Name__________________________
Section_________Date____________
Report for Experiment 11
Aluminum Atoms
Prelaboratory Questions
1.
What measurements will you need in order to determine the thickness of a piece of aluminum
foil?
Length, width and mass of the foil are needed as well as the density of
aluminum
Data/Observations
Data Table
Mass of aluminum block
0.84
g
Volume of water in graduated cylinder
81.5
mL
Volume of water with aluminum block
82.0
mL
Aluminum foil
Width
17.8
cm
Length
18.8
cm
Mass of aluminum foil
2.09
g
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Houghton Mifflin Company.
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Report for Experiment 11
Name__________________
Analysis and Conclusions
1.
Calculate the volume of the aluminum block from the apparent change in the volume of the
water in the cylinder.
82.0 mL – 81.5 mL = 0.5 mL = 0.5 cm3
2.
Since both the aluminum block and the aluminum foil are pure elemental aluminum, we
would expect the ratio of the mass to the volume to be the same for both. That is:
mass of block
volume of block
=
mass of foil
volume of foil
Use this relationship to find the volume of the aluminum foil.
0.84 g
0.5 cm3
=
volume of foil
2.09 g
volume of foil
= 1.24 cm3
Note that this number should only have one significant figure, but we are
including more to be consistent with the dimension measurements
3.
Calculate the thickness of the aluminum foil. (Hint: think about how you would calculate
the volume of a box from its measurement. Think of the piece of aluminum foil as a very
thin box.)
Volume
= l x w x h
1.24 cm3 = 17.8 cm x 18.8 cm x h
h = 3.72 x 10-3 cm
4.
One aluminum atom has a diameter of 0.000000025 cm. How many atoms thick is the
aluminum foil?
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Report for Experiment 11
Name__________________
3.72 x 10-3 cm /(2.5 x 10-8 cm/atom )= 1.5 x 105 atoms
5.
What are the possible sources for error in your experiment?
Answers vary.
Something Extra
Look up the diameters for lithium, sodium, potassium and cesium atoms. What is the relationship
between the atomic number of the element and the diameter of its atoms?
Diameter increases with atomic number.
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Houghton Mifflin Company.
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