Question

1. an element exists as 4 different isotopes. 4.35% have a mass of 49.9461 amu, 83.79% have a mass of 51.9405 amu, 9.50% have a mass of 52.9407 amu, and 2.36% have a mass of 53.9389 amu.

a. what is the average atomic mass of this element?
b. what is the identity of this element?

Explanation:

Step1: Recall average - atomic - mass formula

The average atomic mass ($A_{avg}$) of an element with isotopes is calculated using the formula $A_{avg}=\sum_{i = 1}^{n}m_i\times p_i$, where $m_i$ is the mass of the $i$-th isotope and $p_i$ is the percentage abundance of the $i$-th isotope (expressed as a decimal).

Step2: Convert percentages to decimals

$4.35\%=0.0435$, $83.79\% = 0.8379$, $9.50\%=0.0950$, $2.36\%=0.0236$.

Step3: Calculate the contribution of each isotope

For the first isotope: $m_1 = 49.9461$ amu and $p_1=0.0435$, so $m_1\times p_1=49.9461\times0.0435 = 2.17265535$ amu.
For the second isotope: $m_2 = 51.9405$ amu and $p_2 = 0.8379$, so $m_2\times p_2=51.9405\times0.8379=43.52784495$ amu.
For the third isotope: $m_3 = 52.9407$ amu and $p_3=0.0950$, so $m_3\times p_3=52.9407\times0.0950 = 5.0293665$ amu.
For the fourth isotope: $m_4 = 53.9389$ amu and $p_4=0.0236$, so $m_4\times p_4=53.9389\times0.0236 = 1.27295804$ amu.

Step4: Calculate the average atomic mass

$A_{avg}=2.17265535 + 43.52784495+5.0293665 + 1.27295804=51.99282484\approx52.00$ amu.

Step5: Identify the element

Looking at the periodic table, the element with an average atomic mass close to 52.00 amu is Chromium (Cr).

Answer:

a. 52.00 amu
b. Chromium (Cr)