The Aluminium Nitrate Formula is Al(NO₃)₃, representing a white, water-soluble ionic salt formed between aluminium and nitric acid. Covered in NCERT Chemistry for Class 11 (Chapter 10 — The s-Block Elements and related p-Block content) and Class 12, this compound is also relevant for JEE Main, JEE Advanced, and NEET aspirants who need to understand ionic bonding, oxidation states, and salt chemistry. This article covers the formula, chemical structure, physical and chemical properties, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Aluminium Nitrate Formulas at a Glance
Quick reference for the most important formulas and facts about aluminium nitrate.
- Molecular formula: \( \text{Al(NO}_3)_3 \)
- Molar mass: \( M = 213.00 \text{ g/mol} \) (anhydrous)
- Nonahydrate molar mass: \( M = 375.13 \text{ g/mol} \)
- Ionic dissociation: \( \text{Al(NO}_3)_3 \rightarrow \text{Al}^{3+} + 3\text{NO}_3^{-} \)
- Oxidation state of Al: \( +3 \)
- Oxidation state of N in nitrate: \( +5 \)
- Thermal decomposition: \( 4\text{Al(NO}_3)_3 \rightarrow 2\text{Al}_2\text{O}_3 + 12\text{NO}_2 + 3\text{O}_2 \)
What is the Aluminium Nitrate Formula?
The Aluminium Nitrate Formula is \( \text{Al(NO}_3)_3 \), which represents aluminium(III) nitrate. It is an inorganic salt formed when aluminium reacts with nitric acid. The compound is also called nitric acid aluminium salt or aluminium(III) nitrate.
Aluminium belongs to Group 13 of the periodic table. It has three valence electrons in its outermost shell. Aluminium loses all three electrons to form the \( \text{Al}^{3+} \) cation. The nitrate ion \( \text{NO}_3^{-} \) carries a charge of \( -1 \). Three nitrate ions are therefore needed to balance the \( +3 \) charge of aluminium, giving the formula \( \text{Al(NO}_3)_3 \).
In NCERT Class 11 Chemistry, Chapter 10 discusses the chemistry of aluminium under p-Block elements. Students encounter aluminium salts in reactions involving acids and metals. In Class 12, this compound appears in the context of coordination chemistry and salt hydrolysis. The nonahydrate form, \( \text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O} \), is the most commonly encountered form in laboratories. It is a strong oxidising agent and is used in a variety of industrial and laboratory applications.
Aluminium Nitrate Formula — Expression, Structure and Variables
The molecular formula of aluminium nitrate is written as:
\[ \text{Al(NO}_3)_3 \]
This formula tells us that one aluminium ion is bonded ionically to three nitrate ions. Each nitrate group contains one nitrogen atom and three oxygen atoms. The total atom count per formula unit is: 1 Al + 3 N + 9 O = 13 atoms.
| Symbol / Group | Quantity / Ion | Charge / Unit |
|---|---|---|
| Al | Aluminium cation | \( +3 \) |
| NO₃ | Nitrate anion | \( -1 \) each (3 total) |
| N | Nitrogen (in nitrate) | Oxidation state \( +5 \) |
| O | Oxygen (in nitrate) | Oxidation state \( -2 \) |
| M (anhydrous) | Molar mass | 213.00 g/mol |
| M (nonahydrate) | Molar mass with 9H₂O | 375.13 g/mol |
Derivation of the Formula
Aluminium has an atomic number of 13. Its electronic configuration is \( 2, 8, 3 \). It loses 3 electrons to attain the stable noble gas configuration of neon. This gives the \( \text{Al}^{3+} \) ion.
The nitrate ion \( \text{NO}_3^{-} \) is a polyatomic anion with a charge of \( -1 \). To form a neutral compound, we need three nitrate ions for every one aluminium ion:
\[ \text{Al}^{3+} + 3\text{NO}_3^{-} \rightarrow \text{Al(NO}_3)_3 \]
This cross-multiplication of charges (Al charge = 3, nitrate charge = 1) directly gives the subscript 3 for the nitrate group. The compound is electrically neutral overall.
Chemical Structure of Aluminium Nitrate
Aluminium nitrate is an ionic compound. The \( \text{Al}^{3+} \) cation is surrounded by three \( \text{NO}_3^{-} \) anions in the crystal lattice. Each nitrate ion has a trigonal planar geometry with bond angles of \( 120^\circ \). The N–O bond length in nitrate is approximately 124 pm due to resonance delocalisation across the three oxygen atoms.
