Question

1. iodine is 80% ¹²⁷i, 17% ¹²⁶i, and 3% ¹²⁸i. calculate the average atomic mass of iodine.
2. the natural abundance for boron isotopes is 19.9% ¹⁰b and 80.1% ¹¹b. calculate borons atomic mass.
3. hydrogen is 99% ¹h, 0.8% ²h, and 0.2% ³h. calculate its average atomic mass.
4. rubidium is a soft, silvery - white metal that has two common isotopes, ⁸⁵rb and ⁸⁷rb. if the abundance of ⁸⁵rb is 80.2% and the abundance of ⁸⁷rb is 19.8%, what is the average atomic mass of rubidium?
5. what is the atomic mass of hafnium if, out of every 100 atoms, 5 have a mass of 176, 19 have a mass of 177, 27 have a mass of 178, 14 have a mass of 179, and 35 have a mass of 180.0?

Explanation:

Step1: <Formula for average atomic mass>

The formula for average atomic mass \(M = \sum_{i = 1}^{n} (m_i\times a_i)\), where \(m_i\) is the mass of the \(i -\)th isotope and \(a_i\) is its abundance (in decimal form).

Step2: <Convert abundances to decimals>

For iodine: \(a_1=0.8\), \(a_2 = 0.17\), \(a_3=0.03\); \(m_1 = 127\), \(m_2=126\), \(m_3 = 128\).

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For boron: \(a_1 = 0.199\), \(a_2=0.801\); \(m_1 = 10\), \(m_2 = 11\).

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For hydrogen: \(a_1=0.99\), \(a_2 = 0.008\), \(a_3=0.002\); \(m_1 = 1\), \(m_2=2\), \(m_3 = 3\).

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For rubidium: \(a_1 = 0.802\), \(a_2=0.198\); \(m_1 = 85\), \(m_2 = 87\).

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For hafnium: Calculate abundances. \(a_1=\frac{5}{100}=0.05\), \(a_2=\frac{19}{100}=0.19\), \(a_3=\frac{27}{100}=0.27\), \(a_4=\frac{14}{100}=0.14\), \(a_5=\frac{35}{100}=0.35\); \(m_1 = 176\), \(m_2=177\), \(m_3 = 178\), \(m_4=179\), \(m_5 = 180\).

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Step3: <Sum up the values>

Iodine: \(101.6 + 21.42+3.84=126.86\)
Boron: \(1.99 + 8.811 = 10.801\)
Hydrogen: \(0.99+0.016 + 0.006=1.012\)
Rubidium: \(68.17+17.226=85.396\)
Hafnium: \(8.8+33.63+48.06+25.06+63 = 178.55\)

Answer:

1. The average atomic mass of iodine is \(126.86\).
2. The atomic mass of boron is \(10.801\).
3. The average atomic mass of hydrogen is \(1.012\).
4. The average atomic mass of rubidium is \(85.396\).
5. The atomic mass of hafnium is \(178.55\).