Question

a robot spacecraft returned samples from the planetesimal 98765 aleks, located in the outer solar system. mass - spectroscopic analysis produced the following data on the isotopes of ruthenium in these samples:

isotope	mass (amu)	relative abundance
$^{102}$ru	101.9	8.9%

use these measurements to complete the entry for ruthenium in the periodic table that would be used on 98765 aleks. be sure your answers have the correct number of significant digits.
caution: your correct answer will have the same format but not necessarily the same numbers as the entry for ruthenium in the periodic table we use here on earth.

Explanation:

Step1: Recall the formula for average atomic mass

The average atomic mass ($A$) of an element is calculated by the sum of the product of each isotope's mass ($m_i$) and its relative abundance ($f_i$, expressed as a decimal) for all isotopes. The formula is:
$$A = \sum (m_i \times f_i)$$

Step2: Convert relative abundances to decimals

For $^{100}\text{Ru}$: relative abundance = $91.1\% = \frac{91.1}{100} = 0.911$
For $^{102}\text{Ru}$: relative abundance = $8.9\% = \frac{8.9}{100} = 0.089$

Step3: Calculate the contribution of each isotope

• Contribution of $^{100}\text{Ru}$: $m_1 \times f_1 = 99.9 \, \text{amu} \times 0.911$

$99.9 \times 0.911 = 99.9 \times (0.9 + 0.011) = 99.9 \times 0.9 + 99.9 \times 0.011 = 89.91 + 1.0989 = 91.0089$

• Contribution of $^{102}\text{Ru}$: $m_2 \times f_2 = 101.9 \, \text{amu} \times 0.089$

$101.9 \times 0.089 = 101.9 \times (0.09 - 0.001) = 101.9 \times 0.09 - 101.9 \times 0.001 = 9.171 - 0.1019 = 9.0691$

Step4: Sum the contributions

Add the two contributions to get the average atomic mass:
$A = 91.0089 + 9.0691 = 100.078$

Answer:

The average atomic mass of ruthenium (Ru) for this sample is approximately $\boldsymbol{100.1}$ amu (rounded to the correct number of significant digits, or as calculated: $100.08$ amu, which can be rounded to $100.1$ amu for appropriate precision).