Question

Consider a sample of calcium carbonate in the form of a cube measuring 2.605 in. on each edge.If the sample has a density of 2.70 g/cm3 , how many oxygen atoms does it contain?

Answer 1

Volume of cube is V = a^3

a is edge.

Volume of cube =( 2.605 in)^3

= 17.7 in^3

Now concer this inch to cm

1 in = 2.54 cm

1 in^3 = 16.4 cm^3

17.7 in^3 × ( 16.4 cm^3 / 1 in^3) = 290.28 cm^3

Mass of CaCO3:

290.28 cm^3 × ( 2.70 g / cm^3)

= 783.7 g CaCO3

Number of moles of oxygen:

783.7 g CaCO3 × ( 1 mol CaCO3 / 100 g CaCO3)

= 7.837 mol CaCO3 × ( 3 mol O / 1 mol CaCO3)

= 23.51 mol O

Number of oxygen atoms:

23.51 mol O × ( 6.02×10^23 O atom / 1 mol O)

= 141.5 × 10^23 oxygen atoms

= 1.42×10^25 oxygen atoms.