Question

Question: Use the data from Table 19.1 to predict which is more stable: 16 protons, 16 neutrons and 16 electrons organized as two ${}_{\mathbf{8}}{}^{\mathbf{16}}\mathbf{O}$or one ${}_{\mathbf{16}}{}^{\mathbf{32}}\mathbf{S}$. What is the mass difference?

Short answer

Since the mass of one ${}_{16}{}^{32}\mathrm{S}$atom is less than two ${}_8{}^{16}\mathrm{O}$atoms, so ${}_{16}{}^{32}\mathrm{S}$is more stable.

The mass difference is$0.01775\mathrm{u}$.

Step-by-step solution

Determining the stability

We will take help from Table 19.1 to calculate the total mass of two ${}_8{}^{16}\mathrm{O}$atoms.

Mass of one ${}_8{}^{16}\mathrm{O}$atom$=15.99491\mathrm{u}$

Mass of two${}_8{}^{16}\mathrm{O}$atoms $=2\times 15.99491\mathrm{u}=31.98982\mathrm{u}$

Mass of one ${}_{16}{}^{32}\mathrm{S}$atom$=31.97207\mathrm{u}$

Since the mass of one atom of ${}_{16}{}^{32}\mathrm{S}$is less than mass of two atoms of ${}_8{}^{16}\mathrm{O}$,${}_{16}{}^{32}\mathrm{S}$has more stability.

Mass difference

The mass difference is obtained by subtracting the mass of ${}_{16}{}^{32}\mathrm{S}$atom from the mass of two ${}_8{}^{16}\mathrm{O}$atoms.

$$\begin{gathered} ∆\mathrm{m}=31.98982\mathrm{u}-31.97207\mathrm{u} \\ =0.01775\mathrm{u} \end{gathered}$$