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Thickness In Millimeters Lab Report

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9.) Calculations h = height or thickness
D = M/V
V = M/D
L*W*H = M/D

Trial 1:
This calculation solves for the height (h), or thickness, of the aluminum foil.
(29.97 cm * 30.50 cm * h cm) = (3.82 g/2.70 cm2)
(914.085 cm2 * h cm) = (3.82 g/2.70 cm2) h cm = 0.00155 cm
The following calculations express the height (h)/thickness in millimeters and micrometers.
(0.00155 cm * (10 mm/1 cm)) = 0.0155 mm
(0.00155 cm * (10,000 μm/1 cm)) = 15.5 μm

Trial 2:
This calculation solves for the height (h), or thickness, of the aluminum foil.
(9.60 cm * 6.10 cm * h cm) = (0.23 g/2.70 cm2)
(58.56 cm2 * h cm) = (0.23 g/2.70 cm2) h cm = 0.0015 cm
The following calculations express the height (h)/thickness in millimeters and micrometers.
(0.0015 cm * (10 mm/1 …show more content…

The accepted density of gold is 19.32 g/cm3. In order to calculate the thickness of the gold foil, the following algebraic equations must be used:
M = Mass, D = Density, V = Volume, L = Length, W = Width, H = Height
M/D = V
V/L/W = H
Therefore, M/D/L/W = H
The calculations to determine the thickness of the sample of gold foil is as follows:
29.25 g / 19.32 g/cm3 / 22.41 cm / 20.33 cm = 0.003323 cm
This means that the sample of gold foil is 0.003323 cm thick. In order to convert the thickness into micrometers, the following conversion must be used:
0.003323 cm * 10,000 µm = 33.23 µm
Therefore, the thickness of the sample of gold foil is 33.23 µm.
11.) Error …show more content…

Three different pieces of aluminum foil were used for three different trials in order to determine the thickness. Data was then collected by using an electric balance for mass and a standard ruler for length and width. The formulas used were area = length * width, volume = length * width * height, and density = mass / volume. Thickness (height) was found simply by dividing mass by density, width, and length. The average of our three trials was not only calculated, but also found via the graph and slope.. A graph of area vs. mass was made with data points from our three trials. Using linear fit, the graph showed that the slope, which was then divided by the density. This calculation isolated the height in centimeters. The slope of the graph was 0.004181 g/cm2. This value shows the relationship between the area of the aluminum foil and its mass. The values plotted in this graph were used to calculate the thickness of the aluminum foil via an alternative method. Additionally, the graph had a correlation value of 1.00. This means that area and mass have a strong, direct relationship. According to pure calculations based upon the data alone, the thickness of the aluminum foil was 0.0016 cm. This is identical to the value of thickness provided by the aluminum foil company, Reynold’s. According to calculations using the slope,

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