The Thickness of Aluminum Foil Mass of Al Foil: 0.146 g Length of Al Foil: 5.99 cm Width of Al Foil: [missing value] Data Analysis Use your measured mass and volume of the pellets (in Table I) to calculate the density of aluminum in g/cm^3. Show your work, and report your answer with the correct number of significant figures. Look up the true density of aluminum on the website "www.chemicool.com". Use this to calculate the percent error in your experimentally determined density value. Show your work. Use your measurements for the aluminum foil (in Table 2) along with the true density of aluminum to calculate the foil thickness in cm. Show your work and report your answer in scientific notation. Consider the foil to be a very flat rectangular box, where the volume of the foil is given by the formula: Volume of foil = length x width x thickness.
The Correct Answer and Explanation is:
Given Data:
- Mass of aluminum foil = 0.146 g
- Length of aluminum foil = 5.99 cm
- Width of aluminum foil = Missing (denoted as W)
- True density of aluminum (from www.chemicool.com) = 2.70 g/cm³
Step 1: Rearranging the volume formula to find thickness
The volume of a rectangular solid is given by:Volume=Length×Width×Thickness\text{Volume} = \text{Length} \times \text{Width} \times \text{Thickness}Volume=Length×Width×Thickness
The volume can also be expressed using mass and density:Volume=MassDensity\text{Volume} = \frac{\text{Mass}}{\text{Density}}Volume=DensityMass
Equating the two expressions for volume:MassDensity=Length×Width×Thickness\frac{\text{Mass}}{\text{Density}} = \text{Length} \times \text{Width} \times \text{Thickness}DensityMass=Length×Width×Thickness
Solving for thickness:Thickness=MassDensity×Length×Width\text{Thickness} = \frac{\text{Mass}}{\text{Density} \times \text{Length} \times \text{Width}}Thickness=Density×Length×WidthMass
Substitute known values:Thickness=0.146 g2.70 g/cm3×5.99 cm×W\text{Thickness} = \frac{0.146 \text{ g}}{2.70 \text{ g/cm}^3 \times 5.99 \text{ cm} \times W}Thickness=2.70 g/cm3×5.99 cm×W0.146 gThickness=0.14616.1733×W cm\text{Thickness} = \frac{0.146}{16.1733 \times W} \text{ cm}Thickness=16.1733×W0.146 cm
This is the expression for thickness in terms of the missing width. Without the width value, a final numerical answer cannot be determined.
Step 2: Calculating Percent Error (using hypothetical measured density)
Assuming the experimentally measured density (from pellets) was determined to be 2.59 g/cm³, the percent error is:Percent Error=(∣Measured−True∣True)×100\text{Percent Error} = \left( \frac{|\text{Measured} – \text{True}|}{\text{True}} \right) \times 100Percent Error=(True∣Measured−True∣)×100Percent Error=(∣2.59−2.70∣2.70)×100=(0.112.70)×100≈4.07%\text{Percent Error} = \left( \frac{|2.59 – 2.70|}{2.70} \right) \times 100 = \left( \frac{0.11}{2.70} \right) \times 100 \approx 4.07\%Percent Error=(2.70∣2.59−2.70∣)×100=(2.700.11)×100≈4.07%
Summary (Scientific Notation and Significant Figures)
The thickness of the foil in terms of width:Thickness=0.14616.1733×W cm\text{Thickness} = \frac{0.146}{16.1733 \times W} \text{ cm}Thickness=16.1733×W0.146 cm
The final thickness will be in scientific notation once the width is known. For instance, if the width was 10.0 cm, then:Thickness=0.146161.733=9.03×10−4 cm\text{Thickness} = \frac{0.146}{161.733} = 9.03 \times 10^{-4} \text{ cm}Thickness=161.7330.146=9.03×10−4 cm
This method treats aluminum foil as a thin rectangular solid and applies dimensional and density-based reasoning for volume calculation, expressing the final result to the correct significant figures.
