Pharmaceutical Calculation.pdf

Pharmaceutical calculations

1. Prepared by: Dr.Harshil M Patel, Assistant Professor
Dept. of Pharmaceutics, SNLPCP, UMRAKH.

2. ▪Inourdaytodaylifeaswellasinthepracticeof
Pharmaceuticalprofession,itisalwaysnecessarytohave
someunittomeasureaquantity.
▪Thescience,whichdealswithweightsandmeasure,iscalled
Metrology.Aseachandeverypharmaceuticaloperation
impliesthefundamentalknowledgeofmetrology,itappears
appropriatetouswithvariousaspectsofmetrology.
➢TherearetwosystemsofweightandMeasures:
1.TheImperialSystem
2.TheMetricSystem
2

3. ▪It is an old system of weights and measures.
Measurements of Weights in Imperial System:
▪Weight is a measure of the gravitational force acting on a body and is directly proportional to its mass.
▪The imperial system are of two types:
1.Avoirdupois system
▪In this system pound (lb) is taken as standard of weight (mass).
1 pound avoir (lb) = 16 oz avoir (oz is pronounced as ounce)
1 pound avoir (lb) = 7000 grains (gr)
2.Apothecary or Troy system:
▪In this system grain (gr) is taken as standard of weight (Mass).
1 pound apoth (lb) = 12 ounce (oz)
1 pound apoth (lb) = 5760 Grains (gr)
1 Ounce = 8 drachms (oz)
1 Drachms (oz) = 3 scrupls
1 Scrupls = 20 grains (gr)
3

4. ▪1 gallon ( c ) = 4 Quarts
▪1 Quarts = 2 pint
▪1 pint ( o ) = 20 Fluid ounce
▪1 Fluid ounce = 8 Fluid Drachm
▪1 fluid drachm = 3 Fluid scruple
▪1 fluid scruple = 20 minims
4

5. ▪ThemetricsystemisusedintheIndianPharmacopoeiaforthemeasurementof
weightandcapacity.ThemetricsysteminIndiawasimplantedfrom1
st
April,1964in
Pharmacyprofession.
▪Kilogramistakenasthestandardweight(Mass).
Conversion
1 kilogram (kg) 1000 Grams Kilo=1000 Greek Word
1 hectogram (hg) 100 Grams Hecto=100 Greek Word
1 dekagram (dg) 10 Grams Deka=10 Greek Word
1 gram (g) 1 Grams
1 decigram (dcg) 1/10 Grams Deci=1/10 Latin Word
1 centigram (cg) 1/100 Grams Centi=1/100 Latin Word
1 milligram (mg) 1/1000 Grams Milli=1/1000 Latin Word
1 microgram (mcg) 10
-6
Grams Micro=10
-6
1 nanogram (ng) 10
-9
Grams Nano=10
-9 5

6. Measurement of Volume:
‘Liter’ is a Taken as the standard of volume.
1 liter (L,Lit) = 1000 ml
1 Microliter (mcl) = 1/1000 ml
Conversion
Domestic Measure Metric System Imperial System
1 Drop 0.06 ml 1 minim
1 Teaspoonful 4 ml 1 fluid drachms
1 Desert Spoonful 8 ml 2 fluid drachms
1 Tablespoonful 15 ml 4 fluid drachms
1 wine-glassful 60 ml 2 fluid ounces
1 Teacupful 120 ml 4 fluid ounces
1 Tumblerful 240 ml 8 fluid ounces
6

7. 1 Kilogram 2.2 pounds (lb)
1 Ounce apoth. 30 g
1 Pound avoir. 450 g
1 Grain 65 mg
7

8. PercentageSolution:
▪Theconcentrationofasubstancecanbeexpressedinthefollowing
threetypesofpercentages:
1.Weightinvolume(w/v):Requiredtoexpressconcentrationofa
solidinliquid.
2.Weightinweight(w/w):requiredtoexpressconcentrationofasolid
insolidmixture.
3.VolumeinVolume(v/v):Requirestoexpressconcentrationofa
liquidinanotherliquid.
8

