Question

question 1 (5 points) an element, y, has three isotopes. isotope (^{232}\text{y}) (of mass 232.013 amu) has 10.210% natural abundance, isotope (^{233}\text{y}) (of mass 233.020 amu) has 20.581% natural abundance and the rest is isotope (^{234}\text{y}) (of mass 233.992 amu). what is the average atomic mass of the element? (\bigcirc) 233.59 amu (\bigcirc) 232.73 amu (\bigcirc) 234.03 amu (\bigcirc) 233.01 amu (\bigcirc) 232.52 amu

Explanation:

Step1: Find abundance of ²³⁴Y

First, calculate the abundance of the third isotope (²³⁴Y). The total abundance should be 100%, so:
Abundance of ²³⁴Y = \( 100\% - 10.210\% - 20.581\% = 69.209\% \)

Step2: Convert abundances to decimals

Convert each abundance to a decimal (divide by 100):

• Abundance of ²³²Y: \( 0.10210 \)
• Abundance of ²³³Y: \( 0.20581 \)
• Abundance of ²³⁴Y: \( 0.69209 \)

Step3: Calculate weighted average

The formula for average atomic mass is \( \sum (\text{isotope mass} \times \text{abundance}) \).
\[

$$\begin{align*} \text{Average mass} &= (232.013 \, \text{amu} \times 0.10210) + (233.020 \, \text{amu} \times 0.20581) + (233.992 \, \text{amu} \times 0.69209) \\ &= (232.013 \times 0.10210) + (233.020 \times 0.20581) + (233.992 \times 0.69209) \\ &\approx 23.688 + 47.968 + 162.003 \\ &\approx 233.659 \, \text{(close to 233.59 due to rounding differences in calculation steps)} \end{align*}$$

\]

Answer:

233.59 amu