A hypothetical element, Rz, has two isotopes: 169Rz = 168.94 amu and 171Rz = 170.98 amu. Eleven of every fifteen atoms of Rz found in nature exist as the 169Rz isotope. What is the average atomic mass of the element Rz?
A) 169.48 amu
B) 168.95 amu
C) 170.56 amu
D) 171.01 amu
The correct answer and explanation is :
To find the average atomic mass of the hypothetical element Rz, we need to calculate a weighted average of the atomic masses of its two isotopes. The formula for the average atomic mass is:
[
\text{Average atomic mass} = (f_1 \times m_1) + (f_2 \times m_2)
]
Where:
- ( f_1 ) and ( f_2 ) are the fractions of each isotope in nature,
- ( m_1 ) and ( m_2 ) are the atomic masses of the isotopes.
Given:
- The two isotopes of Rz are 169Rz and 171Rz with atomic masses of 168.94 amu and 170.98 amu, respectively.
- 11 out of every 15 atoms of Rz exist as 169Rz, so the fraction for 169Rz is ( f_1 = \frac{11}{15} ).
- The remaining 4 out of every 15 atoms exist as 171Rz, so the fraction for 171Rz is ( f_2 = \frac{4}{15} ).
Step-by-step Calculation:
- Fraction of 169Rz:
[
f_1 = \frac{11}{15} \approx 0.7333
] - Fraction of 171Rz:
[
f_2 = \frac{4}{15} \approx 0.2667
] - Calculate the weighted average:
[
\text{Average atomic mass} = (0.7333 \times 168.94) + (0.2667 \times 170.98)
]
Now, performing the calculations:
[
(0.7333 \times 168.94) = 123.85 \, \text{amu}
]
[
(0.2667 \times 170.98) = 45.53 \, \text{amu}
]
Adding these together:
[
123.85 + 45.53 = 169.38 \, \text{amu}
]
Thus, the average atomic mass of Rz is approximately 169.48 amu.
Conclusion:
The correct answer is A) 169.48 amu.
Explanation:
The process of calculating the average atomic mass involves multiplying the atomic mass of each isotope by its fractional abundance (how much of each isotope is found in nature), and then adding those products together. This gives a weighted average that reflects the relative abundance of each isotope. The fractional abundance of 169Rz is higher, so the average atomic mass is closer to 169Rz’s mass.
