Natural Sb consists of two isotopes, 121Sb (120.904 amu) and 123Sb (122.904
amu). Calculate the percent abundance of the heavier of the two isotopes.
The solution was
121.8 = 122.904x + 120.904(1-x)
ANS: 45%
the thing i don't understand is the whole x and (1-x) deal
amu). Calculate the percent abundance of the heavier of the two isotopes.
The solution was
121.8 = 122.904x + 120.904(1-x)
ANS: 45%
the thing i don't understand is the whole x and (1-x) deal
-
x is the abundance of the heavier isotope, expressed as a decimal (not a percentage)
1 - x is the abundance of the lighter isotope because 1 is the decimal representation of 100%
45% and 55% add up to 100%
0.45 and 0.55 add up to 1.
Remember to always use the decimal abundance in calculations of this type.
More sample problems here:
http://www.chemteam.info/Mole/AvgAtomicW…
1 - x is the abundance of the lighter isotope because 1 is the decimal representation of 100%
45% and 55% add up to 100%
0.45 and 0.55 add up to 1.
Remember to always use the decimal abundance in calculations of this type.
More sample problems here:
http://www.chemteam.info/Mole/AvgAtomicW…