To find the percent abundance of isotope Z-31, we will use the formula for the average atomic mass:

$\text{Average atomic mass} = (m_1 \times f_1) + (m_2 \times f_2)$

where:

• $m_1$ and $m_2$ are the masses of the isotopes (28.97 amu for Z-29 and 30.99 amu for Z-31).
• $f_1$ and $f_2$ are the fractional abundances of the isotopes.
• The sum of the fractional abundances equals 1: $f_1 + f_2 = 1$.

Let $f_2$ be the fractional abundance of Z-31. Then, the fractional abundance of Z-29 will be $1 - f_2$.

Step 1: Set up the equation

$30.22 = (28.97 \times (1 - f_2)) + (30.99 \times f_2)$

Step 2: Expand the equation

$30.22 = 28.97 - 28.97 f_2 + 30.99 f_2$

Step 3: Combine like terms

$30.22 = 28.97 + (30.99 - 28.97) f_2$

$30.22 = 28.97 + 2.02 f_2$

Step 4: Solve for $f_2$

$30.22 - 28.97 = 2.02 f_2$

$1.25 = 2.02 f_2$

$f_2 = \frac{1.25}{2.02} \approx 0.6188$

So, the fractional abundance of Z-31 is approximately 0.6188, which means the percent abundance is:

$0.6188 \times 100 \approx 61.88\%$

Thus, the percent abundance of isotope Z-31 is approximately 61.88%.

Would you like more details or have any questions?

Related Questions

1. What would be the percent abundance of Z-29 in this scenario?
2. How does the average atomic mass relate to isotopic masses?
3. If a third isotope were present, how would the calculation change?
4. How do you find the mass of an element using isotopic data?
5. What is the role of significant figures in calculating average atomic mass?

Tip

Always check that the sum of fractional abundances equals 1 to avoid calculation errors.