The Magnesium Nitrate Formula is Mg(NO₃)₂, representing an ionic compound formed by one magnesium cation (Mg²⁺) and two nitrate anions (NO₃⁻). This inorganic salt is covered in NCERT Chemistry for Class 10 and revisited in Class 11 under ionic equilibrium and salt chemistry. It is also relevant for NEET and JEE Main aspirants studying chemical bonding and solutions. This article covers the formula, structure, molar mass, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Magnesium Nitrate Formulas at a Glance
Quick reference for the most important formulas related to Magnesium Nitrate.
- Molecular formula: \( \text{Mg(NO}_3)_2 \)
- Molar mass: \( M = 148.31 \text{ g/mol} \)
- Moles formula: \( n = \frac{m}{M} \)
- Molarity: \( C = \frac{n}{V} \)
- Dissociation: \( \text{Mg(NO}_3)_2 \rightarrow \text{Mg}^{2+} + 2\text{NO}_3^- \)
- Percent composition of Mg: \( \%\text{Mg} = \frac{24.31}{148.31} \times 100 \)
- Hydrated form molar mass (hexahydrate): \( M = 256.41 \text{ g/mol} \)
What is Magnesium Nitrate Formula?
The Magnesium Nitrate Formula, written as Mg(NO₃)₂, represents an ionic salt composed of magnesium and nitrate ions. Magnesium (Mg) is an alkaline earth metal in Group 2 of the periodic table. It forms a stable 2+ cation. The nitrate ion (NO₃⁻) carries a single negative charge. To balance the 2+ charge of magnesium, two nitrate ions are needed. This gives the formula Mg(NO₃)₂.
In NCERT Chemistry Class 10 (Chapter 2 — Acids, Bases and Salts), salt formation is discussed in detail. Magnesium nitrate is formed when magnesium or magnesium oxide reacts with dilute nitric acid. In Class 11 (Chapter 7 — Equilibrium), its dissociation in water is studied. The compound is highly soluble in water and is used as a fertiliser and in pyrotechnics. Its IUPAC name is magnesium dinitrate. The anhydrous form has the formula Mg(NO₃)₂, while the common laboratory form is the hexahydrate, Mg(NO₃)₂·6H₂O.
Magnesium Nitrate Formula — Expression and Variables
The molecular formula for magnesium nitrate is:
\[ \text{Mg(NO}_3)_2 \]
This formula tells us that each formula unit contains one magnesium atom, two nitrogen atoms, and six oxygen atoms. The molar mass is calculated as follows:
\[ M = 24.31 + 2(14.01 + 3 \times 16.00) = 24.31 + 2(62.00) = 148.31 \text{ g/mol} \]
| Symbol / Ion | Component | Quantity in Formula | Atomic/Molar Mass (g/mol) |
|---|---|---|---|
| Mg²⁺ | Magnesium cation | 1 | 24.31 |
| NO₃⁻ | Nitrate anion | 2 | 62.00 each |
| N | Nitrogen atoms | 2 | 14.01 each |
| O | Oxygen atoms | 6 | 16.00 each |
| Mg(NO₃)₂ | Anhydrous molar mass | — | 148.31 |
| Mg(NO₃)₂·6H₂O | Hexahydrate molar mass | — | 256.41 |
Structure of Magnesium Nitrate
Magnesium nitrate has an ionic structure. The central Mg²⁺ ion is surrounded by two NO₃⁻ ions through electrostatic attraction. Each nitrate ion has a planar, trigonal geometry. The nitrogen atom is at the centre of the nitrate ion. It is bonded to three oxygen atoms with bond angles of 120°. The overall compound is white and crystalline in its solid state. In solution, it fully dissociates as shown below:
\[ \text{Mg(NO}_3)_2 \xrightarrow{\text{H}_2\text{O}} \text{Mg}^{2+}(aq) + 2\text{NO}_3^-(aq) \]
Derivation of Molar Mass
To find the molar mass of Mg(NO₃)₂, add the atomic masses of all atoms present.
Step 1: Identify atoms — 1 Mg, 2 N, 6 O.
Step 2: Use atomic masses — Mg = 24.31, N = 14.01, O = 16.00.
Step 3: Calculate — \( M = 24.31 + 2(14.01) + 6(16.00) = 24.31 + 28.02 + 96.00 = 148.33 \text{ g/mol} \) (≈ 148.31 g/mol using precise values).
For the hexahydrate Mg(NO₃)₂·6H₂O: \( M = 148.31 + 6(18.02) = 148.31 + 108.12 = 256.43 \text{ g/mol} \).
