Question

Suppose that a hypothetical element consists
of a mixture of two isotopes. One isotope,
having mass 44 amu, is present in 18.5%
abundance, while the other isotope, having
mass 46 amu, accounts for the other 81.5%.
What should be the experimentally determined
atomic weight for this hypothetical element?

Answer 1

Answer on Question#54674 – Chemistry – General Chemistry

Question

Suppose that a hypothetical element consists of a mixture of two isotopes. One isotope, having mass 44 amu, is present in 18.5% abundance, while the other isotope, having mass 46 amu, accounts for the other 81.5%. What should be the experimentally determined atomic weight for this hypothetical element?

Answer:

The experimentally determined atomic weight for this hypothetical element is:

$$A(E) = 44 \cdot \frac{18.5\%}{100\%} + 46 \cdot \frac{81.5\%}{100\%} = 45.63 \text{ amu}$$

Answer: 45.63 amu

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