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Stoichiometry/Demensional Analysis on a hammered sheet of gold

Submitted by Hikari on Wed, 09/03/2008 - 12:48

I am a Sophomore taking chemistry for the first time, and our teacher assigned us some homework problems. Even though it's for a completion grade, I want to try to figure out the right answers, but I need some hints on some of them, particularly this one:

"To the alchemists, gold was the material representation of purity. It is a very soft metal that can be hammered into extremely thin sheets. If a 2g piece of gold is hammered into a sheet whose area is 42.8 sq. ft., what is the average thickness of the sheet? (The density of gold is 19.32 g/cm^3.)"

Okay, so I know that volume = base * height; and I know that density = mass/volume... but the area of the sheet of gold is in feet and the unit for density involves centimeters. How do I do dimensional analysis when one unit is squared and the other is cubed?


 

Hikari wrote:

Okay, so I know that volume = base * height;

volume = area of base * height

the area of base is in squared units, times height, adds the third unit and you end up with length cubed.

From the mass (2g) and the density (19.32 g/cm3) find the volume in cm3

so now you know the volume and the area of the base, solve for height.

2.54 cm = 1 in and 12 in = 1 ft

cube both sides on both equalitites to find the conversions from cm3 -> in3 -> ft3

you can also convert the area to square inches and then to square cm and work everything in cm

 


Oh, I see. I guess I was just over-thinking it. Thank you!


Actually, I still don't think I'm getting the right answer. Please check my work.

v = d/m, so:

v = (2 g) / (19.32 g/cm^3) = 1.035196687 -> 1.035 cm^3

1.035 cm^3 * (2.54^3 in^3 / 1cm^3) * (1^3 ft^3 / 12^3 in^3) =

1.035 cm^3 * (16.39 in^3 / 1cm^3) * (1 ft^3/ 1728 in^3) = .0098169271 -> .0098 ft^3

v = b*h
h = v/b

h = .0098ft^3/42.8ft^2 = 2.29e-4 ft

I feel like I'm doing something wrong.


almost

V = m/d              you have d/m

however, your math is correct in that part because you placed the mass on the numerator

Thus v = 1.035 cm3 is correct

Hikari wrote:

1.035 cm^3 * (2.54^3 in^3 / 1cm^3) * (1^3 ft^3 / 12^3 in^3) =

that part is backwards it should be (13 in3 / 2.543 cm3)

When you correct for that part v = 3.6596 x 10-6 ft3

v = b*h
h = v/b

h = 3.6596 x 10-6 ft3/42.8ft2 = 8.71 x 10-8 ft

in inches = 1.05 x 10-6 thus about 1 micro inch very thin sheet indeed, but in order for 2g to cover over 42 ft2 it would have to be very thin.


Oops! I see now. Thank you so much for the help!