Question

1. element x has two naturally occurring isotopes of masses 106.905 amu and 108.905 amu. the percentage abundance of these two isotopes are 51.83% and 48.17% respectively. what is the average atomic mass of element x ?. a. 106.9 amu b. 107.9 amu c. 108.5 amu d. 107.2 amu e. 108.2 amu

Explanation:

Step1: Recall the formula for average atomic mass

The formula for average atomic mass ($A$) of an element with isotopes is $A = (m_1 \times f_1) + (m_2 \times f_2)$, where $m_1, m_2$ are the masses of isotopes and $f_1, f_2$ are their fractional abundances.
First, convert percentage abundances to fractions: $f_1=\frac{51.83}{100} = 0.5183$, $f_2=\frac{48.17}{100}=0.4817$.

Step2: Calculate the contribution of each isotope

For the first isotope: $m_1\times f_1=106.905\times0.5183$. Let's calculate that: $106.905\times0.5183\approx106.905\times0.5 + 106.905\times0.0183\approx53.4525+1.956\approx55.4085$.
For the second isotope: $m_2\times f_2 = 108.905\times0.4817$. Calculate: $108.905\times0.4817\approx108.905\times0.5 - 108.905\times0.0183\approx54.4525 - 1.993\approx52.4595$.

Step3: Sum the contributions

Add the two contributions: $55.4085 + 52.4595\approx107.868\approx107.9$ amu.

Answer:

B. 107.9 amu