Another recent minilab of ours strove to find the thickness of a piece of aluminum foil. I’ve concluded that the piece my partner and I used was about .046 millimeters, or 46 microns.
I figured that out by using the density of aluminum, the measured mass, the area of our piece, and by solving for the thickness. The density we used was 2.7g/cm3, my partner and I settled on a square that was 7cm by 7cm, though we ultimately had to decide it was 6.98cm by 6.95cm, which gave us an area of 48.511cm2, which through converting to mm and rounding was ultimately made 4,850 mm2. The mass we found the square to be was .603 grams.
If density= mass/volume, then v= m/D, and v= l*w*h, and l*w=4,851.1mm2, then 4,851.1mm2*h= .603g/(2.7g/cm2). Converting units a bit more, 4,851.1mm2*h= 603mm3/2.7. Therefore, h= 603mm3/(2.7*4,851.1mm2), h= 603mm/(2.7*4,851.1), and finally, H= .046037mm, which rounds to .046mm, as seen above in the first paragraph.
My result is reliable up to two significant digits, because our density of 2.7 had the least number of significant digits at two. We also calculated our masses and lengths to the best of our ability, so I would say that the calculation of my aluminum foil’s thickness is very accurate and reliable.