Chapter 3, Problem 84QP

(a)

Interpretation Introduction

Interpretation:

The formula of ammonia is to be determined.

Explanation of Solution

Ammonia is a combination of nitrogen and hydrogen in which a covalent bond is formed by sharing of electrons between nitrogen and hydrogen atoms to form a molecular compound. Therefore, the formula of ammonia is ${\text{NH}}_3$ .

(b)

Interpretation Introduction

Interpretation:

The formula of ammonium phosphate is to be determined.

Explanation of Solution

The symbol and charge of the elements are analyzed on the basis of their position in the periodic table. The ammonium ion carries a $\left(+1\right)$ charge and the phosphate ion carries a $\left(-3\right)$ . So, three ammonium ions $\left({\text{3 NH}}_4^{+}\right)$ are required to neutralize the charge of the phosphate ion $\left({\text{PO}}_4^{3-}\right)$ to form an ionic compound. For writing the formula of ionic compound, the cation is written first followed by the anion. Therefore, the formula of ammonium phosphate is ${\text{NH}}_4{\left({\text{PO}}_4\right)}_3$ .

(c)

Interpretation Introduction

Interpretation:

The formula of hydrochloric acid is to be determined.

Explanation of Solution

Hydrochloric acid contains hydrogen cation and chloride anion in which the hydrogen ion contains a $+1$ charge and the chloride ion contains a $-1$ , which is already balanced. For writing the formula of an acid, the notation “aq� is used after writing the formula of the acid. Therefore, the formula of hydrofluoric acid is $\text{HCl}\left(aq\right)$ .

(d)

Interpretation Introduction

Interpretation:

The formula of potassium hydroxide is to be determined.

Explanation of Solution

The potassium ion carries a $+1$ charge that is determined on the basis of its position in the periodic table. Thus, a hydroxide ion carries a $\left(-1\right)$ charge that neutralizes the charge of the potassium ion $\left({\text{K}}^{+}\right)$ in order to form an ionic compound. For writing the formula of an ionic compound, the cation is written first, followed by the anion. Therefore, the formula of potassium hydroxide is $\text{KOH}$ .

(e)

Interpretation Introduction

Interpretation:

The formula of chromium(III) oxide.

Explanation of Solution

The name chromium(III) indicates that the chromium ion contains a $\left(+3\right)$ charge and the oxide ion $\left({\text{O}}^{2-}\right)$ carries a $\left(-2\right)$ charge, which is analyzed on the basis of their position in the periodic table. Thus, three oxide ions $\left(3{\text{ O}}^{2-}\right)$ are required to neutralize the charge of two chromium ion $\left({\text{2 Cr}}^{3+}\right)$ to form an ionic compound. For writing the formula of an ionic compound, the cation is written first, followed by the anion. Therefore, the formula of chromium(II) oxide is ${\text{Cr}}_2{\text{O}}_3$ .

(f)

Interpretation Introduction

Interpretation:

The formula of magnesium nitrate is to be determined.

Explanation of Solution

The magnesium ion carries a $+2$ charge which is determined on the basis of its position in the periodic table. The nitrate ion carries a $\left(-1\right)$ charge, so two nitrate ions $\left(2{\text{ NO}}_3^{2-}\right)$ are required to neutralize the charge of one magnesium ion $\left({\text{Mg}}^{2+}\right)$ in order to form an ionic compound. For writing the formula of an ionic compound, the cation is written first, followed by the anion. Therefore, the formula of magnesium nitrate is $\text{Mg}{\left({\text{NO}}_3\right)}_2$ .

(g)

Interpretation Introduction

Interpretation:

The formula of manganese(IV) oxide.

Explanation of Solution

The name manganese (IV) indicates that the manganese ion contains a $\left(+4\right)$ charge and the oxide ion $\left({\text{O}}^{2-}\right)$ carries a $\left(-2\right)$ charge, which is analyzed on the basis of their position in the periodic table. Thus, two oxide ions $\left({\text{2 O}}^{2-}\right)$ are required to neutralize the charge of one manganese ion $\left({\text{Mn}}^{4+}\right)$ in order to form an ionic compound. For writing the formula of an ionic compound, the cation is written first, followed by the anion. Therefore, the formula of manganese(IV) oxide is ${\text{MnO}}_{\text{2}}$ .

(h)

Interpretation Introduction

Interpretation:

The formula of nitrogen dioxide is to be determined.

Explanation of Solution

In nitrogen dioxide, the first word indicates that nitrogen is the first element and its symbol is written first, and the second word indicates that dioxide is the second element which takes the subscript two when writing the formula of the molecular compound. Therefore, the formula for nitrogen dioxide is ${\text{NO}}_2$ .