Gold can be hammered into extremely thin sheets called gold leaf. If a 200. mg piece of gold is hammered . . .

Question

into a sheet measuring 2.4 × 1.0 ft, what is the average thickness of the sheet in meters? How might the thickness be expressed without exponential notation, using an appropriate metric prefix.

(thickness = volume/area; density of gold = 19.32 g/cm3)

Okay, I know the answer already, it's 46 nm. But I have no idea how to get to the answer. I've looked at other people's work and I still can't make sense of it. Someone please help!!!

Answer 1

First, to make things a little easier later on, convert the imperial units to metric:

length = 2.4 feet

length = 2.4 * 12 * 2.54

length = 73.152 cm

breadth = 1.0 feet

breadth = 1.0 * 12 * 2.54

breadth = 30.48 cm

Now calculate the volume of the sheet:

volume = length * breadth * height

volume = 73.152 * 30.48 * h

volume = 2230 h

And since:

density = mass / volume

19.32 = 0.2 / 2230h

2230h * 19.32 = 0.2

43077h = 0.2

h = 0.000004643 cm

Converting to nanometres and to two significant figures:

h = 46 nm

Edit: pisgahchemist, I disagree, and was in fact taught to only include units in the final answer (in various subjects including maths, physics and chemistry). Including units in equations only serves to complicate matters. The units of the values used in the equations are already known, and the units of the answer are included for clarity, anything beyond that is unnecessary, and undesirable, fluff.

Furthermore, it is my opinion that you would lose marks for giving an answer with more precision than is able. The dimentions of the sheet were only given to two significant figures, therefore it is not possible to know the average thickness to an accuracy of more than two significant figures, and in your answer you gave three.

Also, that would be neither the minimum, nor maximum, number of gold atoms for the thickness, but the average, and only assuming that they were all neatly stacked upon each other, which of course they wouldn't be.

Answer 2

Density = 19.32g/cm^3 =19320mg/cm^3

Density = Mass/Volume, That implies Volume = Mass/Density

Now mass is 200 mg....

Volume = 200/19320 cm^3

Area = 2.4 X 1.0 feet

But, 1 Feet = 30.48 cm. Hence,

Area = 73.152cm X 30.48cm^2

        = 2229.67296 cm^2 ~ 2230cm^2

Thickness = Volume/Area cm

                = 200/(19320 X 2230) cm

                = 200/(43083600) cm

                = 0.00000464360 cm

                ~ 46 X 10^-7 cm

                = 46 X 10^-9 m

                = 46 nm

Here, you get the answer like that...  

Answer 3

Think about what relates the volume and mass of a substance.  That would be "density".  Density is the ratio of mass to volume.

D = m / V

It even gives you the density of gold in the problem.  Major hint.  Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...

V = L x W x H

Volume = Length x Width x Height

start by converting 200.0 mg into grams

1000 mg = 1 g

200. mg x (1 g / 10^3 mg) = 0.200 g

V = m / D

V = 0.200 g / (19.32 g/cm^3)

V = 0.01035 cm^3

Convert 2.4 ft and 1 ft to cm

2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm

1 ft = 30.48 cm

Compute the height (thickness)

V = LxWxH

H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm

H = 4.64 x 10^-6 cm

Convert to nanometers

4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm

Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold.  This would assume that the gold atoms are all in a row.  This would give the minimum number of gold atoms.

Atomic radius gold = 174 pm

Diameter = 348 pm

46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold

======== Follow up ========

As a chemistry teacher, I know that your teacher is going to want you to put the correct units with EVERY number in the problem.  For instance Lavalamp, while getting the correct answer, failed to put in most of the units.  He would have received only partial credit for his solution.