19. Iridium has two naturally occurring isotopes. \( { }^{191} 1 \mathrm{r} \) and \( { }^{193} 1 \mathrm{r} \) have atomic masses of 190.9609 amu and 192.9633 amu , respectively. The average atomic mass for iridium is 192.22 amu . What is the percent natural abundance for each isotope?

To find the percent natural abundance of each isotopes of iridium, we can set up a system of equations using the average atomic mass. Let \( x \) be the fraction of \( { }^{191} \mathrm{Ir} \) and \( (1 - x) \) be the fraction of \( { }^{193} \mathrm{Ir} \). The equation for the average atomic mass is: \[ 190.9609 \, \text{amu} \cdot x + 192.9633 \, \text{amu} \cdot (1 - x) = 192.22 \, \text{amu} \] Expanding and rearranging gives: \[ 190.9609x + 192.9633 - 192.9633x = 192.22 \] Combining like terms leads to: \[ -2.0024x + 192.9633 = 192.22 \] Subtracting \( 192.9633 \) from both sides results in: \[ -2.0024x = -0.7433 \] Now, solving for \( x \): \[ x = \frac{0.7433}{2.0024} \approx 0.3701 \] Thus, the percent natural abundance for \( { }^{191} \mathrm{Ir} \) is approximately \( 37.01\% \) and for \( { }^{193} \mathrm{Ir} \): \[ 1 - x \approx 0.6299 \quad \text{or} \quad 62.99\% \] In summary: - The percent natural abundance of \( { }^{191} \mathrm{Ir} \) is approximately \( 37.01\% \). - The percent natural abundance of \( { }^{193} \mathrm{Ir} \) is approximately \( 62.99\% \).