Why isn’t the average atomic mass of silver given on the periodic table simply 108?
a. Calculate the average mass in grams of an individual atom of silver.
b. Calculate the mass in grams of 10 trillion silver atoms.
c. Calculate the mass in grams of 5.00 × 1045 silver atoms.
The correct answer and explanation is :
Explanation:
The average atomic mass of silver (Ag) is not exactly 108 amu because silver exists as a mixture of two naturally occurring isotopes: Ag-107 and Ag-109. These isotopes have slightly different atomic masses and exist in different percent abundances. The weighted average of these isotopes, based on their relative natural abundances, results in the reported atomic mass of 107.87 amu, rather than precisely 108 amu.
The average atomic mass is calculated using the formula:
[
\text{Average atomic mass} = (\% \text{ of isotope 1} \times \text{mass of isotope 1}) + (\% \text{ of isotope 2} \times \text{mass of isotope 2})
]
For silver:
[
(51.84\% \times 106.905 amu) + (48.16\% \times 108.905 amu) = 107.87 \text{ amu}
]
This explains why the periodic table lists 107.87 amu rather than 108.
Calculations:
(a) Calculate the average mass in grams of an individual silver atom.
1 atomic mass unit (amu) = 1.66054 × 10⁻²⁴ g
Silver’s atomic mass = 107.87 amu
[
\text{Mass of one Ag atom} = 107.87 \times 1.66054 \times 10^{-24} \text{ g}
]
[
= 1.791 \times 10^{-22} \text{ g}
]
(b) Calculate the mass in grams of 10 trillion (10¹³) silver atoms.
[
\text{Mass} = (1.791 \times 10^{-22} \text{ g}) \times (10^{13})
]
[
= 1.791 \times 10^{-9} \text{ g} \text{ or } 1.791 \text{ nanograms}
]
(c) Calculate the mass in grams of 5.00 × 10⁴⁵ silver atoms.
[
\text{Mass} = (1.791 \times 10^{-22} \text{ g}) \times (5.00 \times 10^{45})
]
[
= 8.955 \times 10^{23} \text{ g}
]
[
= 1.49 \times 10^{22} \text{ moles of silver} \text{ (since 1 mole = 107.87 g)}
]
Thus, the mass of 5.00 × 10⁴⁵ silver atoms is 8.955 × 10²³ g or about 895.5 quintillion grams.