Chapter 6, Problem 27A

(a)

Interpretation Introduction

Interpretation:

The percentage by mass of each element needs to be determined in sodium sulphate.

Concept Introduction:

In any compound, the mass percent of an element can be determined as follows:

  $\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}$

Answer to Problem 27A

Mass percent of Na = 32.37%

Mass percent of S = 22.57%

Mass percent of O = 45.05%

Explanation of Solution

Sodium sulfate: Na₂SO₄

Molar mass of Na₂SO₄ = 142.04 g/mol

Molar mass of Na = 22.99 g/mol

Molar mass of S = 32.065 g/mol

Molar mass of O = 15.999 g/mol

There are 2 moles of Na, 1 mol of S and 4 moles of O in 1 mol of Na₂SO₄

  $\begin{array}{l}\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}\hfill \\ {\text{Total mass of Na in 1 mol of Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\left(\text{2×22}\text{.99}\right)\text{g = 45}\text{.98 g}\hfill \\ {\text{Total mass of S in 1 mol of Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\left(\text{1×32}\text{.065}\right)\text{g = 32}\text{.065 g}\hfill \\ {\text{Total mass of O in 1 mol of Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\left(\text{4×15}\text{.999}\right)\text{g = 63}\text{.996 g}\hfill \\ \text{Mass percent of Na =}\left(\frac{\text{45}\text{.98g}}{\text{142}\text{.04g}}\text{}\text{×100}\right)\text{\% = 32}\text{.37\%}\hfill \\ \text{Mass percent of S =}\left(\frac{\text{32}\text{.065g}}{\text{142}\text{.04g}}\text{}\text{×100}\right)\text{\% = 22}\text{.57\%}\hfill \\ \text{Mass percent of O =}\left(\frac{\text{63}\text{.996g}}{\text{142}\text{.04g}}\text{}\text{×100}\right)\text{\% = 45}\text{.05\%}\hfill \end{array}$

(b)

Interpretation Introduction

Interpretation:

The percentage by mass of each element needs to be determined in sodium sulphite.

Concept Introduction:

In any compound, the mass percent of an element can be determined as follows:

  $\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}$

Answer to Problem 27A

Mass percent of Na = 36.48%

Mass percent of S = 25.44%

Mass percent of O = 38.08%

Explanation of Solution

Sodium sulphite: Na₂SO₃

Molar mass of Na₂SO₃ = 126.043 g/mol

Molar mass of Na = 22.99 g/mol

Molar mass of S = 32.065 g/mol

Molar mass of O = 15.999 g/mol

There are 2 moles of Na, 1 mol of S and 3 moles of O in 1 mol of Na₂SO₃

  $\begin{array}{l}{\text{Total mass of Na in 1 mol of Na}}_{\text{2}}{\text{SO}}_{\text{3}}\text{=}\left(\text{2×22}\text{.99}\right)\text{g = 45}\text{.98 g}\hfill \\ {\text{Total mass of S in 1 mol of Na}}_{\text{2}}{\text{SO}}_{\text{3}}\text{=}\left(\text{1×32}\text{.065}\right)\text{g = 32}\text{.065 g}\hfill \\ {\text{Total mass of O in 1 mol of Na}}_{\text{2}}{\text{SO}}_{\text{3}}\text{=}\left(\text{3×15}\text{.999}\right)\text{g = 47}\text{.997 g}\hfill \\ \text{Mass percent of Na =}\left(\frac{\text{45}\text{.98g}}{\text{126}\text{.043g}}\text{}\text{×100}\right)\text{\% = 36}\text{.48\%}\hfill \\ \text{Mass percent of S =}\left(\frac{\text{32}\text{.065g}}{\text{126}\text{.043g}}\text{}\text{×100}\right)\text{\% = 25}\text{.44\%}\hfill \\ \text{Mass percent of O =}\left(\frac{\text{47}\text{.997g}\text{}}{\text{126}\text{.043g}}\text{×100}\right)\text{\% = 38}\text{.08\%}\hfill \end{array}$

(c)

Interpretation Introduction

Interpretation:

The percentage by mass of each element needs to be determined in sodium sulphide.

