Percentage Strength

“Preparation” is frequently a solution but may be an ointment, a suppository, a powder mixture, and so on.

2. If the solute and the preparation are expressed in the same units, the concentration is dimensionless. It indicates the portion of the preparation represented by solute, and is written as a decimal or fraction. For example, if 10.0 mL of alcohol were dissolved in a sufficient quantity of water to make 40.0 mL of solution, the concentration of alcohol would be

Percent concentration is defined as the number of parts of solute in 100 units of solution. Concentration expressed as a decimal may be converted to percent by multiplying by 100:

This concentration may be written 25.0% by volume or 25.0% volume in volume or 25.0% v/v to indicate that both the solute and solution were measured by volume.

If 12.0 mL of peppermint oil were dissolved in sufficient alcohol to make 80.0 mL of solution, what would be the concentration of peppermint oil in this solution expressed as

A. a decimal?

B. a percent?

Solutions.

A. 0.150 v/v

B. 15.0% v/v

CALCULATIONS

A. concentration =

mL = 0.150 v/v

B. 0.150 v/v × 100 = 15.0% v/v

3. The formula for 1 liter of syrup contains 0.5 mL of orange oil. What is the percentage strength (v/v) of the oil in the syrup?

Solution. 0.05% v/v

CALCULATION

4. If the quantity of solute and of the preparation are expressed in the same units of weight, the concentration is dimensionless. If 10.0 g of charcoal are mixed with 65.0 g of another powder to make a total of 75.0 g, the charcoal concentration is

Note: In a preparation in which all concentrations are w/w, the sum of all contributions must be 100%.

If 12.0 g of lanolin are combined with 2.0 g of white wax and 16.0 g of petrolatum to make an ointment, what is the percentage concentration of lanolin in the ointment?

Solution. 40.0% w/w

CALCULATIONS

5. If 250 g of dextrose are dissolved in 300 mL of water, what is the percentage strength of dextrose in the solution in a w/w basis?

Solution. 45% w/w

CALCULATIONS

300 mL water = 300 g

6. A pharmacist adds 5.25 g of hydrocortisone to 150 g of a 2.5% hydrocortisone ointment. What is the percentage (w/w) of hydrocortisone in the finished product?

Solution. 5.8% w/w

CALCULATIONS

7. When the solute is measured by weight and the solution by volume, concentration is not dimensionless. If 1.25 g of sodium chloride is dissolved in sufficient water to make 55.0 mL of solution, the concentration is

Commonly, this concentration will be stated as a percent, w/v. The appendage “w/v” tells us that the solute is expressed in grams and that the quantity of solution is determined by volume in milliliters. These implied units must not be neglected in calculations.

By multiplying grams per milliliter by 100, we obtain the number of grams in 100 mL, which defines percent w/v:

If 5.75 g of boric acid are dissolved in sufficient alcohol to make a total volume of 120 mL, what is the strength of boric acid in the solution in

A. g/mL?

B. percent w/v?

Solutions.

A. 0.0479 g/mL

B. 4.79% w/v

CALCULATIONS

B. 0.0479 g/mL × 100% = 4.79% w/v

8. If 20 mL of a 3.5% (w/v) xylocaine solution are added to a 100 mL bag of D5W injection,

A. What is the percentage strength (w/v) of xylocaine in the final product?

B. Express the strength of final dilution in mg/mL.

Solutions.

A. 0.58%

B. 5.8 mg/mL

CALCULATIONS

9. To review these definitions of percent, the concentration of substance in a solution or mixture is expressed in terms of the amount of substance and the finished preparation. If the concentration is p percent w/w, then 100 weight units of the preparation contain p weight units of substance. The same weight units must be used for both substance and preparation. If the concentration is q percent v/v, then 100 volume units of the preparation contain q volume units of substance. The same volume units must be used for both substance and preparation. If the concentration is r percent w/v, then 100 mL of the preparation contain r grams of substance.

In each case, determine the percent concentration of glycerin and indicate whether w/w, v/v, or w/v:

A. 4.00 g of glycerin are dissolved in sufficient alcohol to make 25.0 mL.

B. 1.50 g of phenol are dissolved in 8.00 g of glycerin.

C. 10.0 mL of glycerin are dissolved in sufficient water to make 38.0 mL of solution.

Solutions.

A. 16.0% w/v

B. 84.2% w/w

C. 26.3% v/v

CALCULATIONS

= 0.160 g/mL = 16.0% w/v

= 0.842 w/w = 84.2% w/w

= 0.263 v/v = 26.3% v/v

Default Rules if Percentage Type is not Indicated

10. Sometimes the concentration of a solution is indicated without stating whether it is to be w/w, w/v, or v/v. In such cases, solutions are prepared so that solids can be weighed and liquids measured by volume. When the solute and the solvent (and therefore the solution) are liquids, the solution strength is assumed to be percent v/v. If the solute is a solid and the solvent a liquid, solution strength is assumed to be percent w/v. If both the solute and solvent are solid (or in a mixture of solids) percent w/w is assumed. These default rules come into play only when the percentage type is not indicated.

