Question

How many total ions are in 2.95 g of magnesium sulfite?

Answer 1

Answer:

Approximately $5.65 \times 10^{-2}\; \rm mol$, which is approximately $3.41 \times 10^{22}$ particles.

Explanation:

Formula of magnesium sulfite: $\rm MgSO_3$.

Look up the relative atomic mass of $\rm Mg$. $\rm S$, and $\rm O$ on a modern periodic table:

• $\rm Mg$: $24.305$.
• $\rm S$: $32.06$.
• $\rm O$: $15.999$.

The ionic compound $\rm MgSO_3$ consist of magnesium ions $\rm {Mg}^{2+}$ and sulfite ions $\rm {SO_3}^{2-}$.

Notice that $\rm {Mg}^{2+}\!$ and $\rm {SO_3}^{2-}\!$ combine at a one-to-one ratio to form the neural compound $\rm MgSO_3$. Therefore, each $\rm MgSO_3\!$ formula unit would include one $\rm {Mg}^{2+}$ ion and one $\rm {SO_3}^{2-}$ ion (that would be two ions in total).

Calculate the formula mass of one such formula unit:

$\begin{aligned}&\; M(\mathrm{MgSO_3}) \\ = & \; 24.305 + 32.06 + 3 \times 15.999 \\ = & \; 104.362\; \rm g \cdot mol^{-1}\end{aligned}$.

In other words, the mass of one mole of $\rm MgSO_3$ formula units (which includes one mole of $\rm {Mg}^{2+}$ ions and one mole of $\rm {SO_3}^{2-}$ ions) would be $104.362\; \rm g$.

Calculate the number of moles of such formula units in that $2.95\; \rm g$ of $\rm MgSO_3$:

$\begin{aligned}n&= \frac{m}{M}\\ &=\frac{2.95\; \rm g}{104.362\; \rm g \cdot mol^{-1}} \approx 2.82670 \times 10^{-2}\; \rm mol\end{aligned}$.

There are two moles of ions in each mole of $\rm MgSO_3$ formula units. Therefore, that $2.82670 \times 10^{-2}\; \rm mol$ of $\rm MgSO_3\!$ formula units would include approximately $2 \times 2.82670 \times 10^{-2}\; \rm mol \approx 5.65\times 10^{-2}\; \rm mol$ of ions.