Chapter 8, Problem 47QAP

Calculate the percent by mass of the element listed first in the formulas for each of the following compounds.

methane, ${\text{CH}}_4$

sodium nitrate, ${\text{NaNO}}_3$

carbon monoxide, $\text{CO}$

nitrogen dioxide, ${\text{NO}}_2$

1-octanol, ${\text{C}}_8{\text{H}}_{18}\text{O}$

calcium phosphate, ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$

3-phenyiphenol, ${\text{C}}_{12}{\text{H}}_{10}\text{O}$

aluminum acetate, $\text{Al}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_3$

Interpretation Introduction

(a)

Interpretation:

Thepercent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 47QAP

$\text{M}\text{e}\text{t}\text{h}\text{a}\text{n}\text{e},\text{C}{\text{H}}_4 \text{C}\text{\%} =74.88\text{\%}$.

Explanation of Solution

A chemical compound is a collection of several atoms. Molar masses of each and every atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$

$\text{M}\text{e}\text{t}\text{h}\text{a}\text{n}\text{e},\text{C}{\text{H}}_4 \text{C}\text{\%} =$

$\text{Mass of 1 mol of }\text{C} =1\times 12.01\text{g}=12.01\text{g}$

$\text{Mass of 4 mol of H = 4 }\times \text{ 1}\text{.008 g = 4}\text{.032 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{C}{\text{H}}_4\text{= 12}\text{.01+ 4}\text{.032 = }16.04\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}{\text{ mass of CH}}_{\text{4}}\text{= 16}{\text{.04 g mol}}^{\text{-1}}$

Mass of the carbon present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{12}\text{.01 g}}{\text{1 mol}}=12.01\text{ g}$

Mass percent of $\text{C}$ = $\frac{12.01\text{g}\text{C} }{16.04\text{g} \text{C}{\text{H}}_4}\times 100\text{\%}=\underset{\_}{\underset{\_}{74.88\text{\%}}}$.

Interpretation Introduction

(b)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 47QAP

$\text{S}\text{o}\text{d}\text{i}\text{u}\text{m}\text{n}\text{i}\text{t}\text{r}\text{a}\text{t}\text{e},\text{N}\text{a}\text{N}{\text{O}}_3 \text{N}\text{a}\text{\%} = 27.05\text{\%}$.

Explanation of Solution

${\text{Sodium nitrate, NaNO}}_{\text{3}}\text{ Na \% =}$

$\text{Mass of 1 mol of Na}=1\times \text{ 23}\text{.0 }\text{g}=\text{ 23}\text{.0 }\text{g}$

$\text{Mass of 1 mol of N = 1 }\times \text{ 14}\text{.01 g = 14}\text{.01 g}$

$\text{Mass of 3 mol of O = 3 }\times \text{ 16}\text{.00 g = 48}\text{.00 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}{\text{ NaNO}}_{\text{3}}\text{= 23}\text{.0+14}\text{.01+48}\text{.00 = 85}\text{.01 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}{\text{ mass of NaNO}}_{\text{3}}\text{= 85}{\text{.01 g mol}}^{\text{-1}}$

Mass of the sodium present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{23}\text{.0 g}}{\text{1 mol}}=23.0\text{ g}$

Mass percent of $\text{N}\text{a}$ = $\frac{\text{23}\text{.0 g Na }}{\text{85}\text{.01 g NaNO3 }}\times 100\text{\%}=\underset{\_}{\underset{\_}{27.05\text{\%}}}$.

Interpretation Introduction

(c)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 47QAP

$\text{C}\text{a}\text{r}\text{b}\text{o}\text{n}\text{m}\text{o}\text{n}\text{o}\text{x}\text{i}\text{d}\text{e}{,}_{,}\text{C}\text{O} \text{C}\text{\%} = 42.88\text{\%}$.

Explanation of Solution

$\text{C}\text{a}\text{r}\text{b}\text{o}\text{n}\text{m}\text{o}\text{n}\text{o}\text{x}\text{i}\text{d}{\text{e}}_{,}\text{C}\text{O} \text{C}\text{\%} =$

$\text{Mass of 1 mol of C}=1\times \text{ 12}\text{.01 }\text{g}=\text{ 12}\text{.01 }\text{g}$

$\text{Mass of 1 mol of O = 1 }\times \text{ 16}\text{.00 g = 16}\text{.00 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{C}\text{O}\text{ = 12}\text{.01+16}\text{.00 = 28}\text{.01 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of CO = 28}{\text{.01 g mol}}^{\text{-1}}$

Mass of the carbon present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{12}\text{.01 g}}{\text{1 mol}}=12.01\text{ g}$

Mass percent of $\text{C}$ = $\frac{\text{12}\text{.01 g C }}{\text{28}\text{.01 g CO }}\times 100\text{\%}=\underset{\_}{\underset{\_}{42.88\text{\%}}}$.

