Question

vanadium has an atomic mass of 50.9415 amu. it has two common isotopes. One isotope has a mass of 50.9440 amu and a relative abundance of 99.75%.
A. what is the abundance of the other isotope?
B. what is the mass of the other isotope?

Answer 1

To find the abundance and mass of the other isotope of vanadium, we can use the concept of weighted averages based on the relative abundance of the isotopes.

### A. Finding the abundance of the other isotope

Let's denote:
- The mass of the first isotope \( m_1 = 50.9440 \, \text{amu} \)
- The abundance of the first isotope \( a_1 = 99.75\% = 0.9975 \)
- The abundance of the second isotope \( a_2 \)
- We know that the total abundance must equal 100%, so:

\[
a_1 + a_2 = 1
\]

This means:

\[
a_2 = 1 - a_1 = 1 - 0.9975 = 0.0025 \, \text{or} \, 0.25\%
\]

So, the abundance of the other isotope is **0.25%**.

### B. Finding the mass of the other isotope

We can set up the equation for the average atomic mass based on the isotopes and their abundances:

\[
\text{Average atomic mass} = (m_1 \cdot a_1) + (m_2 \cdot a_2)
\]

Where:
- The average atomic mass of vanadium is \( 50.9415 \, \text{amu} \)
- Rearranging for \( m_2 \) (mass of the second isotope) gives us:

\[
50.9415 = (50.9440 \cdot 0.9975) + (m_2 \cdot 0.0025)
\]

Calculating \( (50.9440 \cdot 0.9975) \):

\[
50.9440 \cdot 0.9975 \approx 50.8408
\]

Now we plug this into our equation:

\[
50.9415 = 50.8408 + (m_2 \cdot 0.0025)
\]

Subtract \( 50.8408 \) from both sides:

\[
50.9415 - 50.8408 = m_2 \cdot 0.0025
\]

\[
0.1007 = m_2 \cdot 0.0025
\]

Now, solve for \( m_2 \):

\[
m_2 = \frac{0.1007}{0.0025} \approx 40.28 \, \text{amu}
\]

Thus, the mass of the other isotope is approximately **40.28 amu**.

### Summary:

A. The abundance of the other isotope is **0.25%**.
B. The mass of the other isotope is approximately **40.28 amu**.