You find the molar mass of calcium metal, it is listed as 40.1*g*mol^-140.1⋅g⋅mol−1. You should check your copy of the Periodic Table to see if I have got it right. This means that "Avogadro's number"Avogadro's number of calcium atoms, i.e. 6.022xx10^236.022×1023 individual calcium atoms have a mass of 40.1*g40.1⋅g. And thus we can find the number of calcium atoms in a lump of metal, simply by measuring the mass of the lump and doing a simple calculation. And of course, we can also find the number of calcium atoms given a mass, and a formula for a calcium-containing material.
So "Moles of calcium"Moles of calcium == (197*cancelg)/(40.1*cancelg*mol^-1)
=??mol. We get an answer in "moles", because dimensionally 1/(mol^-1)=1/(1/(mol))=mol as required. Note that an answer that uses N_A to represent the given number would be quite acceptable; of course you could multiply it out.
If I were you I would study the relevant section of your text that deals with this principle. The idea of equivalent mass, the use of mass to represent a NUMBER of combining particles, is fundamental to the study of chemistry, and should not require too much angst to incorporate.
How many sodium atoms (approx.) in 23*g of sodium metal?
