Chapter 11, Problem 11.8P

(a)

Interpretation Introduction

Interpretation:

Mass of oxygen in $13\text{ g}$ calcium acetate has to be determined.

Concept Introduction:

Moleis S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of the total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is g/mol.

The expression for number of moles, mass and molar mass of compound is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

Answer to Problem 11.8P

Mass of oxygen in $13\text{ g}$ calcium acetate is $5.2598\text{ g}$.

Explanation of Solution

The formula to calculate number of moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{.Ca}$ is as follows:

  ${\text{Number of moles of C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{.Ca}=\frac{{\text{Given mass of C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{.Ca}}{{\text{molar mass of C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{.Ca}}$        (1)

Substitute $13\text{ g}$ for given mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{.Ca}$ and $\text{158}\text{.17 g/mol}$ for molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{.Ca}$ in equation (1).

  $\begin{array}{c}{\text{number of moles of C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{.Ca}=\frac{13\text{ g}}{\text{158}\text{.17 g/mol}}\\ =\text{0}\text{.0821 mol}\end{array}$

1 mole ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{.Ca}$ contains 4 moles of $\text{O}$ atoms. Therefore, number of moles of $\text{O}$ will be equal to 4 times number of moles ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{.Ca}$ and is $\text{0}\text{.3287 mol}$.

The formula to calculate number of moles of $\text{O}$ is as follows:

  $\text{Number of moles of O}=\frac{\text{Given mass of O}}{\text{molar mass of O}}$        (2)

Rearrange equation (2) to calculate mass of $\text{O}$.

  $\text{Mass of O}=\left(\text{Number of moles of O}\right)\left(\text{molar mass of O}\right)$        (3)

Substitute $\text{15}\text{.999 g/mol}$ for molar mass of $\text{O}$ and $\text{0}\text{.3287 mol}$ for number of moles of $\text{O}$ in equation (3).

  $\begin{array}{c}\text{Mass of O}=\left(\text{0}\text{.3287 mol}\right)\left(\text{15}\text{.999 g/mol}\right)\\ =5.2598\text{ g}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Mass of fluorine in $0.22\text{ g}$ xenon tetrafluoride has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.8P

Mass of fluorine in $0.22\text{ g}$ xenon tetrafluoride is $0.0805\text{ g}$.

Explanation of Solution

The formula to calculate number of moles of ${\text{XeF}}_{\text{4}}$ is as follows:

  ${\text{Number of moles of XeF}}_{\text{4}}=\frac{{\text{Given mass of XeF}}_{\text{4}}}{{\text{molar mass of XeF}}_{\text{4}}}$        (4)

Substitute $0.22\text{ g}$ for given mass of ${\text{XeF}}_{\text{4}}$ and $\text{207}\text{.2836 g/mol}$ for molar mass of ${\text{XeF}}_{\text{4}}$ in equation (4).

  $\begin{array}{c}{\text{number of moles of XeF}}_4=\frac{0.22\text{ g}}{\text{207}\text{.2836 g/mol}}\\ =\text{0}\text{.00106 mol}\end{array}$

1 mole ${\text{XeF}}_{\text{4}}$ contains 4 moles of $\text{F}$ atoms. Therefore, number of moles of $\text{F}$ will be equal to 4 times number of moles of ${\text{XeF}}_{\text{4}}$ and is $\text{0}\text{.00424 mol}$.

The formula to calculate number of moles of $\text{F}$ is as follows:

  $\text{Number of moles of F}=\frac{\text{Given mass of}\text{F}}{\text{molar mass of F}}$        (5)

Rearrange equation (5) to calculate mass of $\text{F}$.

  $\text{Mass of F}=\left(\text{Number of moles of F}\right)\left(\text{molar mass of F}\right)$        (6)

Substitute $\text{18}\text{.9984 g/mol}$ for molar mass of $\text{F}$ and $\text{0}\text{.00424 mol}$ for number of moles of $\text{F}$ in equation (6).

  $\begin{array}{c}\text{Mass of F}=\left(\text{0}\text{.00424 mol}\right)\left(\text{18}\text{.9984 g/mol}\right)\\ =0.0805\text{ g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Mass of water in $\text{25 g}$ barium chloride dihydrate has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.8P

Mass of water in $\text{25 g}$ barium dichloride dihydrate is $3.6859\text{ g}$.

