Question 2.5: Calculating the Atomic Mass of an Element Problem Silver (Ag......
Calculating the Atomic Mass of an Element
Problem Silver (Ag; Z = 47) has 46 known isotopes, but only two occur naturally, ^{107}Ag and ^{109}Ag. Given the following data (Table 1), calculate the atomic mass of Ag:
Table 1
| Isotope | Mass (amu) | Abundance (%) |
| ^{107}Ag | 106.9051 | 51.84 |
| ^{109}Ag | 108.9048 | 48.16 |
Plan From the mass and abundance of the two Ag isotopes, we have to find the atomic mass of Ag (weighted average of the isotopic masses). We divide each percent abundance by 100 to get the fractional abundance and then multiply that by each isotopic mass to find the portion of the atomic mass contributed by each isotope. The sum of the isotopic portions is the atomic mass (Equation 2.3).
Solution Finding the fractional abundances:
Finding the portion of the atomic mass from each isotope:
= 106.90509 \text{amu} × 0.5184 = 55.42 \text{amu}
\text{Portion of atomic mass from } ^{107}\text{Ag} = 108.90476 \text{amu} × 0.4816 = 52.45 \text{amu}
Finding the atomic mass of silver:
Or, in one step using Equation 2.3:
= (106.90509 \text{amu})(0.5184) + (108.90476 \text{amu})(0.4816)
= 107.87 \text{amu}
Check The individual portions seem right: ∼100 amu × 0.50 = 50 amu. The portions should be almost the same because the two isotopic abundances are almost the same.
We rounded each portion to four significant figures because that is the number of significant figures in the abundance values. This is the correct atomic mass (to two decimal places); in the List of Elements (inside front cover), it is rounded to 107.9 amu.