In the nonahydrate form \( \text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O} \), water molecules coordinate around the \( \text{Al}^{3+} \) ion, forming a hydrated complex. The compound appears as white crystalline solids or powder at room temperature.
Physical Properties of Aluminium Nitrate
| Property | Value |
|---|---|
| Appearance | White crystalline solid |
| Molar mass (anhydrous) | 213.00 g/mol |
| Molar mass (nonahydrate) | 375.13 g/mol |
| Melting point (nonahydrate) | 73.9 °C |
| Boiling point | Decomposes at ~150 °C |
| Solubility in water | Highly soluble (73.9 g/100 mL at 20 °C) |
| Density (nonahydrate) | 1.72 g/cm³ |
| Odour | Odourless |
Chemical Properties of Aluminium Nitrate
Aluminium nitrate is a strong oxidising agent. It decomposes on heating to produce aluminium oxide, nitrogen dioxide, and oxygen:
\[ 4\text{Al(NO}_3)_3 \xrightarrow{\Delta} 2\text{Al}_2\text{O}_3 + 12\text{NO}_2 + 3\text{O}_2 \]
In aqueous solution, it dissociates completely into ions:
\[ \text{Al(NO}_3)_3 \rightarrow \text{Al}^{3+}(aq) + 3\text{NO}_3^{-}(aq) \]
The solution is acidic in nature. The \( \text{Al}^{3+} \) ion undergoes hydrolysis:
\[ \text{Al}^{3+} + 3\text{H}_2\text{O} \rightleftharpoons \text{Al(OH)}_3 + 3\text{H}^+ \]
Aluminium nitrate reacts with sodium hydroxide to give aluminium hydroxide and sodium nitrate:
\[ \text{Al(NO}_3)_3 + 3\text{NaOH} \rightarrow \text{Al(OH)}_3 + 3\text{NaNO}_3 \]
Complete Chemistry Formula Sheet — Aluminium Compounds and Related Salts
| Formula Name | Expression | Key Variables / Notes | Molar Mass (g/mol) | NCERT Chapter Reference |
|---|---|---|---|---|
| Aluminium Nitrate | \( \text{Al(NO}_3)_3 \) | Al: +3, NO₃⁻: –1, ionic salt | 213.00 | Class 11, Ch 10 & 11 |
| Aluminium Chloride | \( \text{AlCl}_3 \) | Lewis acid, covalent character | 133.34 | Class 11, Ch 11 |
| Aluminium Oxide (Alumina) | \( \text{Al}_2\text{O}_3 \) | Amphoteric oxide | 101.96 | Class 11, Ch 11 |
| Aluminium Hydroxide | \( \text{Al(OH)}_3 \) | Amphoteric hydroxide | 78.00 | Class 11, Ch 11 |
| Aluminium Sulphate | \( \text{Al}_2(\text{SO}_4)_3 \) | Used in water purification (alum) | 342.15 | Class 11, Ch 11 |
| Potash Alum | \( \text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O} \) | Double salt | 474.39 | Class 12, Ch 5 |
| Ammonia | \( \text{NH}_3 \) | Formed in Haber process; reacts with Al salts | 17.03 | Class 11, Ch 10 |
| Nitric Acid | \( \text{HNO}_3 \) | Reacts with Al to form Al(NO₃)₃ | 63.01 | Class 12, Ch 7 |
| Silicon Dioxide | \( \text{SiO}_2 \) | Acidic oxide; found with Al in earth’s crust | 60.08 | Class 11, Ch 11 |
| Bleaching Powder | \( \text{Ca(OCl)Cl} \) | Oxidising agent like Al(NO₃)₃ | 127.00 | Class 11, Ch 10 |
Aluminium Nitrate Formula — Solved Examples
Example 1 (Class 9-10 Level) — Calculating Molar Mass
Problem: Calculate the molar mass of aluminium nitrate, \( \text{Al(NO}_3)_3 \), using atomic masses: Al = 27, N = 14, O = 16.
Given: Formula = \( \text{Al(NO}_3)_3 \); Atomic masses: Al = 27 u, N = 14 u, O = 16 u
Step 1: Count each type of atom in the formula.
