A hypothetical element has two main isotopes

with mass numbers of 71 and 74. If 80.00% of
the isotopes have a mass number of 71 amu,
what atomic weight should be listed on the
periodic table for this element?
(answer in AMU).

I would rather see the atomic weight of each isotope; failing that I'm assuming they are 71 and 74.

0.80(71) + 0.20(74) = ?

71.6

To calculate the atomic weight of the element, we need to consider the abundance of each isotope and its corresponding mass.

Given that 80.00% of the isotopes have a mass number of 71 amu, we can assume that the remaining 20.00% have a mass number of 74 amu.

Calculating the weighted average:

(0.80 * 71 amu) + (0.20 * 74 amu)

56.8 amu + 14.8 amu

The atomic weight of the element should be listed as 71.6 amu on the periodic table.

To determine the atomic weight of the element, we need to calculate the weighted average of the masses of its isotopes.

Given that 80.00% of the isotopes have a mass number of 71 amu, we can assume that 80.00% of the element consists of the 71 amu isotope.

To calculate the atomic weight, we can use the following formula:

Atomic Weight = (Percentage of Isotope 1 × Mass of Isotope 1) + (Percentage of Isotope 2 × Mass of Isotope 2)

In this case, the percentages are 80.00% for the isotope with a mass number of 71, and 20.00% for the isotope with a mass number of 74. The masses of the isotopes are 71 amu and 74 amu, respectively.

Now let's calculate:

Atomic Weight = (0.80 × 71 amu) + (0.20 × 74 amu)
Atomic Weight = 56.8 amu + 14.8 amu
Atomic Weight = 71.6 amu

Therefore, the atomic weight (or atomic mass) listed on the periodic table for this hypothetical element should be approximately 71.6 amu.