Problem: What is the nuclear binding energy of a lithium-7 nucleus in units of kJ/mol and eV/nucleus? (Mass of a lithium-7 atom = 7.016003 amu.)

First, calculate mass defect: 

 Li–7 (atomic mass = 7.016003 amu)

Atomic number→ 3 → 3 protons

Mass number → 7 

Neutrons → 7-3 = 4

Mass values needed: 

Proton = 1.007277 amu

electron = 0.000548597 amu

neutron = 1.008665 amu

Lithium = 7.016003 amu

Calculate mass defect: 

$$\begin{gathered} \boxed{\mathbf{∆}\mathbf{m}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{[}\mathbf{Z}(m_p +m_e)\mathbf{+}(A-Z){\mathbf{m}}_{\mathbf{n}}\mathbf{]}\mathbf{-}{\mathbf{m}}_{\mathbf{atom}}} \\ \mathbf{∆}\mathbf{m}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{[}\mathbf{3}(1.007277 \mathrm{amu} +0.000548597 \mathrm{amu})\mathbf{+}(7-3)(1.008665 \mathrm{amu})\mathbf{]} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{016003}\mathbf{ }\mathbf{amu} \\ \mathbf{∆}\mathbf{m}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{[}\mathbf{3}(1.007826)\mathbf{+}\left(4\right)(1.008665 \mathrm{amu})\mathbf{]}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{016003}\mathbf{ }\mathbf{amu} \end{gathered}$$

Δm = 0.042105 amu

Substitute the equation below: 

1 amu c² = 931.49 MeV

$$\begin{gathered} \boxed{\mathbf{E}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{∆}{\mathbf{mc}}^{\mathbf{2}}} \\ \mathbf{E}\mathbf{ }\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{042105}\mathbf{ }\cancel{\mathbf{amu}}\mathbf{)}\cancel{{\mathbf{c}}^{\mathbf{2}}}\mathbf{ }\mathbf{\times }\frac{\mathbf{931}\mathbf{.}\mathbf{49}\cancel{\mathbf{ }\mathbf{MeV}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{amu}\mathbf{ }{\mathbf{c}}^{\mathbf{2}}}\mathbf{ }}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{6}}\mathbf{ }\mathbf{eV}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{ }\mathbf{MeV}}} \end{gathered}$$

E = 3.92 x 10⁷  eV/nucleus

Convert to kJ/mol: