Chapter 5, Problem 5.65AP

(a)

Interpretation Introduction

Interpretation:

The formula weight and molar mass of ${\text{NaNO}}_{\text{2}}$ has to be calculated.

Answer to Problem 5.65AP

The formula weight of ${\text{NaNO}}_{\text{2}}$ is $\text{69}\text{.00 amu}\text{.}$

The molar mass of ${\text{NaNO}}_{\text{2}}$ is $\text{69}\text{.00 g/mol}\text{.}$

Explanation of Solution

The formula weight of ${\text{NaNO}}_{\text{2}}$ is calculated as,

  $\begin{array}{l}\text{1Na}\text{atoms}\text{×}\text{22}\text{.99}\text{amu}\text{=}\text{22}\text{.99}\text{amu}\\ \text{1}\text{N}\text{atoms×14}\text{.01amu}\text{=}\text{14}\text{.01}\text{amu}\\ \text{2}\text{O}\text{atoms×16}\text{.00}\text{amu}\text{=}\underset{\_}{\text{32}\text{.00}\text{amu}}\\ \text{Formula weight}\text{of}{\text{NaNO}}_{\text{2}}\text{=69}\text{.00}\text{amu}\end{array}$

The formula weight of ${\text{NaNO}}_{\text{2}}$ is $\text{69}\text{.00 amu}\text{.}$

Since, the formula weight of ${\text{NaNO}}_{\text{2}}$ is $\text{69}\text{.00 amu,}$ the molar mass of ${\text{NaNO}}_{\text{2}}$ is $\text{69}\text{.00 g/mol}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The molar mass of ${\text{C}}_{\text{2}}{\text{H}}_4$ has to be calculated.

Answer to Problem 5.65AP

The formula weight of ${\text{C}}_{\text{2}}{\text{H}}_4$ is $\text{28}\text{.05 amu}\text{.}$

The molar mass of ${\text{C}}_{\text{2}}{\text{H}}_4$ is $\text{28}\text{.05 g/mol}\text{.}$

Explanation of Solution

The formula weight of ${\text{C}}_{\text{2}}{\text{H}}_4$ is calculated as,

  $\begin{array}{l}\text{2}\text{C}\text{atoms}\text{×}\text{12}\text{.01}\text{amu}\text{=}\text{24}\text{.02}\text{amu}\\ \text{4}\text{H}\text{atoms×1}\text{.008}\text{amu}\text{=}\underset{\_}{\text{4}\text{.032}\text{amu}}\\ \text{Formula weight}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{=28}\text{.05}\text{amu}\end{array}$

The formula weight of ${\text{C}}_{\text{2}}{\text{H}}_4$ is $\text{28}\text{.05 amu}\text{.}$

Since, the formula weight of ${\text{C}}_{\text{2}}{\text{H}}_4$ is $\text{28}\text{.05 amu,}$ the molar mass of ${\text{C}}_{\text{2}}{\text{H}}_4$ is $\text{28}\text{.05 g/mol}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The molar mass of ${\text{C}}_{\text{9}}{\text{H}}_{\text{13}}{\text{NO}}_{\text{2}}$ has to be calculated.

Answer to Problem 5.65AP

The formula weight of ${\text{C}}_{\text{9}}{\text{H}}_{\text{13}}{\text{NO}}_{\text{2}}$ is $\text{167}\text{.20}\text{amu}\text{.}$

The molar mass of ${\text{C}}_{\text{9}}{\text{H}}_{\text{13}}{\text{NO}}_{\text{2}}$ is $\text{167}\text{.20}\text{g/mol}\text{.}$

Explanation of Solution

The formula weight of ${\text{C}}_{\text{9}}{\text{H}}_{\text{13}}{\text{NO}}_{\text{2}}$ is calculated as,

  $\begin{array}{l}\text{9 C}\text{atoms}\text{×}\text{12}\text{.01}\text{amu}\text{=}\text{108}\text{.09}\text{amu}\\ \text{13}\text{H}\text{atoms×1}\text{.008}\text{amu}\text{=}\text{13}\text{.104}\text{amu}\\ \text{1}\text{N}\text{atoms×14}\text{.01}\text{amu}\text{=}\text{14}\text{.01}\text{amu}\\ \text{2}\text{O}\text{atoms×16}\text{.00}\text{amu}\text{=}\underset{\_}{\text{32}\text{.00}\text{amu}}\\ \text{Formula weight}\text{of}{\text{C}}_{\text{9}}{\text{H}}_{\text{13}}{\text{NO}}_{\text{2}}\text{=167}\text{.204}\text{amu}\end{array}$

The formula weight of ${\text{C}}_{\text{9}}{\text{H}}_{\text{13}}{\text{NO}}_{\text{2}}$ is $\text{167}\text{.20}\text{amu}\text{.}$

Since, the formula weight of ${\text{C}}_{\text{9}}{\text{H}}_{\text{13}}{\text{NO}}_{\text{2}}$ is $\text{167}\text{.20}\text{amu,}$ the molar mass of ${\text{C}}_{\text{9}}{\text{H}}_{\text{13}}{\text{NO}}_{\text{2}}$ is $\text{167}\text{.20}\text{g/mol}\text{.}$