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 of isotope 191Ir is 37.3% and of isotope 193Ir is 62.7% based on the given average atomic mass of iridium.

Step-by-step explanation:

To determine the percent natural abundance of the two isotopes of iridium, which are 191Ir with an atomic mass of 190.9609 amu and 193Ir with an atomic mass of 192.9633 amu, we can set up the following equations based on the average atomic mass of iridium (192.22 amu):

• Let x be the fraction of 191Ir in nature.
• Let (1-x) be the fraction of 193Ir in nature.

Using the formula for average atomic mass, we get:

190.9609(x) + 192.9633(1-x) = 192.22

Solving this for x gives us:

x = (192.22 - 192.9633) / (190.9609 - 192.9633) = 0.373

The percent natural abundance of 191Ir is then:

x * 100% = 37.3%

And for 193Ir:

(1 - x) * 100% = (1 - 0.373) * 100% = 62.7%

Therefore, the percent natural abundance of 191Ir is 37.3% and of 193Ir is 62.7%.