Question

Consider a sample of calcium carbonate in the form of a cube measuring 2.205 in. on each edge.
Part A
If the sample has a density of 2.72 g/cm3 , how many oxygen atoms does it contain?

Answer 1

Density of the sample$=2.72\frac{g}{cm^3}$

Edge length$=2.205in=2.205\times2.54=5.6007cm$

So the volume of the cubic sample$=5.6007\times5.6007\times5.6007=175.68cm^3$

Hence the mass of the sample$=2.72\times175.68=477.85g$

Molar mass of $CaCO_3$ $=100\frac{g}{mol}$

Therefore number of moles of $CaCO_3$ in the given sample$\frac{477.85}{100}=4.7785mol$

In $1mol$ $CaCO_3$, number of $O$ atoms$=3\times N_A$

So in $4.7785mol$ $CaCO_3$, number of $O$ atoms$=4.7785\times3\times N_A=8.634\times10^{24}atoms$