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1. In Part A, how would the density calculations change if the volume of the container used was not accurate? For example, what if the volume of the container was smaller than was assumed?
2. Considering mass, volume, and density, which of these variables changed between the two different trials of the film canister (or prescription bottle) in Part A? Which variables were held constant?
Answer:
(Score for Question 3: _ of 2 points)
3. Considering mass, volume, and density, which of these variables changed between the two wooden blocks of different sizes in Part B? Which variables were held constant?
Answer:
(Score for Question 4: _ of 2 points)
4. Imagine that you have a piece of foam. Suppose you squeezed it hard enough to change its shape. Briefly describe how this compression would affect its mass, volume, and density.
Answer:
(Score for Question 5: _ of 2 points)
5. A golf ball and a ping pong ball are approximately the same volume. However, a golf ball has much more mass. Which ball is denser? Explain your answer.
Answer:
(Score for Question 6: _ of 4 points)
6. How does changing the mass of a substance affect the density if the volume is kept the same? How does changing the volume of a substance affect the density if the volume is kept the same? If you prefer, you can use specific numbers and calculations to show the relationship instead.
Answer:
(Score for Question 7: _ of 3 points)
7. A cylindrical can has a mass of 100,000 grams. The radius of the can is 20 centimeters. The height of the can is 50 centimeters. Calculate the volume and density for the cylinder.
Answer:
(Score for Question 8: _ of 2 points)
8. Describe the relationship between mass, volume, and density.
Answer:
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9. Sketch a graph of the relationship between mass, volume, and density.
Answer:
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10. A baseball has a mass of 150 grams. The radius of the ball is 3.6 centimeters. Calculate the volume and density for the ball.
Answer:

The uncertainty in the volume must be determined by error propagation. Mass, length, and diameter measurements contribute to the overall uncertainty.

Volume by pycnometry

Pycnometry is a technique that uses the density relationship between volume and mass, and the vessel used is called a pycnometer .

To perform pycnometry measurements, the mass of the cylinder and the mass of a flask filled with water to a mark (A, Fig. 3) are recorded. The cylinder is then inserted into the flask. Water is displaced when the cylinder is inserted. The volume of water displaced is removed by pipet, thereby restoring the water level to the mark (B). The combined mass of the flask, remaining water, and cylinder is then measured.

Figure 3

The sums of the masses before and after are equal. The massA, the massB, and the masscylinder were all measured on the balance. There is only one unknown in the equation - the mass of the displaced water.

( 4a )

massA + masscylinder = massB + massdisplaced water

massdisplaced water = massA + masscylinder − massB

The volume of water removed is equal to the volume of the cylinder. Masswater can be converted to volume using the density of water.

( 4b )

Vdisplacedwater = Vcylinder = massdisplaced water / densitywater

The density of the cylinder is calculated using mcyl/Vcyl.

The uncertainty calculation requires a few steps and assumptions. The volume of the cylinder was equal to the volume of the water. Vwater was based on the three mass measurements - the mass of the cylinder, of A, and of B.

The uncertainty in masscylinder comes from the balance reading.

The uncertainty associated with massA and massB depends on your ability to precisely adjust the level of the water to the mark at the exactly same place every time (calibration). By repeatedly filling the flask to the mark and taking the mass readings, the average mass of A and the standard deviation (the fluctuation in the mass due to variations in the exact liquid level) can be found.

( 4c )

mA

=

mA,trial1 + mA, trial2 + mA, trial3 +

# trials

( 4d )

σmA = ±

mA

-mA,trial1

2+

mA

-mA,trial2

2+

# trials-1

Assume the uncertainty in the mass of both A and B is the same: mA ± σmA; mB ± σmA.

The uncertainty in the mass of water displaced is determined by error propagation:

( 4e )

σmwater = σmA + σmB + σmcyl = σmA + σmA + σmcyl

The density of water at room temperature is known quite precisely and is assumed to contribute negligible error (see table at the end of the lab), so dividing σm, water by the density of water to give σV, water is adequate. Since σV, water = σV, cyl, the uncertainty in the density can be determined.

( 4f )

σρ = ± ρ

σV

V

+

σm

m

You will use pycnometry in parts 4 and 5 to determine the volume and/or density of a hollow cylinder and of a mixed cylinder.

Volume of a void inside a hollow cylinder

A hollow cylinder has an empty space inside.

Figure 4

The volume of the cylinder is comprised of the volume of metal and the volume of the void inside.

( 5a )

Vcyl = Vmetal + Vvoid → Vvoid = Vcyl - Vmetal

Vcyl is determined by pyncometry. The volume occupied by the metal can be determined using the mass of the cylinder (which is due to only the metal, not the void) and the density of the metal, which was determined previously in the lab (either Al or brass, depending on the cylinder). Use the value for density that is closest to the literature values - 2.70 g/cm3 for Al; between 8 and 9 g/cm3 for brass.

( 5b )

Vmetal =

mcyl

ρmetal

No error propagation is required

Percent composition of a mixed cylinder

The total mass of the cylinder, mcyl, is the sum of the mass of Al and brass (mAl + mbrass). In terms of fractional composition, this would be Xmcyl and (1 - X)mcyl, respectively, where X is the Al fraction and (1-X) is the brass fraction (the remainder).

The cylinder volume is determined by pycnometry and is the sum of the volumes of the two metals:

( 6a )

Vcyl = VAl + Vbrass

Replace each volume by its mass divided by its density using V=m/ρ:

( 6b )

Vcyl =

mAl

ρAl

+

mbrass

ρmass

Replace the masses by the equivalent expressions in terms of X and mcyl:

( 6c )

Vcyl =

X mcyl

ρAl

+

(1−X)mcyl

ρbrass

Divide through by mcyl and replace Vcyl/mcyl with 1/ρcyl:

( 6d )

1

ρcyl

=

X

ρAl

+

(1 − X)

ρbrass

Collect terms on the right-hand side that contain X:

( 6e )

1

ρcyl

= X

1

ρAl

−

1

ρbrass

+

1

ρbrass