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 the isotopes and $f_1, f_2$ are their fractional abundances.
First, convert the percentage abundances to fractional abundances. For the first isotope, $f_1=\frac{51.83}{100} = 0.5183$, and for the second isotope, $f_2=\frac{48.17}{100}=0.4817$.

Step2: Calculate the contribution of each isotope

The mass of the first isotope $m_1 = 106.905$ amu, so its contribution is $m_1\times f_1=106.905\times0.5183$. Let's calculate that: $106.905\times0.5183\approx106.905\times0.5 + 106.905\times0.0183\approx53.4525+1.9564\approx55.4089$ amu.
The mass of the second isotope $m_2 = 108.905$ amu, so its contribution is $m_2\times f_2 = 108.905\times0.4817$. Calculate that: $108.905\times0.4817\approx108.905\times0.5 - 108.905\times0.0183\approx54.4525 - 1.9930\approx52.4595$ amu.

Step3: Sum the contributions

Now, add the two contributions together: $A=55.4089 + 52.4595\approx107.8684\approx107.9$ amu.

Answer:

B. 107.9 amu