9. Weight in Volume (w/v):
In this case the general formula for 1% (w/v) is:
The formula is actually:
Solute: 1 parts by weight solute 1 g
Solvent up to: 100 parts by volume solvent up to 100 ml
Weight in Volume (v/v):
In this case the general formula for 1% (v/v) is:
The formula is actually:
Solute: 1 parts by volume solute 1 ml
Solvent up to: 100 parts by volume solvent up to 100 ml
9

10. ▪These type of calculation involve the mixing of two similar preparations, but of different
strengths, to produce a preparation of intermediate strength.
▪The name is derived from the Latin allegation, meaning the act of attaching and hence
refers to lines drawn during calculation to bind quantities together.
Method:
Higher
Concentration
Parts of Higher
concentration= R-L
Lower
Concentration
Parts of Lower
concentration= H-R
Intermediate
Concentration
10

11. ▪Prepare 600ml of 60 % v/v alcohol from 95 % v/v alcohol.
▪Higher Concentration -95 %
▪Required concentration-60 %
▪Lower concentration-0%
▪so from allegation method it is obtained:
Volume of 60 % alcohol solution=600 ml
The total volume of 95 % alcohol required=
The volume of 95 % alcohol required=
????????????���??????�95%????????????????????????ℎ????????????
????????????���??????�????????????���+????????????���??????�95%????????????????????????ℎ????????????
* 600
=
60
35+60
* 600
= 379 ml
95 % 60-0-=60 Parts
0 % 95-60= 35 Parts
60 %
11

12. ▪Forexcisepurpose,thestrengthofalcoholinindicatedbydegreesproof.
TheUSsystem:Proofspiritis50%alcoholbyvolume(or42.49%byweight).
TheBritish/IndianSystem:Proofspiritis57.1%ethanolbyvolume(or48.24%by
weight).
▪Proofspiritisthatmixtureofalcoholandwater,whichat50
0
Fweights12/13
th
ofan
equalvolumeofwater.
[N.B.Densityofproofspirit=12/13ofdensityofwaterat51
0
F=0.923g/ml]
▪Thismeansthatanyalcoholicsolutionthatcontains57.1%v/valcoholisaproofspirit
andissaidtobe100proofs.
100 degree proof alcohol = 57.1 % alcohol
12

13. ▪Ifthestrengthofthealcoholisabove57.1%alcoholthenthesolutionis
called“OVERPROOF”.
▪Itthestrengthofthealcoholisbelow57.1%alcoholthenthesolutioniscalled
“UNDERPROOF”.
▪InUSsystem:Theproofspiritis50%alcoholbyvolumeor42.49%by
weight.
▪InIndia,theexcisedutyiscalculatedintermsofrupeesperliterofproof
alcohol.Soanystrengthofalcoholisrequiredtobeconvertedtodegree
proof.WeshallfollowtheBritishsystem.
▪Conversionofstrengthofalcoholfrom%v/vtodegreeproofasperIndian
system.
Strength of alcohol=
%??????/??????��������
��.�%??????????????????
* 100
▪Conversionofstrengthofalcoholfromdegreesproofto%v/vasperIndian
System.
Strengthofalcoholin%v/v=
??????���������????????????����??????��������??????����∗��.�
���
13

14. 1.Find the strength of 95 % v/v alcohol in terms of proof spirit.
Strength of alcohol =
%??????/??????��������
��.�%??????????????????
* 100
Strength of alcohol=
??????�%??????/??????��������
��.�%??????????????????
* 100
Strength of alcohol = 166.34 degree Proof = (1.66.34-100) Degrees
Over Proof= 66.34 op
2.Find the strength of 20 % v/v alcohol in terms of proof spirit.
Strength of alcohol =
%??????/??????��������
��.�%??????????????????
* 100
Strength of alcohol=
��%??????/??????��������
��.�%??????????????????
* 100
Strength of alcohol = 35.03degree Proof = (100-35.03) Degrees
Under Proof = 64.97 up
14

