Name: ________________________
Period : ____________
Date:___________
Structure of the Molecule: Calculating its Size
Part 1: Calculating the Diameter of a BB
Purpose
In this experiment the diameter of a BB will be determined without directly measuring it.
(The
Discussion
This activity clarifies the relationship between surface area and volume of a 3dimentional object, and sets the stage for the follow-up experiment: Determining
Diameter of an Oleic Acid Molecule. (In that experiment the size of a single
molecule will be calculated.) To understand the relationship between surface area
volume, consider eight wooden blocks arranged to form a single 2 x 2 x 2
centimeter cube (figure a). Because all cubes have 6 sides, it follows that the
surface area will be 6 times the area of one face (4 square cm) or 24 square cm.
cubical form exposes the minimum surface area (which is the reason buildings in cold
are often cubical in shape). Now, consider the surface area of any other configuration of
the eight blocks. For example, consider an arrangement of a 1 x 2 x 4 cm rectangular
prism (figure b). The outer surface area of the new structure is greater. (28 square cm)
And, finally, if the blocks are spread out to form a stack only one cube thick, 1 x 1 x 8
centimeters (figure c), the surface area is maximized. (34 square cm)
The different configurations have different surface areas, but the volume remains
constant. For each of the examples given, the height of the structure can be calculated
by dividing the volume of the structure by the area of the top surface:
𝑉𝑜𝑙𝑢𝑚𝑒
𝐴𝑟𝑒𝑎
=
(𝑙𝑒𝑛𝑔𝑡ℎ 𝑥 𝑤𝑖𝑑𝑡ℎ 𝑥 ℎ𝑒𝑖𝑔ℎ𝑡)
(𝑙𝑒𝑛𝑔𝑡ℎ 𝑥 𝑤𝑖𝑑𝑡ℎ)
figure a
the
and
outer
The
figure
b
areas
figure c
= 𝐻𝑒𝑖𝑔ℎ𝑡
For example, in figure 1 the volume of the structure = 2 cm x 2 cm x 2 cm = 8 cm3
The area of the top surface of that same structure = 2 cm x2 cm = 4 cm2
Dividing the volume by the area, as in the above equation, gives:
8 cm3
4 cm2
= 2 𝑐𝑚
As can be seen in figure a, the height is indeed 2 cm. It is this concept of the relationship of surface area to
volume that we will use to calculate both the diameter of a BB and the radius of a molecule.
The same concept of surface area to volume can be used for a cylindrical shape also. A pancake is essentially
a cylinder with a very small height. The volume of pancake batter is the same whether it is in the mixing bowl
or spread out on a surface, but the surface area is much different. The volume of a pancake equals the surface
area of one flat side multiplied by the height (thickness of the pancake).
𝑉 = 𝜋𝑟 2 𝑥 ℎ
But, much the same as with the blocks, if both the volume and the surface area are known the thickness can
be calculated:
𝑉𝑜𝑙𝑢𝑚𝑒
𝐴𝑟𝑒𝑎
=
𝜋𝑟 2 𝑥 ℎ
𝜋𝑟 2
= 𝐻𝑒𝑖𝑔ℎ𝑡
Instead of cubical blocks or pancake batter, consider a graduated cylinder that contains BBs. The volume of the
BBs (including the air between the BBs) is easily read on the side of the cylinder. If the BBs are poured into a
tray, their volume remains the same. The procedure below uses the concepts discussed above; follow it and
calculate the diameter of a single BB.
Procedure
Step 1: Use a graduated cylinder to measure out 75 cm3 of the BBs. (Note that 1 mL = 1 cm3).
Volume =
cm3
1
Step 2: Carefully spread the BBs out to make a compact circle one pellet thick on the tray. Use the string
provided to construct a border to help stabilize the BBs. With a ruler, determine the diameter (in centimeters)
of the BB circle in three places and compute the average diameter. Using the average, calculate the area of the
BB circle in the space below (remember: diameter = 2 radii).
1st diameter:
2nd diameter:
3rd diameter:
Average diameter =
=
cm
Area = 𝜋𝑟 2 =
=
cm2
Step 3: Using the area and volume of the BBs, calculate the height (diameter) of a BB. Show your
computations.
Height =
𝑉𝑜𝑙𝑢𝑚𝑒
𝐴𝑟𝑒𝑎
=
=
cm
Step 4: Check your result by using the calipers to measure the height (diameter) of a BB.
