Question

in the chemical closet, you found an unlabeled vial with a solid piece of an unknown element inside (element z). you decided to put it in the mass - spec to figure out its atomic mass. the results showed that it has two naturally occurring isotopes, z - 85, and z - 87. z - 85 has a natural abundance of 72.17% and a mass of 84.912 amu. z - 87 has a natural abundance of 27.83% and a mass of 86.909 amu. calculate the average atomic mass and determine the identity of mystery element z

Explanation:

Step1: Recall the formula for average atomic mass

The average atomic mass ($A$) of an element with isotopes is calculated by the sum of the product of each isotope's mass ($m_i$) and its natural abundance ($f_i$, in decimal form). The formula is:
$$A = \sum (m_i \times f_i)$$

Step2: Convert abundances to decimals

• For Z-85: Abundance $f_1 = 72.17\% = \frac{72.17}{100} = 0.7217$, Mass $m_1 = 84.912\ \text{amu}$
• For Z-87: Abundance $f_2 = 27.83\% = \frac{27.83}{100} = 0.2783$, Mass $m_2 = 86.909\ \text{amu}$

Step3: Calculate the contribution of each isotope

• Contribution of Z-85: $m_1 \times f_1 = 84.912 \times 0.7217$

$84.912 \times 0.7217 \approx 61.28$ (rounded for simplicity, but we'll use precise values)

• Contribution of Z-87: $m_2 \times f_2 = 86.909 \times 0.2783$

$86.909 \times 0.2783 \approx 24.19$ (rounded, but precise calculation follows)

Step4: Sum the contributions

Precise calculation:
$84.912 \times 0.7217 = 84.912 \times 0.7 + 84.912 \times 0.02 + 84.912 \times 0.0017$
$= 59.4384 + 1.69824 + 0.1443504 = 61.2809904$

$86.909 \times 0.2783 = 86.909 \times 0.2 + 86.909 \times 0.07 + 86.909 \times 0.0083$
$= 17.3818 + 6.08363 + 0.7213447 = 24.1867747$

Total average atomic mass: $61.2809904 + 24.1867747 = 85.4677651\ \text{amu}$

Answer:

The average atomic mass of element Z is approximately $\boldsymbol{85.47\ \text{amu}}$. (This matches the atomic mass of Rubidium (Rb), so the mystery element Z is likely Rubidium.)