1. A hypothetical pure element consists entirely of two isotopes. The natural percent abundance of the isotope with mass 25.5 u is 86.1%. The other isotope has mass 29.2 u. Calculate the atomic weight of the naturally occurring element.
2. Suppose that a hypothetical element consists of a mixture of two isotopes. One isotope, having mass 44 amu, is present in 20.9% abundance, while the other isotope, having mass 46 amu, accounts for the other 79.1%. What should be the experimentally determined atomic weight for this hypothetical element?Answer in units of amu.
26
25.5(0.861) + 29.2(1-0.861) = ?
29.6
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To calculate the atomic weight of an element, you need to consider the atomic masses and percent abundances of its isotopes. Here's how you can calculate the atomic weights for the given hypothetical elements:
1. For the first question:
- The percent abundance of the isotope with mass 25.5 u is 86.1%.
- The percent abundance of the isotope with mass 29.2 u is therefore (100% - 86.1%) = 13.9%.
- To calculate the average atomic weight, you multiply the atomic mass of each isotope by its percent abundance, then add the products.
(25.5 u x 0.861) + (29.2 u x 0.139) = 21.9755 u + 4.0468 u = 26.0223 u.
Therefore, the atomic weight of the naturally occurring element is 26.0223 u.
2. For the second question:
- The percent abundance of the isotope with mass 44 amu is 20.9%.
- The percent abundance of the isotope with mass 46 amu is therefore (100% - 20.9%) = 79.1%.
- To calculate the average atomic weight, you multiply the atomic mass of each isotope by its percent abundance, then add the products.
(44 amu x 0.209) + (46 amu x 0.791) = 9.136 amu + 36.386 amu = 45.522 amu.
Therefore, the experimentally determined atomic weight for this hypothetical element is 45.522 amu.