Explanation of Solution

The formula to calculate number of moles of ${\text{BaCl}}_2.{\text{2H}}_{\text{2}}\text{O}$ is as follows:

  ${\text{Number of moles of BaCl}}_2.{\text{2H}}_{\text{2}}\text{O}=\frac{{\text{Given mass of BaCl}}_2.{\text{2H}}_{\text{2}}\text{O}}{{\text{molar mass of BaCl}}_2.{\text{2H}}_{\text{2}}\text{O}}$        (7)

Substitute $\text{25 g}$ for given mass of ${\text{BaCl}}_2.{\text{2H}}_{\text{2}}\text{O}$ and $\text{244}\text{.26 g/mol}$ for molar mass of ${\text{BaCl}}_2.{\text{2H}}_{\text{2}}\text{O}$ in equation (7).

  $\begin{array}{c}{\text{number of moles of BaCl}}_2.{\text{2H}}_{\text{2}}\text{O}=\frac{25\text{ g}}{\text{244}\text{.26 g/mol}}\\ =\text{0}\text{.1023 mol}\end{array}$

1 mole ${\text{BaCl}}_2.{\text{2H}}_{\text{2}}\text{O}$ contains 2 moles of ${\text{H}}_{\text{2}}\text{O}$. Therefore, number of moles of ${\text{H}}_{\text{2}}\text{O}$ will be equal to 2 times number of moles of ${\text{BaCl}}_2.{\text{2H}}_{\text{2}}\text{O}$ and is $\text{0}\text{.2046 mol}$.

The formula to calculate number of moles of ${\text{H}}_{\text{2}}\text{O}$ is as follows:

  ${\text{Number of moles of H}}_{\text{2}}\text{O}=\frac{{\text{Given mass of H}}_{\text{2}}\text{O}}{{\text{molar mass of H}}_{\text{2}}\text{O}}$        (8)

Rearrange equation (8) to calculate mass of ${\text{H}}_{\text{2}}\text{O}$.

  ${\text{Mass of H}}_{\text{2}}\text{O}=\left({\text{Number of moles of H}}_{\text{2}}\text{O}\right)\left({\text{molar mass of H}}_{\text{2}}\text{O}\right)$        (9)

Substitute $\text{18}\text{.0152 g/mol}$ for molar mass of ${\text{H}}_{\text{2}}\text{O}$ and $\text{0}\text{.2046 mol}$ for number of moles of ${\text{H}}_{\text{2}}\text{O}$ in equation (9).

  $\begin{array}{c}{\text{Mass of H}}_{\text{2}}\text{O}=\left(\text{0}\text{.2046 mol}\right)\left(\text{18}\text{.0152 g/mol}\right)\\ =3.6859\text{ g}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Mass of carbon in $2\text{ g}$${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.8P

Mass of carbon in $2\text{ g}$${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is $1.6718\text{ g}$.

Explanation of Solution

The formula to calculate number of moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is as follows:

  ${\text{Number of moles of C}}_{\text{6}}{\text{H}}_{\text{14}}=\frac{{\text{Given mass of C}}_{\text{6}}{\text{H}}_{\text{14}}}{{\text{molar mass of C}}_{\text{6}}{\text{H}}_{\text{14}}}$        (10)

Substitute $2\text{ g}$ for given mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ and $\text{86}\text{.1753 g/mol}$ for molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ in equation (10).

  $\begin{array}{c}{\text{number of moles of C}}_{\text{6}}{\text{H}}_{\text{14}}=\frac{2\text{ g}}{\text{86}\text{.1753 g/mol}}\\ =\text{0}\text{.0232 mol}\end{array}$

1 mole ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ contains 6 moles of $\text{C}$ atoms. Therefore, number of moles of $\text{C}$ will be equal to 6 times number of moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ and is $\text{0}\text{.1392 mol}$.

The formula to calculate number of moles of $\text{C}$ is as follows:

  $\text{Number of moles of C}=\frac{\text{Given mass of C}}{\text{molar mass of C}}$        (11)

Rearrange equation (11) to calculate mass of $\text{C}$.

  $\text{Mass of C}=\left(\text{Number of moles of C}\right)\left(\text{molar mass of C}\right)$        (12)

Substitute $\text{12}\text{.0107 g/mol}$ for molar mass of $\text{C}$ and $\text{0}\text{.1392 mol}$ for number of moles of $\text{C}$ in equation (12).

  $\begin{array}{c}\text{Mass of C}=\left(\text{0}\text{.1392 mol}\right)\left(\text{12}\text{.0107 g/mol}\right)\\ =1.6718\text{ g}\end{array}$