Chapter 7, Problem 27PE

(a)

Interpretation Introduction

Interpretation:

The percent composition of the compound sodium bromide has to be given.

Answer to Problem 27PE

The percent composition of sodium is $\text{22}\text{.33\%}\text{.}$

The percent composition of bromine is $\text{77}\text{.66\%}\text{.}$

Explanation of Solution

Given,

The atomic mass of sodium atom $\text{=23}\text{g}$

The atomic mass of bromine atom $\text{=80}\text{g}$

The molecular weight of $\text{NaBr}$$\text{=23+80=103}\text{g}$

The percent composition for each element can be calculated as,

The percent composition of sodium $\text{=}\left(\frac{\text{23}\text{g}}{\text{103}\text{g}}\right)\text{×100=22}\text{.33\%}$

The percent composition of sodium is $\text{22}\text{.33\%}\text{.}$

The percent composition of bromine $\text{=}\left(\frac{80\text{g}}{\text{103}}\right)\text{×100=77}\text{.66\%}$

The percent composition of bromine is $\text{77}\text{.66\%}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The percent composition of the compound ${\text{KHCO}}_{\text{3}}$  has to be given.

Answer to Problem 27PE

The percent composition of Potassium is $\text{39}\text{.06\%}\text{.}$

The percent composition of Hydrogen is $\text{1}\text{.007\%}\text{.}$

The percent composition of Carbon is $\text{12}\text{.00\%}\text{.}$

The percent composition of Oxygen is $\text{47}\text{.95\%}\text{.}$

Explanation of Solution

Given,

The atomic mass of potassium atom $\text{=39}\text{.10}\text{g}$

The atomic mass of hydrogen atom $\text{=1}\text{.008}\text{g}$

The atomic mass of oxygen atom $\text{=16}\text{g}$

The atomic mass of carbon atom $\text{=12}\text{.01}\text{g}$

The molecular weight of ${\text{KHCO}}_{\text{3}}$$\text{=39}\text{.10}\text{g+1}\text{.008}\text{g+12}\text{g+3×16}\text{g}$

The molecular weight of ${\text{KHCO}}_{\text{3}}$$\text{=100}\text{.1}\text{g}$

The percent composition for each element can be calculated as,

The percent composition of Potassium $\text{=}\left(\frac{\text{39}\text{.10}\text{g}}{\text{100}\text{.1}\text{g}}\right)\text{×100=39}\text{.06\%}$

The percent composition of Potassium is $\text{39}\text{.06\%}\text{.}$

The percent composition of Hydrogen $\text{=}\left(\frac{\text{1}\text{.008g}}{\text{100}\text{.1}\text{g}}\right)\text{×100=1}\text{.007\%}$

The percent composition of Hydrogen is $\text{1}\text{.007\%}\text{.}$

The percent composition of Carbon $\text{=}\left(\frac{\text{12}\text{.01}\text{g}}{\text{100}\text{.1}\text{g}}\right)\text{×100=12}\text{.00\%}$

The percent composition of Carbon is $\text{12}\text{.00\%}\text{.}$

The percent composition of Oxygen $\text{=}\left(\frac{48.00\text{g}}{\text{100}\text{.1}\text{g}}\right)\text{×100=47}\text{.95\%}$

The percent composition of Oxygen is $\text{47}\text{.95\%}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The percent composition of the compound ${\text{FeCl}}_{\text{3}}$ has to be given.

Answer to Problem 27PE

The percent composition of Iron is $\text{34}\text{.43\%}\text{.}$

The percent composition of Chlorine is $\text{65}\text{.56\%}\text{.}$

Explanation of Solution

Given,

The atomic mass of iron atom $\text{=55}\text{.85}\text{g}$

The atomic mass of chlorine atom $\text{=35}\text{.45}\text{g}$

The molecular weight of ${\text{FeCl}}_{\text{3}}$$\text{=55}\text{.85+3}\times \text{35}\text{.45=162}\text{.2}\text{g}$

The percent composition for each element can be calculated as,

The percent composition of iron $\text{=}\left(\frac{\text{55}\text{.85}\text{g}}{\text{162}\text{.2}\text{g}}\right)\text{×100=34}\text{.43\%}$

The percent composition of iron is $\text{34}\text{.43\%}\text{.}$

The percent composition of Chlorine $\text{=}\left(\frac{\text{106}\text{.35}\text{g}}{\text{162}\text{.2}\text{g}}\right)\text{×100=65}\text{.56\%}$

The percent composition of Chlorine is $\text{65}\text{.56\%}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The percent composition of the compound ${\text{SiCl}}_{\text{4}}$ has to be given.

