The atomic mass of Eu is 151.96. There are only two naturally occurring isotopes of europium, 151Eu with a mass of 151.0 and 153Eu with a mass of 153.0. The natural abundance of the 131Eu isotope must be approximately
To determine the natural abundance of the 153Eu isotope, we need to consider the atomic mass of europium and the mass of the two naturally occurring isotopes (151Eu and 153Eu).
The atomic mass of Europium (Eu) is given as 151.96.
Let's assume the natural abundance of the 151Eu isotope as "x" (in percentage). Then, the natural abundance of the 153Eu isotope will be (100 - x) (since the total abundance must add up to 100%).
Now, we need to set up an equation using the atomic masses and natural abundances:
(x/100)(151) + ((100 - x)/100)(153) = 151.96
Simplifying this equation, we get:
(151x + 153(100 - x))/100 = 151.96
Expanding and simplifying further:
151x + 15300 - 153x = 15196
Combine like terms:
-2x = -104
Solving for "x," we divide both sides of the equation by -2:
x = -104 / -2
x = 52
Therefore, the natural abundance of the 151Eu isotope is approximately 52%.
To determine the natural abundance of the 151Eu isotope, subtract the mass of the 153Eu isotope from the atomic mass of europium and then divide by the difference between the masses of the two isotopes.
Atomic mass of Eu = 151.96
Mass of 153Eu = 153.0
Natural abundance of 151Eu = (Atomic mass of Eu - Mass of 153Eu) / (Mass of 151Eu - Mass of 153Eu)
Natural abundance of 151Eu = (151.96 - 153.0) / (151.0 - 153.0)
Natural abundance of 151Eu = (-1.04) / (-2.0)
Natural abundance of 151Eu ≈ 0.52
Therefore, the natural abundance of the 151Eu isotope is approximately 0.52.
Surely you made a typo in the last sentence and meant 151Eu and not 131. I am assuming you meant 151.
Let X = fraction of 153, then
1-X = fraction of 151.
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151.0(1-X) + 153.0(X) = 151.96
Solve for X and 1-X which gives the fraction and that times 100 gives percent. .