Question

Iridium has two naturally occurring isotopes. 1911r and 1931r have atomic masses of 1 90.9609 amu and 1 92 .9633 amu, respectively. The average atomic mass for iridium is 192.22 amu. What is the percent natural abundance for each isotope?

Answer 1

The percent natural abundance for isotope 1911r is 37.71% and for isotope 1931r is 62.29%.

Step-by-step explanation:

For the percent natural abundance of each isotope, we can use the formula:

Percent abundance = (Mass of the isotope * Abundance) / Average atomic mass

For isotope 1911r:

Percent abundance = (190.9609 amu * x) / 192.22 amu = 1.0

For isotope 1931r:

Percent abundance = (192.9633 amu * (1 - x))/ 192.22 amu = 1.0

Solving these equations, we find that x = 0.3771 or 37.71% and (1 - x) = 0.6229 or 62.29%. Therefore, the percent natural abundance for isotope 1911r is 37.71% and for isotope 1931r is 62.29%.