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

  • Thread starter Thread starter domtar
  • Start date Start date
  • Tags Tags
    Atoms
Click For Summary

Discussion Overview

The discussion revolves around calculating the total number of atoms in 1.00 g of calcium carbonate (CaCO3), focusing on the participant's approach and identifying potential errors in their calculations. The context is primarily homework-related, involving stoichiometry and molecular composition.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant calculated the number of molecules in 1.00 g of CaCO3 as 0.01 moles and then multiplied by Avogadro's number, arriving at 6.022x10^21 atoms, but questioned the correctness of this approach.
  • Another participant noted that there are 5 atoms in one molecule of CaCO3, comprising 1 calcium atom, 1 carbon atom, and 3 oxygen atoms.
  • A third participant expressed uncertainty about the calculations and the conclusions drawn from them.
  • A later reply suggested recalculating the total number of atoms based on the correct number of molecules and the composition of CaCO3.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as there are differing views on the calculations and the interpretation of the problem. Some participants clarify the composition of CaCO3, while others express confusion about the calculations.

Contextual Notes

The initial calculation focused on the number of molecules rather than the total number of atoms, which led to a misunderstanding of the problem. There are unresolved steps regarding how to correctly calculate the total number of atoms from the number of molecules.

Who May Find This Useful

This discussion may be useful for students working on stoichiometry problems, particularly those involving molecular composition and conversions between moles and atoms.

domtar
Messages
6
Reaction score
0

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.
 
Physics news on Phys.org
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?
 

Similar threads

Calculating Age of Native American Campfire Using Radioactive Decay of CaCO3
  • · Replies 1 ·
Replies
1
Views
3K
Chemistry Decomposition of C5H6O3 equilibrium
  • · Replies 1 ·
Replies
1
Views
3K
Expected concentration of fluoride released from PFAA
  • · Replies 3 ·
Replies
3
Views
2K
Calculate Total Water Hardness (ppm of Equiv. CaCO3)
  • · Replies 1 ·
Replies
1
Views
15K
Titration Troubles: Where Did We Go Wrong?
  • · Replies 5 ·
Replies
5
Views
3K
Finding a rock porosity change after an acid mix treatment
  • · Replies 1 ·
Replies
1
Views
2K
Chemistry Calculating the volume of individual solution(s) [Mole/Atoms Concept]
  • · Replies 10 ·
Replies
10
Views
2K
Chemistry Where Did the 400g in Enthalpy Change of Neutralization Come From?
  • · Replies 3 ·
Replies
3
Views
3K
How do I calculate the density of calcium from atomic radius?
  • · Replies 11 ·
Replies
11
Views
17K
Calculating Charge on 1 g N3- Ions: Where Did I Go Wrong?"
  • · Replies 6 ·
Replies
6
Views
11K