How many g of caco3 are present in a sample if there are 4.52 x 1024 atoms of carbon in that sample?

mass of CaCO3 = 750.5 grams

Explanation:
One mole of any substance contains Avogadro's number of atoms.
This mean that "x" moles of a substance will contain x * Avogadro's number atoms.
We are given that the sample contains 4.52 * 10^24 carbon atoms. 
Therefore, we can get the number of moles of carbon in the sample as follows:
number of atoms = number of moles * Avogadro's number
Therefore:
number of moles = number of atoms / Avogadro's number
number of moles = (4.52 * 10^24) / (6.022 * 10^23) = 7.505 moles

Now, from the given compound, we can note that one mole of CaCO3 contains one mole of carbon.
This means that 7.505 moles of carbon will be present in 7.505 moles of CaCO3

Final step is to get the mass of 7.505 moles of CaCO3 as follows:
From the periodic table:
molar mass of calcium = 40 grams
molar mass of carbon = 12 grams
molar mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams/mole
This means that:
mass of 7.505 moles of CaCO3 = 7.505 * 100 = 750.5 grams

Hope this helps :)