Question

an element has two naturally occurring isotopes. one is 10.013 amu and is 19.9% abundant. the other is 11.01 amu and is 80.1% abundant. what is the average atomic mass? what element is it?

Explanation:

Step1: Calculate the contribution of the first isotope

The contribution of an isotope to the average atomic mass is calculated by multiplying its mass by its abundance (in decimal form).
The abundance of the first isotope \(19.9\%=0.199\), and its mass \(m_1 = 10.013\) amu.
The contribution \(C_1=m_1\times0.199=10.013\times0.199 = 1.992587\) amu.

Step2: Calculate the contribution of the second isotope

The abundance of the second isotope \(80.1\% = 0.801\), and its mass \(m_2=11.01\) amu.
The contribution \(C_2=m_2\times0.801=11.01\times0.801=8.81901\) amu.

Step3: Calculate the average atomic mass

The average atomic mass \(M = C_1 + C_2\).
\(M=1.992587+8.81901 = 10.811597\approx10.812\) amu.

Answer:

The average atomic mass is approximately \(10.812\) amu. The element is Boron.