The Avogadro constant, represents the number of carbon-12 atoms in exactly 12 g of pure carbon-12. the value of Avogadro constant is 6.0221421 푥 10^23. How many atoms of K-40 (Radioactive isotope) are present in 225 mL of whole milk containing 1.65 mg K/mL?

1. The Mole and Avogadro
Dr. K. Shahzad Baig
Memorial University of Newfoundland
(MUN)
Canada
Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario.
Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.

2. The mass of one mole of atoms of an element is called its molar mass, M.
The Avogadro constant, represents the number of carbon-12 atoms in exactly 12 g of
pure carbon-12. the value of Avogadro constant is
𝑁𝐴 = 6.0221421 𝑥 1023 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚𝑜𝑙−1
The Concept of the Mole and the Avogadro Constant
1 mol 12C = 6.02214 * 1023 12C atoms = 12.0000 g
1 mol 16O = 6.02214 * 1023 16O atoms = 15.9949 g

3. Consider the statement:
1 mol S = 6.022 x 1023 S atoms = 32.06 g S
Molar mass and the Avogadro constant are used
in a variety of calculations involving the mass,
amount (in moles), and number of atoms in a
sample of an element.

4. Problem statement
Convert 9500g of iron to number of atoms in the sample.
9500 𝑔 𝐹𝑒 𝑥
1 𝑚𝑜𝑙𝑒
55.8 𝑔
𝑥
6.02214 1023 𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙𝑒
= 1.0 𝑥 1026 𝑎𝑡𝑜𝑚𝑠
Problem statement
What is the mass of 0.250 moles of aluminum?
0.250 𝑚𝑜𝑙𝑒𝑠 𝑥
27.0 𝑔
1 𝑚𝑜𝑙
= 6.75 𝑔 𝐴𝑙

5. Problem statement
Find the mass, in cg, of 3.25 x 1021 atoms of lithium.
3.25 𝑥 1021 𝑎𝑡𝑜𝑚𝑠 𝑥
1 𝑚𝑜𝑙
6.022𝑥 1023
𝑥
6.9 𝑔
1 𝑚𝑜𝑙
𝑥
100 𝑐𝑔
1 𝑔
= 3.7 𝑐𝑔
Problem statement
How many atoms are in 10.0g of gold?
10 𝑔 𝑔𝑜𝑙𝑑 𝑥
1 𝑚𝑜𝑙
197.0 𝑔
𝑥
6.022 𝑥 1023 𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙
= 3.06 𝑥 1022 𝑎𝑡𝑜𝑚𝑠

6. EXAMPLE 2-8
Relating Number of Atoms, Amount in Moles, and Mass in Grams
In the sample of sulfur weighing 4.07 g,
(a) how many moles of sulfur are present, and
(b) what is the total number of sulfur atoms in the sample?
(b) The conversion mol S atoms S is carried out using the Avogadro constant as a
conversion factor
(a) For the conversion g S → mol S,
The result of this calculation should be stored without rounding it off because it is required
in part (b).

7. PRACTICE EXAMPLE A:
What is the mass of of Cu?
PRACTICE EXAMPLE B:
How many lead-206 atoms are present in a 22.6
g sample of lead metal?

8. EXAMPLE 2-9
Combining Several Factors in a Calculation Molar Mass, the Avogadro
Constant, Percent Abundances
Potassium-40 is one of the few naturally occurring radioactive isotopes of elements of
low atomic number. Its percent natural abundance among K isotopes is 0.012%. How
many atoms of K-40 are present in 225 mL of whole milk containing 1.65 mg K/mL?
First, we convert from mL milk to g K.

9. Next, we convert from g K to mol K
and then we convert from mol K to atoms K.
Finally, we convert from atoms K to atoms 40K.

10. PRACTICE EXAMPLE A:
How many Pb atoms are present in a small piece of lead with a volume of The
density of
PRACTICE EXAMPLE B:
Rhenium-187 is a radioactive isotope that can be used to determine the age of
meteorites. A 0.100 mg sample of Re contains of What is the percent abundance of
rhenium-187 in the sample?

11. Problem statement
How many Pb atoms are present in a small piece of lead with a volume of 1.068 𝑚𝐿
The density of Pb = 11.34 g / cm3
𝑑 = 𝑚/𝑣
𝑚𝑜𝑙𝑒 =
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑃𝑏 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑃𝑏 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑃𝑏
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑃𝑏 = 11.34
𝑔
𝑚𝐿
𝑥 1.068 𝑚𝐿
𝑃𝑏 𝑎𝑡𝑜𝑚𝑠 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑃𝑏 (𝑔)
207.2
𝑔
𝑚𝑜𝑙
𝑥
6.02 𝑥 1023 𝑃𝑏 𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙 𝑃𝑏

12. Problem statement
Rhenium has two isotopes, Ru 185 and Ru 187. Determine the % abundance of Ru-185.
P and (1-P) represent the two percent natural abundance. Mathematically:
[P x 185] + [(1-P) x 187] = 186
P = .373845
37.3845% Re-185 and 62.6155% Re-187