 π
is used to denote pure base, solvent or vehicle.
Example-1: βΆ0 = 0Γ
100 100
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129 0 100
βΆ
=
βΆ
= 0%
π·ππππππ πππ‘ππ‘πππ ππ 0% β’ππ 0 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 as,
βΆ
50
1 20 + 0 50 950
Adding numerator and denominator directly (6), we have βΆ
=
2.5 1000
βΆ
=
1 400
8.1. MEANING OF πΏ π πΆπΉ
πΏ π
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 And it may also imply that the solvent is completely removed.
mixture.
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) sucrose will it contain?[16]
of
Method -1: βΆ 65% π€ π£ =
65 π 100ππ
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. βΆ π .π .,
65π 0 β 100ππ 15ππ
By direct subtraction of numerator and denominator we have the above expression equal to βΆ=
65π 85ππ
βΆ πΉπ’ππ‘
πππ , π‘π ππππ
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π‘ ππ ππππ πππ‘πππ
100 85 = 76.47π = 76.47% , 100 100ππ 85ππ Γ 85 65π Γ
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
Method -2: βΆ
65 100 Γ 85%
βΆ
65 85 To find the percentage strength 100 85 100 85 Γ 85 65 Γ
βΆ
76.47 100
βΆ βΆ
= 76.47%
9. Solution to Miscellaneous problems. Problem No.-1:In acute hypersensitivity reactions, 0.5mL of a 1:1000 (w/v) solution of epinephrine may be administered subcutaneously or intramuscularly, calculate the milligrams of epinephrine given.[17] Solution: :
1 1000
1π
1000 ππ
π€ π£ = 1000 ππ = 1000 ππ =
0.5 ππ 0.5ππ
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: πΏππ‘ βΆ π
π ππ π‘ ππ ππππ’ππππ
π£πππ’ππ
ππ 10% πππππππ
π πππ’π‘πππ
, π‘ π ππ
10 28 = 1800 100 100
π = 5040
βΆ
Problem.No.3: How many grams of a substance should be added to 240 ml of water to make a 4 % (w/w) solution?[18] Solution: βΆ 4%(π€ π€ ) =
4 96 + 4
To make 96 equal to 240 multiply all the elements of RHS by the factor
240 96
.
240 4 96 βΆ= 240 240 96 96 + 4 96
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IOSR Journal of Pharmacy Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
βΆ=
10π 240π + 10π
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 t he solutions more pragmatic and easily understandable. And, an attempt is made to draw solutions from basic datum or from data. A good r elation is maintained between every steps of the solutions.
References Books: [1] [2]
Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010) Don A. Ballington, Tova Wiegand Green, Pharmacy Calculations (EMC Corporation, USA, 2007)
PageNo. and Section No. [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]
Howard C. Ansel,, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.3 Don A. Ballington, Tova Wiegand Green, Pharmacy Calculations (EMC Corporation, USA, 2007),sec.2.2.3 Howard C. Ansel,, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.9 Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.22 Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.30, Pr. No. 1 Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.85 Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.264 Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.95, pr.No.20 Don A. Ballington, Tova Wiegand Green, Pharmacy Calculations (EMC Corporation, USA, 2007),sec.8.3.1 Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.265 Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.266 Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.253 Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p. 90 Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.253 Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),98. pr.No.65 Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.86
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