Concept Introduction:

In any compound, the mass percent of an element can be determined as follows:

  $\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}$

Answer to Problem 27A

Mass percent of Na = 58.91%

Mass percent of S = 41.08%

Explanation of Solution

Sodium sulphide: Na₂S

Molar mass of Na₂S = 78.05 g/mol

Molar mass of Na = 22.99 g/mol

Molar mass of S = 32.065 g/mol

There are 2 moles of Na and 1 mol of S in 1mol of Na₂S

  $\begin{array}{l}{\text{Total mass of Na in 1 mol of Na}}_{\text{2}}\text{S =}\left(\text{2×22}\text{.99}\right)\text{g = 45}\text{.98 g}\hfill \\ {\text{Total mass of S in 1 mol of Na}}_{\text{2}}\text{S =}\left(\text{1×32}\text{.065}\right)\text{g = 32}\text{.065 g}\hfill \\ \text{Mass percent of Na =}\left(\frac{\text{45}\text{.98g}}{\text{78}\text{.05g}}\text{}\text{×100}\right)\text{\% = 58}\text{.91\%}\hfill \\ \text{Mass percent of S =}\left(\frac{\text{32}\text{.065g}}{\text{78}\text{.05g}}\text{}\text{×100}\right)\text{\% = 41}\text{.08\%}\hfill \end{array}$

(d)

Interpretation Introduction

Interpretation:

The percentage by mass of each element needs to be determined in sodium thiosulphate.

Concept Introduction:

In any compound, the mass percent of an element can be determined as follows:

  $\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}$

Answer to Problem 27A

Mass percent of Na = 29.08%

Mass percent of S = 40.56%

Mass percent of O = 30.36%

Explanation of Solution

(d) Sodium thiosulfate: Na₂S₂O₃

Molar mass of Na₂S₂O₃ = 158.11 g/mol

Molar mass of Na = 22.99 g/mol

Molar mass of S = 32.065 g/mol

Molar mass of O = 15.999 g/mol

  $\begin{array}{l}{\text{Total mass of Na in 1 mol of Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{=}\left(\text{2×22}\text{.99}\right)\text{g = 45}\text{.98 g}\hfill \\ {\text{Total mass of S in 1 mol of Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{=}\left(\text{2×32}\text{.065}\right)\text{g = 64}\text{.13 g}\hfill \\ {\text{Total mass of O in 1 mol of Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{=}\left(\text{3×15}\text{.999}\right)\text{g = 47}\text{.997 g}\hfill \\ \text{Mass percent of Na =}\left(\frac{\text{45}\text{.98g}}{\text{158}\text{.11g}}\text{}\text{×100}\right)\text{\% = 29}\text{.08\%}\hfill \\ \text{Mass percent of S =}\left(\frac{\text{64}\text{.13g}}{\text{158}\text{.11g}}\text{}\text{×100}\right)\text{\% = 40}\text{.56\%}\hfill \\ \text{Mass percent of O =}\left(\frac{\text{47}\text{.997g}}{\text{158}\text{.11g}}\text{}\text{×100}\right)\text{\% = 30}\text{.36\%}\hfill \\ \hfill \end{array}$

(e)

Interpretation Introduction

Interpretation:

The percentage by mass of each element needs to be determined in potassium phosphate.

Concept Introduction:

In any compound, the mass percent of an element can be determined as follows:

  $\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}$

Answer to Problem 27A

Mass percent of K = 55.26%

Mass percent of P = 14.59%

Mass percent of O = 30.15%

Mass percent of P = 20.89%

Explanation of Solution

Potassium phosphate: K₃PO₄

Molar mass of K₃PO₄ = 212.27 g/mol

Molar mass of K = 39.098 g/mol

Molar mass of P = 30.97 g/mol

Molar mass of O = 15.999 g/mol

1 mol of K₃PO₄ contains 3 moles of K, 1 mol of P and 4 moles of O.

