# Finding area of a gold leaf

1. Aug 26, 2007

### P.O.L.A.R

1. The problem statement, all variables and given/known data
Gold, which has a density of 19.32 g/cm^3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold, witha mass of 27.63g, is pressed into a leaf 1.000 µm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.500 µm, what is the length of the fiber?

2. Relevant equations
ρ=m/V (density)

3. The attempt at a solution
Im having trouble finding a direction to go in. I think converting to density is the way to go but I am having trouble using the density formula as well.

2. Aug 26, 2007

### rootX

But aren't you given the density: 19.32?
Hint: You need to know volume of a cylinder, and that other thing (cube ?) in order to solve this.

So, I guess both require two steps.

3. Aug 26, 2007

### P.O.L.A.R

Ok so my intial direction was right but the problem I am having now the area. It just doesnt make sense how I can take the density and find the area. The set-up is not making sense.

4. Aug 26, 2007

### rootX

first find the relationship between area and volume
and then between volume and density/mass

and you will get the area

5. Aug 27, 2007

### P.O.L.A.R

I got it. Thanks for the help the relationship is what I was confusing but all I had to do was divde the density by the mass to get rid of the grams and then convert the cm^3 to µm^3 and relate the area and volume as you said and then the volume with the area.

Thanks

6. Jan 27, 2009

### dhago07

How to solve this problem after finding volume using density and mass

7. Jan 27, 2009

### chemisttree

You need to find a relationship between area and volume for the shape you are considering. For gold leaf, assume that you are using the equation for 2 faces of a cube of material and for a rod assume the area is limited to that of a rod minus the ends.

8. Jan 27, 2009

### dhago07

But what kind of formula should i use for 2 face cube and also for the rod

Thanks for replaying back.

9. Jan 28, 2009

### chemisttree

Using the example of the gold leaf, assume you have a collection of finite elements, each of which is a cube of the same dimension as the thickness (w= 1.000 µm) of the foil. The area of each face of the cube will be w2 and you need two of 'em. Next, you need to understand that the total volume of the gold will be N*w3 where 'N' is equal to the number of elements of dimension w X w X w.

Can you get there from here?