Calculation percentage solutions

Calculations involving percentage solutions

Pharmacopoeia expresses the purity of pharmaceuticals in percentage. Percentage of pharmaceuticals is expressed in any one of the following four manners:

1. Percentage weight by weight (Percent w/w): Number of grams of one pharmaceutical present in 100 grams of the product.

General formula to prepare 1% w/w solution

• · Solute 1 part by weight or 1 gram

•  Solvent to produce 100 parts by weight or 100 grams

Dissolve 1 gram of solid in liquid to produce weight 100 grams of liquid. Such type a solution is rarely prepared.

2. Percentage weight by volume (Percent w/v): Number of grams of one pharmaceutical present in 100 milliliters of product.

General formula to prepare 1% w/v solution

• · Solute 1 part by weight or 1 gram

• · Solvent to produce 100 parts by volume or 100 milliliters

1 gram of solute dissolved in purified water to produce a volume of 100 mL will produce 1% w/v solution.

3. Percentage volume by volume (Percent v/v): Number of milliliters of one pharmaceutical present in 100 milliliters of product.

General formula to prepare 1% v/v solution

• · Solute 1part by volume or 1 millilitre

• · Solvent to produce 100 parts by volume or 100 milliliters

Dissolve 1 mL of one liquid into 100 mL of other liquid will produce 1% v/v solution.

4. Percentage volume by weight (Percent v/w): Number of milliliters of one pharmaceutical present in 100 grams of product.

General formula to prepare 1% v/w solution

• · Solute 1 part by weight or 1 gram

• · Solvent to produce 100 parts by volume or 100 milliliters

Exercise

Q. Calculate the quantity of dextrose to produce 500 mL of dextrose solution 5% w/v.

1 gram of dextrose dissolved in 100 mL purified water produces a dextrose solution of 1% w/v.

1 X 5 gram dextrose dissolved in 100 mL purified water will produce a dextrose solution of 1% w/v.

100 mL of dextrose solution 5% w/v requires 1 X 5 grams of dextrose

1 mL of dextrose solution 5% w/v require 5 gram of dextrose 1X 5/100

500 mL of dextrose solution 5% w/v requires 5 grams of dextrose 1 X 5 X 500/100 = 25 gm dextrose.

Formula to prepare 500 mL of dextrose solution 5% w/v.

• · Dextrose 25 gram

• · Purified water sufficient to produce 500 mL

Q. Calculate the quantity of sodium chloride to produce 500 mL of normal saline solution.

The normal, saline solution contains 0.9% w/v sodium chloride.

1 gram of sodium chloride dissolved in 100 mL purified water produces 1% w/v solution of sodium chloride.

1 X 0.9 gram sodium chloride dissolved in 100 mL purified water produces a 0.9% w/v solution of sodium chloride.

100 mL of normal saline solution requires 1 X 0.9 gm sodium chloride.

1 mL of normal saline solution requires 1 X 0.9/100 gm sodium chloride.

500 mL of normal saline solution requires 1 X 0.9 X 500/100 gm sodium chloride = 4.5 gm

Formula to prepare 500 mL of normal saline solution

• · Sodium chloride 4.5 gm

• · Purified water sufficient to produce 500 mL

Q. Prepare 50 mL of 10% v/v solution of methyl salicylate in alcohol.

1 mL of methyl salicylate in 100 mL alcohol produces 1% v/v solution of methyl salicylate.

1 X 10 mL of methyl salicylate in 100 mL will produce 10% v/v solution of methyl alcohol

100 mL of 10% methyl salicylate solution requires 1 X 10 mL methyl salicylate

1 mL of 10% methyl salicylate solution requires 1 X 10/100 mL methyl salicylate

50 mL of 10% methyl salicylate solution requires 1 X 10 X 50/100 mL methyl salicylate = 10 mL

Formula to prepare 50 mL solution of 10% v/v methyl salicylate solution in alcohol

• · Methyl salicylate 10 mL

• · Alcohol to produce a volume of 50 mL.

Q. Prepare 200 mL of phenol glycerine IP

• · Phenol 160 gm

• · Glycerine 840 gm

Phenol glycerine IP contains 16% w/w of phenol.

The density of glycerine is 1.26.

Formula to convert weight into volume: D =M/V

1.26 = M/200

M = 1.26 X 200

Quantity of glycerine required to prepare 200 mL of phenol glycerine = 200 X 1.26 = 252 gm

Calculation for Quantity of phenol required to prepare 200 mL of phenol glycerine.

840 gm glycerine requires 160 gm of phenol

252 gm glycerine requires 160 X 252/840 = 48 gm

The formula to prepare 200 mL phenol glycerine shall be

• · Phenol 48 gm

• · Glycerine 252 gm

General formulas to prepare 1% w/v solution in the imperial system

1. Solid 1 grain

Solvent to produce 110 minims

2. Solid 4.375 grain

Solvent to produce 1 fluid ounce

3. Solid 35 grain

Solvent to produce 8 fluid ounces.

Sometimes the strength of the solution is expressed as 1: 300. 2: 500, 5: 1000, etc. The strengths of these solutions are expressed as below

· 1: 100 = 1 in 100 = 100/100 % = 1%

· 1: 600 = 1 in 600 = 100/600% = 0.166%

· 3:1000 = 3 in 1000 = 3 X 100/1000% = 0.3%

· 2:10000 = 2 in 10000 2 X 100/10000% = 0.02%

Alcohol dilution

There will be a contraction of volume after dilution, an increase in temperature after dilution, and a release of air bubbles after dilution. Air is more soluble in alcohol than in water. Thus air bubble comes out from alcohol after dilution.

Q. Prepare 500 mL 45% alcohol from 95% alcohol.

The volume required 500 mL

% of Alcohol required 45%

5 of Alcohol supplied 95%

Formula for dilution

Volume of Stronger Alcohol = Volume required X % required/ % used

= 500 X 45/95 = 237 mL.

Formulation to prepare 500 mL 45% alcohol from 95% alcohol

· Alcohol 95% 237 mL

· Purified Water q. s. 500 mL

Calculation for percentage solution

Q. Prepare 200 mL of 8% acetic acid solution.

Acetic acid IP contains 33% w/w of acetic acid. The density of acetic acid is 1.04

Thus the weight of 200 mL of acetic acid IP will be calculated by using the formula D = M/V

1.02 = M/200,

M = 1.02 X 2O0 = 204 gm.

Weight of stronger solution required = weight required X % required/ % Used

= 204 X 8/33 = 49.450 gm

Formulation to prepare 200 mL of 8% acetic acid solution

· Acetic acid IP 49.45 gm

· Purified water q.s 200 gm