Aluminium: 1 atom; Nitrogen: 3 atoms; Oxygen: 9 atoms (3 nitrate groups × 3 oxygen each)
Step 2: Multiply each atomic mass by the number of atoms.
\( \text{Mass of Al} = 1 \times 27 = 27 \text{ u} \)
\( \text{Mass of N} = 3 \times 14 = 42 \text{ u} \)
\( \text{Mass of O} = 9 \times 16 = 144 \text{ u} \)
Step 3: Add all masses.
\( M = 27 + 42 + 144 = 213 \text{ g/mol} \)
Answer
Molar mass of Aluminium Nitrate = 213 g/mol
Example 2 (Class 11-12 Level) — Moles and Mass Calculation
Problem: How many moles of \( \text{NO}_3^{-} \) ions are present in 42.6 g of anhydrous aluminium nitrate? Also find the number of nitrate ions.
Given: Mass of \( \text{Al(NO}_3)_3 \) = 42.6 g; Molar mass = 213 g/mol; Avogadro’s number \( N_A = 6.022 \times 10^{23} \)
Step 1: Find moles of \( \text{Al(NO}_3)_3 \).
\( n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{42.6}{213} = 0.2 \text{ mol} \)
Step 2: Each formula unit of \( \text{Al(NO}_3)_3 \) provides 3 nitrate ions.
\( n(\text{NO}_3^{-}) = 0.2 \times 3 = 0.6 \text{ mol} \)
Step 3: Calculate the number of nitrate ions.
\( N = 0.6 \times 6.022 \times 10^{23} = 3.613 \times 10^{23} \text{ ions} \)
Answer
Moles of \( \text{NO}_3^{-} \) = 0.6 mol; Number of nitrate ions = 3.613 × 10²³
Example 3 (JEE/NEET Level) — Stoichiometry and Reaction
Problem: Aluminium reacts with dilute nitric acid to form aluminium nitrate, ammonium nitrate, and water according to the equation:
\( 8\text{Al} + 30\text{HNO}_3(\text{dilute}) \rightarrow 8\text{Al(NO}_3)_3 + 3\text{NH}_4\text{NO}_3 + 9\text{H}_2\text{O} \)
If 5.4 g of aluminium is dissolved completely, find (a) the mass of aluminium nitrate formed, and (b) the volume of \( 2\text{ M} \) HNO₃ solution required.
Given: Mass of Al = 5.4 g; Molar mass of Al = 27 g/mol; Molar mass of Al(NO₃)₃ = 213 g/mol
Step 1: Find moles of Al.
\( n(\text{Al}) = \dfrac{5.4}{27} = 0.2 \text{ mol} \)
Step 2 (a): From stoichiometry, 8 mol Al gives 8 mol Al(NO₃)₃.
So 0.2 mol Al gives 0.2 mol Al(NO₃)₃.
\( m(\text{Al(NO}_3)_3) = 0.2 \times 213 = 42.6 \text{ g} \)
Step 3 (b): From stoichiometry, 8 mol Al requires 30 mol HNO₃.
\( n(\text{HNO}_3) = 0.2 \times \dfrac{30}{8} = 0.75 \text{ mol} \)
Step 4: Volume of 2 M HNO₃ required:
\( V = \dfrac{n}{C} = \dfrac{0.75}{2} = 0.375 \text{ L} = 375 \text{ mL} \)
Answer
(a) Mass of Al(NO₃)₃ formed = 42.6 g
(b) Volume of 2 M HNO₃ required = 375 mL
CBSE Exam Tips 2025-26 for Aluminium Nitrate Formula
- Memorise the formula correctly: Write \( \text{Al(NO}_3)_3 \) — not AlNO₃ or Al₂(NO₃)₃. The subscript 3 outside the bracket is a common source of error.
- Know the molar mass calculation: CBSE frequently asks students to calculate molar mass. Remember: 1 Al (27) + 3 N (42) + 9 O (144) = 213 g/mol.
- Understand the ionic dissociation: In solution, \( \text{Al(NO}_3)_3 \) gives 4 ions total — 1 Al³⁺ and 3 NO₃⁻. This is important for colligative properties questions.
