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Basic Chemistry

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    The total number of atoms in 1.00 g of CaCO3 (MM = 100.0 g/mol) is:

    3. The attempt at a solution

    My solution: 1.00 / 100.0 = 0.01, then 0.01 x 6.022x10^23 = 6.022x10^21 atoms.

    However, the correct answer is 3.01 x 10^22. How was my approach wrong and did I miss any steps? Any help much appreciated.
     
  2. jcsd
  3. Oct 12, 2011 #2
    5 atoms in the molecule of CaCO3
     
  4. Oct 12, 2011 #3
    I don't think that's the conclusion I was suppose to arrive at... I'm unsure about the sort of calculations.
     
  5. Oct 13, 2011 #4

    Borek

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    Staff: Mentor

    Gabriels-horn is perfectly right - you are asked about total number of atoms, so far you have (correctly) calculated number of molecules.

    One molecule of CaCO3 is made of one Ca atom, one C atom and 3 O atoms - five atoms total.

    How many atoms (in total) in 3 molecules of CaCO3?

    How many atoms (in total) in 6.022x1021 molecules of CaCO3?
     
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