Chapter 3: Problem 38
Write formulas for the following ionic compounds: (a) Ammonium hydrogen sulfite (b) Magnesium acetate (c) Strontium dihydrogen phosphate (d) Silver carbonate (e) Strontium chloride (f) Barium permanganate (g) Aluminum perchlorate
Short Answer
Expert verified
(a) (NH鈧)(HSO鈧), (b) Mg(C鈧侶鈧僌鈧)鈧, (c) Sr(H鈧侾O鈧)鈧, (d) Ag鈧侰O鈧, (e) SrCl鈧, (f) Ba(MnO鈧)鈧, (g) Al(ClO鈧)鈧.
Step by step solution
01
Identify the Ions and Their Charges
For each compound, we need to identify both the cation (positive ion) and the anion (negative ion), along with their respective charges.
- **Ammonium hydrogen sulfite**: Ammonium is NH鈧勨伜 and hydrogen sulfite is HSO鈧冣伝.
- **Magnesium acetate**: Magnesium is Mg虏鈦 and acetate is C鈧侶鈧僌鈧傗伝.
- **Strontium dihydrogen phosphate**: Strontium is Sr虏鈦 and dihydrogen phosphate is H鈧侾O鈧勨伝.
- **Silver carbonate**: Silver is Ag鈦 and carbonate is CO鈧兟测伝.
- **Strontium chloride**: Strontium is Sr虏鈦 and chloride is Cl鈦.
- **Barium permanganate**: Barium is Ba虏鈦 and permanganate is MnO鈧勨伝.
- **Aluminum perchlorate**: Aluminum is Al鲁鈦 and perchlorate is ClO鈧勨伝.
02
Determine the Correct Ratio of Ions
For each compound, determine the number of each ion needed to balance the overall charge of the compound to neutral.
- **Ammonium hydrogen sulfite**: 1 NH鈧勨伜 ion pairs with 1 HSO鈧冣伝 ion: (NH鈧)(HSO鈧).
- **Magnesium acetate**: 1 Mg虏鈦 ion pairs with 2 C鈧侶鈧僌鈧傗伝 ions to balance the 2+ charge: Mg(C鈧侶鈧僌鈧)鈧.
- **Strontium dihydrogen phosphate**: 1 Sr虏鈦 ion pairs with 2 H鈧侾O鈧勨伝 ions: Sr(H鈧侾O鈧)鈧.
- **Silver carbonate**: 2 Ag鈦 ions are needed to balance the 2- charge of CO鈧兟测伝: Ag鈧侰O鈧.
- **Strontium chloride**: 1 Sr虏鈦 ion pairs with 2 Cl鈦 ions: SrCl鈧.
- **Barium permanganate**: 1 Ba虏鈦 ion pairs with 2 MnO鈧勨伝 ions: Ba(MnO鈧)鈧.
- **Aluminum perchlorate**: 1 Al鲁鈦 ion pairs with 3 ClO鈧勨伝 ions: Al(ClO鈧)鈧.
03
Formulate the Ionic Compounds
Using the determined ratios from Step 2, write the chemical formulas for each of the compounds.
- **Ammonium hydrogen sulfite**: (NH鈧)(HSO鈧)
- **Magnesium acetate**: Mg(C鈧侶鈧僌鈧)鈧
- **Strontium dihydrogen phosphate**: Sr(H鈧侾O鈧)鈧
- **Silver carbonate**: Ag鈧侰O鈧
- **Strontium chloride**: SrCl鈧
- **Barium permanganate**: Ba(MnO鈧)鈧
- **Aluminum perchlorate**: Al(ClO鈧)鈧
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Cation and Anion Charges
In ionic compounds, atoms either gain or lose electrons to form ions. These ions come with charges, which play a crucial role in creating stable ionic compounds. Cations are positively charged ions. They form when an atom loses one or more electrons. For example:
- Ammonium (\( NH_4^+ \)) is a cation that results from the loss of electrons by nitrogen bonded to hydrogen.
- Silver (\( Ag^+ \)), magnesium (\( Mg^{2+} \)), and strontium (\( Sr^{2+} \)) form cations by losing electrons.
- The acetate (\( C_2H_3O_2^- \)) anion forms when the acetic acid loses a hydrogen ion.
- Chloride (\( Cl^- \)) and hydrogen sulfite (\( HSO_3^- \)) are common anions obtained by gaining electrons.
Crafting Chemical Formulas of Ionic Compounds
Creating chemical formulas for ionic compounds involves determining the correct ratio of cations to anions. This ensures the compound's overall electrical neutrality. The key is to balance the positive and negative charges so they sum up to zero.Start with identifying the charges of the ions involved. For example:
- For aluminum perchlorate, aluminum's charge is \(+3\), while perchlorate's is \(-1\). This means three perchlorate ions are needed to balance one aluminum ion.
- In magnesium acetate, magnesium has a \(+2\) charge, so it pairs with two acetate ions, each with a \(-1\) charge.
Achieving Compound Neutrality
Neutrality in a compound means that the total positive charges equal the total negative charges. This balance is crucial for the stability of the compound. Achieving compound neutrality often requires careful consideration of the ion charges and a systematic approach to balancing them.Here's a breakdown of how this works:
- In silver carbonate, with silver having a charge of \(+1\) and carbonate \(-2\), two silver ions are needed for every carbonate ion. Thus, the formula is \( Ag_2CO_3 \).
- For strontium chloride, since strontium's charge is \(+2\) and chloride's is \(-1\), two chloride ions balance one strontium ion, resulting in \( SrCl_2 \).