Chapter 2, Problem 2.7P

Identify the mass law that each of the following observations demonstrates, and explain your reasoning:

(a) A sample of potassium chloride from Chile contains the same percent by mass of potassium as one from Poland.

(b) A flash bulb contains magnesium and oxygen before use and magnesium oxide afterward, but its mass does not change.

(c) Arsenic and oxygen form one compound that is 65.2 mass % arsenic and another that is 75.8 mass % arsenic.

Interpretation Introduction

Interpretation: Identification of mass law should be done for two potassium chloride samples one from Chile and another one from Poland.

Concept introduction:Chemical compounds contain its constituent elements in a fixed ratio by mass and it is independent of its origin and the way of preparation, it is known as law of definite proportions.

For an isolated system, despite chemical reaction or physical transformation total mass of the system remain constant, it is known as law of conservation of mass.

If fixed mass of one component combine with different masses of second component to form more than one compound between them, then the ratios of masses of the second component that combine with the first component will be the ratios of simple whole numbers. This law is known as e law of multiple proportions.

Explanation of Solution

A sample of potassium chloride $\left(\text{KCl}\right)$ always contains $52.445\%$ of potassium and $47.555\%$ of chlorine by mass and it does not depend on its source. So it follows laws of definite proportion.

Interpretation Introduction

Interpretation: Identification of mass law should be done for a flashbulb that contains magnesium and oxygen.

Concept introduction: Chemical compounds contain its constituent elements in a fixed ratio by mass and it is independent of its origin and the way of preparation, it is known as law of definite proportions.

For an isolated system, despite chemical reaction or physical transformation total mass of the system remain constant, it is known as law of conservation of mass.

If fixed mass of one component combine with different masses of second component to form more than one compound between them, then the ratios of masses of the second component that combine with the first component will be the ratios of simple whole numbers. This law is known as law of multiple proportions.

Explanation of Solution

In flashbulb, magnesium and oxygen react and form magnesium oxide $\left(\text{MgO}\right)$ . The reaction for the formation of $\text{MgO}$ is as follows:

  $\text{2Mg}+{\text{O}}_2\to 2\text{MgO}$

Mass of single atom of an element is equal to mass of $1\text{mol}$ of that element. Therefore, molar mass of $\text{Mg}$ is $\text{24}\text{.305}\text{g/mol}$ .

The total mass of species in the reactant side is calculated as follows:

  $\begin{array}{c}\text{Total mass on reactant side}=\text{2}\left(\text{Molar mass of Mg}\right)\text{+}\left(\text{Molar}\text{mass}\text{of}{\text{O}}_{\text{2}}\right)\\ =\text{2}\left(\text{24}\text{.305}\text{g/mol}\right)+\text{32}\text{g/mol}\\ =\text{80}\text{.16}\text{g/mol}\end{array}$

The total mass of species in the product side is calculated as follows:

  $\begin{array}{c}\text{Total mass on product side}=2\left(\text{Molar mass of MgO}\right)\\ =2\left(40.305\text{g/mol}\right)\\ =80.16\text{g/mol}\end{array}$

Here the total mass of the component of the reactant side and the product side is equal. The reaction in the flashbulb followslaw of conservation of mass.

Interpretation Introduction

Interpretation: Identification of mass law should be done for two compounds of arsenic and oxygen. Among them, one compound has65.2 mass $\%$ arsenic and another one has 75.8 mass $\%$ arsenic.

Concept introduction: Chemical compounds contain its constituent elements in a fixed ratio by mass and it is independent of its origin and the way of preparation, it is known as aw of definite proportions.

For an isolated system, despite chemical reaction or physical transformation total mass of the system remain constant, it is known as law of conservation of mass.

If fixed mass of one component combine with different masses of second component to form more than one compound between them, then the ratios of masses of the second component that combine with the first component will be the ratios of simple whole numbers. This law is known as law of multiple proportions.

