Question

Calculate the average atomic mass for an element that has four naturally occurring isotopes: 5.90 % with a mass of 53.94 amu, 91.72 % with a mass of 55.93 amu, 2.10 % with a mass of 56.94 amu, and 0.280 % with a mass of 57.93 amu. Identify the element.​

Answer 1

The average atomic mass of the element with the given isotopic composition is 55.86734 amu, and by comparing this value to the periodic table, the element is identified as Manganese (Mn).

Step-by-step explanation:

To calculate the average atomic mass of an element with multiple isotopes, you multiply the mass of each isotope by its relative abundance (converted to a decimal), and then add those values together. For the element in question with four naturally occurring isotopes, the calculation is as follows:

• (0.0590 × 53.94 amu) = 3.18246 amu
• (0.9172 × 55.93 amu) = 51.326936 amu
• (0.0210 × 56.94 amu) = 1.19574 amu
• (0.0028 × 57.93 amu) = 0.162204 amu

Adding these products together gives us:

3.18246 amu + 51.326936 amu + 1.19574 amu + 0.162204 amu = 55.86734 amu as the average atomic mass of the element. By comparing this value to the atomic masses listed in the periodic table, we can identify the element as Manganese (Mn), which has an atomic mass close to 55.85 amu.