This is the solution for carbon: (12) (0.9890) + (13) (0.0110) = 12.011 amu mass number percent abundance 1298.90% 131.10% Recall!!! carbon:

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Average Atomic Mass & % Abundance

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Presentation transcript:

This is the solution for carbon: (12) (0.9890) + (13) (0.0110) = amu mass number percent abundance % % Recall!!! carbon:

If you know the element, look up the atomic mass on the periodic table. It's the number with places after the decimal. If there are only two isotopes they have to add up to 100% (or in decimal form 1) so you set it up like this: (atomic weight of one isotope)(x) + (atomic weight of other isotope)(1-x) = the atomic mass you find on the table. Use algebra to find x. Don't forget to use the distributive property.

Example: Cl-35 weighs amu Cl-37 weighs amu (34.969)(x) + (36.966)(1-x) = (on the table for Chlorine) x x = x = x = so Cl-35 abundance is % and Cl-37 is %

The element Rhenium (Re) has two naturally occurring isotopes, 185 Re and 187 Re, with an average atomic mass of amu. Calculate the relative abundance of each isotope.