15. 3.Calculate the real strength of 30 op & 40 up.
30 op= (100+30)=130 degree proof
Strength of alcohol in % v/v=
??????���������????????????����??????��������??????����∗��.�
���
Strength of alcohol in % v/v=
���∗��.�
���
= 74.23 % v/v
40 up= (100-40)= 60 degree proof
Strength of alcohol in % v/v=
??????���������????????????����??????��������??????����∗��.�
���
Strength of alcohol in % v/v=
��∗��.�
���
= 34.26 % v/v
15

16. 4.How many proof gallons are contained in 5 gallon of 70 % v/v
alcohol?
1 proof gallon = 1 gallon proof alcohol = 1 gallon of 100 degree alcohol
Strength of alcohol =
%??????/??????��������
��.�%??????????????????
* 100
Strength of alcohol =
��%??????/??????��������
��.�%??????????????????
* 100
= 122.59 Degree Proof alcohol
=
���.�??????
���
Proof alcohol
= 1.226 proof alcohol
5 gallon 70 % v/v alcohol = 5 gallons of 1.226 proof alcohol
= 6.013 proof gallon
16

17. ▪Isotonicsolutions:
▪Osmosis:ifasolutionisplacedincontactwithasemipermeable
membranethemovementofthesolventmoleculesthroughthe
membraneiscalledosmosis.
▪Anidealsemipermeablemembraneonlyletsthesolventmoleculesto
passthroughitbutnotthesolutemolecules.Thebiologicalmembranes
arenotidealsemipermeablemembrane.
▪Theyareselectivelypermeable;theygivepassagetosomesolutes
whilestopthepassageofothers.
▪Incaseofbiologicalmembranesanothertermtonicityisused.
17

18. ▪Asolutionisisotonicwithalivingcellifthereisnonetgainorlossof
waterbythecell,whenitisincontactwiththissolution.
▪Ifalivingcelliskeptincontactwithasolutionandthereisnolossor
gainofwaterbythecellthenthesolutionissaidtobeisotonicwiththe
cell.
▪Itisfoundthattheosmoticpressureof0.9%w/vNaClsolutionissame
asbloodplasma.So0.9%w/vNaClsolutionisisotonicwithplasma.
18

19. ▪Isotonic:-Whenasolutionhassomeosmoticpressureasthatof0.9%
w/vNaClsolution.
▪Paratonic:-Notisotonic.
a)Hypertonic:-Theosmoticpressureofthesolutionishigherthen
0.9%w/vNaClsolution.
b)Hypertonic:-Theosmoticpressureofthesolutionislowerthan0.9
%w/vNaClsolution.
19

20. Observation Conclusion Mechanism
The shape and size of the cell
remained unchanged.
The solution is isotonic.
Osmotic pressure of the cell
fluid and the solution are
same. No movement of water
occurs across the cell
membrane.
The size of the cell increased
and may burst.
The solution is hypertonic.
Osmotic Pressure of the cell
fluid is more then the solution.
Water molecules moved from
the solution to the interior of
the cell. So the cell swelled.
The size of the cell is reduced
or shrieked.
The solution is hypertonic.
Osmotic pressure of the cell
fluid is less than the solution
outside. Water molecules
moved from the interior of the
cell to the solution.
20
•Iftheredbloodcellburststhenthehemoglobincomesoutthecellandtheplasma
becomesredincolor.Thisphenomenoniscalledhemolysis.

