Answer:
A) Sample B has more calcium carbonate molecules
Explanation:
M = Molar mass of calcium carbonate = 100.0869 g/mol
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}\ \text{mol}^{-1}[/tex]
For the 4.12 g sample
Moles of a substance is given by
[tex]n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}[/tex]
Number of molecules is given by
[tex]nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}[/tex]
For the 19.37 g sample
[tex]n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}[/tex]
Number of molecules is given by
[tex]nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}[/tex]
[tex]1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}[/tex]
So, sample B has more calcium carbonate molecules.
The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.