Consider a sample of calcium carbonate in the form of a cube measuring 2.505 inches on each edge

If the sample has a density of 2.72 g/cm^3 , how many oxygen atoms does it contain?


Please help me. I'm lost and would appreciate it if someone could explain the answer to me.

(2.505inches x 2.54cm/inch)^3 = 257.59cm^3 = volume
257.59cm^3 x 2.72g/cm^3 = 700.64g CaCO3
700.64gCaCO3 / 100g/mole = 7.01moles
7.01moles CaCO3 x (3 O / 1 CaCO3) x 6.022x10^23atoms/mole = 1.26x10^25atoms
or
%O in CaCO3 = 48g / 100g x 100% = 48%
700.64gCaCO3 x 0.48 = 336.31g O
336.31gO / 16g/mole x 6.022x10^23 atoms/mole = 1.26x10^25atoms