21. 1.Solutionforintravenousinjection:Theinjectionmustbeisotonicwithplasmaotherwise
theredbloodcellmaybehemolyzed.
2.Solutionforsubcutaneousinjections:Isotonicityisrequiredbutnotessential,because
thesolutioniscomingincontactwithfattytissueandnotcontactwithblood.
3.Solutionforintramuscularinjection:Theaqueoussolutionmaybeslightlyhypertonic.
Thiswilldrawwaterfromtheadjoiningtissueandincreasetheabsorptionofthedrug.
4.Solutionforintracutaneousinjection:Diagnosticpreparationmustbeisotonic,becausea
paratonicsolutionmaycauseafalsereaction.
5.Solutionforintrathecalinjection:Intrathecalinjectionsareintroducedinthecavitiesof
brainandspinalchord.Itmixeswithcerebrospinalfluid.ThevolumeofCSFisonly60to80
ml.soasmallvolumeofparatonicinjectionmaychangetheosmoticpressureoftheCSF,
whichmayleadtovomitingandothersideeffect.
6.Solutionsfornasaldrops:Aqueoussolutionappliedwithinthenostrilmayproduce
irritationifitisparatonic.Sonasaldropsmustbeisotonicwithplasma.
7.Solutionforophthalmicuse:Onlyoneortwodropsofophthalmicsolutionsaregenerally
used.Soitisnotessentialforeyedropstobeisotonic.SlightPara-tonicitywillnotproduce
greatirritationbecausetheeyedropswillbedilutedwiththeLachrymalfluid.
21

22. ▪Itisdifficultandtimeconsumingtodeterminetheosmoticpressureofasolution.So
someindirectmethodsareadoptedtocomparebetweentwoisotonicsolutions.Two
solutionswillproducesameosmoticpressureifbothcontainthesamenumbersof
ultimateunits.
▪Theseunitsmaybeasfollows:
1.Theseunitsmaybemoleculesincaseofsubstancesthosedonotionize.
2.Theseunitsmaybeionsincaseofsubstancesthoseionize.
3.Theseunitsmaybebothionsandunionizedmoleculesincaseofweakelectrolytes.
▪Somephysicalpropertiesofthesesolutiondependonthisnumberofunitssuchas
osmoticpressure,freezingpointdepression,vaporpressureetc.thesephysical
propertiesarecalledcolligativepropertiesofthesolutions.
▪Sincethesecolligativepropertiesareinterdependent,soosmoticpressureoftwo
solutionscanbecomparedfromtheircolligativepropertieslikefreezingpoint
depression.
22

23. Freezing point depression method
Sodium Chloride equivalent method
Isotonic solution V-Value method
23

24. ▪Freezingpointofpurewateris0
0
C.Whenanyimpuritiesarethere(
likesalt,drugetc.)thewaterfreezesatsomelowertemperatures.In
caseofasolutionthesoluteunitsreducesthefreezingpointofwater.
So the Freezing point depression = Freezing point of pure water –Freezing point of the
solution
▪ThisFreezingpointdepressiontothenumberofunitsofsolutes
presentinthesolution.
▪Itisalsoproportionaltotheosmoticpressureofthesolution.
▪Now,whilepreparinganinjectionorophthalmicsolutionthedrugis
givenincertainpercentage(%w/v).
▪Thissolutiongenerallyishypertonic.Inthissolutionsomeinertsolute
(likeNaClorDextrose)isdissolvedtoraisetheosmoticpressureupto
theosmoticpressureofserumorPlasma.
24

25. ▪Step:11dentifyareferencesolutionandtheassociatedtonicityparameter
▪Step:2Determinethecontributionofthedrugandadditivestothetotaltonicity.
▪Step:3Determinetheamountofsodiumchlorideneededbysubtractingthe
contributionoftheoriginalsolutionfromthereferencesolution.
The freezing point of plasma = -0.52
0
C
Thefreezingpointdepressionofplasma=Freezingpointofpurewater–Freezingpointofplasma
=0
0
C-(-0.52
0
C)
=0.52
0
C
The freezing point depression of sodium chloride = 0.52
0
C
25