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

[domtar]

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.

 

[gabriels-horn]

5 atoms in the molecule of CaCO3

 

[domtar]

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

 

[Borek]

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?

 

[DeeNos]

Your approach was not entirely wrong, but it seems that you may have missed a step in your calculation. Let's break it down:

1.00 g of CaCO3 is equivalent to 0.01 moles of CaCO3 (using the molar mass of 100.0 g/mol).

Now, 1 mole of any substance contains 6.022x10^23 particles (atoms, molecules, etc.), so 0.01 moles of CaCO3 would contain 0.01 x 6.022x10^23 = 6.022x10^21 particles.

However, CaCO3 is a compound made up of multiple atoms, specifically one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O). So, in order to find the total number of atoms in 0.01 moles of CaCO3, we need to multiply our previous answer by the number of atoms in each molecule of CaCO3.

This means that the total number of atoms in 0.01 moles of CaCO3 is 6.022x10^21 x (1 Ca atom + 1 C atom + 3 O atoms) = 6.022x10^21 x 5 = 3.011x10^22 atoms.

So, your initial approach was correct, but you just forgot to account for the fact that CaCO3 is a compound made up of multiple atoms. Remember to always double check your calculations and make sure you are accounting for all factors in a problem.