Bromine has two stable isotopes: 79Br and 81Br with masses of 78.9183371 amu and 80.9162906 amu. If the average atomic mass for Bromine on Earth is 79.904 amu, what is the mass of 79Br in a 10.0 g sample?
The correct answer and explanation is :
To determine the mass of the 79Br isotope in a 10.0 g sample of bromine, we need to understand how the average atomic mass of bromine is calculated based on the relative abundances of its isotopes, 79Br and 81Br.
Step 1: Understanding the Problem
Bromine exists naturally as a mixture of two isotopes:
- 79Br with a mass of 78.9183371 amu.
- 81Br with a mass of 80.9162906 amu.
The average atomic mass of bromine on Earth is 79.904 amu. This average is a weighted average of the masses of the two isotopes, weighted by their relative abundances in nature.
Step 2: Let’s Set Up the Equation
Let’s assume:
- The fraction of 79Br in the sample is x.
- The fraction of 81Br in the sample is 1 – x (since the total fraction must equal 1).
Now, we can set up the equation for the weighted average atomic mass: Average Atomic Mass=(x⋅78.9183371 amu)+((1−x)⋅80.9162906 amu)\text{Average Atomic Mass} = (x \cdot 78.9183371 \, \text{amu}) + ((1 – x) \cdot 80.9162906 \, \text{amu})
We are given that the average atomic mass of bromine is 79.904 amu, so we can plug this value into the equation: 79.904=(x⋅78.9183371)+((1−x)⋅80.9162906)79.904 = (x \cdot 78.9183371) + ((1 – x) \cdot 80.9162906)
Step 3: Solve for x (the Fraction of 79Br)
Expanding and simplifying the equation: 79.904=78.9183371x+80.9162906−80.9162906×79.904 = 78.9183371x + 80.9162906 – 80.9162906x 79.904=80.9162906−(80.9162906−78.9183371)x79.904 = 80.9162906 – (80.9162906 – 78.9183371)x 79.904−80.9162906=−(80.9162906−78.9183371)x79.904 – 80.9162906 = – (80.9162906 – 78.9183371)x −1.0122906=−(1.9979535)x-1.0122906 = -(1.9979535)x x=1.01229061.9979535≈0.5068x = \frac{1.0122906}{1.9979535} \approx 0.5068
Thus, the fraction of 79Br in the sample is approximately 0.5068 (or 50.68%).
Step 4: Determine the Mass of 79Br in 10.0 g
Now that we know the fraction of 79Br, we can calculate the mass of 79Br in a 10.0 g sample of bromine: Mass of 79Br=0.5068×10.0 g=5.068 g\text{Mass of 79Br} = 0.5068 \times 10.0 \, \text{g} = 5.068 \, \text{g}
Final Answer:
The mass of 79Br in the 10.0 g sample is approximately 5.068 g.
Explanation:
The problem involves calculating the proportion of 79Br in a sample of bromine based on its average atomic mass, which is a weighted average of the isotopes’ masses. By using the equation for a weighted average, we solve for the fraction of 79Br, then apply it to the total mass of the sample to find the amount of 79Br. This method allows us to determine the specific amount of each isotope in a sample of bromine based on its natural isotopic distribution.