Magnesium nitrate is an ionic compound composed of magnesium cations, Mg2+, and nitrate anions, NO3−. The central skill associated with magnesium nitrate is writing the correct ionic formula using charge balance while keeping the polyatomic ion intact.
Formula of magnesium nitrate from ionic charges
Magnesium forms a \(+2\) cation in most ionic compounds, written as Mg2+. The nitrate ion is a polyatomic anion with charge \(-1\), written as NO3−.
Step 1: Write the ions with charges
Mg2+ and NO3−
Step 2: Balance total positive and negative charge
One Mg2+ contributes \(+2\). Each nitrate contributes \(-1\). Two nitrates are needed to supply \(-2\), giving an overall neutral formula unit.
\[ (+2) + 2 \times (-1) = 0 \]
Step 3: Use parentheses for a repeated polyatomic ion
Because the nitrate ion contains multiple atoms, parentheses indicate that the subscript 2 applies to the entire NO3 group:
\[ \text{magnesium nitrate} = \mathrm{Mg(NO_3)_2} \]
A frequent error is writing MgN2O6 or MgNO3 2 without parentheses. Parentheses are required whenever more than one copy of a polyatomic ion is present in the formula unit.
Key facts and quick calculations for magnesium nitrate
Molar mass of Mg(NO3)2
The molar mass is obtained by summing atomic masses according to the subscripts:
\[ M\big(\mathrm{Mg(NO_3)_2}\big) = M(\mathrm{Mg}) + 2 \times \big(M(\mathrm{N}) + 3 \times M(\mathrm{O})\big) \]
| Element | Count in Mg(NO3)2 | Contribution to molar mass |
|---|---|---|
| Mg | 1 | \(1 \times M(\mathrm{Mg})\) |
| N | 2 | \(2 \times M(\mathrm{N})\) |
| O | 6 | \(6 \times M(\mathrm{O})\) |
Dissociation (strong electrolyte behavior)
In water, magnesium nitrate separates into ions. The balanced dissociation equation is:
\[ \mathrm{Mg(NO_3)_2(s) \rightarrow Mg^{2+}(aq) + 2\,NO_3^{-}(aq)} \]
This stoichiometry is essential for ion-concentration questions. For example, a solution with concentration \(C\) in \(\mathrm{Mg(NO_3)_2}\) has nitrate concentration \(2C\) in \(\mathrm{NO_3^{-}}\), assuming complete dissociation.
Worked example: mass needed to prepare a magnesium nitrate solution
Example goal: prepare \(250.0\ \mathrm{mL}\) of \(0.200\ \mathrm{M}\) magnesium nitrate solution. Determine the required mass of \(\mathrm{Mg(NO_3)_2}\) (assume anhydrous magnesium nitrate).
Step 1: Convert volume to liters and compute moles
\[ V = 250.0\ \mathrm{mL} = 0.2500\ \mathrm{L} \] \[ n = M \cdot V = 0.200\ \mathrm{mol \cdot L^{-1}} \times 0.2500\ \mathrm{L} = 0.0500\ \mathrm{mol} \]
Step 2: Convert moles to mass using molar mass
With \(M_m = M\big(\mathrm{Mg(NO_3)_2}\big)\) in \(\mathrm{g \cdot mol^{-1}}\), \[ m = n \cdot M_m = 0.0500\ \mathrm{mol} \times M_m \]
Substituting a numerical molar mass value (from a periodic table) completes the calculation in grams.
If a hydrated form (such as a magnesium nitrate hydrate) is used, replace \(M_m\) with the hydrate’s molar mass. The formula-writing logic for magnesium nitrate remains Mg(NO3)2; only the molar mass changes.
Summary
Magnesium nitrate is built from Mg2+ and NO3−; charge neutrality requires two nitrates per magnesium, giving \(\mathrm{Mg(NO_3)_2}\). Parentheses preserve the polyatomic nitrate ion, and the dissociation \(\mathrm{Mg^{2+} + 2\,NO_3^{-}}\) drives many molarity and ion-concentration calculations.