Copper is made up of two isotopes Cu^63 and Cu^65, actually Cu^63 having the mass Cu^62.9296 amu. and Cu^65 having the mass Cu^64.9278 amu. Given coppers atomic weight is 63.546 gram. What is the percent abundance of each isotope?
Answers
Answer:
69.152
%
→
63
Cu
30.848
%
→
65
Cu
Explanation:
As you know, the average atomic mass of an element is determined by taking the weighted average of the atomic masses of its naturally occurring isotopes.
Simply put, an element's naturally occurring isotopes will contribute to the average atomic mass of the element proportionally to their abundance.
avg. atomic mass
=
∑
i
(
isotope
i
×
abundance
x
)
When it comes to the actual calculation, it's easier to use decimal abundances, which are simply percent abundances divided by
100
.
So, you know that copper has two naturally occurring isotopes, copper-63 and copper-65. This means that their respective decimal abundance must add up to give
1
.
If you take
x
to be the decimal abundance of copper-63, you can say that the decimal abundance of copper-65 will be equal to
1
−
x
.
Therefore, you can say that
copper-63
x
⋅
62.9296
u
+
copper-65
(
1
−
x
)
⋅
64.9278
u
=
63.546
u
Solve this equation for
x
to get
62.9296
⋅
x
−
64.9278
⋅
x
=
63.546
−
64.9278
1.9982
⋅
x
=
1.3818
⇒
x
=
1.3818
1.9982
=
0.69152
This means that the percent abundances of the two isotopes will be
69.152
%
→
63
Cu
30.848
%
→
65
Cu
Answer:
The percentage abundance of Cu-63 is 69.15% and of Cu-65 is 30.85%.
Explanation:
1) Write the following equation:
(62.9296) (x) + (64.9278) (1 − x) = 63.546
Notice that 'x' and 'one minus x' add up to one.
2) Solve for x:
x = 0.6915 (the decimal abundance for Cu-63)
Note that this calculation technique works only with two isotopes. If you have three or more, there are too many variables and not enough equations. I hope it's obvious why you wouldn't do this with an element that has only one stable isotope!
By the way, the 'trick' works because the other equation required is:
x + y = 1
We simply went right to y = 1 − x and substitued it immediately without ever writing down the second equation required.