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 determine the percent natural abundance of each iridium isotope, we'll use the given atomic masses and the average atomic mass of iridium. **Given:** - **Ir-191:** - Mass = 190.9609 amu - Abundance = \( x \) (fraction) - **Ir-193:** - Mass = 192.9633 amu - Abundance = \( 1 - x \) (since there are only two isotopes) **Average atomic mass of Iridium:** \[ 192.22 \text{ amu} \] **Setting Up the Equation:** The average atomic mass is calculated by summing the products of each isotope's mass and its abundance: \[ x \times 190.9609 + (1 - x) \times 192.9633 = 192.22 \] **Solving for \( x \):** 1. Distribute the terms: \[ 190.9609x + 192.9633 - 192.9633x = 192.22 \] 2. Combine like terms: \[ (190.9609 - 192.9633)x + 192.9633 = 192.22 \] \[ -2.0024x + 192.9633 = 192.22 \] 3. Isolate \( x \): \[ -2.0024x = 192.22 - 192.9633 \] \[ -2.0024x = -0.7433 \] \[ x = \frac{-0.7433}{-2.0024} \approx 0.3716 \] **Calculating Percent Abundance:** - **Ir-191:** \[ 0.3716 \times 100 \approx 37.2\% \] - **Ir-193:** \[ 100\% - 37.2\% = 62.8\% \] **Final Answer:** - **\( {}^{191}\text{Ir} \):** 37.2% - **\( {}^{193}\text{Ir} \):** 62.8% **Rounded to one decimal place:** - Ir‑191  37.2 % - Ir‑193  62.8 %