Explanation of Solution

Arsenic and oxygen form two compounds. For the first compound, mass $\%$ of arsenic is $\text{65}\text{.2}\text{\%}$ and therefore the percentage of oxygen is calculated as follows:

  $\begin{array}{c}\text{mass \% of oxygen}=\text{100 \%}-\text{mass \% of arsenic}\\ =1\text{00 \%}-\text{65}\text{.2}\text{\%}\\ =34.8\text{\%}\end{array}$

The formula to calculate ratio of molecularity of arsenic to molecularity of oxygen for first compound is as follows:

  ${\left(\frac{\text{Arsenic}}{\text{Oxygen}}\right)}_1=\left(\frac{\left(\text{Mass}\text{\%}\text{of}\text{arsenic}\right)\left(\text{Atomic}\text{mass}\text{of}\text{oxygen}\right)}{\left(\text{Atomic}\text{mass}\text{of}\text{arsenic}\right)\left(\text{Mass}\text{\%of}\text{oxygen}\right)}\right)$

The mass $\%$ of arsenic is $\text{65}\text{.2}\text{\%}$ .

The mass $\%$ of oxygen is $34.8\text{\%}$ .

Atomic mass of oxygen is $16\text{u}$ .

Atomic mass of arsenic is $74.92\text{u}$ .

Substitute the values in above equation.

  $\begin{array}{c}{\left(\frac{\text{Arsenic}}{\text{Oxygen}}\right)}_1=\left(\frac{\left(\text{Mass}\text{\%}\text{of}\text{arsenic}\right)\left(\text{Atomic}\text{mass}\text{of}\text{oxygen}\right)}{\left(\text{Atomic}\text{mass}\text{of}\text{arsenic}\right)\left(\text{Mass}\text{\%of}\text{oxygen}\right)}\right)\\ =\left(\frac{\left(65.2\text{\%}\right)\left(16\text{u}\right)}{\left(74.92\text{u}\right)\left(34.8\text{\%}\right)}\right)\\ =\frac{2}{5}\end{array}$

Hence the molecular formula of the first compound is ${\text{As}}_{\text{2}}{\text{O}}_{\text{5}}$ .

For the second compound, mass $\%$ of arsenic is $\text{75}\text{.8}\text{\%}$ and therefore the percentage of oxygen is calculated as follows:

  $\begin{array}{c}\text{mass \% of oxygen}=\text{100 \%}-\text{mass \% of arsenic}\\ =1\text{00 \%}-\text{75}\text{.8}\text{\%}\\ =24.2\text{\%}\end{array}$

The formula to calculate ratio of molecularity of arsenic to molecularity of oxygen for second compound is as follows:

  ${\left(\frac{\text{Arsenic}}{\text{Oxygen}}\right)}_2=\left(\frac{\left(\text{Mass}\text{\%}\text{of}\text{arsenic}\right)\left(\text{Atomic}\text{mass}\text{of}\text{oxygen}\right)}{\left(\text{Atomic}\text{mass}\text{of}\text{arsenic}\right)\left(\text{Mass}\text{\%of}\text{oxygen}\right)}\right)$

The mass $\%$ of arsenic is $\text{75}\text{.8}\text{\%}$ .

The mass $\%$ of oxygen is $24.2\text{\%}$ .

Atomic mass of oxygen is $16\text{u}$ .

Atomic mass of arsenic is $74.92\text{u}$ .

Substitute the values in above equation.

  $\begin{array}{c}{\left(\frac{\text{Arsenic}}{\text{Oxygen}}\right)}_2=\left(\frac{\left(\text{Mass}\text{\%}\text{of}\text{arsenic}\right)\left(\text{Atomic}\text{mass}\text{of}\text{oxygen}\right)}{\left(\text{Atomic}\text{mass}\text{of}\text{arsenic}\right)\left(\text{Mass}\text{\%of}\text{oxygen}\right)}\right)\\ =\left(\frac{\left(\text{75}\text{.8}\text{\%}\right)\left(16\text{u}\right)}{\left(74.92\text{u}\right)\left(24.2\text{\%}\right)}\right)\\ =\frac{2}{3}\end{array}$

Hence the molecular formula of the second compound is ${\text{As}}_{\text{2}}{\text{O}}_{\text{3}}$ . As the total mass of arsenic is the same for both compounds, so ratio of the number of oxygen present in the first compound to the number of oxygen present in the second compound is calculated as follows:

  $\begin{array}{c}\left(\frac{{\text{Oxygen}}_1}{{\text{Oxygen}}_2}\right)=\frac{5}{3}\\ =5:3\end{array}$

Ratio of oxygen in both compounds is simple whole numbers. Solaw of multiple proportions is applicable for both compounds.