The Thickness of Aluminum Foil � Density Lab

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Density Lab - Honors
The Thickness of Aluminum Foil
The volume of a regular object is found by using the formula V = L x W x H, where L = length, W =
width, and H = height. Imagine that the regular object is a rectangular-shaped piece of foil. Then the formula
might be revised to V = L x W x T, where T = thickness of the foil. Going one step further, the area of the foil
can be expressed as A = L x W, so the original formula for volume can be restated as V = A x T. Since this
experiment involves finding the thickness, it would be better to rearrange the formula again. Dividing both
sides of the equation by A, the new equation is: T = V / A.

The next problem will be to find the volume and area of a piece of aluminum foil. Remember that
density is a property that is expressed as D = M / V, where M = mass, and V = volume. The density of
aluminum is known, and the mass of a piece of aluminum foil can be measured with a balance. The volume of
the aluminum can then be calculated by using the rearranged equation V = M / D.

Procedures:

1) Measure three rectangular pieces of aluminum foil (you may wish to cut them or use the squares that have
been previously prepared).

2) Using a centimeter ruler, carefully measure the longest length and longest width of each piece of foil.
Record the measurements on the data table. How precise can the measurements be, keeping in mind the use of
significant figures? Think carefully before recording the results.

3) Using a balance find the mass of each piece of aluminum foil. Record the mass on the data table. Again, be
careful to be as precise as possible.

Title:

Length                                                    Density       Volume       Thickness
Foil Piece                  Width (cm)     Area (cm2)     Mass (g)
(cm)                                                     (g/cm3)        (cm3)         (cm)

Foil #1                                                                  2.6989

Foil #2                                                                  2.6989

Foil #3                                                                  2.6989

Analysis

1) Calculate the area for each piece of foil. Show all calculations. Area = L x H.

2) Calculate the volume for each piece of foil. Show all calculations. Volume = Mass/Density.

3) Calculate the thickness of each piece of foil. Show all calculations. Thickness = Volume/Area.

Chemistry: Unit 1 – Matter                                                                        Page 1 of 2
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Conclusions: (To be completed on a separate piece of paper and attached)

Accuracy is the closeness of an experimental value to an accepted value. For example, the accepted
value for the length of a football field is 91.4 meters (100 yards). If an individual were to measure the field and
come close to 91.4 meters, then that measurement would be considered accurate. Precision is the closeness of
the repeated measurements to each other. Using the same example, if the individual measured the field three
times and each time found the length to be 89.0 meters, the measurements would be precise to the tenths, but
not accurate.

In this experiment, the accepted value for the thickness of aluminum foil is available from the instructor.
The closeness to these accepted values will determine the accuracy of the measurements. If more than one trial
is performed with the same piece of aluminum foil, then the precision of the measurements can be calculated.

1) Calculate your percent error (and your percent correctness) for this experiment and determine at least three
sources of error for each lab (this determines your accuracy).

2) How precise were the calculated values and your actual measurements?

3) A very thin layer of gold plating was placed on a metal tray that measured 25.22 cm by 13.22 cm. The gold
plating increased the mass of the plate by 0.0512 g. Calculate the thickness of the plating. The density of gold
is 19.32 g/cm3. Show all work.

4) By mistake, 1.00 quart of oil was dumped into a swimming pool that measures 25.0 m x 30.0 m. The density
of the oil was 0.750 g/cm3. How thick was the resulting oil slick? Some helpful conversions are that 1.06
quarts = 1.00 liter = 1000.00 cm3 and 100.0 cm = 1.00 m. Be careful with significant figures and exponential
notation. Show all work. Hint: Convert your volume of 1.00 quart into liters, then cm3, and convert your
meters into cm before you begin.

5) In the determination of the density of a rectangular metal bar, a student made the following measurements:
length, 8.53 cm; width, 2.4 cm; height, 1.0 cm; mass, 52.7064 g. Calculate the density of the metal to the
correct number of significant figures.

Extra Credit:
The following procedure was carried out to determine the volume of a flask. The flask was massed dry and
then filled with water. If the masses of the empty flask and filled flask were 56.12 g and 87.39 g, respectively,
and the density of water is 0.9976 g/cm3, calculate the volume of the flask in cubic centimeters.

Chemistry: Unit 1 – Matter                                                                         Page 2 of 2

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