Chapter 10, Problem 73A

(a)

Interpretation Introduction

Interpretation: To determine the percentage composition of ${\text{H}}_{\text{2}}\text{S}$ .

Concept Introduction: Mass percentage can be determined by the ratio of mass of element and the mass of the compound which can be written as:

  $\text{Mass}\text{\%=}\frac{\text{Mass}\text{of}\text{}\text{element}}{\text{Mass}\text{of}\text{}\text{compound}}\text{×100}$

Answer to Problem 73A

The mass % of S will be 94.12 g/mol and mass % of hydrogen 5.88 g/mol.

Explanation of Solution

The given compound is ${\text{H}}_{\text{2}}\text{S}$ .

Mass of hydrogen atom = 2 g/mol

Mass of sulphur atom = 32 g/mol

Mass of ${\text{H}}_{\text{2}}\text{S}$ = 34 g/mol

Mass of percentage can be determined by using the formula:

  $\text{Mass}\text{\%=}\frac{\text{Mass}\text{of}\text{}\text{element}}{\text{Mass}\text{of}\text{}\text{compound}}\text{×100}$

Put the given data in the above equation.

  $\begin{array}{c}\text{\%}\text{H}\text{=}\frac{2}{\text{34}}\text{×100}\\ \text{=5}\text{.88}\%\end{array}$

  $\begin{array}{c}\text{\%}\text{S}\text{=}\frac{32}{\text{34}}\text{×100}\\ \text{=94}\text{.12}\%\end{array}$

Therefore, mass % of S will be 94.12 g/mol and mass % of hydrogen 5.88 g/mol.

(b)

Interpretation Introduction

Interpretation: To determine the percentage composition of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ .

Concept Introduction: Mass percentage can be determined by the ration of mass of element and mass of the compound which can be written as:

  $\text{Mass}\text{\%=}\frac{\text{Mass}\text{of}\text{}\text{element}}{\text{Mass}\text{of}\text{}\text{compound}}\text{×100}$

Answer to Problem 73A

mass % of N will be $\text{22}\text{.5}\%$ and mass % of hydrogen $\text{6}\text{.4}\%$ , mass % of C will be $\text{19}\text{.3}\%$ , and mass % of O will be $\text{51}\text{.5}\%$ .

Explanation of Solution

The given compound is ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ .

The molecular mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ is $124.1 \text{g/mol}$

The molar mass of hydrogen atom = 1 g/mol

The molar mass of oxygen atom = 16 g/mol

The molar mass of nitrogen atom = 14 g/mol

The molar mass of carbon atom = 12 g/mol

Mass of the percentage can be determined by using the formula:

  $\text{Mass}\text{\%=}\frac{\text{Mass}\text{of}\text{}\text{element}}{\text{Mass}\text{of}\text{}\text{compound}}\text{×100}$

Put the given data in the above equation.

  $\begin{array}{c}\text{\%}\text{H}\text{=}\frac{1\times 8}{124.1 \text{g/mol}}\text{×100}\\ \text{=6}\text{.4}\%\end{array}$

  $\begin{array}{c}\text{\%}\text{N}\text{=}\frac{2\times 14}{124.1 \text{g/mol}}\text{×100}\\ \text{=22}\text{.5}\%\end{array}$

  $\begin{array}{c}\text{\%}\text{C}\text{=}\frac{2\times 12}{124.1 \text{g/mol}}\text{×100}\\ \text{=19}\text{.3}\%\end{array}$

  $\begin{array}{c}\text{\%}\text{O}\text{=}\frac{4\times 16}{124.1 \text{g/mol}}\text{×100}\\ \text{=51}\text{.5}\%\end{array}$

Therefore, mass % of N will be $\text{22}\text{.5}\%$ and mass % of hydrogen $\text{6}\text{.4}\%$ , mass % of C will be $\text{19}\text{.3}\%$ , and mass % of O will be $\text{51}\text{.5}\%$ .

(c)

Interpretation Introduction

Interpretation: To determine the percentage composition of the given compound.

Concept Introduction: Mass percentage can be determined by the ration of mass of element and mass of the compound which can be written as:

  $\text{Mass}\text{\%=}\frac{\text{Mass}\text{of}\text{}\text{element}}{\text{Mass}\text{of}\text{}\text{compound}}\text{×100}$

Answer to Problem 73A

The mass % of Mg will be 41.67 %and mass % of hydrogen 3.41% , mass % of O will be 54.87%.

Explanation of Solution

The given compound is ${\text{Mg(OH)}}_{\text{2}}$ .

Mass of $\text{Mg}$ atom = 24 g/mol

Mass of oxygen atom =32 g/mol

Mass of hydrogen atom = 2 g/mol

Mass of ${\text{Mg(OH)}}_{\text{2}}$ = 58.31 g/mol

Mass of percentage can be determined by using the formula:

Put the given data in the above equation.

  $\begin{array}{c}\text{\%}\text{H}\text{=}\frac{2.016}{58.31}\text{×100}\\ \text{=}3.45\%\end{array}$

  $\begin{array}{c}\text{\%}\text{Mg}\text{=}\frac{24.30}{58.31}\text{×100}\\ \text{=41}\text{.67}\%\end{array}$

  $\begin{array}{c}\text{\%}\text{O}\text{=}\frac{32}{58.31}\text{×100}\\ \text{=54}\text{.87}\%\end{array}$

Therefore, mass % of Mg will be 41.67 %and mass % of hydrogen 3.45% , mass % of O will be 54.87%.

(d)

Interpretation Introduction

Interpretation: To determine the percentage composition of the given compound.

Concept Introduction: Mass percentage can be determined by the ration of mass of element and mass of the compound which can be written as:

Answer to Problem 73A

The mass % of Na will be 42.1 %and mass % of P will be 18 % , mass % of O will be 39%.

Explanation of Solution

The given compound is ${\text{Na}}_{\text{3}}{\text{(PO}}_{\text{4}}\text{)}$ .

Mass of Na atom = 69 g/mol

Mass of oxygen atom =64 g/mol

Mass of P atom = 31 g/mol

Mass of ${\text{Na}}_{\text{3}}{\text{(PO}}_{\text{4}}\text{)}$ = 164 g/mol

Mass of the percentage can be determined by using the formula:

  $\text{Mass}\text{\%=}\frac{\text{Mass}\text{of}\text{}\text{element}}{\text{Mass}\text{of}\text{}\text{compound}}\text{×100}$

Put the given data in the above equation.

  $\begin{array}{c}\text{\%}\text{Na}\text{=}\frac{69}{164}\text{×100}\\ \text{=}42.1\%\end{array}$

  $\begin{array}{c}\text{\%}\text{P}\text{=}\frac{31}{164}\text{×100}\\ \text{=18}\%\end{array}$

  $\begin{array}{c}\text{\%}\text{O}\text{=}\frac{64}{164}\text{×100}\\ \text{=39}\%\end{array}$

Therefore, mass % of Na will be 42.1 %and mass % of P will be 18 % , mass % of O will be 39%.