Interpretation Introduction

(d)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 47QAP

$\text{N}\text{i}\text{t}\text{r}\text{o}\text{g}\text{e}\text{n}\text{d}\text{i}\text{o}\text{x}\text{i}\text{d}\text{e},\text{N}{\text{O}}_2 \text{N}\text{\%} = 30.44\text{\%}$.

Explanation of Solution

$\text{Nitrogen dioxide,} {\text{NO}}_{\text{2}}\text{ N \% = }$

$\text{Mass of 1 mol of N}=1\times \text{ 14}\text{.01 }\text{g}=\text{ 14}\text{.01 }\text{g}$

$\text{Mass of 2 mol of O = 2 }\times \text{ 16}\text{.00 g = 32}\text{.00 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}{\text{ NO}}_{\text{2}}\text{ = 14}\text{.01+32}\text{.00 = 46}\text{.01 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}{\text{ mass of NO}}_{\text{2}}\text{= 46}{\text{.01 g mol}}^{\text{-1}}$

Mass of the nitrogen present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{14}\text{.01 g}}{\text{1 mol}}=14.01\text{ g}$

Mass percent of $\text{N}$ = $\frac{\text{14}\text{.01 g N }}{\text{46}\text{.01 g NO2 }}\times 100\text{\%}=\underset{\_}{\underset{\_}{30.44\text{\%}}}$.

Interpretation Introduction

(e)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 47QAP

$1-\text{o}\text{c}\text{t}\text{a}\text{n}\text{o}\text{l},{\text{C}}_8{\text{H}}_{18}{\text{O}}_{} \text{C}\text{\%} = 73.76\text{\%}$.

Explanation of Solution

$1-\text{o}\text{c}\text{t}\text{a}\text{n}\text{o}\text{l},{\text{C}}_8{\text{H}}_{18}{\text{O}}_{} \text{C}\text{\%} =$

$\text{Mass of 8 mol of C}=8\times \text{ 12}\text{.01 }\text{g}=\text{ 96}\text{.08 }\text{g}$

$\text{Mass of 18 mol of H = 18 }\times \text{ 1}\text{.01 g = 18}\text{.18 g}$

$\text{Mass of 1 mol of O = 1 }\times \text{ 16}\text{.00 g = 16}\text{.00 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}{\text{C}}_8{\text{H}}_{18}\text{O}\text{= 96}\text{.08+18}\text{.18+16}\text{.00 = 130}\text{.26 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }{\text{C}}_8{\text{H}}_{18}\text{O}\text{= 130}{\text{.26 g mol}}^{\text{-1}}$

Mass of the carbon present in $1\text{m}\text{o}\text{l}$ of compound = $\text{8 mol × }\frac{\text{12}\text{.01 g}}{\text{1 mol}}=96.08\text{ g}$

Mass percent of $\text{C}$ = $\frac{\text{96}\text{.08 g C }}{\text{130}\text{.26 g }{\text{C}}_8{\text{H}}_{18}\text{O}}\times 100\text{\%}=\underset{\_}{\underset{\_}{73.76\text{\%}}}$.

Interpretation Introduction

(f)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 47QAP

$\text{C}\text{a}\text{l}\text{c}\text{i}\text{u}\text{m}\text{p}\text{h}\text{o}\text{s}\text{p}\text{h}\text{a}\text{t}\text{e},\text{C}{\text{a}}_3\text{P}{\text{O}}_4 \text{C}\text{a}\text{\%} =55.87\text{\%}$.

Explanation of Solution

$\text{C}\text{a}\text{l}\text{c}\text{i}\text{u}\text{m}\text{p}\text{h}\text{o}\text{s}\text{p}\text{h}\text{a}\text{t}\text{e},\text{C}{\text{a}}_3\text{P}{\text{O}}_4 \text{C}\text{a}\text{\%} =$

$\text{Mass of 3 mol of Ca}=3\times \text{ 40}\text{.08 }\text{g}=\text{ 120}\text{.24 }\text{g}$

