Given information and chemical model
Aluminium ascorbate is treated as an ionic compound between \(\mathrm{Al^{3+}}\) and the ascorbate anion derived from ascorbic acid (vitamin C).
Step 1: Write the ion formulas
- Aluminium ion: \(\mathrm{Al^{3+}}\).
- Ascorbate ion (monoanion): \(\mathrm{C_6H_7O_6^-}\).
Step 2: Enforce charge neutrality to find the subscripts
An electrically neutral formula unit must have total charge \(0\). Let \(x\) be the number of ascorbate ions that combine with one \(\mathrm{Al^{3+}}\):
\[ (+3) + x(-1) = 0 \]Solving gives:
\[ 3 - x = 0 \;\;\Rightarrow\;\; x = 3 \]Therefore, the formula unit is:
\[ \mathrm{Al(C_6H_7O_6)_3} \]Step 3: Expand to a single chemical formula and reduce to the empirical formula
Expanding \(\mathrm{(C_6H_7O_6)_3}\) multiplies each subscript by \(3\):
\[ \mathrm{Al(C_6H_7O_6)_3 = AlC_{18}H_{21}O_{18}} \]The empirical formula is the simplest whole-number ratio of atoms. Because aluminium appears as \(\mathrm{Al_1}\), no common factor greater than \(1\) divides all subscripts simultaneously. Thus the empirical formula is the same as the expanded formula:
\[ \mathrm{AlC_{18}H_{21}O_{18}} \]Quick check table (atoms per formula unit)
| Component | Count | Contribution to total |
|---|---|---|
| \(\mathrm{Al^{3+}}\) | 1 | \(\mathrm{Al_1}\) |
| \(\mathrm{C_6H_7O_6^-}\) | 3 | \(\mathrm{C_{18}H_{21}O_{18}}\) |
| Total | — | \(\mathrm{AlC_{18}H_{21}O_{18}}\) |
Visualization: charge-balance schematic
Final result
The empirical formula for aluminium ascorbate (under the monoascorbate assumption) is \(\mathrm{AlC_{18}H_{21}O_{18}}\), equivalently written as \(\mathrm{Al(C_6H_7O_6)_3}\).