Chapter 10.3, Problem 36PP

Interpretation Introduction

(a)

Interpretation:

The molar mass of ${\text{Sr(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ needs to be determined. The compound needs to be identified as ionic or molecular compound.

Concept introduction:

A molecule is composed of more than type of atoms of different elements. On the basis of bonding with atoms, they can be ionic and covalent or molecular compounds.

The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound.

Answer to Problem 36PP

${\text{Molar mass of Sr(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{= 211}\text{.5 g/mol}$

It is an ionic compound.

Explanation of Solution

An ionic compound is composed of oppositely charged ions whereas a covalent or molecular compound is formed by equal sharing of valence electrons between bonded atoms. Ionic compounds are usually formed between metals and non-metals whereas covalent compounds usually consist of non-metals. Since ${\text{Sr(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ consists of metal with non-metal therefore it must be ionic compound.

Given compound =

${\text{Sr(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Molar mass of Sr = 87.6 g/mol

Molar mass of O = 15.99 g/mol

Molar mass of N = 14.0 g/mol

Calculate the molar mass of ${\text{Sr(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ = $\left(\text{1×87}\text{.6 }\right)\text{+}\left(\text{6× 15}\text{.99}\right)\text{+}\left(\text{2× 14}\text{.0}\right)\text{ = 211}\text{.5 g/mol}$

Interpretation Introduction

(b)

Interpretation:

The molar mass of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ needs to be determined. The compound needs to be identified as ionic or molecular compound.

Concept introduction:

A molecule is composed of more than type of atoms of different elements. On the basis of bonding with atoms, they can be ionic and covalent or molecular compounds.

The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound.

Answer to Problem 36PP

${\text{Molar mass of (NH}}_{\text{4}}{\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}\text{= 149 g/mol}$

Explanation of Solution

An ionic compound is composed of oppositely charged ions whereas a covalent or molecular compound is formed by equal sharing of valence electrons between bonded atoms. Ionic compounds are usually formed between metals and non-metals whereas covalent compounds usually consist of non-metals. Since ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ consists of two oppositely charged ions therefore it must be ionic compound.

Given compound = ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$

Molar mass of P = 31.0 g/mol

Molar mass of O = 15.99 g/mol

Molar mass of N = 14.0 g/mol

Molar mass of H = 1.0 g/mol

Calculate the molar mass of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ = $\left(\text{3×14}\text{.0 }\right)\text{+}\left(\text{12× 1}\text{.0}\right)+\left(\text{1× 31}\text{.0 }\right)\text{+}\left(\text{4× 15}\text{.99}\right)\text{ = 149 g/mol}$

Interpretation Introduction

(c)

Interpretation:

The molar mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ needs to be determined. The compound needs to be identified as ionic or molecular compound.

Concept introduction:

A molecule is composed of more than type of atoms of different elements. On the basis of bonding with atoms, they can be ionic and covalent or molecular compounds.

The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound.

Answer to Problem 36PP

${\text{Molar mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{= 342 g/mol}$

Explanation of Solution

An ionic compound is composed of oppositely charged ions whereas a covalent or molecular compound is formed by equal sharing of valence electrons between bonded atoms. Ionic compounds are usually formed between metals and non-metals whereas covalent compounds usually consist of non-metals. Since ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ consists of non-metals therefore it must be molecular compound.

Given compound = ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$

Molar mass of O = 15.99 g/mol

Molar mass of C = 12.0 g/mol

Molar mass of H = 1.0 g/mol

Calculate the molar mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ = $\left(\text{12×12}\text{.0 }\right)\text{+}\left(\text{22× 1}\text{.0 }\right)\text{+}\left(\text{11× 15}\text{.99 }\right)\text{ = 342 g/mol}$