Question

question 9 (2 points) an element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99 amu, and 8.82% abundance with 21.99 amu. calculate the average atomic mass of this element. write your answer to 2 decimal places and include units separated by a space. your answer: answer units

Explanation:

Step1: Convert percentages to decimals

$90.92\% = 0.9092$, $8.82\%=0.0882$, $0.26\% = 0.0026$

Step2: Use the formula for average atomic mass

The formula for average atomic mass $A=\sum_{i}m_i\times x_i$, where $m_i$ is the isotopic mass and $x_i$ is the relative - abundance.
$A=(20.99\times0.0026)+(21.99\times0.0882)+(19.99\times0.9092)$

Step3: Calculate each product

$20.99\times0.0026 = 0.054574$
$21.99\times0.0882=1.939518$
$19.99\times0.9092 = 18.174908$

Step4: Sum up the products

$A=0.054574 + 1.939518+18.174908=20.168998\approx20.17$ amu

Answer:

20.17 amu