Chapter 2, Problem 83QP

Interpretation Introduction

Interpretation: The empirical formulafor thegiven compounds are to be written.

Concept introduction:

When the formula of an ionic compound is indicated using the smallest whole number ratios without altering the proportion of the atoms of an individual element to form an electrically neutral substance, it is called an empirical formula.

Electrically neutral compounds are formedwhen positive and negative charges on constituent ions balance each other.

While writing the formula of any compound first the valences of the cation and anions are to be written. Then the valency of the cation will be written as the subscript of the anion and vice versa.

Answer to Problem 83QP

Solution:

a) ${\text{RbNO}}_2$

b) ${\text{K}}_{\text{2}}\text{S}$

c) $\text{NaHS}$

d) ${\text{Mg}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$

e) ${\text{CaHPO}}_4$

f) ${\text{PbCO}}_3$

g) ${\text{SnF}}_{\text{2}}$

h) ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$

i) ${\text{AgClO}}_4$

j) ${\text{BCl}}_3$

Explanation of Solution

a) Rubidium nitrite.

The rubidium ion $\left({\text{Rb}}^{\text{+}}\right)$ and the nitrite ion $\left({\text{NO}}_2{}^{\text{-}}\right)$ combine to form the ionic compoundrubidium nitrite. The sum of the charges is $1+\left(-1\right)=0$. So, no subscripts are necessary. The formula for the compound rubidium nitrite is ${\text{RbNO}}_2$.

b) Potassium sulfide.

The potassium ion $\left({\text{K}}^{\text{+}}\right)$ and the sulfide ion $\left({\text{S}}^2{}^{\text{-}}\right)$ combine to form the ionic compoundpotassium sulfide. The sum of the charges of one $\left({\text{K}}^{\text{+}}\right)$ ion and one $\left({\text{S}}^2{}^{\text{-}}\right)$ ion is $1+\left(-2\right)=-1$. To make the charges add up to zero, multiply the $+1$ charge of the cation by $2$ and add the subscript $2$ to the symbol of potassium. Thus, the formula for the compound potassium sulfide is ${\text{K}}_{\text{2}}\text{S}$.

c) Sodium hydrogen sulfide.

The sodium ion $\left({\text{Na}}^{\text{+}}\right)$, the hydrogen ion $\left({\text{H}}^{\text{+}}\right)$, and the sulfide ion $\left({\text{S}}^2{}^{\text{-}}\right)$ combine to form the ionic compoundsodium hydrogen sulfide. The sum of the charges is $1+1+\left(-2\right)=0$. So, no subscripts are necessary. Thus, the formula for the compound sodium hydrogen sulfide is $\text{NaHS}$.

d) Magnesium phosphate.

The magnesium ion $\left({\text{Mg}}^{\text{2+}}\right)$ and the phosphate ion $\left({\text{PO}}_{\text{4}}{{}^{\text{3}}}^{\text{-}}\right)$ combine to form the ionic compoundmagnesium phosphate. The sum of the charges of one $\left({\text{Mg}}^{\text{2+}}\right)$ ion and one $\left({\text{PO}}_{\text{4}}{{}^{\text{3}}}^{\text{-}}\right)$ ion is $+2+\left(-3\right)=-1$. To make the charges add up to zero, multiply the $+2$ charge of the cation by $3$ and add the subscript $3$ to the symbol of magnesium. Multiply the $-3$ charge of the anion by $2$ and add the subscript $2$ to the symbol of phosphate Thus, the formula for the compound magnesium phosphate is ${\text{Mg}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$.

e)Calcium hydrogen phosphate.

The calcium ion $\left({\text{Ca}}^{2+}\right)$, the hydrogen ion $\left({\text{H}}^{\text{+}}\right)$, and the phosphate ion $\left({\text{PO}}_{\text{4}}{{}^{\text{3}}}^{\text{-}}\right)$ combine to form the ionic compoundcalcium hydrogen phosphate. The sum of the charges is $2+1+\left(-3\right)=0$. So, no subscripts are necessary. Thus, the formula for the compound calcium hydrogen phosphate is ${\text{CaHPO}}_4$.

f) Lead (II) carbonate.

The lead (II) ion $\left({\text{Pb}}^{\text{2+}}\right)$ and the carbonate ion $\left({\text{CO}}_3{{}^2}^{\text{-}}\right)$ combine to form the ionic compoundlead (II) carbonate. The sum of the charges is $-2+\left(-2\right)=0$. So, no subscripts are necessary. Thus, the formula for the compound lead (II) carbonate is ${\text{PbCO}}_3$.

g) Tin $\left(\text{II}\right)$ fluoride.

The tin(II) ion $\left({\text{Sn}}^{\text{2+}}\right)$ and the fluoride ion $\left({\text{F}}^{\text{-}}\right)$ combine to form the ionic compoundtin $\left(\text{II}\right)$ fluoride. The sum of the charges is $2+\left(-1\right)=1$. To make the charges add up to zero, multiply the $-1$ charge of the anion by $2$ and add the subscript $2$ to the symbol of fluoride. Thus, the formula for the compound tin $\left(\text{II}\right)$ fluoride is ${\text{SnF}}_{\text{2}}$.

h) Ammonium sulfate.

The ammonium ion $\left({\text{NH}}_4{}^{\text{+}}\right)$ and the sulfate ion $\left({\text{SO}}_4{}^{\text{2-}}\right)$ combine to form the ionic compoundammonium sulfate. The sum of the charges is $1+\left(-2\right)=-1$. To make the charges add up to zero, multiply the $+1$ charge of the cation by $2$ and add the subscript $2$ to the symbol of ammonium. The formula for the compound ammonium sulfate is ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$.

i) Silver perchlorate.

The silver ion $\left({\text{Ag}}^{\text{+}}\right)$ and the perchlorate ion $\left({\text{ClO}}_{\text{4}}{}^{\text{-}}\right)$ combine to form the ionic compoundsilver perchlorate. The sum of the charges is $1+\left(-1\right)=0$. So, no subscripts are necessary. The formula for the compound silver perchlorate is ${\text{AgClO}}_4$.

j) Boron trichloride.

The boron ion $\left({\text{B}}^{\text{3+}}\right)$ and the chloride ion $\left({\text{Cl}}^{\text{-}}\right)$ combine to form the ionic compoundboron trichloride. The sum of the charges is $3+\left(-1\right)=2$. To make the charges add up to zero, multiply the $-1$ charge of the anion by $3$ and add the subscript $3$ to the symbol of chloride. The formulafor the compound boron trichlorideis ${\text{BCl}}_3$.