$\text{Mass of 1 mol of P = 1 }\times \text{ 30}\text{.97 g = 30}\text{.97 g}$

$\text{Mass of 4 mol of O = 4 }\times \text{ 16}\text{.00 g = 64}\text{.00 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{C}{\text{a}}_3\text{P}{\text{O}}_4\text{= 120}\text{.24+30}\text{.97+64}\text{.00 = 215}\text{.21 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{C}{\text{a}}_3\text{P}{\text{O}}_4\text{= 215}{\text{.21 g mol}}^{\text{-1}}$

Mass of the calcium present in $1\text{m}\text{o}\text{l}$ of compound = $\text{3 mol × }\frac{\text{40}\text{.08 g}}{\text{1 mol}}=120.24\text{ g}$

Mass percent of $\text{C}\text{a}$ = $\frac{\text{120}\text{.24 g Ca }}{\text{215}\text{.21 g }\text{C}{\text{a}}_3\text{P}{\text{O}}_4}\times 100\text{\%}=\underset{\_}{\underset{\_}{55.87\text{\%}}}$.

Interpretation Introduction

(g)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 47QAP

$3-\text{p}\text{h}\text{e}\text{n}\text{y}\text{l}\text{p}\text{h}\text{e}\text{n}\text{o}\text{l},{\text{C}}_{12}{\text{H}}_{10}{\text{O}}_{} \text{C}\text{\%} =84.67\text{\%}$.

Explanation of Solution

$3-\text{p}\text{h}\text{e}\text{n}\text{y}\text{l}\text{p}\text{h}\text{e}\text{n}\text{o}\text{l},{\text{C}}_{12}{\text{H}}_{10}{\text{O}}_{} \text{C}\text{\%} =$

$\text{Mass of 12 mol of C}=12\times \text{ 12}\text{.01 }\text{g}=\text{ 144}\text{.12 }\text{g}$

$\text{Mass of 10 mol of H = 10 }\times \text{ 1}\text{.01 g = 10}\text{.1 g}$

$\text{Mass of 1 mol of O = 1 }\times \text{ 16}\text{.00 g = 16}\text{.00 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}{\text{C}}_{12}{\text{H}}_{10}\text{O}\text{= 144}\text{.12+10}\text{.1+16}\text{.00 = 170}\text{.22 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }{\text{C}}_{12}{\text{H}}_{10}\text{O}\text{= 170}{\text{.22 g mol}}^{\text{-1}}$

Mass of the carbon present in $1\text{m}\text{o}\text{l}$ of compound = $\text{12 mol × }\frac{\text{12}\text{.01 g}}{\text{1 mol}}=144.12\text{ g}$

Mass percent of $\text{C}$ = $\frac{\text{144}\text{.12 g C }}{\text{170}\text{.22 g }{\text{C}}_{12}{\text{H}}_{10}\text{O}}\times 100\text{\%}=\underset{\_}{\underset{\_}{84.67\text{\%}}}$.

Interpretation Introduction

(h)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 47QAP

$\text{A}\text{l}\text{u}\text{m}\text{i}\text{n}\text{u}\text{m}\text{a}\text{c}\text{e}\text{t}\text{a}\text{t}\text{e},\text{A}\text{l}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_{3 }\text{A}\text{l}\text{\%}=13.19\text{\%}$.

Explanation of Solution

$\text{A}\text{l}\text{u}\text{m}\text{i}\text{n}\text{u}\text{m}\text{a}\text{c}\text{e}\text{t}\text{a}\text{t}\text{e},\text{A}\text{l}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_{3 }\text{A}\text{l}\text{\%}=$

$\text{Mass of 1 mol of Al}=1\times \text{ 26}\text{.92 }\text{g}=\text{ 26}\text{.92 }\text{g}$

$\text{Mass of 6 mol of C = 6 }\times \text{ 12}\text{.01 g = 72}\text{.06 g}$

$\text{Mass of 9 mol of H = 9 }\times \text{ 1}\text{.01 g = 9}\text{.09 g}$

$\text{Mass of 6 mol of O = 6 }\times \text{ 16}\text{.00 g = 96}\text{.00 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{A}\text{l}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_{3 }\text{= 26}\text{.92+72}\text{.06+9}\text{.09+96}\text{.00 = 204}\text{.07 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{A}\text{l}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_{3 }\text{= 204}{\text{.07 g mol}}^{\text{-1}}$

Mass of the aluminum present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{26}\text{.92 g}}{\text{1 mol}}=26.92\text{ g}$

Mass percent of $\text{A}\text{l}$ = $\frac{\text{26}\text{.92 g Al }}{\text{204}\text{.07 g }\text{A}\text{l}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_{3 }}\times 100\text{\%}=\underset{\_}{\underset{\_}{13.19\text{\%}}}$.