Complete Chemistry Formula Sheet — Ionic Salts and Related Formulas
| Formula Name | Expression | Variables | Molar Mass (g/mol) | NCERT Chapter |
|---|---|---|---|---|
| Magnesium Nitrate | \( \text{Mg(NO}_3)_2 \) | Mg²⁺ + 2NO₃⁻ | 148.31 | Class 10, Ch 2 |
| Magnesium Nitrate Hexahydrate | \( \text{Mg(NO}_3)_2 \cdot 6\text{H}_2\text{O} \) | Mg²⁺ + 2NO₃⁻ + 6H₂O | 256.41 | Class 10, Ch 2 |
| Barium Acetate | \( \text{Ba(CH}_3\text{COO)}_2 \) | Ba²⁺ + 2CH₃COO⁻ | 255.43 | Class 10, Ch 2 |
| Zinc Bromide | \( \text{ZnBr}_2 \) | Zn²⁺ + 2Br⁻ | 225.18 | Class 10, Ch 2 |
| Ammonium Acetate | \( \text{NH}_4\text{CH}_3\text{COO} \) | NH₄⁺ + CH₃COO⁻ | 77.08 | Class 11, Ch 7 |
| Magnesium Chloride | \( \text{MgCl}_2 \) | Mg²⁺ + 2Cl⁻ | 95.21 | Class 10, Ch 2 |
| Magnesium Sulphate | \( \text{MgSO}_4 \) | Mg²⁺ + SO₄²⁻ | 120.37 | Class 10, Ch 2 |
| Magnesium Oxide | \( \text{MgO} \) | Mg²⁺ + O²⁻ | 40.30 | Class 10, Ch 1 |
| Nitric Acid | \( \text{HNO}_3 \) | H⁺ + NO₃⁻ | 63.01 | Class 11, Ch 8 |
| Moles Formula | \( n = \frac{m}{M} \) | n = moles, m = mass (g), M = molar mass | — | Class 11, Ch 1 |
| Molarity Formula | \( C = \frac{n}{V} \) | C = molarity (mol/L), V = volume (L) | — | Class 11, Ch 1 |
Magnesium Nitrate Formula — Solved Examples
Example 1 (Class 9-10 Level) — Finding Molar Mass
Problem: Calculate the molar mass of magnesium nitrate, Mg(NO₃)₂.
Given: Atomic masses — Mg = 24, N = 14, O = 16 (approximate values used in CBSE exams)
Step 1: Write the formula: \( \text{Mg(NO}_3)_2 \)
Step 2: Count each atom — 1 Mg, 2 N, 6 O
Step 3: Multiply atomic mass by number of atoms for each element:
\( \text{Mg}: 1 \times 24 = 24 \)
\( \text{N}: 2 \times 14 = 28 \)
\( \text{O}: 6 \times 16 = 96 \)
Step 4: Add all values: \( M = 24 + 28 + 96 = 148 \text{ g/mol} \)
Answer
Molar mass of Mg(NO₃)₂ = 148 g/mol
Example 2 (Class 11-12 Level) — Calculating Moles and Molarity
Problem: 29.6 g of magnesium nitrate is dissolved in water to make 500 mL of solution. Calculate (a) the number of moles and (b) the molarity of the solution.
Given: Mass (m) = 29.6 g, Molar mass (M) = 148 g/mol, Volume (V) = 500 mL = 0.5 L
Step 1: Use the moles formula: \( n = \frac{m}{M} \)
Step 2: Substitute values: \( n = \frac{29.6}{148} = 0.2 \text{ mol} \)
Step 3: Use the molarity formula: \( C = \frac{n}{V} \)
Step 4: Substitute values: \( C = \frac{0.2}{0.5} = 0.4 \text{ mol/L} \)
Answer
(a) Number of moles = 0.2 mol
(b) Molarity = 0.4 M
Example 3 (JEE/NEET Level) — Percentage Composition and Ion Concentration
Problem: A 0.5 M aqueous solution of Mg(NO₃)₂ is prepared. Calculate (a) the percentage by mass of magnesium in the compound and (b) the concentration of NO₃⁻ ions in the solution.
Given: Molar mass of Mg(NO₃)₂ = 148 g/mol, Atomic mass of Mg = 24 g/mol, Molarity = 0.5 M
Part (a) — Percentage Composition:
Step 1: Use the formula: \( \%\text{Mg} = \frac{\text{Atomic mass of Mg}}{\text{Molar mass of Mg(NO}_3)_2} \times 100 \)
Step 2: Substitute: \( \%\text{Mg} = \frac{24}{148} \times 100 = 16.22\% \)
Part (b) — Nitrate Ion Concentration:
Step 3: Write the dissociation equation: \( \text{Mg(NO}_3)_2 \rightarrow \text{Mg}^{2+} + 2\text{NO}_3^- \)
Step 4: Each mole of Mg(NO₃)₂ gives 2 moles of NO₃⁻. So: \( [\text{NO}_3^-] = 2 \times 0.5 = 1.0 \text{ mol/L} \)
Answer
(a) Percentage of Mg = 16.22%
(b) Concentration of NO₃⁻ ions = 1.0 M
CBSE Exam Tips 2025-26 for Magnesium Nitrate Formula
- Write the formula correctly: Always use parentheses — Mg(NO₃)₂, not MgNO₃₂. The parentheses show that the subscript 2 applies to the entire nitrate group.
