Question

1. bromine has two isotopes, bromine-79 and bromine-81, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%) were determined by experiments.

a. how many protons, neutrons, and electrons does each isotope have? (complete on your paper)
table: isotope, # protons, # neutrons, # electrons. bromine-79: 35, 44, 35; bromine-81: 35, 46, 35
b. calculate the average atomic mass of bromine based on these experiments. round your answer to the nearest hundredth.
79.88 amu
*hint: do not round until the very end

Explanation:

Part A: Protons, Neutrons, Electrons Calculation

Step 1: Protons (Atomic Number)

Bromine has an atomic number of 35, so both isotopes have 35 protons.
For Bromine - 79: \( \text{Protons} = 35 \)
For Bromine - 81: \( \text{Protons} = 35 \)

Step 2: Neutrons (Mass Number - Atomic Number)

Mass number of Bromine - 79 is 79, so neutrons: \( 79 - 35 = 44 \)
Mass number of Bromine - 81 is 81, so neutrons: \( 81 - 35 = 46 \)

Step 3: Electrons (Equal to Protons in Neutral Atoms)

Both isotopes are neutral, so electrons equal protons.
For Bromine - 79: \( \text{Electrons} = 35 \)
For Bromine - 81: \( \text{Electrons} = 35 \)

Part B: Average Atomic Mass Calculation

Step 1: Recall the Formula for Average Atomic Mass

The formula for average atomic mass (\( A \)) is:
\[ A = (m_1 \times f_1) + (m_2 \times f_2) \]
where \( m_1, m_2 \) are the masses of isotopes, and \( f_1, f_2 \) are their fractional abundances.

Step 2: Convert Percent Abundances to Fractions

Bromine - 79: \( f_1 = \frac{50.69}{100} = 0.5069 \), \( m_1 = 78.9183 \, \text{amu} \)
Bromine - 81: \( f_2 = \frac{49.31}{100} = 0.4931 \), \( m_2 = 80.9163 \, \text{amu} \)

Step 3: Calculate the Contribution of Each Isotope

Contribution of Bromine - 79: \( 78.9183 \times 0.5069 \)
\[ 78.9183 \times 0.5069 \approx 78.9183 \times 0.5069 \approx 40.00 \, (\text{precise: } 78.9183 \times 0.5069 = 40.003 \, \text{(approx)}) \]

Contribution of Bromine - 81: \( 80.9163 \times 0.4931 \)
\[ 80.9163 \times 0.4931 \approx 39.88 \, (\text{precise: } 80.9163 \times 0.4931 = 39.882 \, \text{(approx)}) \]

Step 4: Sum the Contributions

\[ A = 40.003 + 39.882 = 79.885 \approx 79.89 \, \text{amu} \]

Final Answers

Part A:

• Bromine - 79: Protons = 35, Neutrons = 44, Electrons = 35
• Bromine - 81: Protons = 35, Neutrons = 46, Electrons = 35

Part B:

The average atomic mass of Bromine is \(\boldsymbol{79.89}\) amu (rounded to the nearest hundredth).

Answer:

Part A: Protons, Neutrons, Electrons Calculation

Step 1: Protons (Atomic Number)

Bromine has an atomic number of 35, so both isotopes have 35 protons.
For Bromine - 79: \( \text{Protons} = 35 \)
For Bromine - 81: \( \text{Protons} = 35 \)

Step 2: Neutrons (Mass Number - Atomic Number)

Mass number of Bromine - 79 is 79, so neutrons: \( 79 - 35 = 44 \)
Mass number of Bromine - 81 is 81, so neutrons: \( 81 - 35 = 46 \)

Step 3: Electrons (Equal to Protons in Neutral Atoms)

Both isotopes are neutral, so electrons equal protons.
For Bromine - 79: \( \text{Electrons} = 35 \)
For Bromine - 81: \( \text{Electrons} = 35 \)

Part B: Average Atomic Mass Calculation

Step 1: Recall the Formula for Average Atomic Mass

The formula for average atomic mass (\( A \)) is:
\[ A = (m_1 \times f_1) + (m_2 \times f_2) \]
where \( m_1, m_2 \) are the masses of isotopes, and \( f_1, f_2 \) are their fractional abundances.

Step 2: Convert Percent Abundances to Fractions

Bromine - 79: \( f_1 = \frac{50.69}{100} = 0.5069 \), \( m_1 = 78.9183 \, \text{amu} \)
Bromine - 81: \( f_2 = \frac{49.31}{100} = 0.4931 \), \( m_2 = 80.9163 \, \text{amu} \)

Step 3: Calculate the Contribution of Each Isotope

Contribution of Bromine - 79: \( 78.9183 \times 0.5069 \)
\[ 78.9183 \times 0.5069 \approx 78.9183 \times 0.5069 \approx 40.00 \, (\text{precise: } 78.9183 \times 0.5069 = 40.003 \, \text{(approx)}) \]

Contribution of Bromine - 81: \( 80.9163 \times 0.4931 \)
\[ 80.9163 \times 0.4931 \approx 39.88 \, (\text{precise: } 80.9163 \times 0.4931 = 39.882 \, \text{(approx)}) \]

Step 4: Sum the Contributions

\[ A = 40.003 + 39.882 = 79.885 \approx 79.89 \, \text{amu} \]

Final Answers

Part A:

• Bromine - 79: Protons = 35, Neutrons = 44, Electrons = 35
• Bromine - 81: Protons = 35, Neutrons = 46, Electrons = 35

Part B:

The average atomic mass of Bromine is \(\boldsymbol{79.89}\) amu (rounded to the nearest hundredth).