Questions: Element X has an average atomic mass of 44.68574 amu and has two isotopes. One isotope of element X has an atomic mass of 45.57456 amu and a percent abundance was measured to be 14.29%. What is the atomic mass of the other isotope? Your answer MUST have the correct number of significant figures.

Solution Steps

We are given an element \( X \) with an average atomic mass of 44.68574 amu. It has two isotopes. One isotope has an atomic mass of 45.57456 amu with a percent abundance of 14.29%. We need to find the atomic mass of the other isotope.

The average atomic mass of an element is calculated using the formula:

\[ \text{Average Atomic Mass} = (m_1 \times f_1) + (m_2 \times f_2) \]

where \( m_1 \) and \( m_2 \) are the atomic masses of the isotopes, and \( f_1 \) and \( f_2 \) are their respective fractional abundances. Given:

• \( m_1 = 45.57456 \, \text{amu} \)
• \( f_1 = 0.1429 \) (since 14.29% = 0.1429)
• Average Atomic Mass = 44.68574 amu

We need to find \( m_2 \), the atomic mass of the other isotope. The fractional abundance of the second isotope is \( f_2 = 1 - f_1 = 0.8571 \).

Substitute the known values into the equation:

\[ 44.68574 = (45.57456 \times 0.1429) + (m_2 \times 0.8571) \]

Calculate the contribution of the first isotope:

\[ 45.57456 \times 0.1429 = 6.5075 \]

Substitute back into the equation:

\[ 44.68574 = 6.5075 + (m_2 \times 0.8571) \]

Solve for \( m_2 \):

\[ 44.68574 - 6.5075 = m_2 \times 0.8571 \]

\[ 38.1782 = m_2 \times 0.8571 \]

\[ m_2 = \frac{38.1782}{0.8571} = 44.5556 \]

Final Answer