﻿ Testing Density of Various Soap Products
Testing Density of Various Soap Products
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## Testing Density of Various Soap Products

Developers:

Joseph J. Leo
Log College Middle School
Centennial School District

Allen P. Marks
Research Scientist
Rohm and Haas Company
Spring House, PA

Level:

6 to 8

Discipline:

Physical Science, General Science

Objectives:

Students will have the opportunity to observe, measure and analyze the density of aerosol shaving creams and foams generated from shampoo and dishwashing liquids or detergents.

Background:

The experiment is designed to measure the density (mass per unit volume) generated by the foam(ing) content of each product.

Materials:

1/2 - or 1-gallon plastic jugs
250- or 300-mL beakers (2 for each product tested)
25 mL and 50 mL graduated cylinders
safety goggles or glasses
shampoo (Head and Shoulders � , Prell � )
shaving cream (Barbasol � , Gillette � )
liquid soap - (Dawn � , Ivory � , Palmolive � )
triple beam balance
spatula or wooden tongue depressors

Procedure:

1. Weigh and record the tare weight of each beaker to be used. (Mass) (Grams)

2. Fill each beaker with water and find the volume of the beaker by subtracting the tare weight from the beaker and water weight. Since the density of water is 1.0 g/mL, the weight of the water in grams is equal to the volume of the beaker in milliliters. The volume of water within the beaker will be the actual volume of foam that the beaker will hold. (Note that a 250-mL to 300-mL beaker has a volume greater than noted). (Volume) (Milliliters)

3. Prepare a 5% by volume solution of the soap products or shampoo to be used:

A 5% solution is made up in a plastic jug by adding 5 milliliters of soap to 95 milliliters of water. Gently swirl the contents to disperse the soap in the water. Then, shake the jug to develop the foam.

Aerosol shaving cream may be used in place of preparing foam, but it doesn�t work as well as soap foam because it does not pack well into the beaker without trapping air pockets.

4. Fill two empty beakers with foam, by inverting the jug and squeezing.

5. Weigh the beakers and subtract the weight of the beaker from the beaker packed with foam. The remainder is the mass of the foam.

6. Remember that 1 gram of water is equal to one milliliter.

7. Calculate the density of the foam in the following manner:

Example:

 Density = Mass of foam in beaker Volume of water held by beaker

 Density = Wt of the beaker and foam minus Wt of beaker Wt of beaker and water minus Wt of beaker

 Density = Weight of Foam (Grams) Volume of beaker (mL)

 Density = 1379 - 1109 = 27g Foam 270 mL 270 mL water

 Density = 0.10 gm/mL

Problems:

1. What is the density of the foam produced by 2% solution by volume if the container holds 270 mL. and its tare weight is 110 grams? The full container is 130 grams. Calculate the density.

2. Calculate the density of a 5% Dawn� detergent solution using the same container with the full container weighing 148 grams.

Extensions:

1. The density of foam generated may be varied by preparing several different solutions of 2%, 10% and 20% by volume. Also, the solution�s density will vary by the type of product you use. Soap detergents and shampoo were found to be the best test products.

2. An economical analysis can be made by testing several different brands of dishwashing liquids or shampoos at a constant volume (2%, 5%, etc.). Calculate the densities of each product and then calculate the cost/volume (milliliters or ounces) as to price of product.

Economic Analysis

(Data made up but you can supply real data.)

Product A:
 Cost: \$.50 for 16 fl. oz. ( a fl. oz. is about 29.6 mL, I think, so 16 fl. oz. - 47.4 mL) 5 % solution makes foam of density 0.10 g/mL (The 5% solution will have a density of about 1.0 before foaming since it's mainly water.)

 So 100 g of 5% solution will generate 100 = 100 mL of foam .07

 The whole container contains 47.4 = 95 aliquots of 5 mL samples .50

 So the bottle will make 95,000 mL or 95 liters of foam for \$.50, or .50 = 0.0053 cents/liter 95

Product B:
 Cost: \$ .60 for 16 fl. oz. 5 vol % solution makes foam of density 0.07 g/cc

 So 100 g of 5% solution will generate 100 = 1429 mL of foam .07

 The whole container makes 1429 x 95 = 136,000 m = 136 liters 136 liters

 So the cost is .60 = 0.0044 cents/liter 136

Please note that this experiment works well in cooperative learning groups, with every member having a task to perform.

 This experiment is courtesy of