Chapter 2, Problem 87QRT

(a)

Interpretation Introduction

Interpretation:

The molar mass of the compound $\text{PbS}$, lead (II) sulfide (galena) and mass percent of each element present in the compound has to be calculated.

Explanation of Solution

The given compound is $\text{PbS}$, lead (II) sulfide (galena).

Mass of lead is $\text{207}\text{.2}\text{g/mol}$.

Mass of sulfur is $\text{32}\text{.065}\text{g/mol}$.

Molar mass of $\text{PbS}$ $\text{=}$ $\text{207}\text{.2}\text{g/mol}\text{Pb}\text{+}\text{32}\text{.065}\text{g/mol}\text{S}\text{=}\text{239}\text{.3}\text{g/mol}\text{PbS}$.

Then,

Mass percent of $\text{Pb}$

  $\text{=}\frac{\text{mass}\text{of}\text{Pb}\text{per}\text{mol}\text{of}\text{PbS}}{\text{mass}\text{of}\text{PbS}\text{per}\text{mol}\text{PbS}}\text{×}\text{100}\text{\%}\text{=}\frac{\text{207}\text{.2}\text{g}\text{Pb}}{\text{239}\text{.3}\text{g}\text{PbS}}\text{×}\text{100}\text{\%}\text{=}\text{86}\text{.58}\text{\%}\text{Pb}\text{in}\text{PbS}$.

Mass percent of $\text{S}$ $\text{=}\text{100}\text{\%}\text{-}\text{86}\text{.58}\text{\%}\text{=}\text{13}\text{.41}\text{\%}\text{S}\text{in}\text{PbS}$.

(b)

Interpretation Introduction

Interpretation:

The molar mass of the given compound ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ and mass percent of each element present in the compound has to be calculated.

Explanation of Solution

The given compound is ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$, ethane, a hydrocarbon fuel.

Mass of carbon is $12.0107\text{g/mol}$.

Mass of hydrogen is $\text{1}\text{.0079}\text{g/mol}$.

Molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ $\text{=}$ $\text{(2×12}\text{.0171)}\text{g/mol}\text{C}\text{+}\text{(1×1}\text{.0079)}\text{g/mol}\text{H}\text{=}\text{30}\text{.688}\text{g/mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$.

Then,

Mass percent of $\text{C}$

$\text{=}\frac{\text{mass}\text{of}\text{C}\text{per}\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}{\text{mass}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{per}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{×}\text{100}\text{\%}\text{=}\frac{\text{24}\text{.0214}\text{g}\text{C}}{\text{30}\text{.0688}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{×}\text{100}\text{\%}\text{=}\text{79}\text{.8881}\text{\%}\text{C}\text{in}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$.

Mass percent of $\text{C}$ $\text{=}\text{100}\text{\%}\text{-}\text{79}\text{.8881}\text{\%}\text{=}\text{20}\text{.1119}\text{\%}\text{C}$.

(c)

Interpretation Introduction

Interpretation:

The molar mass of the given compound ${\text{CH}}_{\text{3}}\text{COOH}$ and mass percent of each element present in the compound has to be calculated.

Explanation of Solution

The given compound is ${\text{CH}}_{\text{3}}\text{COOH}$, acetic, an important ingredient in vinegar.

Mass of carbon is $12.0107\text{g/mol}$.

Mass of hydrogen is $\text{1}\text{.0079}\text{g/mol}$.

Mass of oxygen is $\text{15}\text{.9994}\text{g/mol}$

Molar mass of ${\text{CH}}_{\text{3}}\text{COOH}$ $\text{=}$

$\text{(2×12}\text{.0171)}\text{g/mol}\text{C}\text{+}\text{(4×1}\text{.0079)}\text{g/mol}\text{H}\text{+}\text{(2}\text{×}\text{15}\text{.9994}\text{)}\text{g/mol}\text{O}\text{=}\text{60}\text{.0518}\text{g/mol}{\text{CH}}_{\text{3}}\text{COOH}$.

Then,

Mass percent of $\text{C}$

$\text{=}\frac{\text{mass}\text{of}\text{C}\text{per}\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}{\text{mass}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{per}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{×}\text{100}\text{\%}\text{=}\frac{\text{24}\text{.0214}\text{g}\text{C}}{\text{60}\text{.0518}\text{g}{\text{CH}}_{\text{3}}\text{COOH}}\text{×}\text{100}\text{\%}\text{=}\text{40}\text{.0011}\text{\%}\text{C}\text{in}{\text{CH}}_{\text{3}}\text{COOH}$.

Mass percent of $\text{H}$

$\text{=}\frac{\text{mass}\text{of}\text{H}\text{per}\text{mol}\text{of}{\text{CH}}_{\text{3}}\text{COOH}}{\text{mass}\text{of}{\text{CH}}_{\text{3}}\text{COOH}\text{per}\text{mol}{\text{CH}}_{\text{3}}\text{COOH}}\text{×}\text{100}\text{\%}\text{=}\frac{\text{4}\text{.0316}\text{g}\text{H}}{\text{60}\text{.0518}\text{g}{\text{CH}}_{\text{3}}\text{COOH}}\text{×}\text{100}\text{\%}\text{=}\text{6}\text{.7135}\text{\%}\text{H}\text{in}{\text{CH}}_{\text{3}}\text{COOH}$.

Mass percent of $\text{O}$ $\text{=}\text{100}\text{\%}\text{-}\text{79}\text{.8881}\text{\%}\text{-}\text{6}\text{.7135\%}\text{=}\text{53}\text{.284}\text{\%}\text{O}$.

(d)

Interpretation Introduction

Interpretation:

The molar mass of the given compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ and mass percent of each element present in the compound has to be calculated.

Explanation of Solution

The given compound is ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$, ammonium nitrate, a fertilizer.

Mass of nitrogen is $14.0067\text{g/mol}$.

Mass of hydrogen is $\text{1}\text{.0079}\text{g/mol}$.

Mass of oxygen is $\text{15}\text{.9994}\text{g/mol}$

Molar mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ $\text{=}$

$\text{(2×14}\text{.0067)}\text{g/mol}\text{N}\text{+}\text{(4×1}\text{.0079)}\text{g/mol}\text{H}\text{+}\text{(3}\text{×}\text{15}\text{.9994}\text{)}\text{g/mol}\text{O}\text{=}\text{80}\text{.0432}\text{g/mol}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$.

Then,

Mass percent of $\text{N}$

$\text{=}\frac{\text{mass}\text{of}\text{N}\text{per}\text{mol}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{\text{mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{per}\text{mol}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\text{×}\text{100}\text{\%}\text{=}\frac{4.0316\text{g}\text{N}}{80.0432\text{g}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\text{×}\text{100}\text{\%}\text{=}34.9919\text{\%}\text{N}\text{in}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$.

Mass percent of $\text{H}$

$\text{=}\frac{\text{mass}\text{of}\text{H}\text{per}\text{mol}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{\text{mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{per}\text{mol}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\text{×}\text{100}\text{\%}\text{=}\frac{\text{4}\text{.0316}\text{g}\text{H}}{\text{80}\text{.0432}\text{g}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\text{×}\text{100}\text{\%}\text{=}\text{5}\text{.0368}\text{\%}\text{H}\text{in}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$.

Mass percent of $\text{O}$ $\text{=}\text{100}\text{\%}\text{-}34.9979\text{\%}\text{-}5.0368\text{\%}\text{=}59.9654\text{\%}\text{O}$.