Chapter 1, Problem 112QRT

Interpretation Introduction

Interpretation:

From the given, the mass of oxygen in each sample has to be calculated.  Also, the mass of iron in the sample of rust and the mass of aluminum in the final sample has to be calculated.

Explanation of Solution

The mass of each atom to the mass of sample is given as

  $\begin{array}{l}\text{35}\text{.48}\text{g}\text{rust}\text{=}{\text{m}}_{\text{Fe}}\text{+}{\text{m}}_{\text{O}}\\ \\ \text{22}\text{.65}\text{g}\text{aluminum oxide}\text{=}{\text{m}}_{\text{AI}}\text{+}{\text{m}}_{\text{O}}\\ \\ {\text{m}}_{\text{Fe}}\text{=}\text{2}\text{.069}{\text{m}}_{\text{Al}}\end{array}$

Elimination of ${\text{m}}_{\text{O}}$ from the above expression:

  $\text{35}\text{.48}\text{g}\text{-}\text{22}\text{.65}\text{g}\text{=}{\text{m}}_{\text{Fe}}{\text{+m}}_{\text{O}}\text{-}{\text{(m}}_{\text{Al}}\text{+}{\text{m}}_{\text{O}}\text{)}\text{=}{\text{m}}_{\text{Fe}}\text{-}{\text{m}}_{\text{Al}}$

The value of ${\text{m}}_{\text{Fe}}$ is substituted to get ${\text{m}}_{\text{Al}}$.

  $\begin{array}{l}\text{35}\text{.48}\text{g}\text{-}\text{22}\text{.65}\text{g}\text{=}{\text{m}}_{\text{Fe}}\text{-}{\text{m}}_{\text{Al}}\\ \\ \text{12}\text{.83}\text{g}\text{=}\text{2}\text{.069}{\text{m}}_{\text{Al}}\text{-}{\text{m}}_{\text{Al}}\text{=}\text{1}\text{.069}{\text{m}}_{\text{Al}}\\ \\ {\text{m}}_{\text{Al}}\text{=}\text{12}\text{.00}\text{g}\text{Al}\end{array}$

The ${\text{m}}_{\text{O}}$ is calculated as

  $\begin{array}{l}{\text{m}}_{\text{O}}\text{=}\text{22}\text{.65}\text{g}\text{aluminum oxide}\text{-}{\text{m}}_{\text{Al}}\\ \\ \text{=}\text{22}\text{.65}\text{g}\text{-}\text{12}\text{.00}\text{g}\text{Al}\\ \\ \text{=}\text{10}\text{.65}\text{g}\text{O}\end{array}$

The ${\text{m}}_{\text{Fe}}$ is calculated as

  $\begin{array}{l}{\text{m}}_{\text{Fe}}\text{=}\text{35}\text{.48}\text{g}\text{rust}\text{-}{\text{m}}_{\text{O}}\\ \\ \text{=}\text{35}\text{.48}\text{g}\text{rust}\text{-}\text{10}\text{.65}\text{g}\text{O}\\ \\ \text{=}\text{24}\text{.83}\text{g}\text{Fe}\end{array}$

The ratio of calculated mass of Iron to Aluminum is

  $\frac{\text{24}\text{.83}\text{g}}{\text{12}\text{.00}\text{g}}\text{=}\text{2}\text{.069}$

The mass of iron in rust is $\text{24}\text{.83}\text{g}$.

The mass of aluminum in final sample is $\text{12}\text{.00}\text{g}$.

The mass of oxygen in each sample is $\text{10}\text{.65}\text{g}$.