Chapter 8, Problem 16E

Interpretation Introduction

(a)

Interpretation:

The molar mass of iron $\left(\text{II}\right)$ acetate, $\text{Fe}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is to be stated.

Concept introduction:

The mass of one mole of any substance is known as its molar mass. The molar mass of an element is equal to the atomic mass of an element. Molar mass is expressed in $\text{g}/\text{mol}$. The molar mass of a compound is the sum of all the atoms of each element present in the compound.

Answer to Problem 16E

The molar mass of iron $\left(\text{II}\right)$ acetate, $\text{Fe}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is $173.94\text{g}/\text{mol}$.

Explanation of Solution

The molar mass of iron $\left(\text{II}\right)$ acetate, $\text{Fe}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is computed by the formula written below.

$\text{Molar mass of iron}\left(\text{II}\right)\text{acetate }\left(\text{Fe}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\right)=\text{Fe}+\text{6H}+\text{4C}+\text{4O}$...(1)

The value of molar mass of iron, hydrogen, carbon and oxygen are $\text{55}\text{.84 }\text{g}/\text{mol}$, $\text{1}\text{.01 }\text{g}/\text{mol}$, $\text{12}\text{.01 }\text{g}/\text{mol}$ and $\text{16}\text{.00 }\text{g}/\text{mol}$ respectively.

Substitute the vales of molar mass of iron, hydrogen, carbon and oxygen in equation number (1).

$\begin{array}{rll}\text{Molar massof iron}\left(\text{II}\right)\text{ acetate}& =& \left(\begin{array}{l}\text{55}\text{.84 }\text{g}/\text{mol}+\text{6}\times \text{1}\text{.01 }\text{g}/\text{mol}+\text{4}\times \text{12}\text{.01 }\text{g}/\text{mol}+\\ \text{4}\times \text{16}\text{.00 }\text{g}/\text{mol}\end{array}\right)\\ & =& \text{173}\text{.94 }\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of the iron $\left(\text{II}\right)$ acetate, $\text{Fe}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is $173.94\text{g}/\text{mol}$.

Conclusion

The molar mass of the iron $\left(\text{II}\right)$ acetate, $\text{Fe}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is $173.94\text{g}/\text{mol}$.

Interpretation Introduction

(b)

Interpretation:

The molar mass of iron $\left(\text{II}\right)$ phosphate, ${\text{Fe}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is to be stated.

Concept introduction:

The mass of one mole of any substance is known as its molar mass. The molar mass of an element is equal to the atomic mass of an element. Molar mass is expressed in $\text{g}/\text{mol}$. The molar mass of a compound is the sum of all the atoms of each element present in the compound.

Answer to Problem 16E

The molar mass of iron $\left(\text{II}\right)$ phosphate, ${\text{Fe}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{357}\text{.49}\text{ g}/\text{mol}$.

Explanation of Solution

The molar mass of iron $\left(\text{II}\right)$ phosphate, ${\text{Fe}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is computed by the formula written below.

$\text{Molar mass}\text{of iron}\left(\text{II}\right){\text{ phosphate Fe}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}=3\text{Fe}+2\text{P}+8\text{O}$  …(2)

The value of molar mass of iron, phosphorus and oxygen are $\text{55}\text{.85 }\text{g}/\text{mol}$, $\text{30}\text{.97 }\text{g}/\text{mol}$ and $\text{16}\text{.00 }\text{g}/\text{mol}$ respectively.

Substitute the values of molar mass of iron, phosphorus and oxygen in equation number (2).

$\begin{array}{rll}\text{Molar massof iron}\left(\text{II}\right)\text{ phosphate}& =& \left(\begin{array}{l}\text{3}\times \text{55}\text{.85 }\text{g}/\text{mol}+\text{2}\times \text{30}\text{.97 }\text{g}/\text{mol}+\\ \text{8}\times \text{16}\text{.00 }\text{g}/\text{mol}\end{array}\right)\\ & =& \text{357}\text{.49}\text{ g}/\text{mol}\end{array}$

Therefore, the molar mass of iron $\left(\text{II}\right)$ phosphate, ${\text{Fe}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{357}\text{.49}\text{ g}/\text{mol}$.

