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Average Atomic Mass | Introduction to Chemistry

Calculate the average atomic mass of an element given its isotopes and their ... natural abundanceThe abundance of a particular isotope naturally found on the ... mass numberThe total number of protons and neutrons in an atomic nucleus.

Pravalika
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a)

Given that

Mass of ^{6}_{3} Li, m_1 = 6.01512

Mass of ^{7}_{3} Li, m_2 = 7.01600

Abundances isotopes ^{6}_{3} Li, \eta_1 = 7.5 %

Abundances isotopes ^{3}_{7} Li, \eta_2 = 92.5 %

Step 1: Find the atomic mass of an isotope with abundance.

Relationship between isotope abundance and average atomic mass

Atomic mass of lithium = \frac{m_1 \eta_1 + m_2 \eta_2}{\eta_1 + \eta_2}

= \frac{6.01512 * (7.5) + 7.01600 * (92.5)}{7.5 + 92.5}

= \frac{45.1134 + 648.98}{100}

= 6.941 u

b)

Mass of ^{10}_{5} B, m_1 = 10.01294

Mass of ^{11}_{5} B, m_2 = 11.00931

Atomic mass of boron = 10.811 u

Abundances isotopes ^{10}_{5} B, \eta_1 = ?

Abundances isotopes ^{11}_{5} B, \eta_2 = ?

Step 1: Relation between abundance of isotopes and average atomic mass.

Lets say, Abundances isotopes ^{10}_{5} B, \eta_1 = x %

Abundances isotopes  _5^{11}B,\ \eta_2=\left(100\ -x\right)\%

Atomic mass of boron = \frac{m_1 \eta_1 + m_2 \eta_2}{\eta_1 + \eta_2}

10.811 = \frac{ 10.01294 * x + 11.00931 * (100 - x)}{x + (100 - x)}

10.811 * 100 = 10.01294x + 1100.931 - 11.00931x

1081.11 - 1100.931 = - 0.99637x

x = \frac{19.821}{0.99637}

x = 19.89 %

Abundances isotopes ^{10}_{5} B, \eta_1 = 19.89 %

Abundances isotopes  _5^{11}B,\ \eta_2=\left(100\ -x\right)\% = 100 - 19.89 = 80.11 %

Hence, Abundances isotopes \eta_1=19.89 % and Abundances isotopes \eta_2 =\ 80.11%