To find the percent abundance of lithium-6, we will use the formula for the average atomic mass:

$\text{Atomic Mass} = (f_{\text{Li-6}} \times \text{mass of Li-6}) + (f_{\text{Li-7}} \times \text{mass of Li-7})$

Where:

• $f_{\text{Li-6}}$ = fraction of lithium-6 isotope
• $f_{\text{Li-7}} = 1 - f_{\text{Li-6}}$ (since the total abundance must equal 100%)
• Atomic mass of lithium = 6.9412 amu
• Mass of lithium-6 (\text{Li-6}) = 6.015123 amu
• Mass of lithium-7 (\text{Li-7}) = 7.016005 amu

Step 1: Set up the equation

$6.9412 = (f_{\text{Li-6}} \times 6.015123) + [(1 - f_{\text{Li-6}}) \times 7.016005]$

Step 2: Expand the equation

$6.9412 = f_{\text{Li-6}} \times 6.015123 + (1 - f_{\text{Li-6}}) \times 7.016005$

$6.9412 = f_{\text{Li-6}} \times 6.015123 + 7.016005 - f_{\text{Li-6}} \times 7.016005$

Step 3: Combine like terms

$6.9412 = f_{\text{Li-6}} \times (6.015123 - 7.016005) + 7.016005$

$6.9412 = f_{\text{Li-6}} \times (-1.000882) + 7.016005$

Step 4: Subtract 7.016005 from both sides

$6.9412 - 7.016005 = f_{\text{Li-6}} \times (-1.000882)$

$-0.074805 = f_{\text{Li-6}} \times (-1.000882)$

Step 5: Solve for $f_{\text{Li-6}}$

$f_{\text{Li-6}} = \frac{-0.074805}{-1.000882}$

$f_{\text{Li-6}} = 0.074772$

So, the fraction of lithium-6 is approximately 0.074772.

Step 6: Convert to percentage

$\text{Percent abundance of Li-6} = 0.074772 \times 100 = 7.48\%$

Thus, the percent abundance of lithium-6 is 7.48%.

Would you like a further breakdown of any steps?

Related Questions:

1. What is the percent abundance of lithium-7?
2. How does atomic mass affect the properties of elements?
3. Why is lithium used in high-temperature lubricants?
4. What is the difference between atomic mass and mass number?
5. How do isotopes influence the average atomic mass of an element?

Tip:

When solving problems with multiple isotopes, always remember the total abundance sums to 100%, and isolate variables step-by-step for clarity.