  $\begin{array}{l}{\text{Total mass of K in 1 mol of K}}_{\text{3}}{\text{PO}}_{\text{4}}\text{=}\left(\text{3×39}\text{.098}\right)\text{g = 117}\text{.29 g}\hfill \\ {\text{Total mass of P in 1 mol of K}}_{\text{3}}{\text{PO}}_{\text{4}}\text{=}\left(\text{1×30}\text{.97}\right)\text{g = 30}\text{.97 g}\hfill \\ {\text{Total mass of O in 1 mol of K}}_{\text{3}}{\text{PO}}_{\text{4}}\text{=}\left(\text{4×15}\text{.999}\right)\text{g = 63}\text{.996 g}\hfill \\ \text{Mass percent of K =}\left(\frac{\text{117}\text{.29g}}{\text{212}\text{.27g}}\text{}\text{×100}\right)\text{\% = 55}\text{.26\%}\hfill \\ \text{Mass percent of P =}\left(\frac{\text{30}\text{.97g}\text{}}{\text{212}\text{.27g}}\text{×100}\right)\text{\% = 14}\text{.59\%}\hfill \\ \text{Mass percent of O =}\left(\frac{\text{63}\text{.996g}}{\text{212}\text{.27g}}\text{}\text{×100}\right)\text{\% = 30}\text{.15\%}\hfill \end{array}$

(f)

Interpretation Introduction

Interpretation:

The percentage by mass of each element needs to be determined in potassium hydrogen phosphate.

Concept Introduction:

In any compound, the mass percent of an element can be determined as follows:

  $\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}$

Answer to Problem 27A

Mass percent of K = 44.89%

Mass percent of H = 0.58%

Mass percent of P =17.78%

Mass percent of O = 36.74%

Explanation of Solution

Potassium hydrogen phosphate: K₂HPO₄

Molar mass of K₂HPO₄ = 174.2 g/mol

Molar mass of K = 39.098 g/mol

Molar mass of P = 30.97 g/mol

Molar mass of O = 15.999 g/mol

Molar mass of H = 1.008 g/mol

1 mol of K₂HPO₄ contains 2 moles of K, 1 mol of H, 1 mol of P and 4 moles of O

  $\begin{array}{l}{\text{Total mass of K in 1 mol of K}}_{\text{2}}{\text{HPO}}_{\text{4}}\text{=}\left(\text{2×39}\text{.098}\right)\text{g = 78}\text{.196 g}\hfill \\ {\text{Total mass of H in 1 mol of K}}_{\text{2}}{\text{HPO}}_{\text{4}}\text{=}\left(\text{1×1}\text{.008}\right)\text{g = 1}\text{.008 g}\hfill \\ {\text{Total mass of P in 1 mol of K}}_{\text{2}}{\text{HPO}}_{\text{4}}\text{=}\left(\text{1×30}\text{.97}\right)\text{g = 30}\text{.97 g}\hfill \\ {\text{Total mass of O in 1 mol of K}}_{\text{2}}{\text{HPO}}_{\text{4}}\text{=}\left(\text{4×15}\text{.999}\right)\text{g = 63}\text{.996 g}\hfill \\ \text{Mass percent of K =}\left(\frac{\text{78}\text{.196g}}{\text{174}\text{.2g}}\text{}\text{×100}\right)\text{\% = 44}\text{.89\%}\hfill \\ \text{Mass percent of H =}\left(\frac{\text{1}\text{.008g}}{\text{174}\text{.2g}}\text{}\text{×100}\right)\text{\% = 0}\text{.58\%}\hfill \\ \text{Mass percent of P =}\left(\frac{\text{30}\text{.97g}}{\text{174}\text{.2g}}\text{}\text{×100}\right)\text{\% = 17}\text{.78\%}\hfill \\ \text{Mass percent of O =}\left(\frac{\text{63}\text{.996g}}{\text{174}\text{.2g}}\text{}\text{×100}\right)\text{\% = 36}\text{.74\%}\hfill \end{array}$

(g)

Interpretation Introduction

Interpretation:

The percentage by mass of each element needs to be determined in potassium dihydrogen phosphate.