Conclusion

The molar mass of iron $\left(\text{II}\right)$ phosphate, ${\text{Fe}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{357}\text{.49}\text{ g}/\text{mol}$.

Interpretation Introduction

(c)

Interpretation:

The molar mass of tribromine octaoxide, ${\text{Br}}_{\text{3}}{\text{O}}_{\text{8}}$ is to be stated.

Concept introduction:

The mass of one mole of any substance is known as its molar mass. The molar mass of an element is equal to the atomic mass of an element. Molar mass is expressed in $\text{g}/\text{mol}$. The molar mass of a compound is the sum of all the atoms of each element present in the compound.

Answer to Problem 16E

The molar mass of tribromine octaoxide, ${\text{Br}}_{\text{3}}{\text{O}}_{\text{8}}$ is $\text{367}\text{.70 }\text{g}/\text{mol}$.

Explanation of Solution

The molar mass of tribromine octaoxide, ${\text{Br}}_{\text{3}}{\text{O}}_{\text{8}}$ is computed by the formula written below.

$\text{Molar mass of tribromine octaoxide }\left({\text{Br}}_{\text{3}}{\text{O}}_{\text{8}}\right)=3\text{Br}+8\text{O}$  …(3)

The value of molar mass of bromine and oxygen is $\text{79}\text{.90 }\text{g}/\text{mol}$ and $\text{16}\text{.00 }\text{g}/\text{mol}$ respectively.

Substitute the values of molar mass of bromine and oxygen in the equation number (3).

$\begin{array}{rll}\text{Molar}\text{mass}\text{of}\text{tribromine octaoxide}& =& \text{3}\times \text{79}\text{.90 }\text{g}/\text{mol}+\text{8}\times \text{16}\text{.00 }\text{g}/\text{mol}\\ & =& \text{367}\text{.70 }\text{g}/\text{mol}\end{array}$

Therefore the molar mass of tribromine octaoxide, ${\text{Br}}_{\text{3}}{\text{O}}_{\text{8}}$ is $\text{367}\text{.70 }\text{g}/\text{mol}$.

Conclusion

The molar mass of tribromine octaoxide, ${\text{Br}}_{\text{3}}{\text{O}}_{\text{8}}$ is $\text{367}\text{.70 }\text{g}/\text{mol}$.

Interpretation Introduction

(d)

Interpretation:

The molar mass of trinitroglycerin, ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{3}}$ is to be stated.

Concept introduction:

The mass of one mole of any substance is known as its molar mass. The molar mass of an element is equal to the atomic mass of an element. Molar mass is expressed in $\text{g}/\text{mol}$. The molar mass of a compound is the sum of all the atoms of each element present in the compound.

Answer to Problem 16E

The molar mass of trinitroglycerin, ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{3}}$ is $\text{227}\text{.098 }\text{g}/\text{mol}$.

Explanation of Solution

The molar mass of trinitroglycerin, ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{3}}$ is computed by the formula written below.

$\text{Molar mass of trinitro glycerin }\left({\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{3}}\right)=\text{3C}+9\text{O}+5\text{H}+\text{3N}$  …(4)

The value of molar mass of carbon, oxygen, hydrogen and nitrogen are $\text{12}\text{.01 }\text{g}/\text{mol}$, $\text{16}\text{.00 }\text{g}/\text{mol}$, $\text{1}\text{.01 }\text{g}/\text{mol}$ and $\text{14}\text{.006 }\text{g}/\text{mol}$ recpectively.

Substitute the values of molar mass of carbon, oxygen, hydrogen and nitrogen in equation (4).

$\begin{array}{rll}\text{Molar mass of trinitro glycerin}& =& \left(\begin{array}{l}\text{3}\times \text{12}\text{.01 }\text{g}/\text{mol}+\text{9}\times \text{16}\text{.00 }\text{g}/\text{mol}+\text{5}\times \text{1}\text{.01 }\text{g}/\text{mol}+\\ \text{3}\times \text{14}\text{.006 }\text{g}/\text{mol}\end{array}\right)\\ & =& \text{227}\text{.098 }\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of the trinitroglycerin, ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{3}}$ is $\text{227}\text{.098 }\text{g}/\text{mol}$.

Conclusion

The molar mass of trinitroglycerin, ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{3}}$ is $\text{227}\text{.098 }\text{g}/\text{mol}$.