Chapter 3, Problem 60E

(a)

Interpretation Introduction

Interpretation: The number of moles of compound is given. By using the number of moles, the mass of phosphorous in each compound given in exercise 52 is to be determined.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole. One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ atoms.

Hence,

$\begin{array}{l}\left(\text{6}\text{.022}\times {\text{10}}^{\text{23}} \text{atoms}\right)\left(\frac{\text{12}\text{u}}{\text{1}\text{atom}}\right)=\text{12}\text{g}\\ \Rightarrow \text{1}\text{u}=\frac{\text{1}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}} }\text{g}\end{array}$

To determine: The mass of phosphorous $\left(\text{P}\right)$ in $\text{5}\text{.00}$ moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$.

(b)

Interpretation Introduction

Interpretation: The number of moles of compound is given. By using the number of moles, the mass of phosphorous in each compound given in exercise 52 is to be determined.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole. One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ atoms.

Hence,

$\begin{array}{l}\left(\text{6}\text{.022}\times {\text{10}}^{\text{23}} \text{atoms}\right)\left(\frac{\text{12}\text{u}}{\text{1}\text{atom}}\right)=\text{12}\text{g}\\ \Rightarrow \text{1}\text{u}=\frac{\text{1}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}} }\text{g}\end{array}$

(c)

Interpretation Introduction

Interpretation: The number of moles of compound is given. By using the number of moles, the mass of phosphorous in each compound given in exercise 52 is to be determined.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole. One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ atoms.

Hence,

$\begin{array}{l}\left(\text{6}\text{.022}\times {\text{10}}^{\text{23}} \text{atoms}\right)\left(\frac{\text{12}\text{u}}{\text{1}\text{atom}}\right)=\text{12}\text{g}\\ \Rightarrow \text{1}\text{u}=\frac{\text{1}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}} }\text{g}\end{array}$

To determine: The mass of phosphorous $\left(\text{P}\right)$ in $\text{5}\text{.00}$ moles of ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$.