- Memorise the ion charges: Mg²⁺ has a 2+ charge and NO₃⁻ has a 1− charge. Two nitrate ions are needed to balance one magnesium ion. This is a common 1-mark question.
- Molar mass calculation: In CBSE board exams, use Mg = 24, N = 14, O = 16 for quick calculation. The molar mass comes to 148 g/mol. We recommend memorising this value directly.
- Distinguish anhydrous and hydrated forms: Mg(NO₃)₂ is anhydrous (M = 148 g/mol). Mg(NO₃)₂·6H₂O is the hexahydrate (M ≈ 256 g/mol). CBSE questions often specify which form to use.
- Reaction with nitric acid: Know the equation — \( \text{MgO} + 2\text{HNO}_3 \rightarrow \text{Mg(NO}_3)_2 + \text{H}_2\text{O} \). This appears in Class 10 acid-base reactions.
- Solubility: Magnesium nitrate is highly soluble in water. It is a strong electrolyte. It fully dissociates in aqueous solution. This is important for Class 11 electrolyte questions.
Common Mistakes to Avoid with Magnesium Nitrate Formula
- Mistake 1 — Missing parentheses: Writing MgNO₃₂ instead of Mg(NO₃)₂ is a very common error. Without parentheses, the formula is chemically incorrect. Always enclose NO₃ in brackets before adding the subscript.
- Mistake 2 — Wrong ion charges: Some students write Mg⁺ (1+ charge) instead of Mg²⁺ (2+ charge). Magnesium always forms a 2+ ion. This error leads to writing MgNO₃ as the formula, which is wrong.
- Mistake 3 — Incorrect atom count: Students sometimes count 2 oxygen atoms instead of 6. The NO₃ group has 3 oxygens. With 2 nitrate groups, there are 6 oxygen atoms in total.
- Mistake 4 — Confusing anhydrous and hexahydrate molar mass: Using 256 g/mol when the question asks about anhydrous Mg(NO₃)₂ leads to wrong answers. Always check whether the question specifies the hydrated form.
- Mistake 5 — Incorrect nitrate ion concentration: When calculating ion concentrations in solution, students often forget to multiply the molarity by 2 for NO₃⁻. One formula unit gives two nitrate ions upon dissociation.
JEE/NEET Application of Magnesium Nitrate Formula
In our experience, JEE and NEET aspirants encounter the Magnesium Nitrate Formula in several key topic areas. Understanding the formula deeply — beyond just memorising it — gives a significant advantage.
1. Mole Concept and Stoichiometry (JEE Main / NEET)
The mole concept is one of the most heavily tested topics. Questions often give a mass of Mg(NO₃)₂ and ask for moles, number of molecules, or number of ions. The key formula is \( n = \frac{m}{M} \). Since Mg(NO₃)₂ dissociates into 3 ions (1 Mg²⁺ and 2 NO₃⁻), the total ion concentration is 3 times the molarity. This is tested in JEE Main questions on colligative properties and osmotic pressure.
2. Ionic Equilibrium and Electrolytes (JEE Advanced / NEET)
Magnesium nitrate is a salt of a strong acid (HNO₃) and a strong base (Mg(OH)₂). It does not hydrolyse in water. Its aqueous solution is therefore neutral (pH ≈ 7). JEE Advanced tests this concept in hydrolysis and buffer problems. NEET questions may ask whether the solution is acidic, basic, or neutral — and the answer is neutral for Mg(NO₃)₂.
3. Percentage Composition and Empirical Formula (JEE Main)
Calculating the percentage by mass of each element in Mg(NO₃)₂ is a standard JEE Main question type. Our experts suggest practising percentage composition problems with all common ionic compounds. For Mg(NO₃)₂: %Mg ≈ 16.2%, %N ≈ 18.9%, %O ≈ 64.8%. These values can appear as data in reverse calculation problems where the empirical formula must be determined.
4. Colligative Properties (JEE Main)
Since Mg(NO₃)₂ is a strong electrolyte, it dissociates completely into 3 ions. The van’t Hoff factor \( i = 3 \). This value is used in calculating boiling point elevation, freezing point depression, and osmotic pressure. A common JEE question compares the freezing point depression of equimolar solutions of Mg(NO₃)₂ and NaCl. Since i = 3 for Mg(NO₃)₂ and i = 2 for NaCl, magnesium nitrate causes greater depression.
FAQs on Magnesium Nitrate Formula
We hope this comprehensive guide on the Magnesium Nitrate Formula has helped you understand the formula, structure, molar mass, and applications clearly. For more chemistry formula resources, explore our complete guide on Chemistry Formulas. You may also find our articles on the Barium Acetate Formula, the Zinc Bromide Formula, and the Ammonium Acetate Formula helpful for your exam preparation. For official NCERT resources, visit the NCERT official website.