Calculate the average atomic mass of element. What is the identity of the element?
The atomic masses of and are 6.0151 u and 7.0160 u, respectively. Calculate the natural percentage abundances of these two isotopes. The average atomic mass of Li is 6.941 u.
How many atoms of Au are in an ounce of gold? (Use the atomic mass unit (u) approach
The Correct Answer and Explanation is:
Let’s solve these problems step-by-step using atomic mass unit (u) concepts and the definition of average atomic mass.
Part 1: Identity and Average Atomic Mass of Lithium
We are given:
- Mass of isotope 1 (⁶Li): 6.0151 u
- Mass of isotope 2 (⁷Li): 7.0160 u
- Average atomic mass of Li: 6.941 u
Let x be the fractional abundance of ⁶Li, so the fractional abundance of ⁷Li is (1 − x).
Using the formula for average atomic mass:
$$
6.941 = (6.0151)(x) + (7.0160)(1 – x)
$$
Expanding and simplifying:
$$
6.941 = 6.0151x + 7.0160 – 7.0160x
$$
$$
6.941 = 7.0160 – (7.0160 – 6.0151)x
$$
$$
6.941 = 7.0160 – 1.0009x
$$
$$
1.0009x = 7.0160 – 6.941 = 0.0750
$$
$$
x = \frac{0.0750}{1.0009} ≈ 0.0749
$$
So, ⁶Li is about 7.49%, and ⁷Li is about 92.51% abundant.
Part 2: Identity of the Element
The element described is Lithium (Li), since its average atomic mass is 6.941 u, which is listed on the periodic table as the atomic mass of lithium. This confirms the isotope data pertains to lithium.
Part 3: Number of Atoms of Gold (Au) in 1 Ounce
Given:
- Mass of 1 ounce = 28.3495 grams
- Atomic mass of Au (gold) = 196.97 u
- 1 mole of atoms = 6.022 × 10²³ atoms
- 1 mole of Au weighs 196.97 grams
Now, calculate the number of moles in 1 ounce of gold:
$$
\text{Moles of Au} = \frac{28.3495 \text{ g}}{196.97 \text{ g/mol}} ≈ 0.1438 \text{ mol}
$$
Now, multiply by Avogadro’s number to find the number of atoms:
$$
\text{Atoms of Au} = 0.1438 \text{ mol} × 6.022 × 10^{23} \text{ atoms/mol} ≈ 8.66 × 10^{22} \text{ atoms}
The average atomic mass of an element reflects the weighted average of all naturally occurring isotopes based on their relative abundances. In the case of lithium (Li), it has two stable isotopes: lithium-6 and lithium-7. Their respective atomic masses are 6.0151 u and 7.0160 u. The weighted average formula allows us to solve for the natural abundances using the known average atomic mass of 6.941 u. Solving the equation yields about 7.49% for ⁶Li and 92.51% for ⁷Li, showing ⁷Li is the dominant isotope in nature.
Next, we explored the number of atoms in an ounce of gold (Au). One mole of gold has a mass of 196.97 grams and contains Avogadro’s number of atoms, 6.022 × 10²³. By converting ounces to grams (1 oz = 28.3495 g) and using dimensional analysis, we calculated the number of moles of gold in an ounce to be approximately 0.1438. Multiplying this by Avogadro’s number gives about 8.66 × 10²² atoms in one ounce of gold.
These calculations illustrate how atomic mass units (u), isotopic abundances, and mole-based conversions interconnect. They are essential in chemistry for understanding atomic structure, isotopic distribution, and real-world applications, such as determining the atomic quantity in a given sample.
