1. The Size and Mass of an Oleic Acid Molecule

2. 1. Estimate the Width of the drop

3. Example : Width of drop = 0.03cm = 3 X 10 -2 cm

4. 2. Volume of the drop = (width of the drop) 3 (width of the drop) 3

5. Example: (3 X 10 -2 cm) 3 = 3 X 10 -5 cm 3

6. 3. Estimate the diameter of the Floating Circle Diameter

7. Example : Diameter = 17cm

8. 4. Calculate the Area of the Floating Circle A = πr 2 A = πr 2 (r = d / 2 ) ( r = d / 2 )

9. Example: Diameter = 17cm ; radius = 17/2 = 8.5cm Area = π r 2 = 3.14 X 8.5 X 8.5 = 226.9 = 2.3 X 10 2 cm 2

10. 5. Divide the Volume of the drop by the area of the circle to find the thickness of the floating circle The maximum size of the oleic acid molecule = the thickness of the floating film

11. Example: Thickness = volume/area = 3 X 10 -5 cm 3 / 2.3 X10 2 cm 2 = 1 X 10 -7 cm

12. 6. Since the oleic acid molecule has a height that is ten times the width of the base of the molecule, divide the thickness by ten to get the width of the molecule’s base 7. Square the width of the base of the molecule to find the area of the base of the molecule.

13. Example: (1 x 10 -7 cm) / 10 = 1 X 10 -8 cm Area of molecule’s base = (1 X 10 -8 cm) 2 = 1 X 10 -16 cm 2

14. 8. Divide the area of the floating circular film by the area of the base of one molecule to find the number of molecules in the drop.

15. Example: (2.3 X 10 2 cm 2 ) / ( 1 X 10 -16 cm 2 ) = 2 X 10 18 molecules

16. 9. The mass of the drop = volume X density volume X density

17. Example: ( 3 X 10 -5 cm 3 ) X (1 g/cm 3 )= 3 X 10 -5 g

18. 10. The mass of a molecule = the mass of all the molecules in the drop divided by the number of molecules in the drop

19. Example: ( 3 X 10 -5 g) / ( 2 X 10 18 molecules) = 2 X 10 -23 g