Answer to Problem 27PE

The percent composition of Silicon is $\text{16}\text{.53\%}\text{.}$

The percent composition of Chlorine is $\text{83}\text{.46\%}\text{.}$

Explanation of Solution

Given,

The atomic mass of silicon atom $\text{=28}\text{.09}\text{g}$

The atomic mass of chlorine atom $\text{=35}\text{.45}\text{g}$

The molecular weight of ${\text{SiCl}}_{\text{4}}$$\text{=28}\text{.09}\text{g+4}\times \text{35}\text{.45}\text{g=169}\text{.9}\text{g}$

The percent composition for each element can be calculated as,

The percent composition of Silicon $\text{=}\left(\frac{28.09\text{g}}{\text{169}\text{.9}\text{g}}\right)\text{×100=16}\text{.53\%}$

The percent composition of Silicon is $\text{16}\text{.53\%}\text{.}$

The percent composition of Chlorine $\text{=}\left(\frac{\text{141}\text{.8}\text{g}}{\text{169}\text{.9}\text{g}}\right)\text{×100=83}\text{.46\%}$

The percent composition of Chlorine is $\text{83}\text{.46\%}\text{.}$

(e)

Interpretation Introduction

Interpretation:

The percent composition of the compound ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{3}}\right)}_{\text{3}}$ has to be given.

Answer to Problem 27PE

The percent composition of aluminium is $\text{15}\text{.77\%}\text{.}$

The percent composition of sulfur is $\text{28}\text{.12\%}\text{.}$

The percent composition of oxygen is $\text{56}\text{.11\%}\text{.}$

Explanation of Solution

Given,

The atomic mass of aluminium atom $\text{=26}\text{.98}\text{g}$

The atomic mass of sulfur atom $\text{=32}\text{g}$

The atomic mass of oxygen atom $\text{=16}\text{g}$

The molecular weight of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{3}}\right)}_{\text{3}}$$\text{=2×26}\text{.98+3×}\left(\text{32+3×16}\right)\text{=342}\text{.2}\text{g}$

The percent composition for each element can be calculated as,

The percent composition of aluminium $\text{=}\left(\frac{53.96\text{g}}{342.2\text{g}}\right)\text{×100=15}\text{.77\%}$

The percent composition of aluminium is $\text{15}\text{.77\%}\text{.}$

The percent composition of sulfur $\text{=}\left(\frac{96.21\text{g}}{342.2\text{g}}\right)\text{×100=28}\text{.12\%}$

The percent composition of sulfur is $\text{28}\text{.12\%}\text{.}$

The percent composition of oxygen $\text{=}\left(\frac{192\text{g}}{342.2\text{g}}\right)\text{×100=56}\text{.11\%}$

The percent composition of oxygen is $\text{56}\text{.11\%}\text{.}$

(f)

Interpretation Introduction

Interpretation:

The percent composition of the compound ${\text{AgNO}}_{\text{3}}$ has to be given.

Answer to Problem 27PE

The percent composition of $\text{Ag}$ is $\text{63}\text{.51\%}\text{.}$

The percent composition of nitrogen is $\text{8}\text{.246\%}\text{.}$

The percent composition of oxygen is $\text{28}\text{.25\%}\text{.}$

Explanation of Solution

Given,

The atomic mass of $\text{Ag}$ atom $\text{=107}\text{.9}\text{g}$

The atomic mass of nitrogen atom $\text{=14}\text{.01}\text{g}$

The atomic mass of oxygen atom $\text{=16}\text{g}$

The molecular weight of ${\text{AgNO}}_{\text{3}}$$\text{=107}\text{.9+14}\text{.01+48=169}\text{.9}\text{g}$

The percent composition for each element can be calculated as,

The percent composition of $\text{Ag}$$\text{=}\left(\frac{107.9\text{g}}{169.9\text{g}}\right)\text{×100=63}\text{.51\%}$

The percent composition of $\text{Ag}$ is $\text{63}\text{.51\%}\text{.}$

The percent composition of nitrogen $\text{=}\left(\frac{14.01\text{g}}{169.9\text{g}}\right)\text{×100=8}\text{.246\%}$

The percent composition of nitrogen is $\text{8}\text{.246\%}\text{.}$

The percent composition of oxygen $\text{=}\left(\frac{48\text{g}}{169.9\text{g}}\right)\text{×100=28}\text{.25\%}$

The percent composition of oxygen is $\text{28}\text{.25\%}\text{.}$