Density problem

Question

(A) Atoms are very small compared to objects on the macroscopic scale. The radius of a gold atom is 144 pm. What is this value in meters and in centimeters?

(b) The mass of a single gold atom is 3.27×10^-22 g. Suppose enough Au atoms were lined up like beads on a string to span a distance of 49.0 cm ( 19 inches). How many atoms would be required?

(C)What mass in grams of Au would be used?

(D)Could you weigh out this amount of gold using a typical laboratory balance?

(E) Taking the density of gold metal to be 19.3 g/cm^3, calculate the mass of metal needed to form a piece of Au wire with the same length as the distance in b, but with a diameter of 1.50 mm. Hint: The volume of a cylinder is a times its radius squared times its height.

(F) How many gold atoms does this represent?

Answer 1

A) 144 pm x 1 m / 1x10¹² pm = 1.44x10^-10 m = radius of 1 gold atom

1.44x10^-10 m x 1 cm / 0.01 m = 1.44x10^-8 cm = radius of 1 gold atom

B) From (A) above, we can find the diameter of the gold atom as 1.44x10^-8cm x 2 = 2.88x10^-8 cm

49.0 cm x 1 atom / 2.88x10^-8 cm = 1.70x10⁹ atoms required

C) 1.70x10⁹ atoms x 3.27x10^-22 g / atom = 6.29x10^-13 g

D) No. A typical lab balance (depending on type) might weigh 10^-3 g quantities, not 10^-13g.

E) We need to first find the volume of the wire (in cm³) using π r² h

π = 3.14

r = 1/2 x 1.50 mm = 0.75 mm x 1 cm / 10 mm = 0.075 cm

h = length of wire = 49.0 cm

V = (3.14)(0.075 cm)²(49.0 cm)

V = 865 cm³

Next, we use the density provided, to calculate the mass:

865 cm³ x 19.3 g / cm³ = 44.8 g

F) 44.8 g x 1 atom / 3.27x10^-22 g = 1.37x10²³ gold atoms