Let us look at an example. Say that you make up a solution of peppermint oil in alcohol by dissolving 2.0 mL of oil in enough alcohol to make 100 mL of solution. The concentration is

If your product is labeled “Peppermint Oil Solution, 2.0%,” the default rules apply. Since both peppermint oil and alcohol are liquids, it is assumed that both will be measured by volume. Therefore, “2.0%” means 2.0% v/v in this case, so that your label describes the product accurately.

However, if you were to make up a solution by dissolving 2.0 g of peppermint oil in sufficient alcohol to make up 100 mL of solution? Then “2.0%” would not accurately describe the product. It would have to be labeled 2.0% w/v.

If 4.0 g of peppermint oil are dissolved in 96.0 g of alcohol, how would you describe the concentration of peppermint oil in this solution?

Solution. 4% w/w

CALCULATIONS

Peppermint oil and alcohol are liquids and would normally be measured by volume. But in this example, the solution was prepared by weight. The “percent w/w” designation transmits this information. If the concentration were written “4.0%,” the default rules would apply and we would be (incorrectly) led to believe that the solution was prepared by volume.

11. If 500 mg of sodium bicarbonate are dissolved in water to make 50 mL of solution, what would be the percentage concentration of sodium bicarbonate in the solution?

Solution. 1% w/v

CALCULATIONS

12. Decide whether each of the following systems is w/w, v/v, or w/v, using the default rules:

A. 1% solution of zinc sulfate (a solid) in distilled water

B. An ointment containing 3% sulfur in petrolatum (both solids)

C. 10% solution of sugar in alcohol

D. 15% solution of alcohol in distilled water

Solutions.

A. w/v

B. w/w

C. w/v

D. v/v

Prescriptions with Ingredients Listed as Percentage

13. In a moment we will work with some prescriptions in which the ingredients are listed by percent. Of course, in order to prepare the medication, it is necessary to convert the percentages to weights and volumes. In w/w prescriptions, we calculate the weight of each material, since each must be weighed individually. But in w/v and v/v solutions, the volume of the solvent is usually not determined explicitly. There are two reasons:

(a) The solution is usually completed in a graduate by adding sufficient solvent to make the desired final volume. It is thus not necessary to know the volume of solvent.

(b) It is often impossible to calculate exactly what the volume of solvent should be. In w/v solutions, the volume occupied by the dissolved solute is not known. In very dilute solutions, the volume of solute may be negligible. But in general, there is no way to tell how much solvent to use. In v/v solutions, there may be shrinkage of volume when certain liquids are mixed: 50 mL of alcohol plus 50 mL of water yield less than 100 mL of solution. However, when chemically similar liquids are combined (e.g., mixing one fixed oil with another or one hydrocarbon with another), the total volume is usually the sum of the volumes of the components.

	Zinc sulfate (a solid)	1/2%
	Aq. pur. qs ad	60.0 mL

In this prescription the drag content is specified on the basis of concentration. We must calculate the quantities needed to prepare this solution. Zinc sulfate is a solid and the final product is a liquid, so the concentration is taken to be w/v. Thus the zinc sulfate concentration is 0.5 g/100 mL or 0.005 g/mL.

The most direct way of calculating the quantity of zinc sulfate is to rearrange the definition of concentration:

to quantity of solute = concentration × quantity of solution

Dissolve 0.3 g of zinc sulfate in sufficient purified water to make 60 mL.

Now consider this prescription:

	Eucalyptus oil	2.5% v/v
	Mineral oil	qs ad 30.0

What quantities should be used for this prescription?

Solution. Eucalyptus oil: 0.75 mL

Mineral oil: sufficient to make 30.0 mL of solution

CALCULATION

30.0 mL × 0.025 = 0.75 mL

14.		Sulfur 10.0%
		Benzoic acid 1.5%
		Petrolatum qs ad 30.0
		Ft ung

All ingredients are solids. What quantity of each should be used to make this ointment? (Remember that all three ingredients must be weighed.)

Solution.	Sulfur: 3.00 g
	Benzoic acid: 0.450 g
	Petrolatum: 26.6 g

CALCULATIONS

All ingredients are solids. According to the default rules, we treat the concentrations as percent w/w:

Sulfur: 30.0 g × 0.1 = 3.00 g
Benzoic acid: 30.0 g × 0.015 = 0.450 g

The sum of all percentages must be 100%. This is necessarily true only in a preparation in which all concentrations are w/w. From the prescription, we can see that 88.5% of the ointment is petrolatum.

Petrolatum: 30.0 g × 0.885 = 26.6 g

To check, verify that the sum of all contributions is 30.0 g.

How much boric acid (a solid) and camphor water (a liquid) are needed for this prescription? (Here is a situation in which two concentration types appear in the same solution. The boric acid must be considered on a w/v basis, whereas the camphor water is handled on a v/v basis. Treat each substance separately in doing your calculations.)

Solution.	Boric acid: 300 mg
	Camphor water: 5.25 mL

CALCULATIONS

Boric acid: 15.0 mL × 0.02 g/mL = 0.300 g

Camphor water: 15.0 mL × 0.35 = 5.25 mL

16. Iodine tincture is a 2% w/v solution of iodine.

A. How many grams of iodine will 40.0 mL of the tincture contain?

B. How many milliliters of the tincture contain 0.600 g of iodine? (Use proportion or rearrange the concentration definition to solve for amount of preparation.)

Solutions.