Chapter 3.4, Problem 28SSC

Interpretation Introduction

Interpretation:

Whether the compound $\text{I}$ and compound $\text{II}$ are the same or different needs to be explained.

Concept introduction:

The mass percentage is defined as the percentage of component present in it. The mass percentage of the element is the mass of element divided by the total mass of the compound and multiplied by 100.

$\text{Mass percentage =}\frac{\text{mass}\text{of}\text{element}}{\text{mass}\text{of}\text{total}\text{compound}}\text{×100}$

Answer to Problem 28SSC

For sample $I$ : The mass percentage of the Fe is $69.05$ and mass percentage of O is $30.05$ informed compounds.

For sample $II$ : The mass percentage of the Fe is $77.73$ and mass percentage of O is $22.27$ informed compounds.

So, the compound or sample is not same.

Explanation of Solution

Given information:

The table:

Analysis Data of Two Iron Compounds
compound	Total mass (g)	Mass Fe (g)	Mass O (g)	Mass percentage Fe	Mass percentage O
$I$	$75.00$	$52.46$	$22.54$
$II$	$56.00$	$43.53$	$12.47$

It is known that the mass of compound $=75\text{g}$

The mass value of Fe for sample I is $52.46$ and the mass value of O is $22.54$ from the given table. So

The percentage by mass for sample $I$calculated as −

Percentage by mass of element, $\text{Fe}=\frac{\text{52}\text{.46 g}}{\text{75 g}}\text{×100}\text{=}\text{69}\text{.94\%}$

Percentage by mass of element, $\text{O}=\frac{\text{22}\text{.54 g}}{\text{75 g}}\text{×100}\text{=}\text{30}\text{.05\%}$

Themass percentage of Fe and O is calculated for sample $II$-

It is known that the mass of compound $=56\text{g}$

From the given table −

Percentage by mass of element, $\text{Fe}=\frac{\text{43}\text{.54 g}}{\text{56 g}}\text{×100}\text{=}\text{77}\text{.73\%}$

Percentage by mass of element, $\text{O}=\frac{\text{12}\text{.47 g}}{\text{56 g}}\text{×100}\text{=}\text{22}\text{.26\%}$

Thus, the required table is:

Analysis Data of Two Iron Compounds
compound	Total mass (g)	Mass Fe (g)	Mass O (g)	Mass percentage Fe	Mass percentage O
$\text{I}$	$\begin{array}{l}75.00\hfill \end{array}$	$\begin{array}{l}52.46\hfill \end{array}$	$\begin{array}{l}22.54\hfill \end{array}$	$\begin{array}{l}69.05\hfill \end{array}$	$\begin{array}{l}30.05\hfill \end{array}$
$\text{II}$	$\begin{array}{l}56.00\hfill \end{array}$	$43.53$	$\begin{array}{l}12.47\hfill \end{array}$	$77.73$	$\begin{array}{l}22.27\hfill \end{array}$

Now for the law of multiple proportion -

Sample $\text{I,}\text{Fe:O=}\frac{\text{mass}\text{of}\text{Fe}}{\text{mass}\text{of}\text{O}}\text{=}\frac{\text{52}\text{.46}}{\text{22}\text{.54}}\text{=2}\text{.327}$

Sample $\text{II,}\text{Fe:O}\text{=}\frac{\text{mass}\text{of}\text{Fe}}{\text{mass}\text{of}\text{O}}\text{=}\frac{\text{43}\text{.53}}{\text{12}\text{.47}}\text{=}\text{3}\text{.491}$

The relative mass ratio of Fe in the sample $\text{I}$ and sample $\text{II}$ are $2.327:3.491=2:3$

Hence, the compound or sample is not same.

Conclusion

The calculated percentage by mass for sample $\text{I}$ and sample $\text{II}$ are not the same.