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sample lab report - chemistry

Date of Lab
Experiment 3A: The Thickness Of A Thin Aluminum Sheet
The purpose of this experiment is to ultimately find the thickness of the aluminum foil sheets by
using formulas such as, but are not limited to, V= L x W x H, and V= m / D. In addition, this
experiment also demonstrates the correct use of applying significant figures and exponential
notation. The experiment will be conducted by weighing the pieces of aluminum, measuring
their dimensions, and then conducting the calculations to figure out the thickness of the
aluminum foil.
4 pieces of rectangular aluminum foil (at least 10cm long on each side)
Centimetre Ruler
Dial balance
Please refer to the “ Experiment 3A: The Thickness Of A Thin Aluminum Sheet ” lab
worksheet for the procedure.
Data and Observation:
Qualitative Data:
Aluminum foil was very wrinkly and thus made it hard to measure and may be
The pieces of aluminum foil were folded to be weighed.
Quantitative Data:
Table 1: Quantitative Data for Aluminum Foil
Sheet Number
Length (cm)
Width (cm)
Mass (g)
23.0 ± 0.1cm
15.2 ± 0.1cm
1.40 ± 0.05g
11.2 ± 0.1cm
11.2 ± 0.1cm
0.55 ± 0.05g
10.1 ± 0.1cm
9.6 ± 0.1cm
0.38 ± 0.05g
30.0 ± 0.1cm
22.7 ± 0.1cm
2.84 ± 0.05g
Uncertainty is calculated by dividing the smallest unnumbered difference by 2.
1. For each of the pieces of aluminum foil, find the following:
a. Area (A)
b. Volume (V)
c. Thickness (T)
Table 2: Calculations for Aluminum Foil, Area, Volume, and Thickness
a. Area (A)
b. Volume (V)
c. Thickness (T)
350 cm2
0.52 cm3
1.5 x 10-3 cm
125 cm2
0.20 cm3
1.6 x 10-3 cm
97.0 cm2
0.14 cm3
1.4 x 10-3 cm
681 cm2
1.05 cm3
1.54 x 10-3 cm
All calculations are based on the fact that the density of aluminum foil is 2.70 g/cm3.
Sample Calculations: Area: 23.0cm x 15.2cm = 350cm2
Volume: 1.4g ÷ 2.70 g/cm3 = 0.52 cm3
Thickness: 0.52cm3 ÷ 350 cm2 = 1.5 x 10-3 cm
2. Omitted.
3. How precise were your answers? Recall the definition of precision.
The definition of precision is a reproducible measurement, and generally, the more significant
digits it has, the more precise the measurement is.
Given that the uncertainty of the density of the aluminum foil was not used, the results of the
thickness was fairly consistent, and therefore precise. In addition, the answers to the
calculations have many digits that had to be rounded due to significant figures. The large
amount of digits in each calculation results in a more precise answer.
4. If you had used a crude balance that allowed only one significant figure, how would
this have affected your results for
a. Area
b. Volume
c. Thickness
Area: No effect since it does not use a scale and uses a ruler to the nearest tenth of a number.
Volume and Mass: It would affect the volume and thickness and be less precise since the
answer is largely based on the mass of the aluminum foil, which requires the use of the crude
scale. It will have less sig figs, and less sig figs means less precise.
5. Could this method be used to determine the thickness of an oil spill? What information
would be needed?
The area would need to be known, as well as the volume. Then, to find the thickness, the
volume would need to be divided by the area.
6. A very thin layer of gold plating was placed on a metal tray that measured 25.22cm by
13.22cm. The gold plating increased the mass of the plate by 0.0512g. Calculate the
thickness of the plating. The density of gold is 19.32 g/cm3.
Area: 25.22cm x 13.22 cm = 333.4 cm2
Volume: 0.512g / 19.32 g/cm3 = 2.65 x10-3 cm3
Thickness: 2.65 x10-3 / 333.4 cm2 = 7.95 x 10-6 cm
7. By mistake, one litre (1000cm3) of oil was dumped into a swimming pool that measures
25.0m by 30.0m. The density of the oil was 0.750 g/cm3. How thick was the resulting oil
slick? Be careful with significant figures and exponential notation.
Area: 2500cm x 3000 cm = 7.5 x106 cm2
Volume: 1000mL / 0.750 g/cm3 = 1330 cm3
Thickness: 1330cm3 / 7.5 x105 cm2 = 1.773 x10-4cm
8. How might this method of finding thickness be used in finding the size of molecules?
Finding the mass of the molecule as well as the dimensions are measurable using instruments.
Using mathematical calculations and formulas used above, the thickness of the molecules can
be derived.
In conclusion, the purpose of the experiment was achieved, by finding the thickness of the
aluminum foil at a fairly consistent rate. In addition, it allowed the correct use of sig figs and
exponential notation. The thickness of the aluminum sheets 1, 2, 3, 4, were 1.40 ± 0.05g, 0.55
± 0.05g, 0.38 ± 0.05g, 2.84 ± 0.05g. Although the results were fairly consistent, there
were a couple of systematic errors. All the aluminum sheets were previously crumpled
which can result in an inaccurate measurement of dimensions. As well, the scale
seemed to be calibrated improperly, which could lead to potential errors in the mass and
the calculations that were based on the mass. If this experiment were to be conducted
again, the scales should be double checked if they are calibrated properly before using
them to weigh, as well new pieces of aluminum foil should be cut out to avoid
inaccuracy of dimensions.