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How do we detect isotopes? With a Mass Spectrometer
Isotopes of Chlorine
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Isotopes of Hydrogen
protium
deuterium
tritium
“heavy water�
radioactive
Isotope Chart
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Calculating Average Atomic Mass Need: 1. Mass numbers of each isotope. 2. Relative abundance of each isotope. (percentage/fractional amount of each isotope)
Substitute into the average mass equation and solve‌
Average Atomic Mass Equation AAM= (mass #)(%) + (mass #)(%) + ‌ Isotope 1
Isotope 2
Calculate the AAM for chlorine From the mass spectrometer we know that Cl35 isotope is 3x as abundant as Cl37
So we can say that Cl35 has an abundance of 75% and Cl37 has an abundance of 25%
Calculate the AAM for chlorine AAM= (mass #)(%) + (mass #)(%) + ‌ AAM= (35 amu)(.75) + (37 amu)(.25) = (26.25 amu) + (9.25 amu) = 35.5 amu
Calculating the Percentage of Isotopes in a sample Need: 1. Mass numbers of each isotope. 2. Average atomic mass of the element. 3. A little mathematical setup (algebra magic). Substitute into the total mass equation and solve‌
Total Mass Equation total mass + total mass + ‌ = total mass of sample Isotope 1
Isotope 2
(mass #)(# of atoms) + (mass #)(# of atoms) + ‌ = (AAM)(total # of atoms)
Calculate the % of each isotope for boron
From the periodic table we know that boron has an AAM of 10.8 amu and we are told that there are 2 isotopes: B10 isotope and B11 isotope.
A little mathematical setup first‌ Consider that the sample you are working with contains 100 atoms of the element.
Let the number of atoms for B10 isotope be = x atoms. Therefore the number of the B11 isotope = (100 - x) atoms
(mass #)(# of atoms) + (mass #)(# of atoms) + ‌ = (AAM)(total # of atoms)
(10 amu)(x atoms) + (11 amu)(100 - x) = (10.8 amu)(100)
10x + 11(100 - x) = (10.8)(100) 10x + 1100 - 11x = 1080 -1x + 1100 = 1080 -1x = 1080 - 1100 -1x = -20 x = 20
B10 isotope = x atoms = 20 atoms
= 20/100 = 20% B11 isotope = (100 - x) atoms = (100 - 20) atoms
= 80 atoms = 80/100 = 80%