Perform the classic general chemistry density lab experiment online. Calculate foil thickness from length, width, and mass using the density of aluminum with instant conversions to microns, millimeters, and atomic layers.
Enter your rectangular foil dimensions and measured balance mass.
One of the quintessential experiments in university General Chemistry 1 and AP Chemistry laboratories is calculating the physical thickness of a thin sheet of aluminum foil. Because aluminum foil is far too thin to be measured directly with an ordinary metric ruler, scientists use indirect dimensional analysis utilizing mass and the known physical density of aluminum.
Aluminum foil is shaped as a very thin rectangular box (rectangular prism). Its three-dimensional volume ($V$) is:
By physical definition, density ($d$) equals mass ($m$) divided by volume ($V$):
Equating both volume expressions and solving for thickness ($t$):
Area = 15.2 cm × 10.5 cm = 159.6 cm²
Volume = 0.692 g ÷ 2.70 g/cm³ = 0.2563 cm³
Thickness = 0.2563 cm³ ÷ 159.6 cm² = 0.0016059 cm (1.61 × 10⁻³ cm)
| Commercial Foil Grade | Thickness in Microns (μm) | Thickness in Millimeters (mm) | Thickness in Mils (1/1000") | Common Kitchen Applications |
|---|---|---|---|---|
| Standard Household Foil | 14 to 18 μm | 0.014 to 0.018 mm | 0.55 to 0.70 mils | Wrapping sandwiches, covering leftovers, light baking. |
| Heavy Duty Foil | 20 to 25 μm | 0.020 to 0.025 mm | 0.80 to 1.00 mils | Roasting meats, lining baking pans, freezer storage. |
| Extra Heavy / Grill Foil | 30 to 38 μm | 0.030 to 0.038 mm | 1.20 to 1.50 mils | Outdoor barbecue grilling, high-heat smoking, heavy duty wrapping. |
(1.60 × 10⁻³) ÷ (2.86 × 10⁻⁸) ≈ 55,900 atoms thick.
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