Empirical Formula Of Sr2 And O2-

Empirical Formula of Sr²⁺ and O²⁻: How to Derive SrO from Ionic Charges

When chemists talk about the empirical formula of a substance, they are looking for the simplest whole‑number ratio of atoms that makes up the compound. For ionic substances formed from a metal cation and a non‑metal anion, this ratio is dictated by charge balance. The pair Sr²⁺ (strontium ion) and O²⁻ (oxide ion) is a classic example: each strontium ion carries a +2 charge, while each oxide ion carries a –2 charge. Because the magnitudes are equal and opposite, one Sr²⁺ exactly neutralizes one O²⁻, giving the empirical formula SrO. The following sections walk through the concept, the step‑by‑step procedure, the underlying science, and common questions that arise when students first encounter this topic.

Introduction The empirical formula represents the most reduced ratio of elements in a chemical compound. For molecular compounds, this ratio is often found by converting mass percentages to moles and then simplifying. For ionic compounds, the process is even more straightforward: the formula emerges directly from the requirement that the total positive charge equals the total negative charge. Understanding how to derive SrO from Sr²⁺ and O²⁻ not only reinforces the concept of charge neutrality but also lays the groundwork for predicting formulas of more complex salts, such as Sr(OH)₂ or Sr₃(PO₄)₂.

Key term: Empirical formula – the simplest integer ratio of atoms in a compound.

Understanding Ions and Charge Balance ### What Are Sr²⁺ and O²⁻?

• Sr²⁺ is the strontium cation formed when a strontium atom (atomic number 38) loses two electrons from its 5s orbital.
• O²⁻ is the oxide anion formed when an oxygen atom (atomic number 8) gains two electrons to fill its 2p subshell.

Both ions achieve a noble‑gas electron configuration (Sr²⁺ resembles Kr; O²⁻ resembles Ne), which makes them particularly stable in a solid lattice.

The Charge‑Balance Principle

In any neutral ionic solid, the sum of all positive charges must equal the sum of all negative charges. Mathematically:

[ \sum (\text{cation charge} \times \text{number of cations}) + \sum (\text{anion charge} \times \text{number of anions}) = 0 ]

For a binary compound containing only Sr²⁺ and O²⁻, let x be the number of Sr²⁺ ions and y be the number of O²⁻ ions. The charge‑balance equation becomes:

[ (+2)x + (-2)y = 0 \quad \Rightarrow \quad 2x - 2y = 0 \quad \Rightarrow \quad x = y]

Thus, the smallest whole‑numbers satisfying this relation are x = 1 and y = 1, giving the empirical formula SrO.

Determining the Empirical Formula: Step‑by‑Step Below is a practical workflow that students can follow whenever they are given the charges of a cation and an anion and asked to find the empirical formula.

1. Write down the ionic charges

   • Cation: Sr²⁺ → +2
   • Anion: O²⁻ → –2

2. Set up the charge‑balance equation
   [ (+2)(\text{# Sr}) + (-2)(\text{# O}) = 0 ]

3. Assign variables to the unknown quantities
   Let a = number of Sr²⁺ ions, b = number of O²⁻ ions.

4. Solve for the simplest integer ratio
   [ 2a - 2b = 0 \implies a = b ]
   The smallest positive integers that satisfy a = b are a = 1, b = 1.

5. Write the empirical formula using the obtained ratio

   • Sr: 1 → Sr
   • O: 1 → O
   • Result: SrO

6. Verify charge neutrality

   • Total positive charge = (+2) × 1 = +2
   • Total negative charge = (–2) × 1 = –2
   • Net charge = 0 → formula is correct.

This procedure can be generalized: divide the magnitude of the anion charge by the greatest common divisor (GCD) of the cation and anion charge magnitudes to get the cation subscript, and do the reverse for the anion subscript.

Scientific Explanation Behind SrO Formation

Lattice Energy and Stability

When Sr²⁺ and O²⁻ come together in the solid state, they arrange themselves in a crystalline lattice (specifically, the rock‑salt structure, SrO adopts the NaCl type). The electrostatic attraction between oppositely charged ions releases a considerable amount of energy, known as lattice energy. For SrO, the lattice energy is approximately –3220 kJ mol⁻¹, which compensates for the energy required to remove two electrons from strontium and add two electrons to oxygen. The net process is exothermic, making SrO thermodynamically favorable under standard conditions.

Role of Polarizability

Although both ions are relatively small, the oxide anion is highly polarizable due to its diffuse electron cloud. This polarizability enhances covalent character in the Sr–O bond, contributing to the compound’s high melting point (~2430 °C) and insolubility in water (SrO reacts with water to form Sr(OH)₂, but the oxide itself does not dissolve).

Comparison with Other Alkaline‑Earth Oxides

• MgO (Mg²⁺ + O²⁻) also adopts the rock‑salt structure, but its lattice energy is larger (≈ –3795 kJ mol⁻¹) because Mg²⁺ is smaller, leading to shorter interionic distances.
• CaO (Ca²⁺ + O²⁻) lies between

Scientific Explanation Behind SrO Formation (Continued)

Comparison with Other Alkaline-Earth Oxides

The trend in lattice energy across the alkaline-earth metals (Be, Mg, Ca, Sr, Ba) is a direct consequence of the inverse relationship between ionic radius and lattice energy. As the cation size increases down the group (Sr²⁺ > Ca²⁺ > Mg²⁺ > Be²⁺), the distance between the cation and anion increases, weakening the electrostatic attraction. Consequently, lattice energy decreases:

• MgO: Lattice energy ≈ –3795 kJ/mol (smallest cation, highest energy).
• CaO: Lattice energy ≈ –3414 kJ/mol (larger cation than MgO).
• SrO: Lattice energy ≈ –3220 kJ/mol (even larger cation than CaO).

This decreasing lattice energy down the group explains why SrO, while stable, is less energetically favorable than MgO. However, the stability of SrO is still significant due to the substantial lattice energy released when the highly charged Sr²⁺ and O²⁻ ions form the rock-salt lattice. The large lattice energy compensates for the lower ionization energy of strontium compared to magnesium.

The Rock-Salt Structure: Maximising Stability

Both MgO and SrO adopt the rock-salt (NaCl-type) structure. This arrangement is highly efficient for ionic compounds with equal numbers of cations and anions. In this structure, each ion is surrounded by six nearest neighbors of the opposite charge, forming a three-dimensional network of alternating Sr²⁺ and O²⁻ ions. This close packing maximizes the number of attractive ion-ion interactions (electrostatic bonds) and minimizes repulsive interactions between like-charged ions, contributing significantly to the high lattice energy and overall stability of the compound.

Conclusion

The empirical formula SrO is derived directly from the ionic charges of strontium (Sr²⁺) and oxide (O²⁻) through a straightforward charge-balancing process. This yields a 1:1 ratio of Sr²⁺ to O²⁻ ions, resulting in the formula SrO. The formation of SrO is thermodynamically and kinetically favorable, driven by the substantial lattice energy released when these highly charged ions arrange themselves in the stable rock-salt structure. This structure, characterized by each ion being surrounded by six nearest neighbors of opposite charge, maximizes electrostatic attraction and minimizes repulsion. While lattice energy decreases down the group (MgO > CaO > SrO), SrO remains a stable ionic compound, exemplifying the fundamental principles of ionic bonding and charge neutrality that govern the formation of countless other compounds. The empirical formula SrO serves as a foundational example of how ionic charges dictate the simplest ratio of atoms in a compound.