Finding the Atoms in 1.00 g of CaCO3: Where Did I Go Wrong?

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To find the total number of atoms in 1.00 g of CaCO3, the initial calculation correctly determines the number of molecules as 6.022 x 10^21. However, since each molecule of CaCO3 contains five atoms (one calcium, one carbon, and three oxygen), the total number of atoms is actually 3.01 x 10^22. The error was in only calculating the number of molecules rather than multiplying by the number of atoms per molecule. Understanding the composition of CaCO3 is crucial for accurate calculations. The discussion emphasizes the importance of recognizing the total atom count in molecular compounds.
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Homework Statement



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

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.
 
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5 atoms in the molecule of CaCO3
 
I don't think that's the conclusion I was suppose to arrive at... I'm unsure about the sort of calculations.
 
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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