Antimony-121 has a mass of 120.9038 u
antimony-123 has a mass of 122.9042 u
the atomic mass of antimony from the periodic table is 121.760 amu.
i need to know the percent abundance of the two isotopes.
please show all steps and formulas
thank you
2007-01-26 17:56:00 · 8 answers · asked by I Like Cheese 2 in Science & Mathematics ➔ Chemistry
The atomic mass of an element is the weighted average of the masses of its isotopes.
You know that:
Antimony-121 has a mass of 120.9038 u, x% abundance
Antimony-123 has a mass of 122.9042 u, y% abundance
There are only 2 isotopes for antimony and their percent abundances should add up to 100%. In other words:
x% + y% = 100%
y = 1-x
(percentages written as decimals)
So, now let's put everything together. In order to calculate the atomic mass, multiply the percent abundance of an isotope by its atomic mass; then add the product of all the isotopes:
(Atomic Mass of Antimony-121)(Percent Abundance of Antimony-121) + (Atomic Mass of Antimony-123)(Percent Abundance of Antimony-123) = Atomic Mass of Element Antimony
(120.9038 amu)(x) + (122.9042 amu)(y) = 121.760 amu
Replacing 1-x for y gives:
(120.9038 amu)(x) + (122.9042 amu)(1-x) = 121.760 amu
Solve for x:
120.9038x + 122.9042 -122.9042x = 121.760 amu
-2.0040x = -1.1442
x = 0.57096 = 57.096%
Solve for y:
y = 1 - x
y = 1 - 0.57096 = 0.42904 = 42.904%
2007-01-26 18:56:20 · answer #1 · answered by ybdogsct 2 · 1⤊ 1⤋
This Site Might Help You.
RE:
how do i calculate the percent abundance of two isotopes of antimoney when i am given their masses?
Antimony-121 has a mass of 120.9038 u
antimony-123 has a mass of 122.9042 u
the atomic mass of antimony from the periodic table is 121.760 amu.
i need to know the percent abundance of the two isotopes.
please show all steps and formulas
thank you
2015-08-10 06:26:09 · answer #2 · answered by ? 1 · 0⤊ 0⤋
Antimony Isotopes
2016-10-22 11:29:59 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Antimony 121
2016-12-28 05:41:08 · answer #4 · answered by byro 4 · 0⤊ 0⤋
one way of doing this is:
120.9038x + 122.9042(1-x) = 121.760
120.9038x + 122.9042-122.9042x = 121.760
-122.9042 -122.9042
120.9038x - 122.9042x = -1.1442
-2.0004 = -1.1442
x = 0.571986
the x is the percentage value. If we multiply the atomic masses by the x and (1-x) and then add them, we should get the average atomic mass.
120.9038(0.571986)+122.9042(1-0.571986)=121.759999...
the (1-x) is so that we use the same variable and do not have to do 2 equations.
so, this value works out
now, multiply x by 100 and subtract 1 from x to get the other value
Antimony-121 is 57.1986%
Antimony-123 is 42.8014%
There are six significant figures because the least number of significant figures in the problem are six
2007-01-26 18:19:57 · answer #5 · answered by Fit B 1 · 1⤊ 0⤋
Easy!
122.9042 - 120.9038 = (delta) 2.0004
(the difference in weight of the two isotopes)
121.760-120.9038 = .8562 (difference in 100% antimony-121 and "wild" antimony
.8562 / 2.0004 = about .428 or 42.8%
conversely 1-.428 = .572 or 57.2%
I'll leave the rest to you.
2007-01-26 18:07:50 · answer #6 · answered by Anonymous · 0⤊ 0⤋
let the % of ant-121 be = "a" so that of ant-123 will be = "100-a"
thus: a/100 x 120.9038 + (100-a)/100 x 122.9042 = 121.760
simplifying we get:
1.209038a + 122.9042 - 1.229042a = 121.760
0.020004a = 1.1442
thus a = 57.2%
So % abundance of anti-121 = 57.2% and that of anti-123 = 42.8%
2007-01-26 18:15:37 · answer #7 · answered by Southpaw 5 · 0⤊ 0⤋
lets x be the fraction of antimony 121. so, the fraction of antimony 123 is (1-x)
you can write
120.9038 x +(1-x) 122.9042 = 121.76
120.9038 x - 122.9042 x= 121.76-122.9042
-2.0004x =-1.1442
x= 0.572 = 57.2% of Sb 121
and 100-57.2 = 42.8% of Sb 123 (Sb =antimony)
2007-01-26 18:23:23 · answer #8 · answered by maussy 7 · 0⤊ 0⤋