How many atoms in an #197*g# mass of calcium metal?

1 Answer
Jan 22, 2017

Answer:

Approx. #5xxN_A#, where #N_A# is #"Avogadro's number"#.

Explanation:

You find the molar mass of calcium metal, it is listed as #40.1*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"# of calcium atoms, i.e. #6.022xx10^23# individual calcium atoms have a mass of #40.1*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"# #=# #(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.

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