The density of gold is 19.3 g/cm^3. How many atoms are in a cube of gold which measures 1.50 inches on each side?

Answer 1

To get the number of atoms, we must first find the number of moles and then apply Avogadro's number.

density (ρ) = m/V (mass divided by volume)

In this case, since we have a cube that is 1.50 inches on a side, we can calculate volume. I'll chose to switch the units to centimeters by applying the conversion factor 1 inch = 2.54 cm. (1.50 inches)(2.54 cm/inch) = 3.81 cm. The volume is (3.81)³ = 55.306 cm³.

Using ρ = m/V or m = ρV we get m = (19.3)(55.306) = 1067.4 g

Now, we look the atomic mass of gold from the periodic table and find it is 196.967 g/mol.

To calculate the number of moles divide the mass by the atomic mass so 1067.4/196.967 = 5.42 moles

To get the number of atoms, multiply by Avogadro's number so (5.42)(6.022 x 10²³) = 3.26 x 10²⁴ atoms