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# How many atoms of chalk in my name?

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## Ashley Palmer

on 28 April 2015

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#### Transcript of How many atoms of chalk in my name?

Procedure
First- Mass a piece of chalk
Second- Write names with chalk
Third- Remass chalk
Fourth- Subtract the second mass of chalk from the first mass of chalk
Fifth- Make some calculations!
Calculations:
1. Percent composition of calcium carbonate.
2. Percent (out of one) multiplied by grams of chalk used
3. Converting grams to moles
4. Converting moles to molecules
Data collected
Mass of unused chalk- 9.17 grams
What was written- Ashley Palmer Domenica Wardle
Mass after writing- 9.07 grams
0.10 grams used
Calcium Carbonate
Formula- CaCO
Conclusion
The first thing we did when starting the experiment was mass the original unused chalk. We then wrote our names on a chalkboard and re-massed the piece of chalk we used. After that we calculated the percent composition of the CaCO4 and each element. Then we divided that by 100 and multiplied the number by the total mass of chalk we used. Finally, we converted grams to moles and then to atoms.
Purpose
The purpose of this experiment is to find out the number of atoms of calcium, carbon, and oxygen it takes to write your name with chalk, with the assumption that the piece of chalk is 100% calcium carbonate.
How many atoms of chalk in my name?
3
Mass of calcium- 40.08 g/mol
Mass of carbon- 12.01 g/mol
Mass of oxygen x3- 48.00 g/mol
Total mass of calcium carbonate:
100.09 g/mol
Percent composition

CaCO3= 100.09 grams
% times grams of chalk used
grams of chalk used- 0.10 grams
Grams ---> Moles
Calcium= 40.08 g/mol
Carbon= 12.01 g/mol
Oxygen= 16.00 g/mol
Moles ---> Molecules
6.014x 10 atoms of Calcium
6.015x 10 atoms of Carbon
1.805x 10 atoms of Oxygen
(out of one)
Calcium- 40.08 g/100.09 g
Carbon- 12.01 g/100.09 g
Oxygen- 48.00 g/100.09 g
= 0.4004 % Ca
= 0.1200 % C
= 0.4796 % O
0.10 g x 0.4004 % Ca
0.10 g x 0.1200 % C
0.10 g x 0.4796 % O
= 0.04004g Ca
= 0.01200g C
= 0.04796g O
0.04004g/40.08g Ca
0.01200g/ 12.01g C
0.04796g/ 16.00g O
= 0.0009990 mol Ca
=0.0009992 mol C
= 0.002998 mol O
23
0.0009990mol x 6.02x10
0.0009992mol x 6.02x10
0.002998mol x 6.02x10
23
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