1st measured diameter:
2nd measured diameter:
3rd measured diameter:
Average diameter =
=
cm
Step 5: Calculate the % error between your experimental diameter value and the caliper measurement.
Show your work.
2
Name: ________________________
Period : ____________
Date:___________
Structure of the Molecule: Calculating its Size
Part 2: Calculating the Length of a Molecule of Oleic Acid
Purpose
In this experiment the diameter of a single molecule of oleic acid will be calculated.
Required Equipment and Supplies
tray
10-mL graduated cylinder
water
oleic acid solution (5mL oleic acid in 995 mL of ethanol)
eyedropper
calcium carbonate (chalk) dust
Discussion
In this experiment the length of a single molecule of oleic acid will be calculated! The procedure for measuring
the diameter of a molecule will be much the same as that of measuring the diameter of a BB in the previous
activity. The diameter is calculated by dividing the volume of a drop of oleic acid by the area of the monolayer
film of molecules that is formed on the surface of the water. The height of the monolayer is equal to the length
of the molecule .
Procedure
Step 1: Pour water into the tray to a depth of about 1 cm. Wait for the water to settle and then spread chalk
dust evenly over the surface of the water. This will ensure that the oleic acid film is visible. The dust layer
should be sufficient so the shape of the film will be seen easily, but not so thick as to hem in the oleic acid.
Step 2: Holding the eyedropper vertically, gently add a single drop of the oleic acid solution to the surface of
the water. Be sure that the drop does not include an air bubble. When the drop touches the water, the alcohol
in it will dissolve in the water, but the oleic acid will not. The oleic acid spreads out to form a nearly circular
patch on the water. Measure the diameter of the oleic acid patch in three places each 60 degrees from the
previous measurement. Compute the average diameter of the circular patch.
Trial 1 1st diameter:
Trial 2 1st diameter:
2nd diameter:
2nd diameter:
3rd diameter:
3rd diameter:
Average diameter =
Average diameter =
Average diameter of Trial 1 and Trial 2:
=
cm
Area = 𝜋𝑟 2 =
=
cm2
Step 3: Count the number of drops of solution needed to occupy 3 mL (or 3 cm3) of a 10 mL graduated
cylinder. Take care in making this measurement and read the bottom of the solution meniscus at eye level.
Number of drops in 3.0 mL of solution:
Average number of drops in 1 cm3 =
=
Divide 1 cm3 by the average number of drops in 1 cm3 to determine the volume of a single drop.
Volume of single drop =
=
3
cm3
Step 4: In actuality the volume of the oleic acid in the drop is much less than the volume of the drop itself. This
is because the concentration of oleic acid in the solution is only 5 mL of acid per liter (1000 mL) of solution.
5
Therefore, every cubic centimeter of the solution contains just 1000 cm3, or 0.005 cm3 of oleic acid.
Consequently, the volume of oleic acid in one drop is 0.0050 of the volume of one drop. Multiply the volume of
a single drop by 0.0050 to find the volume of oleic acid in the drop. This represents the volume of the layer of
oleic acid in the tray.
Volume of oleic acid =
=
cm3
Step 5: Calculate the height of the monolayer of oleic acid molecules by dividing the volume of oleic acid by
the area of the circle. This height is also the length of a single molecule. As we learned in Part A of this lab:
𝑉𝑜𝑙𝑢𝑚𝑒
𝐴𝑟𝑒𝑎
=
𝜋𝑟 2 𝑥 ℎ
𝜋𝑟 2
= 𝐻𝑒𝑖𝑔ℎ𝑡
Height of monolayer = length of molecule =
=
cm
Convert your value to nanometers (10-9):
=
nm
The size of an oleic acid molecule as obtained by this method is
good, but the actually shape and characteristics of the molecule
must be considered.
An oleic acid molecule is not spherical, but rather elongated like a hot dog. One
end is attracted to water (hydrophilic), and the other end points away from the
water surface (hydrophobic). The molecules actually stand up vertically. So the
calculated height is more accurately thought of as the length of an oleic acid
molecule.
Questions
1. What is meant by a monolayer?
2. Why is it necessary to dilute the oleic acid for this experiment?
3. Why use alcohol for the dilution?
4. The shape of oleic acid molecule is more like that of a cylinder than a sphere. Furthermore, one end is
attracted to water (hydrophilic) so that the molecule stands up on the surface of water. Assume an
oleic molecule is 10 times longer than it is wide. Calculate the approximate volume of one oleic acid
molecule. Show all work below.
4