Chapter 9, Problem 16CR

Interpretation Introduction

(a)

Interpretation:

The percent by mass of the element whose symbol occurs first in the following compound’s formula should be calculated.

Concept Introduction:

Percent by mass of an element can be calculated by the following equation.

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$.

Answer to Problem 16CR

Percentage of C per mole of C₆ H_{6 ( l )} by mass = $92.26\text{\%}$.

Explanation of Solution

Molar mass of C₆ H_{6 ( l )} = $\left\{\left(6\times 12.01\right)+\left(6\times 1.008\right)\right\}\text{ g/mol}$ = $78.108\text{ g/mol}$

Mass of C present per mole of C₆ H₆ = $\left(6\times 12.01\right)\text{ g/mol}$ = $72.06\text{ g/mol}$

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$

Percentage of C per mole of C₆ H_{6 ( l )} by mass = $\frac{\text{72}\text{.06 g/mol}}{\text{78}\text{.108 g/mol}}\times 100$

= $92.26\text{\%}$.

Interpretation Introduction

(b)

Interpretation:

The percent by mass of the element whose symbol occurs first in the following compound’s formula should be calculated.

Concept Introduction:

Percent by mass of an element can be calculated by the following equation.

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$.

Answer to Problem 16CR

Percentage of Na per mole of Na₂ SO₄ by mass = $32.36\text{\%}$.

Explanation of Solution

Molar mass of Na₂ SO_{4 ( s )} = $\left\{\left(2\times 22.99\right)+32.07+\left(4\times 16.00\right)\right\}\text{ g/mol}$ = $\text{142}\text{.05 g/mol}$

Mass of Na present per mole of Na₂ SO₄ = $\left(2\times 22.99\right)\text{ g/mol}$ = $45.98\text{ g/mol}$

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$

Percentage of Na per mole of Na₂ SO₄ by mass = $\frac{\text{45}\text{.98 g/mol}}{\text{142}\text{.05 g/mol}}\times 100$

= $32.36\text{\%}$.

Interpretation Introduction

(c)

Interpretation:

The percent by mass of the element whose symbol occurs first in the following compound’s formula should be calculated.

Concept Introduction:

Percent by mass of an element can be calculated by the following equation.

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$.

Answer to Problem 16CR

Percentage of C per mole of CS_{2 ( l )} by mass = $\text{15}\text{.77 }\text{\%}$.

Explanation of Solution

Molar mass of CS_{2 ( l )} = $\left\{\text{ 12}\text{.01 }+\left(\text{2 }\times \text{ 32}\text{.07}\right)\right\}\text{ g/mol}$ = $\text{76}\text{.15 g/mol}$

Mass of C present per mole of CS_{2 ( l )} = $\left(\text{1 }\times \text{ 12}\text{.01}\right)\mathrm{g<mo>/</mo><mtext>m</mtext><mtext>o</mtext><mtext>l</mtext>}$ = $\text{12}\text{.01 g/mol}$

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$

Percentage of C per mole of CS_{2 ( l )} by mass = $\frac{\text{12}\text{.01 g/mol}}{\text{76}\text{.15 g/mol}}\times 100$

= $\text{15}\text{.77 }\text{\%}$.

Interpretation Introduction

(d)

Interpretation:

The percent by mass of the element whose symbol occurs first in the following compound’s formula should be calculated.

Concept Introduction:

Percent by mass of an element can be calculated by the following equation.

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$.

Answer to Problem 16CR

Percentage of Al per mole of AlCl_{3 ( s )} by mass = $\text{20}\text{.24 }\text{\%}$.

Explanation of Solution

Molar mass of AlCl_{3 ( s )} = $\left\{\text{ 26}\text{.98 }+\left(\text{3 }\times \text{ 35}\text{.45}\right)\right\}\text{ g/mol}$ = $\text{133}\text{.33 g/mol}$

Mass of Al present per mole of AlCl_{3 ( s )} = $\left(\text{1 }\times \text{ 26}\text{.98}\right)\text{ g/mol}$ = $\text{26}\text{.98 g/mol}$

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$

Percentage of Al per mole of AlCl_{3 ( s )} by mass = $\frac{\text{26}\text{.98 g/mol}}{\text{133}\text{.33 g/mol}}\times 100$

= $\text{20}\text{.24 }\text{\%}$.

Interpretation Introduction

(e)

Interpretation:

The percent by mass of the element whose symbol occurs first in the following compound’s formula should be calculated.

Concept Introduction:

Percent by mass of an element can be calculated by the following equation.

