Question

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

Answer 1

Approximately
`5.65 * 10^(-2)\; \rm mol`, which is approximately
`3.41 * 10^(22)` particles.

Step-by-step 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 * 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&= (m)/(M)\\ &=(2.95\; \rm g)/(104.362\; \rm g \cdot mol^(-1)) \approx 2.82670 * 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 * 10^(-2)\; \rm mol` of
`\rm MgSO_3\!` formula units would include approximately
`2 * 2.82670 * 10^(-2)\; \rm mol \approx 5.65* 10^(-2)\; \rm mol` of ions.