Chapter 12, Problem 98A

(a)

Interpretation Introduction

Interpretation: The formula of aluminum carbonate is to be written.

Concept Introduction: A chemical formula is an expression that shows the components that make up a compound as well as the proportions of those elements that make up the compound.

Answer to Problem 98A

The formula of aluminum carbonate is ${\text{Al}}_{\text{2}}{\text{CO}}_{\text{3}}$ .

Explanation of Solution

The given compound is aluminum carbonate.

Aluminum is found in Group 3A of the periodic table. Its ion has a charge of $+3$ and is denoted by the symbol ${\text{Al}}^{\text{3+}}$ . The carbonate ion is denoted by the chemical formula ${\text{CO}}_{\text{3}}^{\text{2-}}$ , found in the salt form of carbonic acid.

Therefore, the formula of aluminum carbonate is ${\text{Al}}_{\text{2}}{\text{CO}}_{\text{3}}$ .

(b)

Interpretation Introduction

Interpretation: The formula of nitrogen dioxide is to be written.

Concept Introduction: A chemical formula is an expression that shows the components that make up a compound as well as the proportions of those elements that make up the compound.

Answer to Problem 98A

The formula of nitrogen dioxide is ${\text{NO}}_2$ .

Explanation of Solution

The given compound is nitrogen dioxide.

Due to the presence of the prefix di-, the anion of oxygen known as dioxide includes two atoms of oxygen and has a total charge of $-4$ . The chemical formula of oxide is ${\text{O}}^{\text{2-}}$ . The total charge that is associated with nitrogen is $+4$ . The chemical formula of nitrogen cation is ${\text{N}}^{4+}$ .

Therefore, the formula of nitrogen dioxide is ${\text{NO}}_2$ .

(c)

Interpretation Introduction

Interpretation: The formula of potassium sulfide is to be written.

Concept Introduction: A chemical formula is an expression that shows the components that make up a compound as well as the proportions of those elements that make up the compound.

Answer to Problem 98A

The formula of potassium sulfide is ${\text{K}}_{\text{2}}\text{S}$ .

Explanation of Solution

The given compound is potassium sulfide.

Potassium is found in Group 1A of the periodic table. A potassium ion $\left({\text{K}}^{\text{+}}\right)$ is the anion of potassium that carries a positive charge of one. Sulfide $\left({\text{S}}^{\text{2-}}\right)$ is the anion of sulfur, and it has a $-2$ charge since sulfur is placed in Group 6A.

Therefore, the formula of potassium sulfide is ${\text{K}}_{\text{2}}\text{S}$ .

(d)

Interpretation Introduction

Interpretation: The formula of manganese(II) chromate is to be written.

Concept Introduction: A chemical formula is an expression that shows the components that make up a compound as well as the proportions of those elements that make up the compound.

Answer to Problem 98A

The formula of manganese(II) chromate is ${\text{MnCrO}}_4$ .

Explanation of Solution

The given compound is manganese(II) chromate.

Manganese(II) Chromate is an ionic molecule whose overall charge is neutral (zero). Manganese is a transition metal containing multiple charges, indicated by the Roman number. The number indicates that the ion has a charge of $2+$ and is represented as ${\text{Mn}}^{\text{2+}}$ . Chromate is a $2-$ charged polyatomic anion and is represented as ${\text{CrO}}_{\text{4}}^{\text{2-}}$ .

Therefore, the formula of manganese(II) chromate is ${\text{MnCrO}}_4$ .

(e)

Interpretation Introduction

Interpretation: The formula of sodium bromide is to be written.

Concept Introduction: A chemical formula is an expression that shows the components that make up a compound as well as the proportions of those elements that make up the compound.

Answer to Problem 98A

The formula of sodium bromide is $\text{NaBr}$ .

Explanation of Solution

The given compound is sodium bromide.

The overall charge of the ionic substance known as sodium bromide is considered to be neutral or zero. In the periodic table, sodium is found in Group 1A with the other alkali metals. As a result, the sodium has a charge of $1+$ and is represented as ${\text{Na}}^{\text{+}}$ . Bromine is found in Group 7A, and the ion of bromine has a charge of $1-$ . It is represented as ${\text{Br}}^{\text{-}}$

Therefore, the formula of sodium bromide is $\text{NaBr}$ .