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$.

Answer to Problem 16CR

Percentage of Cu per mole of Cu₂ O_{( s )} by mass = $\text{88}\text{.819 }\text{\%}$.

Explanation of Solution

Molar mass of Cu₂ O_{( s )} = $\left\{(2\times 63.55)+16.00\right\}\text{ g/mol}$ = $\text{143}\text{.10 g/mol}$

Mass of Cu present per mole of Cu₂ O_{( s )} = $\left(\text{2 }\times \text{ 63}\text{.55}\right)\text{ g/mol}$ = $\text{127}\text{.10 g/mol}$

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$

Percentage of Cu per mole of Cu₂ O_{( s )} by mass = $\frac{\text{127}\text{.10 g/mol}}{\text{143}\text{.10 g/mol}}\times 100$

= $\text{88}\text{.819 }\text{\%}$.

Interpretation Introduction

(f)

Interpretation:

The percent by mass of the element whose symbol occurs first in the following compound’s formula should be calculated.

Concept Introduction:

Percent by mass of an element can be calculated by the following equation.

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$.

Answer to Problem 16CR

Percentage of Cu per mole of CuO_{( s )} by mass = $\text{79}\text{.89 }\text{\%}$.

Explanation of Solution

Molar mass of CuO_{( s )} = $\left\{63.55+16.00\right\}\text{ g/mol}$ = $\text{79}\text{.55 g/mol}$

Mass of Cu present per mole of CuO_{( s )} = $\left(\text{1 }\times \text{ 63}\text{.55}\right)\text{ g/mol}$ = $\text{63}\text{.55 g/mol}$

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$

Percentage of Cu per mole of CuO_{( s )} by mass = $\frac{\text{63}\text{.55 g/mol}}{\text{79}\text{.55 g/mol}}\times 100$

= $\text{79}\text{.89 }\text{\%}$.

Interpretation Introduction

(g)

Interpretation:

The percent by mass of the element whose symbol occurs first in the following compound’s formula should be calculated.

Concept Introduction:

Percent by mass of a element can be calculated by the following equation.

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$.

Answer to Problem 16CR

Percentage of Co per mole of Co₂ O_{3 ( s )} by mass = $\text{71}\text{.059 }\text{\%}$.

Explanation of Solution

Molar mass of Co₂ O_{3 ( s )} = $\left\{(2\times 58.93)+(3\times 16.00\right\}\text{ g/mol}$ = $\text{165}\text{.86 g/mol}$

Mass of Co present per mole of Co₂ O_{3 ( s )} = $\left(\text{2 }\times \text{ 58}\text{.93}\right)\text{ g/mol}$ = $\text{117}\text{.86 g/mol}$

$\text{Percent by mass \%=}\frac{\text{Mass of particular element per mole of compound}}{\text{Molar mass of the compound}}\text{× 100}$

Percentage of Co per mole of Co₂ O_{3 ( s )} by mass = $\frac{\text{117}\text{.86 g/mol}}{\text{165}\text{.86 g/mol}}\times 100$

= $\text{71}\text{.059 }\text{\%}$.

Interpretation Introduction

(h)

Interpretation:

The percent by mass of the element whose symbol occurs first in the following compound’s formula should be calculated.

Concept Introduction:

Percent by mass of a element can be calculated by the following equation.

$Percent by mass %=Mass of particular element per mole of compoundMolar mass of the compound× 100$.

Answer to Problem 16CR

Percentage of C per mole of C₆ H_{1 2} O_{6 ( s )} by mass = $\text{39}\text{.99 }\text{\%}$.

Explanation of Solution

Molar mass of C₆ H_{1 2} O_{6 ( s )} = $\left\{(6\times 12.01)+(12\times 1.008)+(6\times 16.00)\right\}\text{ g/mol}$ = $\text{180}\text{.16 g/mol}$

Mass of C present per mole of C₆ H_{1 2} O_{6 ( s )} = $\left(\text{6 }\times \text{ 12}\text{.01}\right)\text{ g/mol}$ = $\text{72}\text{.06 g/mol}$

$\text{Percent by mass \%=}\frac{\text{Mass of particular element per mole of compound}}{\text{Molar mass of the compound}}\text{× 100}$

Percentage of C per mole of C₆ H_{1 2} O_{6 ( s )} by mass = $\frac{\text{72}\text{.06 g/mol}}{\text{180}\text{.16 g/mol}}\times 100$

= $\text{39}\text{.99 }\text{\%}$.