Write down the empirical formula for the ionic compound tin (II) bromide and then answer the following questions: What is the charge of the cation? What is the charge of the anion? What is the subscript of the cation in the empirical formula? What is the subscript of the anion in the empirical formula?
The Correct Answer and Explanation is:
ChatGPT said:
Empirical Formula of Tin(II) Bromide: SnBr₂
Answers:
- Charge of the cation: +2
- Charge of the anion: –1
- Subscript of the cation in the empirical formula: 1
- Subscript of the anion in the empirical formula: 2
Textbook-Style Explanation
When writing the empirical formula for an ionic compound, it is essential to first identify the ions involved and their respective charges. Ionic compounds are composed of positively charged cations (usually metals) and negatively charged anions (usually nonmetals). The charges of the ions determine how many of each ion are needed to form a neutral compound.
In the case of tin(II) bromide, the Roman numeral “II” indicates that the tin ion has a charge of +2. This is written as Sn²⁺. Tin can form multiple ions (Sn²⁺ and Sn⁴⁺), so the (II) specifies which oxidation state is being used.
Bromide is derived from the element bromine, which belongs to group 17 (halogens) on the periodic table. Halogens typically form ions with a –1 charge because they gain one electron to achieve a full outer shell. Therefore, the bromide ion is Br⁻.
To write the empirical formula, the total positive charge must balance the total negative charge. One Sn²⁺ ion has a charge of +2. Each Br⁻ ion has a charge of –1, so two Br⁻ ions are needed to balance one Sn²⁺ ion. This results in the neutral compound SnBr₂.
Thus, the empirical formula is SnBr₂, where:
- There is 1 tin ion (Sn²⁺) (subscript of cation = 1),
- There are 2 bromide ions (Br⁻) (subscript of anion = 2).
The formula shows the simplest whole-number ratio of ions that results in a neutral compound, which is the definition of an empirical formula for an ionic substance.
