Questions: Calculate the number of cations and anions in 4.32 g of Na2SO4.

Solution Steps

The molar mass of $\mathrm{Na}_2\mathrm{SO}_4$ is calculated by summing the atomic masses of its constituent elements:

• Sodium (Na): $2 \times 22.9898 \, \text{g/mol} = 45.9796 \, \text{g/mol}$
• Sulfur (S): $1 \times 32.065 \, \text{g/mol} = 32.065 \, \text{g/mol}$
• Oxygen (O): $4 \times 15.999 \, \text{g/mol} = 63.996 \, \text{g/mol}$

Adding these together gives the molar mass of $\mathrm{Na}_2\mathrm{SO}_4$:

$$M = 45.9796 + 32.065 + 63.996 = 142.0406 \, \text{g/mol}$$

Using the given mass of $\mathrm{Na}_2\mathrm{SO}_4$ (4.32 g), we calculate the number of moles:

$$\text{moles of } \mathrm{Na}_2\mathrm{SO}_4 = \frac{4.32 \, \text{g}}{142.0406 \, \text{g/mol}} = 0.0304 \, \text{mol}$$

Each formula unit of $\mathrm{Na}_2\mathrm{SO}_4$ contains:

• 2 sodium cations ($\mathrm{Na}^+$)
• 1 sulfate anion ($\mathrm{SO}_4^{2-}$)

Thus, the number of cations is twice the number of moles of $\mathrm{Na}_2\mathrm{SO}_4$, and the number of anions is equal to the number of moles of $\mathrm{Na}_2\mathrm{SO}_4$.

• Number of $\mathrm{Na}^+$ cations: $2 \times 0.0304 \, \text{mol} = 0.0608 \, \text{mol}$
• Number of $\mathrm{SO}_4^{2-}$ anions: $0.0304 \, \text{mol}$

Using Avogadro's number ($6.022 \times 10^{23} \, \text{mol}^{-1}$), we convert moles to the number of ions:

• Number of $\mathrm{Na}^+$ cations: $$0.0608 \, \text{mol} \times 6.022 \times 10^{23} \, \text{mol}^{-1} = 3.664 \times 10^{22} \, \text{cations}$$

• Number of $\mathrm{SO}_4^{2-}$ anions: $$0.0304 \, \text{mol} \times 6.022 \times 10^{23} \, \text{mol}^{-1} = 1.832 \times 10^{22} \, \text{anions}$$

Final Answer