Notation check: \( \mathrm{NaNO_3} \) is sodium nitrate. Writing “nano3” in lowercase is not a standard chemical formula, so the chemically meaningful interpretation is the properly capitalized \( \mathrm{NaNO_3} \).
Step 1: Identify the ions in \( \mathrm{NaNO_3} \)
Sodium (Na) is a Group 1 metal and forms \( \mathrm{Na^+} \). The nitrate group is a known polyatomic anion with charge \(-1\): \( \mathrm{NO_3^-} \).
Therefore, sodium nitrate is composed of: \[ \mathrm{Na^+} \quad \text{and} \quad \mathrm{NO_3^-} \]
Step 2: Use charge balance to confirm the formula
An ionic compound formula is electrically neutral. Since \( \mathrm{Na^+} \) has charge \(+1\) and \( \mathrm{NO_3^-} \) has charge \(-1\), the simplest neutral ratio is \(1:1\):
\[ (+1) + (-1) = 0 \quad \Rightarrow \quad \mathrm{NaNO_3} \]
Step 3: Name the compound from the ions
Ionic naming uses the cation name followed by the anion name:
- Cation: sodium
- Anion: nitrate
The name is sodium nitrate.
Step 4: Assign oxidation states in \( \mathrm{NaNO_3} \)
For common oxidation-state rules, sodium is \(+1\) and oxygen is \(-2\). Let the oxidation state of nitrogen be \(x\). The sum of oxidation states in a neutral compound is zero:
\[ (+1) + x + 3(-2) = 0 \quad \Rightarrow \quad 1 + x - 6 = 0 \quad \Rightarrow \quad x = +5 \]
Oxidation states: Na \(= +1\), N \(= +5\), O \(= -2\).
Step 5: Dissociation in water
Sodium nitrate is treated as a strong electrolyte in water, separating into ions:
\[ \mathrm{NaNO_3(s)} \rightarrow \mathrm{Na^+(aq)} + \mathrm{NO_3^-(aq)} \]
Step 6: Molar mass and percent composition
Using standard atomic masses \(M(\mathrm{Na}) = 22.98976928\), \(M(\mathrm{N}) = 14.0067\), and \(M(\mathrm{O}) = 15.999\) in \( \mathrm{g\cdot mol^{-1}} \):
\[ M(\mathrm{NaNO_3}) = 22.98976928 + 14.0067 + 3(15.999) = 84.99346928\ \mathrm{g\cdot mol^{-1}} \]
| Element | Count | Atomic mass \((\mathrm{g\cdot mol^{-1}})\) | Contribution \((\mathrm{g\cdot mol^{-1}})\) |
|---|---|---|---|
| Na | 1 | 22.98976928 | 22.98976928 |
| N | 1 | 14.0067 | 14.0067 |
| O | 3 | 15.999 | 47.997 |
| Total | — | — | 84.99346928 |
Percent composition (mass percent) follows from each element’s contribution divided by the molar mass:
\[ \%\mathrm{Na} = \frac{22.98976928}{84.99346928}\times 100 = 27.05\% \] \[ \%\mathrm{N} = \frac{14.0067}{84.99346928}\times 100 = 16.48\% \] \[ \%\mathrm{O} = \frac{47.997}{84.99346928}\times 100 = 56.47\% \]