Question

How many atoms of oxygen are present in 300 grams of CaCO3?

Answer 1

the formula unit mass can be calculated as :- 

atomic mass of :- calcium + carbon + 3 X atomic mass of oxygen

= 40 +12+ 3 X 16
= 100 g.

thus, the molar mass of CaCO3 is i.e. , (6.022 X 10 to the power 23 particles) is 100 g.
so, 100 g of CaCO3 = 3 X 6.022 X 10(23) atoms of oxygen 
       1 g. of  CaCO3 = [(18.066 X 10 (23) / 100] atoms of oxygen 
     300 g. of CaCO3 = [(18.066 X 10(23) X 300) / 100] atoms of oxygen 
 = 54.198 X 10(23) atoms of oxygen 

Answer 2

1 mole contains 3 oxygen atoms
3 moles contain 9 oxygen
so answer is 9*6.023*10^23
                 =5.4207*10^24