- Thermal decomposition equation: We recommend memorising the balanced equation \( 4\text{Al(NO}_3)_3 \rightarrow 2\text{Al}_2\text{O}_3 + 12\text{NO}_2 + 3\text{O}_2 \) for 2025-26 board exams. It frequently appears in 3-mark questions.
- Amphoteric nature of Al(OH)₃: Questions on aluminium salts often lead to Al(OH)₃. Know that it dissolves in both acid and base — this is a common 2-mark question.
- Distinguish anhydrous vs nonahydrate: When a question mentions the laboratory form, it refers to \( \text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O} \) with molar mass 375.13 g/mol. Always check which form the question specifies.
Common Mistakes to Avoid with the Aluminium Nitrate Formula
- Wrong formula writing: Many students write AlNO₃ or Al(NO₃)₂. The correct formula is \( \text{Al(NO}_3)_3 \) because Al has a +3 oxidation state and each nitrate carries –1 charge. Three nitrate groups are needed.
- Forgetting the brackets: Writing AlNO₃₃ instead of Al(NO₃)₃ is a common error. The brackets indicate that the subscript 3 applies to the entire nitrate group \( \text{NO}_3 \), not just oxygen.
- Incorrect molar mass calculation: Students sometimes count only 3 oxygen atoms instead of 9. Each of the 3 nitrate groups has 3 oxygen atoms, giving 9 oxygen atoms in total.
- Confusing oxidation states: The oxidation state of nitrogen in \( \text{NO}_3^{-} \) is +5, not +3. Do not confuse it with the oxidation state of aluminium (+3).
- Ignoring the hydrated form: In stoichiometry problems, always check whether the question refers to anhydrous Al(NO₃)₃ (213 g/mol) or the nonahydrate Al(NO₃)₃·9H₂O (375.13 g/mol). Using the wrong molar mass will give an incorrect answer.
JEE/NEET Application of the Aluminium Nitrate Formula
In our experience, JEE aspirants encounter the Aluminium Nitrate Formula in several key topic areas. Understanding these patterns helps you score more efficiently in MCQ-based papers.
1. Stoichiometry and Mole Concept
JEE Main frequently tests mole calculations involving ionic compounds. Questions ask for the number of ions, moles of a specific ion, or the mass of product formed. For \( \text{Al(NO}_3)_3 \), remember it produces 4 ions per formula unit upon dissociation. This directly affects van’t Hoff factor \( i = 4 \) in colligative property calculations:
\[ \Delta T_b = i \cdot K_b \cdot m = 4 K_b m \]
2. Redox Chemistry and Oxidising Nature
NEET and JEE Advanced test the oxidising behaviour of nitrate salts. Aluminium nitrate is a strong oxidiser. In reactions with reducing agents, the nitrate ion (N in +5 state) gets reduced. Students must balance redox reactions involving \( \text{Al(NO}_3)_3 \) using the ion-electron method or oxidation number method — both are tested in Class 12 NCERT Chapter 8.
3. Salt Hydrolysis and pH
The \( \text{Al}^{3+} \) ion is a small, highly charged cation. It undergoes significant hydrolysis in water, making the solution acidic. JEE Advanced questions on ionic equilibrium ask students to calculate the pH of \( \text{Al(NO}_3)_3 \) solutions using the hydrolysis constant. The relevant equilibrium is:
\[ \text{Al}^{3+} + \text{H}_2\text{O} \rightleftharpoons \text{AlOH}^{2+} + \text{H}^+ \]
Our experts suggest practising at least 5 problems on salt hydrolysis involving Group 13 metal ions before your JEE Advanced exam. This topic carries 2-3 marks in most years.
4. Coordination Chemistry (JEE Advanced)
In JEE Advanced, aluminium nitrate appears in the context of coordination compounds. The \( \text{Al}^{3+} \) ion acts as a Lewis acid and can form complexes with water, hydroxide, and other ligands. Understanding the coordination number (typically 6 for Al³⁺) and geometry (octahedral) is essential for scoring in this section.
FAQs on Aluminium Nitrate Formula
We hope this comprehensive guide on the Aluminium Nitrate Formula has helped you understand the structure, properties, and applications of Al(NO₃)₃. For more related chemistry formulas, explore our Complete Chemistry Formulas Hub. You may also find these articles useful: Ammonia Formula, Silicon Dioxide Formula, and Bleaching Powder Formula. For the official NCERT syllabus reference, visit ncert.nic.in.