Key Concepts
- The relative atomic mass, or atomic weight, of an element is the weighted average of the masses of the isotopes in the naturally occurring element relative to the mass of an atom of the carbon-12 isotope which is taken to be exactly 12.
- For an element E with the naturally occurring isotopes aE, bE, cE, and with the respective abundances of A%, B%, C% etc,
the relative atomic mass (r.a.m.) = (A/100 x a) + (B/100 x b) + (C/100 x c)
- Given the relative atomic mass (r.a.m.) of an element and the mass of each of its isotopes, the relative abundance of each isotope can be calculated:
let y/100 = abundance of isotope 1
and (100 - y)/100 = abundance of isotope 2
then, r.a.m = [y/100 x mass isotope 1] + [(100 - y)/100 x mass isotope 2]
and solve for y
The Abundance of Some Naturally Occurring Isotopes
| Element |
Isotopes |
Mass Number |
Relative Abundance (%) |
|---|
| hydrogen |
1H |
1 |
99.99 |
|---|
| 2H |
2 |
0.01 |
| lithium |
6Li |
6 |
7.59 |
|---|
| 7Li |
7 |
92.41 |
| boron |
10B |
10 |
19.6 |
|---|
| 11B |
11 |
80.4 |
| carbon |
12C |
12 |
98.89 |
|---|
| 13C |
13 |
1.11 |
| nitrogen |
14N |
14 |
99.63 |
|---|
| 15N |
15 |
0.37 |
| oxygen |
16O |
16 |
99.76 |
|---|
| 17O |
17 |
0.04 |
| 18O |
18 |
0.20 |
| magnesium |
24Mg |
24 |
78.99 |
|---|
| 25Mg |
25 |
10.00 |
| 26Mg |
26 |
11.01 |
| silicon |
28Si |
28 |
92.23 |
|---|
| 29Si |
29
| 4.68 |
| 30Si |
30
| 3.09 |
| chlorine |
35Cl |
35 |
75.78 |
|---|
| 37Cl |
37 |
24.22 |
| argon |
36Ar |
36 |
0.34 |
|---|
| 38Ar |
38 |
0.06 |
| 40Ar |
40 |
99.60 |
| copper |
63Cu |
63 |
69.17 |
|---|
| 65Cu |
65 |
30.83 |
| silver |
107Ag |
107 |
51.84 |
|---|
| 109Ag |
109 |
48.16 |
| barium |
130Ba |
130 |
0.11 |
|---|
| 132Ba |
132 |
0.10 |
| 134Ba |
134 |
2.42 |
| 135Ba |
135 |
6.59 |
| 136Ba |
136 |
7.85 |
| 137Ba |
137 |
11.23 |
| 138Ba |
138 |
71.70 |
| lead |
204Pb |
204 |
1.5 |
|---|
| 206Pb |
206 |
23.6 |
| 207Pb |
207 |
22.6 |
| 208Pb |
208 |
52.3 |
| uranium |
234U |
234 |
0.01 |
|---|
| 235U |
235 |
0.72 |
| 238U |
238 |
99.27 |
Calculating Relative Atomic Mass
Naturally occurring silver is 51.84% silver-107 and 48.16% silver-109.
Calculate the relative atomic mass of silver.
r.a.m. (Ag) = (51.84/100 x 107) + (48.16/100 x 109)
= 55.469 + 52.494
= 107.96
Calculating the Abundance of an Isotope
Copper consists of two isotopes, copper-63 and copper-65.
Its relative atomic mass is 63.62.
Find the abundance of each isotope.
Let y/100 = the abundance of copper-63
and (100 - y)/100 = the abundance of copper-65
63.62 = (y/100 x 63) + [(100 - y)/100 x 65]
63.62 = 63y/100 + 6500/100 - 65y/100
6362 = 63y + 6500 - 65y
-135 = -2y
y = 69
Abundance of copper-63 = 69/100 = 69%
Abundance of copper-65 = 100 - 69 = 31%
| 
to
Bookmark this site!