Question

How many formula units make up 24.2 g of magnesium chloride (MgCl2)?

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Answer 1

Approximately
`1.53 * 10^(23)` formula units (
`0.254\; \rm mol`).

Step-by-step explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium (
`\rm Mg`) and chlorine (
`\rm Cl`):

• 
  `\rm Mg`:
  `24.305`.
• 
  `\rm Cl`:
  `35.45`.

In other words, the mass of
`1\; \rm mol` of
`\rm Mg` atoms would be (approximately)
`24.305\; \rm g`.

Likewise, the mass of
`1\; \rm mol` of
`\rm Cl` atoms would be approximately
`35.45\; \rm g`.

One formula unit of the ionic compound
`\rm MgCl_(2)` includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of
`1\; \rm mol` of the formula units of this compound.

The formula
`\rm MgCl_(2)` includes one
`\rm Mg` atom and two
`\rm Cl` atoms.

Hence, every formula unit of
`\rm MgCl_(2) \!` would include the same number of atoms: one
`\rm Mg\!` atom and two
`\rm Cl\!` atoms. There would be
`1\; \rm mol` of
`\rm Mg` atoms and
`2\; \rm mol` of
`\rm Cl` atoms in
`1\; \rm mol\!` of
`\rm MgCl_(2)` formula units.

Thus, the mass of
`1\; \rm mol\!` of
`\rm MgCl_(2)` formula units would be equal to the mass of
`1\; \rm mol` of
`\rm Mg` atoms plus the mass of
`2\; \rm mol` of
`\rm Cl` atoms. (The mass of
`1\; \rm mol\!\!` of each atom could be found from the relative atomic mass of each element.)

`\begin{aligned}& M({\rm MgCl_(2)}) \\ =\; & 24.305\; {\rm g \cdot mol^(-1)} + 2* {\rm 35.45 \; \rm g \cdot mol^(-1)} \\ =\; & 95.205\; \rm g \cdot mol^(-1)\end{aligned}`.

In other words, the formula mass of
`\rm MgCl_(2)` is
`95.205\; \rm g \cdot mol^(-1)`.

Therefore, the number of formula units in
`m = 24.2\; \rm g` of
`\rm MgCl_(2)` would be:

`\begin{aligned}n &= \frac{m({\rm MgCl_(2)})}{M({\rm MgCl_(2)})} \\ &= (24.2\; \rm g)/(95.205\; \rm g\cdot mol^(-1)) \\ & \approx 0.254\; \rm mol\end{aligned}`.

Multiple
`n` by Avogadro's Number
`N_(A) \approx 6.022 * 10^(23)\; \rm mol^(-1)` to estimate the number of formula units in
`0.254\; \rm mol`:

`\begin{aligned}N &= n \cdot N_(A) \\ &\approx 0.254\; \rm mol * 6.022 * 10^(23)\; \rm mol^(-1) \\ &\approx 1.53* 10^(23)\end{aligned}`.