Concept Introduction:

In any compound, the mass percent of an element can be determined as follows:

  $\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}$

Concept Introduction:

  $\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}$

Answer to Problem 27A

Mass percent of K = 28.73%

Mass percent of H = 1.48%

Mass percent of P = 22.76%

Mass percent of O = 47.03%

Explanation of Solution

  $\begin{array}{l}\left(\text{g}\right){\text{ Potassium dihydrogen phosphate: KH}}_{\text{2}}{\text{PO}}_{\text{4}}\hfill \\ {\text{Molar mass of KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{= 136}\text{.086 g/mol}\hfill \\ \text{Molar mass of K = 39}\text{.098 g/mol}\hfill \\ \text{Molar mass of P = 30}\text{.97 g/mol}\hfill \\ \text{Molar mass of O = 15}\text{.999 g/mol}\hfill \\ \text{Molar mass of H = 1}\text{.008 g/mol}\hfill \\ {\text{1 mol of KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{contains 1 mol of K, 2 moles of H, 1 mol of P and 4 moles of O}\hfill \\ {\text{Total mass of K in 1 mol of KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{=}\left(\text{1×39}\text{.098}\right)\text{g = 39}\text{.098 g}\hfill \\ {\text{Total mass of H in 1 mol of KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{=}\left(\text{2×1}\text{.008}\right)\text{g = 2}\text{.016 g}\hfill \\ {\text{Total mass of P in 1 mol of KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{=}\left(\text{1×30}\text{.97}\right)\text{g = 30}\text{.97 g}\hfill \\ {\text{Total mass of O in 1 mol of KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{=}\left(\text{4×15}\text{.999}\right)\text{g = 63}\text{.996 g}\hfill \\ \text{Mass percent of K =}\left(\frac{\text{39}\text{.098g}}{\text{136}\text{.086g}}\text{}\text{×100}\right)\text{\% = 28}\text{.73\%}\hfill \\ \text{Mass percent of H =}\left(\frac{\text{2}\text{.016g}}{\text{136}\text{.086g}}\text{}\text{×100}\right)\text{\% = 1}\text{.48\%}\hfill \\ \text{Mass percent of P =}\left(\frac{\text{30}\text{.97g}}{\text{136}\text{.086g}}\text{}\text{×100}\right)\text{\% = 22}\text{.76\%}\hfill \\ \text{Mass percent of O =}\left(\frac{\text{63}\text{.996g}}{\text{136}\text{.086g}}\text{}\text{×100}\right)\text{\% = 47}\text{.03\%}\hfill \\ \hfill \end{array}$

(h)

Interpretation Introduction

Interpretation:

The percentage by mass of each element needs to be determined in potassium phosphide.

Concept Introduction:

In any compound, the mass percent of an element can be determined as follows:

  $\text{Mass percent of an element = }\left[\frac{\left(\text{Total mass of an element in 1 mol of compound}\right)}{\left(\text{Molar mass of the compound}\right)}\right]\text{×100\%}$

Answer to Problem 27A

Mass percent of K = 79.11%

Mass percent of P = 20.89%

Explanation of Solution

  $\begin{array}{l}\left(\text{h}\right){\text{ Potassium phosphide: K}}_{\text{3}}\text{P}\hfill \\ {\text{Molar mass of K}}_{\text{3}}\text{P = 148}\text{.27 g/mol}\hfill \\ \text{Molar mass of K = 39}\text{.098 g/mol}\hfill \\ \text{Molar mass of P = 30}\text{.97 g/mol}\hfill \\ {\text{1 mol of K}}_{\text{3}}\text{P contains 3 moles of K and 1 mol of P}\hfill \\ {\text{Total mass of K in 1 mol of K}}_{\text{3}}\text{P =}\left(\text{3×39}\text{.098}\right)\text{g = 117}\text{.294 g}\hfill \\ {\text{Total mass of P in 1 mol of K}}_{\text{3}}\text{P =}\left(\text{1×30}\text{.97}\right)\text{g = 30}\text{.97 g}\hfill \\ \text{Mass percent of K =}\left(\frac{\text{117}\text{.294g}}{\text{148}\text{.27g}}\text{}\text{×100}\right)\text{\% = 79}\text{.11\%}\hfill \\ \text{Mass percent of P =}\left(\frac{\text{30}\text{.97g}}{\text{148}\text{.27g}}\text{}\text{×100}\right)\text{\% = 20}\text{.89\%}\hfill \end{array}$